[R] graph label greek symbol failure
Readers, Previous questions about this requirement have been for m$ users, my failure occurs using linux. I have tried to add the delta (δ) symbol to the y axis label and the result is &D, using the command: ...ylab="δt"... Any advice please? rh...@conference.jabber.org mandriva 2008 r 251 (27-06-07) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] postscript, options
Readers,
I am not able to create a postscript file with the following command:
postscript("/pathto/filename.eps,horizontal=FALSE,onefile="FALSE")
A file is not created, instead the command terminal shows the plus
sign(+) on a new line:
+
What does this mean please?
rh...@conference.jabber.org
mandriva 2008
r 251 (27-06-07)
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Re: [R] graph label greek symbol failure
On 17/08/2009, Patrick Connolly wrote: > On Mon, 17-Aug-2009 at 11:51AM +0100, e-letter wrote: > > |> Readers, > |> > |> Previous questions about this requirement have been for m$ users, my > |> failure occurs using linux. > |> > |> I have tried to add the delta (δ) symbol to the y axis label and the > |> result is &D, using the command: > |> > |> ...ylab="δt"... > |> > |> Any advice please? > > > ?plotmath > Is this an answer? Totally meaningless to me... __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] graph label greek symbol failure
On 17/08/2009, Michael Knudsen wrote: > On Mon, Aug 17, 2009 at 12:51 PM, e-letter wrote: > >> I have tried to add the delta (δ) symbol to the y axis label and the >> result is &D, using the command: >> >> ...ylab="δt"... > > Try ylab = expression(delta*t) instead. > This does not work, the result is expression(delta*t) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] graph label greek symbol failure
On 18/08/2009, Gavin Simpson wrote: > On Mon, 2009-08-17 at 11:51 +0100, e-letter wrote: >> Readers, >> >> Previous questions about this requirement have been for m$ users, my >> failure occurs using linux. >> >> I have tried to add the delta (δ) symbol to the y axis label and the >> result is &D, using the command: >> >> ...ylab="δt"... > > Works for me on Fedora 11 with the font packages mentioned in R > Installation and Admin manual. > >> sessionInfo() > R version 2.9.1 Patched (2009-08-07 r49104) > x86_64-unknown-linux-gnu > > locale: > LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C > > attached base packages: > [1] stats graphics grDevices utils datasets methods base > >> >> Any advice please? > > plot(1:10, ylab = expression(delta*t)) > > ?plotmath > > If your question is rather why doesn't the "δ" glyph you enter at the > keyboard show up in the plot, I'm afraid I can't help. Perhaps something > to do with UTF-8 and locales and the available symbol set on the device > you are plotting on, but you'll need to provide more information (of the > sort requested in the posting guide) on your system and set-up. > my locale: locale: LC_CTYPE=en_GB.UTF-8;LC_NUMERIC=C;LC_TIME=en_GB.UTF-8;LC_COLLATE=en_GB.UTF-8;LC_MONETARY=en_GB.UTF-8;LC_MESSAGES=en_GB.UTF-8;LC_PAPER=en_GB.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_GB.UTF-8;LC_IDENTIFICATION=C Doesn't seem that the locale is at fault? I was plotting on the default x11 window. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] postscript, options
On 18/08/2009, Rodrigo Aluizio wrote:
> I'm not shure but I guess that you miss a " and putted it in the wrong
> place!
> Try this: postscript("/pathto/filename.eps",horizontal=FALSE,onefile=FALSE)
>
> Hope It helps.
Thank you, that did help a little, but the eps file was faulty and I
could not open in either imagemagick or gimp. I found it easier and
quicker to copy the graph into gimp, save in png and use imagemagick
to convert to eps format.
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Re: [R] postscript, options
On 18/08/2009, Stefan Grosse wrote:
> On Tue, 18 Aug 2009 13:52:58 +0100 e-letter wrote:
>
> EL> On 18/08/2009, Rodrigo Aluizio wrote:
> EL> > I'm not shure but I guess that you miss a " and putted it in the
> EL> > wrong place!
> EL> > Try this:
> EL> > postscript("/pathto/filename.eps",horizontal=FALSE,onefile=FALSE)
> EL> >
> EL> > Hope It helps.
> EL>
> EL> Thank you, that did help a little, but the eps file was faulty and I
> EL> could not open in either imagemagick or gimp. I found it easier and
> EL> quicker to copy the graph into gimp, save in png and use imagemagick
> EL> to convert to eps format.
>
> What an effort which also yields bad results regarding quality! EPS is
> vector oriented and I doubt that PNG to EPS gives nice results besides
> if you prefer it as such why not directly create a png using the png
> device?
I want eps for latex. Png not enabled in my installation of R.
Conversion to eps was good with imagemagick, poor with gimp.
>
> What do you mean with faulty? I suspect it was to large because you
> forgot the paper="special" option. Otherwise it creates an A4 plot.
>
The image was blank, on my screen a small window about 50 mm each side
(using imagemagick. In gimp, error dialogue window received, unable to
open file). This did not occur when I used gimp and imagemagick.
> postscript("/pathto/filename.eps",horizontal=FALSE,onefile=FALSE,paper="special")
>
This command created a small blank eps file.
> see
> ?grDevices::postscript
>
I tried that command, which shows content about double and triple
colon operators? Does not show anything about postscript?
> for the options.
>
> Stefan
>
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Re: [R] graph label greek symbol failure
On 18/08/2009, Gavin Simpson wrote: > On Tue, 2009-08-18 at 13:06 +0100, e-letter wrote: >> On 17/08/2009, Michael Knudsen wrote: >> > On Mon, Aug 17, 2009 at 12:51 PM, e-letter wrote: >> > >> >> I have tried to add the delta (δ) symbol to the y axis label and the >> >> result is &D, using the command: >> >> >> >> ...ylab="δt"... >> > >> > Try ylab = expression(delta*t) instead. >> > >> This does not work, the result is >> expression(delta*t) > > It works for the rest of us who suggested this. > > plot(1:10, ylab = expression(delta*t)) > True, but the following commands fails: plot(1:10,ylab="temperature expression(delta*t)") plot(1:10,ylab="temperature" expression(delta*t)) Error: syntax error, unexpected SYMBOL, expecting ',' in "plot(1:10,ylab="temperature" expression" So I want to be able to have 'δt' and 'temperature δt' as a y-axis label. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] graph label greek symbol failure
On 19/08/2009, Gavin Simpson wrote: > On Wed, 2009-08-19 at 14:20 +0100, e-letter wrote: >> On 18/08/2009, Gavin Simpson wrote: >> > On Tue, 2009-08-18 at 13:06 +0100, e-letter wrote: >> >> On 17/08/2009, Michael Knudsen wrote: >> >> > On Mon, Aug 17, 2009 at 12:51 PM, e-letter wrote: >> >> > >> >> >> I have tried to add the delta (δ) symbol to the y axis label and the >> >> >> result is &D, using the command: >> >> >> >> >> >> ...ylab="δt"... >> >> > >> >> > Try ylab = expression(delta*t) instead. >> >> > >> >> This does not work, the result is >> >> expression(delta*t) >> > >> > It works for the rest of us who suggested this. >> > >> > plot(1:10, ylab = expression(delta*t)) >> > >> True, but the following commands fails: >> plot(1:10,ylab="temperature expression(delta*t)") >> plot(1:10,ylab="temperature" expression(delta*t)) >> Error: syntax error, unexpected SYMBOL, expecting ',' in >> "plot(1:10,ylab="temperature" expression" >> >> So I want to be able to have 'δt' and 'temperature δt' as a y-axis label. > > Ah, but you never said that. We aren't mind readers you know ;-) > > What you need is an expression that will, when used, give you a text > label containing "Temperature δt". What you have done is create a > character string of the literal "...expression(delta*t)" which is of > course why it is printed as the label - after all , you asked R to do > this. > > I suggest you read the plotmath help page, accessed via: > > ?plotmath > > executed at the prompt. > > I found this expression stuff complicated when I first started out, > until I realised that whatever gets passed to expression() has to be > a syntactically valid R command. So for your particular example we need: > > plot(1:10, ylab = expression(Temperature ~ delta*t)) > > Temperature is treated as the character string "Temperature"; the "~" > means leave some space between the bits either side of "~"; delta is a > special name that will get replaced by the relevant glyph; the "*" > juxtaposes delta and "t", ie places them next to one another without any > space in between. > Being familiar with latex, I interpret your description of the tilde (~) as non-breaking space, therefore I used the command: plot(1:10,ylab=expression(temperature~delta*t)) to give me the result I wanted; in other words the space characters are not needed between temperature,~ and delta in your suggestion. Thank you. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] change default measurement unit to millimetre
Readers, How do I configure R so that all units of measurement are in millimetres? For example if I want to set the width of a graph, I want to write: width=100 and R interprets this value as millimetres. Yours, rh...@conference.jabber.org r 251 mandriva 2008 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] postscript, options
On 18/08/2009, Stefan Grosse wrote:
> On Tue, 18 Aug 2009 16:08:36 +0100 e-letter wrote:
>
> EL> >
> postscript("/pathto/filename.eps",horizontal=FALSE,onefile=FALSE,paper="special")
> EL> >
> EL> This command created a small blank eps file.
>
> Its hard to say why without any example code.
>
> Does
> postscript("/pathto/filename.eps",width=8,height=8,horizontal=FALSE,onefile=FALSE,paper="special")
> hist(rnorm(100))
> dev.off()
> work?
>
Yes, using the following commands:
postscript("/pathto/filename.eps",width=4,height=4,horizontal=FALSE,onefile=FALSE,paper="special")
hist(rnorm(100))
dev.off()
Command terminal response:
X11
2
I opened successfully the file. Thank you.
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Re: [R] change default measurement unit to millimetre
On 20/08/2009, Patrick Connolly wrote: > On Thu, 20-Aug-2009 at 10:25PM +1000, Jim Lemon wrote: > >> e-letter wrote: >>> Readers, >>> >>> How do I configure R so that all units of measurement are in >>> millimetres? For example if I want to set the width of a graph, I want >>> to write: >>> width=100 >>> and R interprets this value as millimetres. >>> >>> >> Hi e-letter, >> If you just want a quick hack: >> >> mm2in<-function(mm) return(mm/25.4) >> >> then >> >> width=mm2in(100) > > As a dedicated lazy typist, I've discovered that this is less typing: > > width=100/25.4 > Thanks, but really I want to change the entire R installation to accept millimetre as the default unit of measurement. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] linear regression, exclude a datum
Readers, I have been reading the r book (Crawley) and tried to use the influence measures function for linear regression, as described. I have one datum that I wish to show in the graph but exclude from the regression and ab line. x y 0 5 10 9 20 10 30 19 40 4 With the influence measures function I plot the graph but linear regression includes the datum (40,4), which I want to show on the graph but exclude from the linear regression ab line. Is there an alternative package to perform this task please? rhelp at conference.jabber.org r 251 mandriva 2008 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] linear regression, exclude a datum
On 14/09/2009, Steve Lianoglou wrote: > Hi, > > On Sep 14, 2009, at 9:47 AM, e-letter wrote: > >> Readers, >> >> I have been reading the r book (Crawley) and tried to use the >> influence measures function for linear regression, as described. I >> have one datum that I wish to show in the graph but exclude from the >> regression and ab line. >> >> xy >> 05 >> 10 9 >> 20 10 >> 30 19 >> 40 4 >> >> With the influence measures function I plot the graph but linear >> regression includes the datum (40,4), which I want to show on the >> graph but exclude from the linear regression ab line. >> >> Is there an alternative package to perform this task please? > > Please post the code you're using to try and do the regression. > > I think you simply want to remove the data point when you build the > regression model, then plot it later after you plot the regression. > Correct; below are my commands: x<-c(0,10,20,30) y<-c(5,9,12,19) fit<-lm(y~x) plot(y~x) abline(fit) I try these commands as explained in section 12 (graphics) of the introduction manual: points(40,4) plot(40,4,add=TRUE) Warning messages: 1: "add" is not a graphical parameter in: plot.window(xlim, ylim, log, asp, ...) 2: "add" is not a graphical parameter in: plot.xy(xy, type, pch, lty, col, bg, cex, lwd, ...) 3: "add" is not a graphical parameter in: axis(side, at, as.graphicsAnnot(labels), tick, line, pos, outer, 4: "add" is not a graphical parameter in: axis(side, at, as.graphicsAnnot(labels), tick, line, pos, outer, 5: "add" is not a graphical parameter in: box(which = which, lty = lty, ...) 6: "add" is not a graphical parameter in: title(main, sub, xlab, ylab, line, outer, ...) The datum is not added to the original graph. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] linear regression, exclude a datum
On 15/09/2009, Uwe Ligges wrote: > > > e-letter wrote: >> On 14/09/2009, Steve Lianoglou wrote: >>> Hi, >>> >>> On Sep 14, 2009, at 9:47 AM, e-letter wrote: >>> >>>> Readers, >>>> >>>> I have been reading the r book (Crawley) and tried to use the >>>> influence measures function for linear regression, as described. I >>>> have one datum that I wish to show in the graph but exclude from the >>>> regression and ab line. >>>> >>>> x y >>>> 0 5 >>>> 10 9 >>>> 20 10 >>>> 30 19 >>>> 40 4 >>>> >>>> With the influence measures function I plot the graph but linear >>>> regression includes the datum (40,4), which I want to show on the >>>> graph but exclude from the linear regression ab line. >>>> >>>> Is there an alternative package to perform this task please? >>> Please post the code you're using to try and do the regression. >>> >>> I think you simply want to remove the data point when you build the >>> regression model, then plot it later after you plot the regression. >>> >> Correct; below are my commands: >> >> x<-c(0,10,20,30) >> y<-c(5,9,12,19) >> fit<-lm(y~x) >> plot(y~x) >> abline(fit) >> >> I try these commands as explained in section 12 (graphics) of the >> introduction manual: > > > > Hope this example is not in Crawley's book as you cite it, because you > need to make the plot large enough to include the value (40,4) already > when you create the plot, e.g., by specifying xlim and ylim. > No, it is an example I made quickly to post to the mailing list. Which section of the manual explains xlim please? > Anyway, I'd rather explude the value from a data.frame as fol#ows: > > # whole data: > dat <- data.frame(x = c(0,10,20,30,40), y = c(5,9,12,19,4)) > dat > plot(y~x, data=dat) > # fit the regression without the 5th observation: > fit <- lm(y ~ x, data=dat[-5,]) > abline(fit) > Thank you. Please advise which section of the introduction manual explains how to exclude a datum (the instruction '...dat[-5,]'. One of my biggest difficulties is finding the relevant section of the manual(s). __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] margin text warning message NAs coercion
Readers, I tried to the following commands: plot(y~x,ylab=expression(A[1]~B[2],xlab=expression(C~D)) mtext(expression(A[1]~B[2]),"additional text",side=3,line=1) I receive the text that I want, but the command terminal shows the following response: Warning message: NAs introduced by coercion in: mtext(expression(A[1]~B[2]),"additional text", side = 3, What is my mistake please? Yours, r251 rhelpatconference.jabber.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] margin text warning message NAs coercion
On 16/02/2010, Peter Ehlers wrote: > On 2010-02-16 9:21, e-letter wrote: >> Readers, >> >> I tried to the following commands: >> >> plot(y~x,ylab=expression(A[1]~B[2],xlab=expression(C~D)) >> mtext(expression(A[1]~B[2]),"additional text",side=3,line=1) > > Your plot() call is not reproducible. > > Anyway, try > > mtext(expression(A[1]~B[2]~~"additional text"),side=3,line=1) > Thank you. I had seen the command (~~) in the help guide for the package plotmath(grdevices), but there is no statement that this command can be used to join the expression to literal text. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] assessing data variation
I have data like so: timedatum 30 12 60 24 90 37 120 41 150 8 In addition to standard deviation, I want to measure the average of the differences in data for each time interval, i.e. average of 24-12, 37-24, 41-37, 8-41. Is there a statistical term for this task? Which package should I use please? rh...@conference.jabber.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Merge failure using zoo package
Readers,
Please refer to attached example data files. It seems that the merge
function fails for the latter section of the data set. Command
terminal output:
> library(chron)
> library(zoo)
> x<-read.zoo("test1.csv",header=TRUE,sep=",",FUN=times)
> y<-read.zoo("test2.csv",header=TRUE,sep=",",FUN=times)
> z<-(na.approx(merge(x[,2],y[,2]),time(z1)))
> z
x[, 2]y[, 2]
01:01:01 0.5418645 0.1755847
01:01:30 0.3486081 0.2068249
01:01:42 0.4808362 0.2380651
01:02:00 0.6130642 0.4983712
01:02:23 0.3140116 0.7586773
01:19:00 0.8545863 0.8927112
01:24:00 0.965 0.1490374
To overcome this behaviour the files test3 and test4 were created by
removing data that had been merged previously. Command terminal output
below:
> x<-read.zoo("test3.csv",header=TRUE,sep=",",FUN=times)
> y<-read.zoo("test4.csv",header=TRUE,sep=",",FUN=times)
> z<-(na.approx(merge(x[,2],y[,2]),time(z1)))
> z
x[, 2]y[, 2]
01:03:06 0.4827475 0.7350236
01:03:30 0.6951390 0.8376028
01:03:50 0.5798283 0.9401821
01:04:00 0.4645176 0.8330635
01:04:30 0.6167257 0.7259450
01:19:00 0.8545863 0.8927112
01:24:00 0.965 0.1490374
The only way to obtain a more complete merge of the data sets is to
create manually new files where previously merged data is removed and
then put all the merged data into a new file. Surely this package
should merge the data sets completely?
yours,
rh...@conference.jabber.org
r251
mandriva2008
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Merge failure using zoo package
Data files test1, ...2, ...3, ...4 respectively. time1,dataset1 01:01:00,0.73512097 01:01:30,0.34860813 01:02:00,0.61306418 01:02:30,0.01495898 01:03:00,0.27035612 01:03:30,0.69513898 01:04:00,0.46451758 01:04:30,0.61672569 01:05:00,0.82496122 01:05:30,0.34766154 01:06:00,0.69618714 01:06:30,0.39035214 01:07:00,0.01680143 01:07:30,0.28576967 01:08:00,0.01205416 01:08:30,0.89637254 01:09:00,0.63147653 01:09:30,0.01522139 01:10:00,0.27661960 01:10:30,0.50974124 01:11:00,0.68141977 01:11:30,0.90725854 01:12:00,0.83823443 01:12:30,0.53360241 01:13:00,0.17769196 01:13:30,0.83438616 01:14:00,0.67248807 01:14:30,0.09991933 01:15:00,0.03334966 01:15:30,0.93292355 01:16:00,0.15990837 01:16:30,0.05354050 01:17:00,0.55281203 01:17:30,0.37845690 01:18:00,0.89051365 01:18:30,0.16674292 01:19:00,0.85458626 01:19:30,0.19278550 01:20:00,0.73240405 01:20:30,0.16417524 01:21:00,0.73878212 01:21:30,0.51790118 01:22:00,0.83076438 01:22:30,0.4704 01:23:00,0.02108640 01:23:30,0.82911053 01:24:00,0.9646 01:24:30,0.14493657 01:25:00,0.84422332 01:25:30,0.41589974 01:26:00,0.67606367 01:26:30,0.00606434 01:27:00,0.59951991 01:27:30,0.43949260 01:28:00,0.66297385 01:28:30,0.33131298 01:29:00,0.06102041 01:29:30,0.84722118 01:30:00,0.46841491 01:30:30,0.34200755 01:31:00,0.87386578 01:31:30,0.70737403 01:32:00,0.23978781 01:32:30,0.11787278 01:33:00,0.14679814 01:33:30,0.65217063 01:34:00,0.81355908 01:34:30,0.31583482 01:35:00,0.92167666 01:35:30,0.55931271 01:36:00,0.13641271 01:36:30,0.35048575 01:37:00,0.17243584 01:37:30,0.93645686 01:38:00,0.85356548 01:38:30,0.61399352 01:39:00,0.05910707 01:39:30,0.01721605 01:40:00,0.94845557 01:40:30,0.48117810 01:41:00,0.34752402 01:41:30,0.59295472 01:42:00,0.64267429 01:42:30,0.57859933 01:43:00,0.00201441 01:43:30,0.32530995 01:44:00,0.25474645 01:44:30,0.93187534 01:45:00,0.99361033 01:45:30,0.16591641 time2,dataset2 01:01:01,0.17558467 01:01:42,0.23806514 01:02:23,0.75867726 01:03:06,0.73502357 01:03:50,0.94018206 01:04:35,0.61882643 01:05:21,0.68417492 01:06:08,0.05744461 01:06:55,0.33344394 01:07:44,0.68752593 01:08:33,0.17270469 01:09:23,0.81522124 01:10:03,0.68304352 01:10:43,0.38774082 01:11:23,0.84176890 01:12:04,0.0936 01:12:44,0.13431965 01:13:25,0.92210721 01:14:06,0.33630635 01:14:47,0.56690294 01:15:29,0.09870816 01:16:11,0.77864105 01:16:53,0.61803441 01:17:35,0.09133728 01:18:17,0.08925487 01:19:00,0.89271117 01:19:42,0.56605742 01:20:25,0.98520534 01:21:08,0.66104843 01:21:51,0.96948589 01:22:34,0.05692690 01:23:17,0.71887456 01:24:00,0.14903741 01:24:43,0.86569445 01:25:26,0.27923513 01:26:09,0.98365033 01:26:53,0.08308399 01:27:36,0.87071027 01:28:19,0.26475705 01:29:03,0.76409811 01:29:47,0.59563256 01:30:31,0.23995054 01:31:14,0.00951054 01:31:59,0.21367270 time1,dataset1 01:02:30,0.01495898 01:03:00,0.27035612 01:03:30,0.69513898 01:04:00,0.46451758 01:04:30,0.61672569 01:05:00,0.82496122 01:05:30,0.34766154 01:06:00,0.69618714 01:06:30,0.39035214 01:07:00,0.01680143 01:07:30,0.28576967 01:08:00,0.01205416 01:08:30,0.89637254 01:09:00,0.63147653 01:09:30,0.01522139 01:10:00,0.27661960 01:10:30,0.50974124 01:11:00,0.68141977 01:11:30,0.90725854 01:12:00,0.83823443 01:12:30,0.53360241 01:13:00,0.17769196 01:13:30,0.83438616 01:14:00,0.67248807 01:14:30,0.09991933 01:15:00,0.03334966 01:15:30,0.93292355 01:16:00,0.15990837 01:16:30,0.05354050 01:17:00,0.55281203 01:17:30,0.37845690 01:18:00,0.89051365 01:18:30,0.16674292 01:19:00,0.85458626 01:19:30,0.19278550 01:20:00,0.73240405 01:20:30,0.16417524 01:21:00,0.73878212 01:21:30,0.51790118 01:22:00,0.83076438 01:22:30,0.4704 01:23:00,0.02108640 01:23:30,0.82911053 01:24:00,0.9646 01:24:30,0.14493657 01:25:00,0.84422332 01:25:30,0.41589974 01:26:00,0.67606367 01:26:30,0.00606434 01:27:00,0.59951991 01:27:30,0.43949260 01:28:00,0.66297385 01:28:30,0.33131298 01:29:00,0.06102041 01:29:30,0.84722118 01:30:00,0.46841491 01:30:30,0.34200755 01:31:00,0.87386578 01:31:30,0.70737403 01:32:00,0.23978781 01:32:30,0.11787278 01:33:00,0.14679814 01:33:30,0.65217063 01:34:00,0.81355908 01:34:30,0.31583482 01:35:00,0.92167666 01:35:30,0.55931271 01:36:00,0.13641271 01:36:30,0.35048575 01:37:00,0.17243584 01:37:30,0.93645686 01:38:00,0.85356548 01:38:30,0.61399352 01:39:00,0.05910707 01:39:30,0.01721605 01:40:00,0.94845557 01:40:30,0.48117810 01:41:00,0.34752402 01:41:30,0.59295472 01:42:00,0.64267429 01:42:30,0.57859933 01:43:00,0.00201441 01:43:30,0.32530995 01:44:00,0.25474645 01:44:30,0.93187534 01:45:00,0.99361033 01:45:30,0.16591641 time2,dataset2 01:03:06,0.73502357 01:03:50,0.94018206 01:04:35,0.61882643 01:05:21,0.68417492 01:06:08,0.05744461 01:06:55,0.33344394 01:07:44,0.68752593 01:08:33,0.17270469 01:09:23,0.81522124 01:10:03,0.68304352 01:10:43,0.38774082 01:11:23,0.84176890 01:12:04,0.0936 01:12:44,0.13431965 01:13:25,0.92210721 01:14:06,0.33630635 01:14:47,0.56690294 01:15:29,0.09870816 01:16:11,0.77864105 01:16:53,0.61803441 01:17:35,0.09133728 01:18:17,0.08925487 01:19:00,0.89271117 01:19:
Re: [R] Merge failure using zoo package
On 02/04/2010, Gabor Grothendieck wrote: > The files only have one data column. What is the meaning of x[,2], > etc. ? What is z1? > I only want to merge one column from one file with one column from another file. With [x,2], I am trying to select the column of data. > Please provide reproducible code and data all in a single file using > this style so its clear what is what. Also please cut down the size of > your data to the smallest size that will still illustrate the problem. > See other posting; each file that I used is separated by an empty line. This error seems to occur with the data set size as shown in the files. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Merge failure using zoo package
On 02/04/2010, Gabor Grothendieck wrote:
> The code does not run with the files. I need the requested
> information, namely a single file containing code and data and that I
> can just copy into a session without editing and see the result you
> see.
I don't understand how I can combine the four csv files into a single
file of data and terminal commands? Anyway, further terminal output.
The following also occurs with correction of the commands. The data
merge is incomplete (to 1:27:30); data set 1 ends at time 1:45:30;
data set 2 1:31:59
> library(chron)
> library(zoo)
> z1<-read.zoo("test1.csv",header=TRUE,sep=",",FUN=times)
> z2<-read.zoo("test2.csv",header=TRUE,sep=",",FUN=times)
> z3<-(na.approx(merge(z1[,2],z2[,2]),time(z1)))
> z3
z1[, 2]z2[, 2]
01:01:01 0.54186455 0.17558467
01:01:30 0.34860813 0.20682491
01:01:42 0.48083615 0.23806514
01:02:00 0.61306418 0.49837120
01:02:23 0.31401158 0.75867726
01:02:30 0.01495898 0.75079270
01:03:00 0.27035612 0.74290813
01:03:06 0.48274755 0.73502357
01:03:30 0.69513898 0.83760282
01:03:50 0.57982828 0.94018206
01:04:00 0.46451758 0.83306352
01:04:30 0.61672569 0.72594497
01:04:35 0.72084346 0.61882643
01:05:00 0.82496122 0.65150068
01:05:21 0.58631138 0.68417492
01:05:30 0.34766154 0.47526482
01:06:00 0.69618714 0.26635471
01:06:08 0.54326964 0.05744461
01:06:30 0.39035214 0.19544428
01:06:55 0.20357679 0.33344394
01:07:00 0.01680143 0.45147127
01:07:30 0.28576967 0.56949860
01:07:44 0.14891191 0.68752593
01:08:00 0.01205416 0.51591885
01:08:30 0.89637254 0.34431177
01:08:33 0.76392454 0.17270469
01:09:00 0.63147653 0.49396296
01:09:23 0.32334896 0.81522124
01:09:30 0.01522139 0.77116200
01:10:00 0.27661960 0.72710276
01:10:03 0.39318042 0.68304352
01:10:30 0.50974124 0.53539217
01:10:43 0.59558051 0.38774082
01:11:00 0.68141977 0.61475486
01:11:23 0.79433915 0.84176890
01:11:30 0.90725854 0.59232742
01:12:00 0.83823443 0.34288594
01:12:04 0.68591842 0.0936
01:12:30 0.53360241 0.11388206
01:12:44 0.35564718 0.13431965
01:13:00 0.17769196 0.52821343
01:13:25 0.50603906 0.92210721
01:13:30 0.83438616 0.72684026
01:14:00 0.67248807 0.53157330
01:14:06 0.38620370 0.33630635
01:14:30 0.09991933 0.45160464
01:14:47 0.06663450 0.56690294
01:15:00 0.03334966 0.33280555
01:15:29 0.48313660 0.09870816
01:15:30 0.93292355 0.32535246
01:16:00 0.15990837 0.55199675
01:16:11 0.10672443 0.77864105
01:16:30 0.05354050 0.69833773
01:16:53 0.30317627 0.61803441
01:17:00 0.55281203 0.44246870
01:17:30 0.37845690 0.26690299
01:17:35 0.63448528 0.09133728
01:18:00 0.89051365 0.09029608
01:18:17 0.52862829 0.08925487
01:18:30 0.16674292 0.49098302
01:19:00 0.85458626 0.89271117
01:19:30 0.19278550 0.72938430
01:19:42 0.46259477 0.56605742
01:20:00 0.73240405 0.77563138
01:20:25 0.44828965 0.98520534
01:20:30 0.16417524 0.87715304
01:21:00 0.73878212 0.76910073
01:21:08 0.62834165 0.66104843
01:21:30 0.51790118 0.81526716
01:21:51 0.67433278 0.96948589
01:22:00 0.83076438 0.66529956
01:22:30 0.4704 0.36111323
01:22:34 0.24832072 0.05692690
01:23:00 0.02108640 0.38790073
01:23:17 0.42509847 0.71887456
01:23:30 0.82911053 0.43395599
01:24:00 0.9646 0.14903741
01:24:30 0.14493657 0.50736593
01:24:43 0.49457995 0.86569445
01:25:00 0.84422332 0.57246479
01:25:26 0.63006153 0.27923513
01:25:30 0.41589974 0.51404020
01:26:00 0.67606367 0.74884526
01:26:09 0.34106401 0.98365033
01:26:30 0.00606434 0.53336716
01:26:53 0.30279212 0.08308399
01:27:00 0.59951991 0.34562608
01:27:30 0.43949260 0.60816818
>
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Re: [R] Merge failure using zoo package
I tried your instructions, as shown.
> library(chron)
> library(zoo)
> z1<-read.zoo("test1.csv",header=TRUE,sep=",",FUN=times)
> z2<-read.zoo("test2.csv",header=TRUE,sep=",",FUN=times)
> z12<-merge(z1,z2)
> z3<-na.approx(z12,xout=time(z1))
Error in approx(along[!na], y[!na], along[na], ...) :
invalid interpolation method
In addition: Warning message:
the condition has length > 1 and only the first element will be used
in: if (is.na(method)) stop("invalid interpolation method")
> source("http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php/*checkout*/pkg/zoo/R/na.approx.R?rev=676&root=zoo";)
> z3<-na.approx(z12,xout=time(z1))
> z3
z1 z2
01:01:30 0.34860813 0.21977817
01:02:00 0.61306418 0.46662656
01:02:30 0.01495898 0.75482666
01:03:00 0.27035612 0.73832408
...
It looks that the data of z3 shows only values corresponding to times
in z1. In the previous merge, z3 consisted of values for times in both
z1 and z2, e.g.:
z1[, 2]z2[, 2]
01:01:01 0.54186455 0.17558467
01:01:30 0.34860813 0.20682491
01:01:42 0.48083615 0.23806514
01:02:00 0.61306418 0.49837120
With respect to advice about seeking help, I think it would be very
good if your comments about using the command 'textConnection()' were
added to the FAQ section or the posting guide, to help novices improve
the quality of asking for help.
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[R] How to specify only minimum ylim parameter for boxplot
Readers, For a command: x<-a data matrix y<-another data matrix z<-and another data matrix boxplot(c(x,y,z),las=1,xaxt="n",ylab="text",xlab="text",ylim=c(0,..?)) is it possible to specify only that the boxplot graph starts with value 0 at the bottom of the ordinate (y axis) and the maximum value is determined by data in x,y and z? Yours, rh...@conference.jabber.org r251 mandriva2008 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] synchronisation of time series data using interpolation
Readers, I have data with different time stamps that I wish to plot (for example): data set 1 time(hh:mm:ss),datum 01:00:00,500 01:00:15,600 01:00:30,750 01:00:45,720 01:01:00,700 01:01:15,725 01:01:30,640 01:01:45,710 data set 2 time,datum 01:00:12,20 01:01:01,55 01:01:55,22 The time interval in data set 1 does not change, but the time interval in data set 2 does change, such that for a specific total time range (e.g. 60 minutes) there will be more data in data set 1 than in data set 2. I thought I could solve this problem using interpolation, to create a new data set using data from data set 2, interpolated to the time stamps in data set 1: data set 3 time,datum 01:00:00,18 01:00:15,23 01:00:30,30 01:00:45,41 01:01:00,53 01:01:15,46 01:01:30,38 01:01:45,29 Therefore I would then be able to plot the data in data set 1 against the interpolated data in data set 3, because there would be equal quantities of data in both data sets. I've looked at the interp function in the help manual, but I don't understand if this function can perform the task I want. Any advice please? Yours, rhelp at conference.jabber.org r 251 (27-06-07) mandriva 2008 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] synchronisation of time series data using interpolation
I created separate text files for the 2 data sets. I enter the
following comands:
library(zoo)
library(chron)
z1<-read.zoo(textConnection("/path/to/test1.txt"),header=FALSE,sep=",",FUN=times)
z2<-read.zoo(textConnection("/path/to/test2.txt"),header=FALSE,sep=",",FUN=times)
z3<-window(na.approx(merge(z1,z2)),time(z1))
plot(z3$z1,z3$z2)
Error in plot.window(xlim, ylim, log, asp, ...) :
need finite 'xlim' values
In addition: Warning messages:
1: no non-missing arguments to min; returning Inf in: min(x)
2: no non-missing arguments to max; returning -Inf in: max(x)
3: no non-missing arguments to min; returning Inf in: min(x)
4: no non-missing arguments to max; returning -Inf in: max(x)
The resultant graph window was blank, so I entered the following command
plot(z3$z1,z3$z2,xlim=c(0,100),ylim=c(0,100))
The graph window showed y axis (labelled 'z3$z1') and x axis (labelled 'Index').
I do not understand the instruction "...to use window to pick off..."
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Re: [R] synchronisation of time series data using interpolation
Test1 file contained data set 1, test2 contained data set 2 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] synchronisation of time series data using interpolation
I saved the data sets as files and then tried to refer to those files.
Therefore the instruction:
> z1<-read.zoo(textConnection("/path/to/test1.txt")
means that I wanted to replace the manual data entry for Lines1 with a
file containing the data. It seems that your instructions only work
when data is typed from the command terminal. I cannot use data stored
in separated files. Command terminal output below:
> lines1<-read.table("/path/to/test1.txt")
> lines2<-read.table("/path/to/test2.txt")
Warning message:
incomplete final line found by readTableHeader on '/path/to/test2.txt'
in: read.table("/path/to/test2.txt")
This is test1 file:
01:00:00,500
01:00:15,600
01:00:30,750
01:00:45,720
01:01:00,700
01:01:15,725
01:01:30,640
01:01:45,710
This is test2 file:
01:00:12,20
01:01:01,55
01:01:55,22
Then I added a carriage return in test2 file and saved it:
01:00:12,20
01:01:01,55
01:01:55,22
Then I repeated the command terminal entry for lines2:
> lines2<-read.table("/path/to/test2.txt")
> library(zoo)
> library(chron)
> z1<-read.zoo(textConnection(lines1),header=FALSE,sep=",",FUN=times)
Error in textConnection(lines1) : invalid 'text' argument
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[R] plot ᵒ C in graph axis label
Readers, I have tried to use a plotmath command to add the temperature degree sign (i.e. ᵒ C) to the axis label of a graph: > x<-(1:10) > y<-(200:191) > plot(x~y,ylab=expression(*degree~C)) Error: syntax error, unexpected '*', expecting ',' in "plot(x~y,ylab=expression(*" According to plotmath manual, the syntax is '*degree' and I am able to use this command when a number precedes the asterisk (e.g. ylab=expression(0*degree~C)). But I want to omit the need for a number before the degree sign. Any advice please? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] plot subscript text and percentage symbol in graph label axis
Readers, I am unable to plot a label consisting of both subscript text and percentage (%) symbol: x<-(1:10) y<-(200:191) plot(x~y,ylab=expression(~degree~C),xlab=expression(x[2]~%)) Error: syntax error, unexpected ERROR in "plot(x~y,ylab=expression(~degree~C),xlab=expression(x~%)" It seems that % is a special character in R? I have looked in the introductory manual but can see no reference to reserved characters. Yours, rhelpatconference.jabber.org r251 mandriva2008 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] exclude data for boxplot stats using mathematical operator
Readers, I have a data set as follows: 1,1 2,2 3,3 4,4 5,3 6,2 7,-10 8,-9 9,-3 10,2 11,3 12,4 13,5 14,4 15,3 16,2 17,1 I entered this data set using the command 'read.csv'. I want to exclude values fewer than zero in column 2 so then I tried the following command: boxplot.stats(x[>0,2],do.conf=FALSE) Error: syntax error, unexpected GT, expecting ',' in "boxplot.stats(x[>" Any help please? Yours, rhelp at conference.jabber.org r 251 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] whitespace with subscript text
Readers, I have the following command: expression(A[1]B~"%") but that causes an error so I changed to: expression(A[1]~B~"%") but the result is too much whitespace between subscript and B. Is there a way to reduce this whitespace? Yours, rhelp at conference.jabber.org r 251 mandriva 2008 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] unable to use coplot function with zoo package
Readers,
I have tried to use the zoo package to merge datasets and then use the
coplot function, but the graph is not fully created. Only the panel
data is shown. Command terminal output below, with csv files. What is
the meaning of the warning message? Can anyone help please?
rhelpatconference.jabber.org
r 251
mandriva 2008
> library(chron)
> library(zoo)
> z1<-read.zoo("test1.csv",header=FALSE,sep=",",FUN=times)
> z2<-read.zoo("test2.csv",header=FALSE,sep=",",FUN=times)
> z3<-read.zoo("test3.csv",header=FALSE,sep=",",FUN=times)
> z4<-(na.approx(merge(z1,z2,z3),time(z1)))
> coplot(z4$z2~z4$z1|z4$z3)
Warning messages:
1: Index vectors are of different classes: integer times in:
merge.zoo(e1, e2, all = FALSE, retclass = NULL)
2: Index vectors are of different classes: integer times in:
merge.zoo(e1, e2, all = FALSE, retclass = NULL)
3: Index vectors are of different classes: integer times in:
merge.zoo(e1, e2, all = FALSE, retclass = NULL)
4: Index vectors are of different classes: integer times in:
merge.zoo(e1, e2, all = FALSE, retclass = NULL)
5: Index vectors are of different classes: integer times in:
merge.zoo(e1, e2, all = FALSE, retclass = NULL)
6: Index vectors are of different classes: integer times in:
merge.zoo(e1, e2, all = FALSE, retclass = NULL)
> z4$z1
09:50:01 09:50:30 09:50:44 09:51:00 09:51:27 09:51:30 09:52:00 09:52:10
320.61 325.43 320.48 315.53 314.63 313.73 316.11 323.88
09:52:30 09:52:54
331.65 328.48
> z4$z2
09:50:01 09:50:30 09:50:44 09:51:00 09:51:27 09:51:30 09:52:00 09:52:10
404.7500 402.7450 400.7400 300.2450 199.7500 266.7867 333.8233 400.8600
09:52:30 09:52:54
402.8350 404.8100
> z4$z3
09:50:01 09:50:30 09:50:44 09:51:00 09:51:27 09:51:30 09:52:00 09:52:10
166.120 175.770 185.420 198.395 211.370 214.820 218.270 221.720
09:52:30 09:52:54
228.370 235.020
>
test1.csv:
09:50:00,315.79
09:50:30,325.43
09:51:00,315.53
09:51:30,313.73
09:52:00,316.11
09:52:30,331.65
09:53:00,325.31
09:53:30,334.33
09:54:00,334.54
09:54:30,336.55
09:55:00,339.34
09:55:30,341.93
test2.csv:
09:50:01,404.75
09:50:44,400.74
09:51:27,199.75
09:52:10,400.86
09:52:54,404.81
09:53:41,404.36
09:54:20,406.11
09:55:03,408.85
test3.csv:
09:50:01,166.12
09:50:44,185.42
09:51:27,211.37
09:52:10,221.72
09:52:54,235.02
09:53:41,262.85
09:54:20,306.64
09:55:03,346.3
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[R] write to file append by column
Readers, Scenario: data x consists of one column; 1 2 3 data y; 4 5 6 Is it possible to write to file such that the file is: 1,4 2,5 3,6 using the write.file function? I have tried the command: write(x,file="file.csv",ncolumns=1,append=TRUE,sep=",") write(y,file="file.csv",ncolumns=1,append=TRUE,sep=",") but the result is: 1 2 3 4 5 6 yours, rhelpatconference.jabber.org r 251 mandriva 2008 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] write to file append by column
On 24/11/2009, jim holtman wrote: > You can not append a column. Best bet, read the old file in, do a > 'cbind', write the object back out. > > On Tue, Nov 24, 2009 at 5:59 AM, e-letter wrote: >> Readers, >> >> Scenario: data x consists of one column; >> 1 >> 2 >> 3 >> >> data y; >> 4 >> 5 >> 6 >> >> Is it possible to write to file such that the file is: >> 1,4 >> 2,5 >> 3,6 >> >> using the write.file function? I have tried the command: >> >> write(x,file="file.csv",ncolumns=1,append=TRUE,sep=",") >> write(y,file="file.csv",ncolumns=1,append=TRUE,sep=",") >> >> but the result is: >> >> 1 >> 2 >> 3 >> 4 >> 5 >> 6 >> >> yours, >> >> rhelpatconference.jabber.org >> r 251 >> mandriva 2008 >> >> __ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > > > > -- > Jim Holtman > Cincinnati, OH > +1 513 646 9390 > > What is the problem that you are trying to solve? > This is the requested format: 1,4 2,5 3,6 The write functions described previously produce the following format: 1 2 3 4 5 6 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] selection of multiple subscripts
Readers, For a data set 'x': 1 a 2 b 3 c 4 d 5 e 6 f 7 g 8 h 9 i How to select multiple subscripts to plot? For example to plot values 1:3 and 9:10: plot(x[1:3,1],x[,2]) and plot(x[9:10,1],x[,2]) into one plot? Yours, rhelpatconference.jabber.org r251 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] selection of multiple subscripts
On 13/01/2010, Duncan Murdoch wrote: > On 13/01/2010 7:36 AM, e-letter wrote: >> Readers, >> >> For a data set 'x': >> >> 1 a >> 2 b >> 3 c >> 4 d >> 5 e >> 6 f >> 7 g >> 8 h >> 9 i >> >> How to select multiple subscripts to plot? For example to plot values >> 1:3 and 9:10: >> >> plot(x[1:3,1],x[,2]) >> >> and >> >> plot(x[9:10,1],x[,2]) >> >> into one plot? > > Neither of those will work, because your x[,2] vector is longer than the > other vector. > > What you want is something like this: > > plot(col2 ~ col1, data=x[c(1:3, 9:10),]) > Thanks, I now understand the concatenate function would help but forgot the syntax. Anyway I've just realised that the search database for R yields no result for '?concatenate' which is surprising. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] selection of multiple subscripts
On 13/01/2010, e-letter wrote: > On 13/01/2010, Duncan Murdoch wrote: >> On 13/01/2010 7:36 AM, e-letter wrote: >>> Readers, >>> >>> For a data set 'x': >>> >>> 1 a >>> 2 b >>> 3 c >>> 4 d >>> 5 e >>> 6 f >>> 7 g >>> 8 h >>> 9 i >>> >>> How to select multiple subscripts to plot? For example to plot values >>> 1:3 and 9:10: >>> >>> plot(x[1:3,1],x[,2]) >>> >>> and >>> >>> plot(x[9:10,1],x[,2]) >>> >>> into one plot? >> >> Neither of those will work, because your x[,2] vector is longer than the >> other vector. >> >> What you want is something like this: >> >> plot(col2 ~ col1, data=x[c(1:3, 9:10),]) >> > Thanks, I now understand the concatenate function would help but > forgot the syntax. Anyway I've just realised that the search database > for R yields no result for '?concatenate' which is surprising. > For the benefit of other novices: for the data set, the subscripts should have read: 1:3 and 8:9 Alternatively, the data set should have included: 10 j :) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] use zoo package with multiple column data sets
Readers,
I am trying to use the zoo package with an array of data:
file1:
hh:mm:ss1
hh:mm:ss2
hh:mm:ss3
hh:mm:ss4
file2:
hh:mm:ss11 55
hh:mm:ss22 66
hh:mm:ss33 77
hh:mm:ss44 88
I wanted to merge these data set so I tried the following commands:
library(chron)
library(zoo)
z1<-read.zoo("path/to/file1.csv",header=TRUE,sep=",",FUN=times)
z2<-read.zoo("path/to/file2.csv",header=TRUE,sep=",",FUN=times)
z3<-(na.approx(merge(z1,z2),time(z1)))
plot(z3$z1,z3$z2[3])
Warning message:
some methods for “zoo” objects do not work if the index entries in
‘order.by’ are not unique in: zoo(rval, ix)
> z3<-(na.approx(merge(z1,z2),time(z1)))
Error in merge.zoo(z1, z2) : series cannot be merged with non-unique
index entries in a series
I have read the document vignette 'zoo' but section 2.1 does not show
an example of the syntax for the command 'order.by'.
How to use the zoo package where the variable z2 contains many columns of data?
Yours,
rhelpatconference.jabber.org
r251
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Re: [R] use zoo package with multiple column data sets
Assuming my documentation is correct, my version shows faq 1 to refer
to duplicate times but if file2 is:
01:01:0111 55
01:01:0422 66
01:01:0733 77
01:01:1044 88
I cannot see what is duplicate? If I create two new files:
file3:
01:01:0111
01:01:0422
01:01:0733
01:01:1044
file4:
01:01:0155
01:01:0466
01:01:0777
01:01:1088
The previous commands work:
z1<-read.zoo("path/to/file1.csv",header=TRUE,sep=",",FUN=times)
z2<-read.zoo("path/to/file3.csv",header=TRUE,sep=",",FUN=times)
z3<-(na.approx(merge(z1,z2),time(z1)))
plot(z3$z1,z3$z2)
and:
z1<-read.zoo("path/to/file1.csv",header=TRUE,sep=",",FUN=times)
z2<-read.zoo("path/to/file4.csv",header=TRUE,sep=",",FUN=times)
z3<-(na.approx(merge(z1,z2),time(z1)))
plot(z3$z1,z3$z2)
Shouldn't I be able to have one file containing all the columns I want
to make graphs, instead of having to create numerous files of only two
columns of data?
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Re: [R] use zoo package with multiple column data sets
> library(chron)
> library(zoo)
> z1<-read.zoo("~/path/to/test1.csv",header=TRUE,sep=",",FUN=times)
> z2<-read.zoo("~/path/to/test2.csv",header=TRUE,sep=",",FUN=times)
> z3<-(na.approx(merge(z1,z2),time(z1)))
Error in approx(along[!na], y[!na], along[na], ...) :
need at least two non-NA values to interpolate
> z3<-(na.approx(merge(z1,z2[,2]),time(z1)))
Error in approx(along[!na], y[!na], along[na], ...) :
need at least two non-NA values to interpolate
> z3<-(na.approx(merge(z1,z2[,3]),time(z1)))
Error in `[.zoo`(z2, , 3) : subscript out of bounds
> z4<-read.zoo(textConnection(z2),FUN=times)
Error in textConnection(z2) : invalid 'text' argument
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[R] syntax for very simple derivative function
Readers, Two variables 'M', 'T', each contain a column of numbers. Have tried to create a derivative function to see how 'M' varies with respect to 'T': dm<-D(m)(dm/dt)~dt dm(m=M,t=T) but R returns an error: Error: could not find function "dm" What is my mistake please? -- R2151 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] function approx interpolation of time series data sets
Readers, Am trying to use the function 'approx' to interpolate time series data sets: data1: 01:23:40 5 01:23:45 10 01:23:50 12 01:23:55 7 data2: 01:23:42 01:23:47 01:23:51 01:23:54 The objective is to obtain interpolated values of 'data1' column 2 (5, 10, 12, 7) for the times shown in data2. Tried the following command but received the error shown: data3<-approx(data1,xout=data2) Error in approx(data1, xout = data2) : (list) object cannot be coerced to type 'double' What is my mistake please? -- r2151 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can strptime handle milliseconds or AM/PM?
Readers,
Responding to an old post
(http://tolstoy.newcastle.edu.au/R/e2/help/07/06/18850.html), and
using the example in the manual:
monthextract<-strptime("20/2/06 11:16:16.683", "%m")
monthextract
[1] NA
Why is the result 'NA' and not '2'?
--
r2151
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Re: [R] function approx interpolation of time series data sets
On 16/01/2013, Rui Barradas wrote: > Hello, > > Like this? > > > data1 <- read.table(text = " > 01:23:40 5 > 01:23:45 10 > 01:23:50 12 > 01:23:55 7 > ") > > data2 <- read.table(text = " > 01:23:42 > 01:23:47 > 01:23:51 > 01:23:54 > ") > > approx(as.POSIXct(data1$V1, format = "%H:%M:%S"), y = data1$V2, xout = > as.POSIXct(data2$V1, format = "%H:%M:%S")) > Thanks. I have later realised that if data frames are unequal (i.e. data2edit 01:23:42 01:23:47 01:23:51 01:23:54 01:23:58 01:23:59 the result for 'y' is $y [1] NA NA NA NA NA NA Similar error occurs with the 'zoo' package. Is it better to do interpolation/further manipulation in another software tool? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] function approx interpolation of time series data sets
On 18/01/2013, Gabor Grothendieck wrote: > On Fri, Jan 18, 2013 at 7:31 AM, e-letter wrote: >> On 16/01/2013, Rui Barradas wrote: >>> Hello, >>> >>> Like this? >>> >>> >>> data1 <- read.table(text = " >>> 01:23:40 5 >>> 01:23:45 10 >>> 01:23:50 12 >>> 01:23:55 7 >>> ") >>> >>> data2 <- read.table(text = " >>> 01:23:42 >>> 01:23:47 >>> 01:23:51 >>> 01:23:54 >>> ") >>> >>> approx(as.POSIXct(data1$V1, format = "%H:%M:%S"), y = data1$V2, xout = >>> as.POSIXct(data2$V1, format = "%H:%M:%S")) >>> >> >> Thanks. I have later realised that if data frames are unequal (i.e. >> data2edit >> >> 01:23:42 >> 01:23:47 >> 01:23:51 >> 01:23:54 >> 01:23:58 >> 01:23:59 >> >> the result for 'y' is >> >> $y >> [1] NA NA NA NA NA NA >> >> Similar error occurs with the 'zoo' package. > > > its not clear precisely what you tried with zoo but have a look at this: > >> data1 <- " > + 01:23:40 5 > + 01:23:45 10 > + 01:23:50 12 > + 01:23:55 7" >> >> data2 <- " > + 01:23:42 > + 01:23:47 > + 01:23:51 > + 01:23:54" >> >> library(zoo) >> library(chron) >> z1 <- read.zoo(text = data1, FUN = times) >> z2 <- read.zoo(text = data2, FUN = times) >> na.approx(z1, xout = time(z2)) > 01:23:42 01:23:47 01:23:51 01:23:54 > 7.0 10.8 11.0 8.0 > With respect to zoo package, looks like I tried to apply zoo package to the vectors directly, not converting them to zoo objects using 'read.zoo ...'. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] function approx interpolation of time series data sets
On 18/01/2013, Gabor Grothendieck wrote:
> On Fri, Jan 18, 2013 at 7:31 AM, e-letter wrote:
>> On 16/01/2013, Rui Barradas wrote:
>>> Hello,
>>>
>>> Like this?
>>>
>>>
>>> data1 <- read.table(text = "
>>> 01:23:40 5
>>> 01:23:45 10
>>> 01:23:50 12
>>> 01:23:55 7
>>> ")
>>>
>>> data2 <- read.table(text = "
>>> 01:23:42
>>> 01:23:47
>>> 01:23:51
>>> 01:23:54
>>> ")
>>>
>>> approx(as.POSIXct(data1$V1, format = "%H:%M:%S"), y = data1$V2, xout =
>>> as.POSIXct(data2$V1, format = "%H:%M:%S"))
>>>
>>
>> Thanks. I have later realised that if data frames are unequal (i.e.
>> data2edit
>>
>> 01:23:42
>> 01:23:47
>> 01:23:51
>> 01:23:54
>> 01:23:58
>> 01:23:59
>>
>> the result for 'y' is
>>
>> $y
>> [1] NA NA NA NA NA NA
>>
>> Similar error occurs with the 'zoo' package.
>
>
> its not clear precisely what you tried with zoo but have a look at this:
>
>> data1 <- "
> + 01:23:40 5
> + 01:23:45 10
> + 01:23:50 12
> + 01:23:55 7"
>>
>> data2 <- "
> + 01:23:42
> + 01:23:47
> + 01:23:51
> + 01:23:54"
>>
>> library(zoo)
>> library(chron)
>> z1 <- read.zoo(text = data1, FUN = times)
>> z2 <- read.zoo(text = data2, FUN = times)
>> na.approx(z1, xout = time(z2))
> 01:23:42 01:23:47 01:23:51 01:23:54
> 7.0 10.8 11.0 8.0
>
Latest error:
> data1
V1 V2
1 01:23:40 5
2 01:23:45 10
3 01:23:50 12
4 01:23:55 7
> data2
V1
1 01:23:42
2 01:23:47
3 01:23:51
4 01:23:54
5 01:23:58
6 01:23:59
> data1zoo<-read.zoo(text=data1,FUN=times)
Error in textConnection(text) : invalid 'text' argument
> data2zoo<-read.zoo(text=data2,FUN=times)
Error in textConnection(text) : invalid 'text' argument
Then I tried to create objects differently:
data1zoo<-read.zoo('test1.txt',FUN=times)
> data1zoo
01:23:40 01:23:45 01:23:50 01:23:55
5 10 127
data2zoo<-read.zoo('test2.txt',FUN=times)
> data2zoo
01:23:42
01:23:47
01:23:51
01:23:54
01:23:58
01:23:59
> data3<-(na.approx(merge(data1zoo,data2zoo),time(data1zoo)))
Error in na.approx.default(object, x = x, xout = xout, na.rm = FALSE, :
x and index must have the same length
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[R] remove margin between plot and axis
Readers, Am trying to plot a graph with type 'h' and want to remove the white space between the plot lines and the x axis. The help section 'par' suggests that the option 'mai' controls this feature, but the syntax plot(...mai=c(0,1,1)) is ineffective Any advice please? -- r2151 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] remove margin between plot and axis
On 29/01/2013, Pascal Oettli wrote: > Hi, > > Please provide a reproducible example. > test<-seq(10:50) plot(test,type='h',mai=c(0,1,1)) Tried plot(test,type='h',yaxs='i') but this has the non-wanted effect of removing white space from between the highest peak and the upper (top) axis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] remove label from specific axis
Readers, For a graph plot instruction: plot(seq(10:50),type='h',yaxt='n',yaxs='i',lab=c(20,2,2),xlab='x axis label',bty='l',main='graph title') how to remove y-axis label and keep the x-axis label? _ r2151 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] convert 12 time stamp to 24 hour
Readers,
For a 12 hour time stamp:
> testtime<-("2013-01-01 01:00:01 PM")
> testtime
[1] "2013-01-01 01:00:01 PM"
> testtime24hour<-strftime(testtime,'%H:%M:%S')
> testtime24hour
[1] "01:00:01"
how to convert to 24 hour format so that the object 'testtime24hour' is:
'13:00:01'
Thanks.
--
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[R] add data symbol to axis label
Readers, Is it possible to add a data symbol (e.g. pch='21') to an axis label? The objective is to plot a graph with two (2) y-axes and the plotting character for data set 1 is added to the label of y1 axis (left); plot character for data set 2 is added to label of y2 (right) axis. Thanks. -- r2151 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] convert 12 time stamp to 24 hour
On 08/02/2013, e-letter wrote:
> Readers,
>
> For a 12 hour time stamp:
>
>> testtime<-("2013-01-01 01:00:01 PM")
>> testtime
> [1] "2013-01-01 01:00:01 PM"
>> testtime24hour<-strftime(testtime,'%H:%M:%S')
>> testtime24hour
> [1] "01:00:01"
>
Have realised that the object is of type 'character'. When this
conversion is applied to a vector and then the 'plot' function applied
(i.e. plot another vector against the vector of 24-hour time series
values), receive an error.
How to convert the values "01:00:01" to 'ts' objects?
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Re: [R] convert 12 time stamp to 24 hour
Readers, Have since tried to plot converted 24 hour data: testtimedataset V1 1 13:01:41 2 13:02:10 3 13:02:38 4 13:03:05 5 13:03259 > testdata<-seq(1:5) > plot(testdata~testtimedataset) Error in function (formula, data = NULL, subset = NULL, na.action = na.fail, : invalid type (list) for variable 'testtimedataset' Was expecting to see a graph with ordinate values 1:5 for the time series values on the abscissa. What is the mistake please? -- r2151 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] convert 12 time stamp to 24 hour
On 11/02/2013, PIKAL Petr wrote: > > str(testtimedataset) With my real data set, revealed that after conversion, the time series data consists of 'chr' which I guess means character type data, reporting the following error: Error in plot.window(...) : need finite 'xlim' values In addition: Warning messages: 1: In xy.coords(x, y, xlabel, ylabel, log) : NAs introduced by coercion 2: In min(x) : no non-missing arguments to min; returning Inf 3: In max(x) : no non-missing arguments to max; returning -Inf Does this mean I must convert the 24-hour time series values to POSIX*t type objects; if so how please? > > Anyway, by > > plot(testdata~testtimedataset$V1) > > you will get rid of error message but depending on testtimedataset mode you > can get different one. > Confirmed, thanks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] approxfun values
Readers,
According to the help '?approxfun', the function can be used to obtain
the interpolated values. The following test was tried:
> testinterpolation<-read.csv('test.csv',header=FALSE)
> testinterpolation
V1 V2
1 10 2
2 20 NA
3 30 5
4 40 7
5 50 NA
6 60 NA
7 70 2
8 80 6
9 90 9
10 100 NA
> testinterpolationvalues<-approxfun(testinterpolation,y=NULL)
> testinterpolationvalues
function (v)
.C(C_R_approxfun, as.double(x), as.double(y), as.integer(n),
xout = as.double(v), as.integer(length(v)), as.integer(method),
as.double(yleft), as.double(yright), as.double(f), NAOK = TRUE,
PACKAGE = "stats")$xout
> testinterpolationvalues<-approx(testinterpolation,y=NULL)
> testinterpolationvalues
$x
[1] 10.0 11.63265 13.26531 14.89796 16.53061 18.16327 19.79592 21.42857
[9] 23.06122 24.69388 26.32653 27.95918 29.59184 31.22449 32.85714 34.48980
[17] 36.12245 37.75510 39.38776 41.02041 42.65306 44.28571 45.91837 47.55102
[25] 49.18367 50.81633 52.44898 54.08163 55.71429 57.34694 58.97959 60.61224
[33] 62.24490 63.87755 65.51020 67.14286 68.77551 70.40816 72.04082 73.67347
[41] 75.30612 76.93878 78.57143 80.20408 81.83673 83.46939 85.10204 86.73469
[49] 88.36735 90.0
$y
[1] 2.00 2.244898 2.489796 2.734694 2.979592 3.224490 3.469388 3.714286
[9] 3.959184 4.204082 4.448980 4.693878 4.938776 5.244898 5.571429 5.897959
[17] 6.224490 6.551020 6.877551 6.829932 6.557823 6.285714 6.013605 5.741497
[25] 5.469388 5.197279 4.925170 4.653061 4.380952 4.108844 3.836735 3.564626
[33] 3.292517 3.020408 2.748299 2.476190 2.204082 2.163265 2.816327 3.469388
[41] 4.122449 4.775510 5.428571 6.061224 6.551020 7.040816 7.530612 8.020408
[49] 8.510204 9.00
How to obtain a vector consisting _only_ of the interpolated values?
It was expected that 'approx' would return both original and
interpolated values (which the above shows) and that 'approxfun' would
not show the original values.
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r2151
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[R] plot custom x axis ticks values
Readers, For this data set: testvalues<-c(10,20,30,40) How to amend the plot instruction: plot(testvalues,ann=FALSE,type='l',yaxt='n',xaxt='n') so that x axis ticks labels can be added to existing graph with arbitrary value such as 0,100,200,300)? Thanks in advance. -- r2151 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot custom x axis ticks values
On 14/02/2013, PIKAL Petr wrote: > Hi > > much quicker and better for you is to inspect help pages. > > ?axis > > axis(1, at=1:4, labels=c(letters[1:4])) Had tried, but noticed that the labels created were not positioned equally across the entire axis, but occupied about 50 % (the left half) of the axis. How to change this please? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot custom x axis ticks values
On 14/02/2013, MacQueen, Don wrote: >> plot(testvalues,ann=FALSE,type='l',yaxt='n',xaxt='n') >> par()$usr > [1] 0.88 4.12 8.80 41.20 > > The x axis range is from 0.88 to 4.12, so tick labels at 0, 100, 200, 300 > makes no sense. > True per se, but the purpose of the tick labels is to indicate approximate correlation with another data set. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot custom x axis ticks values
On 15/02/2013, Jim Lemon wrote: > On 02/14/2013 09:41 PM, e-letter wrote: >> Readers, >> >> For this data set: >> >> testvalues<-c(10,20,30,40) >> >> How to amend the plot instruction: >> >> plot(testvalues,ann=FALSE,type='l',yaxt='n',xaxt='n') >> >> so that x axis ticks labels can be added to existing graph with >> arbitrary value such as 0,100,200,300)? >> > Hi r2151, > If you want the labels to fit on the axis you will have to include this > information in the call to "plot": > > plot(testvalues,ann=FALSE,type='l',yaxt='n',xaxt='n',ylim=c(0,300)) > axis(2,at=c(0,100,200,300)) > plot(testvalues,ann=FALSE,type='l',yaxt='n',xaxt='n',xlim=c(0,300)) axis(1,at=c(0,100,200,300)) The use of 'xlim' makes the graph unacceptable. Instead, it would be better to plot: plot(testvalues,ann=FALSE,type='l',yaxt='n') The x-axis shows tick marks '1.0', '1.5', ... It is required to replace these values with custom values (e.g. 10, 20... etc.) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot custom x axis ticks values
On 15/02/2013, Duncan Murdoch wrote: > On 13-02-15 3:28 AM, e-letter wrote: >> On 15/02/2013, Jim Lemon wrote: >>> On 02/14/2013 09:41 PM, e-letter wrote: >>>> Readers, >>>> >>>> For this data set: >>>> >>>> testvalues<-c(10,20,30,40) >>>> >>>> How to amend the plot instruction: >>>> >>>> plot(testvalues,ann=FALSE,type='l',yaxt='n',xaxt='n') >>>> >>>> so that x axis ticks labels can be added to existing graph with >>>> arbitrary value such as 0,100,200,300)? >>>> >>> Hi r2151, >>> If you want the labels to fit on the axis you will have to include this >>> information in the call to "plot": >>> >>> plot(testvalues,ann=FALSE,type='l',yaxt='n',xaxt='n',ylim=c(0,300)) >>> axis(2,at=c(0,100,200,300)) >>> >> >> plot(testvalues,ann=FALSE,type='l',yaxt='n',xaxt='n',xlim=c(0,300)) >> axis(1,at=c(0,100,200,300)) >> >> The use of 'xlim' makes the graph unacceptable. Instead, it would be >> better to plot: >> >> plot(testvalues,ann=FALSE,type='l',yaxt='n') >> >> The x-axis shows tick marks '1.0', '1.5', ... >> >> It is required to replace these values with custom values (e.g. 10, 20... >> etc.) > > Why not rescale the values before plotting, and use the automatic ticks? > You can lie about the user scale, but it doesn't always give a helpful > plot, e.g. > > axis(1, at=1:4, labels=c(100, 300, 400, 200)) > Tried, but realised that this works when the vectors for 'at' and 'labels' are equal size. For a data set of e.g. 10 values, syntax error returns. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] mtext unicode failure
Readers,
How to solve this unicode input error please?
> postscript("~/tmp/test.eps",width=100/25.4,height=100/25.4,horizontal=FALSE,onefile=TRUE,paper="special")
> testx<-seq(1:5)
> testy<-seq(1:5)
> plot(testy~testx)
> mtext('text (O₂)\n more text',side=3,line=1)
Warning messages:
1: In mtext("text (O₂)\n more text", side = 3, line = 1) :
conversion failure on 'text (O₂)' in 'mbcsToSbcs': dot substituted for
2: In mtext("text (O₂)\n more text", side = 3, line = 1) :
conversion failure on 'text (O₂)' in 'mbcsToSbcs': dot substituted for <82>
3: In mtext("text (O₂)\n more text", side = 3, line = 1) :
conversion failure on 'text (O₂)' in 'mbcsToSbcs': dot substituted for <82>
4: In mtext("text (O₂)\n more text", side = 3, line = 1) :
conversion failure on 'text (O₂)' in 'mbcsToSbcs': dot substituted for
5: In mtext("text (O₂)\n more text", side = 3, line = 1) :
conversion failure on 'text (O₂)' in 'mbcsToSbcs': dot substituted for <82>
6: In mtext("text (O₂)\n more text", side = 3, line = 1) :
conversion failure on 'text (O₂)' in 'mbcsToSbcs': dot substituted for <82>
If the command to print to postscript is omitted, a graph is viewed
successfully, with the title text:
text(O₂)
more text
Thanks.
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Re: [R] mtext unicode failure
On 19/02/2013, r-help-requ...@r-project.org wrote: > -- > > Message: 22 > Subject: Re: [R] mtext unicode failure > > On 18/02/2013 14:15, e-letter wrote: >> Readers, >> >> How to solve this unicode input error please? > > On what system, in what locale? And where is Unicode mentioned? > GNU/Linux; obtained locale settings like so: Sys.localeconv() decimal_point thousands_sep grouping int_curr_symbol ".""""""GBP " currency_symbol mon_decimal_point mon_thousands_sep mon_grouping "£" "." ",""\003\003" positive_sign negative_sign int_frac_digits frac_digits "" "-" "2" "2" p_cs_precedesp_sep_by_space n_cs_precedesn_sep_by_space "1" "0" "1" "0" p_sign_posn n_sign_posn "1" "1" The unicode was in the 'mtext...' command, but it seems that the mailing list server does not accept UTF-8. Tried to set this encoding: postscript([filename],encoding='UTF-8',...) Error in postscript([filename], encoding = "UTF-8", : failed to load encoding file in postscript() In addition: Warning message: In postscript([filename], encoding = "UTF-8", : failed to load encoding file 'UTF-8' > If you intended a subscript 2, use plotmath. That is not a character in > the standard Postscript fonts, nor is in the encoding you selected. We > don't know what that is since it depends on your locale: see ?postscript. > > Or use the cairo_ps device. Thanks, the result is adequate, although the position of the subscript doesn't look pleasing personally. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] create vector from indices interpolated values
Readers, Is it possible to create a plot command based upon the indices of missing values in a data set? dataset1<-read.table(text=' 10 2 20 NA 30 5 40 7 50 NA 60 NA 70 2 80 6 90 NA 100 9 ') dataset2<-read.table(text=' 0.2 0.4 0.1 0.9 0.2 0.3 1.1 0.7 0.9 0.6 0.4 ') The 'approx' function is used to obtain the interpolated values for 'NA' in dataset1. dataset1interpolatedvalues<-approx(dataset1,y=NULL,xout=dataset1$V1[is.na(dataset1$V2)]) dataset1interpolatedvalues $x [1] 20 50 60 90 $y [1] 3.50 5.33 3.67 7.50 x<-dataset1interpolatedvalues$y How to create a vector 'y' by selecting the values in 'dataset2' using the indices equivalent to interpolated values in 'dataset1' (i.e. indices 2, 5, 6, 9 in dataset1)? The result of creating 'y' should be y 0.4 0.2 0.3 0.9 Then the desired plot would be plot(y~x) Thanks. -- r2151 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] create function to solve derivative
On 16/11/2012, Rolf Turner wrote: > > Your question makes little sense. Functions have derivatives --- at > least some of them do. Data sets do not have derivatives. The > functions D(), deriv() etc. work on specified analytic expressions > for functions --- data sets do not come into the picture. > Is the following procedure correct: Plot data from data set. Suppose the resultant graph was linear. The function could be y=mx+c, so the R function 'lm' could be applied to fit a linear line. The example in the manual for the help file '?D' shows: dx2x <- deriv(~ x^2, "x") ; dx2x So for this example it would be correct to write: functionname<-deriv(~mx+c,"x");dx2x What is the significance of ';dx2x'? It would be appreciated to be referred to the relevant manual sections, thanks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] apply calculations to elements in a vector
Readers, If a vector consists of: 10 20 30 how to create a new vector based upon the results of calculations to the elements, e.g. addition of successive elements, so that a new vector would be: 30 50 i.e. 10+20, then 20+30, etc.? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] column name prefix 'x' after data import via read.delim
Readers, The function 'read.delim' was used to import data into R: columnnamea columnnameb columnnamec 1 2 3 2 3 4 3 4 5 After import, the column names were: X.columnnamea columnnameb columnnamec Why did this occur? -- r2151 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] remove last row of a data frame
Readers, For a data set 'a': 1 2 3 4 Please what is the syntax to remove the last row and create a new object 'b': 1 2 3 Thanks. -- R2151 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] remove last row of a data frame
On 12/12/2012, Daniel Nordlund wrote: >> -Original Message- >> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] >> On Behalf Of e-letter >> Sent: Tuesday, December 11, 2012 11:45 PM >> To: r-help@r-project.org >> Subject: [R] remove last row of a data frame >> >> Readers, >> >> For a data set 'a': >> >> 1 >> 2 >> 3 >> 4 >> >> Please what is the syntax to remove the last row and create a new object >> 'b': >> >> 1 >> 2 >> 3 >> >> Thanks. >> > > If by data set a you mean a data frame called a, then something like this > should work: > > b <- a[-nrow(a),] > > If you haven't already read the manual, "An Introduction to R", that ships > with every copy of R, then now is the time. > Thanks, couldn't find quickly the relevant section in the html document, but within R, the command 'nrow' is relevant. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] abbreviate function using 'with'
Readers,
A csv file was created:
column1,column2,column3,column4
1,10,3,2
2,20,6,4
3,30,12,16
4,40,24,256
The csv was imported:
testcsv<-read.csv('/path/to/test.csv')
testsum<-testcsv[2,2]+testcsv[2,3]+testcsv[2,4]
What is the correct syntax to abbreviate the following command using
the function 'with', to avoid repetitive use of the declaration
'testcsv'? Tried the following, but received the error shown:
with(testcsv,testsum<-[2,2]+[2,3]+[2,4])
Error: unexpected '[' in "with(testcsv,testsum<-["
and:
with(testcsv,{testsum<-[2,2]+[2,3]+[2,4]})
Error: unexpected '[' in "with(testcsv,{testsum<-["
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Re: [R] abbreviate function using 'with'
On 13/02/2014, Rolf Turner wrote:
>
> What you've written is simply not (anything like!) R syntax. You should
> learn to speak R if you are going to use R.
>
Agree; was reviewing the help text examples invoked by '?with'.
> In this particular instance
>
> testsum <- sum(testcsv[2,2:4])
>
> should give what you want. The use of with() is uncalled for in this
> context. The with() function allows you to refer to (e.g.) columns of
> a data frame by name, as if these columns were objects in your workspace
> ("global environment"). That is *not* what you are doing, or need to do
> here.
>
Forgive the example. The objective is to use the function 'with' to
refer to specific indices of a dataframe. In retrospect, the example
given was poor because actually, I want to understand the syntax to
specify particular indices within a single dataframe, without having
to state the dataframe repeatedly.
>
> P. S.: Please read fortune("people who don't exist") and change your
> modus operandi.
>
Never heard of fortune, but understand the latin:
http://www.michaeleisen.org/blog/?p=1554
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[R] Use generalised additive model to plot curve
Readers,
I have been reading 'the r book' by Crawley and think that the
generalised additive model is appropriate for this problem. The
package 'gam' was installed using the command (as root)
install.package("gam")
...
library(gam)
> library(gam)
Loading required package: splines
Loading required package: akima
> library(mgcv)
This is mgcv 1.3-25
Attaching package: 'mgcv'
The following object(s) are masked from package:gam :
gam,
gam.control,
gam.fit,
plot.gam,
predict.gam,
s,
summary.gam
> x<-c(30,50,80,90,100)
> y<-c(160,180,250,450,300)
> model<-gam(y~s(x))
Error in smooth.construct.tp.smooth.spec(object, data, knots) :
A term has fewer unique covariate combinations than specified
maximum degrees of freedom
The objective is to plot y against x, finally to produce a graph with
a smooth curve (and then remove the data points). What is my mistake
please?
yours,
r251
gnu/linux mandriva2008
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[R] dotchart for matrix data
Readers, I am trying to use the function dotchart. The data is: > testdot category values1 values2 values3 values4 1a 10 27 56 709 2b 4 46 47 208 3c 5 17 18 109 4d 6 50 49 308 The following error occurs > dotchart(testdot,groups=testdot[,2]) Error in dotchart(testdot, labels = testdot[, 1], groups = testdot[, 2]) : 'x' must be a numeric vector or matrix According to my understanding (clearly wrong!) of the documentation for dotchart (accessed from the manual in section 'graphics'), columns of data can be selected by 'groups' for subsequent plotting. The objective is to be able to create a dot chart where each row is labelled according to the row names in the 'category' column and two columns can be selected, e.g. 'values1' and 'values2'. Then I tried: > testdot1<-testdot[,1] > testdot2<-testdot[,2] > testdot3<-testdot[,3] > dotchart(c(testdot2,testdot3),labels=testdot1) A graph is produced, but not as expected. Instead of 4 rows labelled (descending order from top row) a,b,c,d, and each row containing two data points, the graph shows 8 rows (?) with the top 4 rows un-labelled and the bottom 4 rows labelled (descending order) d,c,b,a and each row shows only 1 datum point. How do I specify the order of labelling of the rows? How do I write correct commands to obtain a dot chart with 4 rows, each row containing the two (or if required three) data points? Thanks in advance. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] dotchart for matrix data
On 18/12/2010, David Winsemius wrote: > > On Dec 18, 2010, at 7:01 AM, e-letter wrote: > >> Readers, >> >> I am trying to use the function dotchart. The data is: >> >>> testdot >> category values1 values2 values3 values4 >> 1a 10 27 56 709 >> 2b 4 46 47 208 >> 3c 5 17 18 109 >> 4d 6 50 49 308 >> >> The following error occurs >> >>> dotchart(testdot,groups=testdot[,2]) >> Error in dotchart(testdot, labels = testdot[, 1], groups = testdot[, >> 2]) : >>'x' must be a numeric vector or matrix >> >> According to my understanding (clearly wrong!) of the documentation >> for dotchart (accessed from the manual in section 'graphics'), columns >> of data can be selected by 'groups' for subsequent plotting. > > The misunderstanding is in how you see the grouping information versus > how R expects it. R generally expects such data in what is called > "long" format, i.e. there will be one values columns and a category > column. There are various ways to change the arrangement of your data. > The function stack(), the function reshape(), or probably most > commonly the function melt from reshape2 being the typical chosen > routes. > Reshape and melt are not installed (version251) so for this task manual rearrangement data is easier. > require(reshape) Loading required package: reshape [1] FALSE Warning message: there is no package called 'reshape' in: library(package, lib.loc = lib.loc, character.only = TRUE, logical = TRUE, > library(reshape) Error in library(reshape) : there is no package called 'reshape' > mdot<-melt(dot) Error: could not find function "melt" However before doing so why this is relevant because of the alternative creation objects 'testdot1'. Aren't these objects suitable, since a (undesireable) graph was produced? >> The >> objective is to be able to create a dot chart where each row is >> labelled according to the row names in the 'category' column and two >> columns can be selected, e.g. 'values1' and 'values2'. Then I tried: >> >>> testdot1<-testdot[,1] >>> testdot2<-testdot[,2] >>> testdot3<-testdot[,3] >>> dotchart(c(testdot2,testdot3),labels=testdot1) > > See if this is more to your liking: > > require(reshape) # I'm not sure why I have reshape_0.8.3 rather than > reshape2 loaded > # I'm pretty sure Hadley would prefer that people > use pkg:reshape2 > mdot <- melt(dot) > dotchart(mdot$value, groups=mdot$category, labels=mdot$variable) > # OR more readable > with(mdot, dotchart(value, groups=category, labels=variable) ) > > I'm not sure I got the roles of "values" and "category" correct, but > it should be a simple matter to switch them in the dotcghart call if > that is your pleasuRe. > I don't have dotcghart either. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] dotchart for matrix data
>Ben Bolker
>Sat, 18 Dec 2010 07:07:24 -0800
>David Winsemius comcast.net> writes:
>
>>
>>
>> On Dec 18, 2010, at 7:01 AM, e-letter wrote:
>>
>> > Readers,
>> >
>> > I am trying to use the function dotchart. The data is:
>> >
>> >> testdot
>> > category values1 values2 values3 values4
>> > 1a 10 27 56 709
>> > 2b 4 46 47 208
>> > 3c 5 17 18 109
>> > 4d 6 50 49 308
>> >
>> > The following error occurs
>> >
>> >> dotchart(testdot,groups=testdot[,2])
>> > Error in dotchart(testdot, labels = testdot[, 1], groups = testdot[,
>> > 2]) :
>> >'x' must be a numeric vector or matrix
>> >
>> > According to my understanding (clearly wrong!) of the documentation
>> > for dotchart (accessed from the manual in section 'graphics'), columns
>> > of data can be selected by 'groups' for subsequent plotting.
>>
>
> Following up on David's response:
>
>
>d <- read.table(textConnection("category values1 values2 values3 values4
>1a 10 27 56 709
>2b 4 46 47 208
>3c 5 17 18 109
>4d 6 50 49 308"),
>header=TRUE)
>
>## Something like this is probably as close as you can get with
>## stock 'dotchart' -- it does *not* (as far as I can tell) put
>## different points on the same line, just groups lines
>
I am trying to create a chart like this
(http://www.b-eye-network.com/images/content/Fig4_3.jpg); so this is
not possible using R?
>dotchart(as.matrix(d[,-1]),labels=as.character(d[,1]))
>dotchart(as.matrix(d[,c("values1","values2")]),labels=as.character(d[,1]))
>
>## reshaping data:
>library(reshape)
>mdot <- melt(d)
>
>## using the lattice package
>
>library(lattice)
>dotplot(value~category,groups=variable,data=mdot)
>dotplot(value~variable,groups=category,data=mdot,auto.key=TRUE,
>scales=list(y=list(log=10)))
>
>## you could also use ggplot2 ...
> ?ggplot2
No documentation for 'ggplot2' in specified packages and libraries:
you could try 'help.search("ggplot2")'; seems I need to retrieve this
package first. Thanks for the suggestion.
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Re: [R] dotchart for matrix data
On 18/12/2010, Peter Ehlers wrote: > On 2010-12-18 07:50, e-letter wrote: >>> Ben Bolker >>> Sat, 18 Dec 2010 07:07:24 -0800 > > [... snip ...] > >> I am trying to create a chart like this >> (http://www.b-eye-network.com/images/content/Fig4_3.jpg); so this is >> not possible using R? > > That looks an awful lot like what lattice's dotplot would > produce. So: have you tried dotplot() as Ben has suggested? > Unfortunately I have been unable to use the "melt" package; 3 mirrors have failed to install. What I don't understand is why the objects created are not in the correct format, otherwise how would the plot be created? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] dotchart for matrix data
On 18/12/2010, Peter Ehlers wrote:
> On 2010-12-18 07:50, e-letter wrote:
>>> Ben Bolker
>>> Sat, 18 Dec 2010 07:07:24 -0800
>
> [... snip ...]
>
>> I am trying to create a chart like this
>> (http://www.b-eye-network.com/images/content/Fig4_3.jpg); so this is
>> not possible using R?
>
> That looks an awful lot like what lattice's dotplot would
> produce. So: have you tried dotplot() as Ben has suggested?
>
For the benefit of other novices, this is what I did (gnu/linux):
sign-in to a command terminal as root
start R
install.packages("reshape2")
In another terminal as normal user
require(reshape2)
mdot<-melt(testdot)
dotplot(value~category,groups=variable,data=mdot)
The resultant graph shows the categories on the abscissa and there
does not seem to be a way of selecting two columns from the original
matrix. It is not like the graph cited in the hyperlink quoted above,
but at least I have successfully followed instructions!
I changed the axes with the command:
dotplot(category~value,groups=variable,data=mdot)
I believe there is a command to select only certain rows of data which
I think will achieve the desired graph, if unable I'll ask again, so
thank you all.
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Re: [R] dotchart for matrix data
On 18/12/2010, e-letter wrote: > On 18/12/2010, Peter Ehlers wrote: >> On 2010-12-18 07:50, e-letter wrote: >>>> Ben Bolker >>>> Sat, 18 Dec 2010 07:07:24 -0800 >> >> [... snip ...] >> >>> I am trying to create a chart like this >>> (http://www.b-eye-network.com/images/content/Fig4_3.jpg); so this is >>> not possible using R? >> >> That looks an awful lot like what lattice's dotplot would >> produce. So: have you tried dotplot() as Ben has suggested? >> > If one set of value ranges for 10-20 and another set ranges form 1000-1500, how to adjust the graph such that: with categories on the ordinate (y-axis), can the bottom abscissa (x-axis bottom) be set with a scale suitable for the data set of range 1000-1500? can the top abscissa (x-top) be set with a scale 10-20? or is it better practice to change the scale of the data set 1000-1500 by two orders of magnitude? how is it possible to control the order of the category variables? For example, if the graph shows: a b c d Is it possible to change to a d b c Thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] dotchart for matrix data
On 20/12/2010, e-letter wrote: > On 18/12/2010, e-letter wrote: >> On 18/12/2010, Peter Ehlers wrote: >>> On 2010-12-18 07:50, e-letter wrote: >>>>> Ben Bolker >>>>> Sat, 18 Dec 2010 07:07:24 -0800 >>> >>> [... snip ...] >>> >>>> I am trying to create a chart like this >>>> (http://www.b-eye-network.com/images/content/Fig4_3.jpg); so this is >>>> not possible using R? >>> >>> That looks an awful lot like what lattice's dotplot would >>> produce. So: have you tried dotplot() as Ben has suggested? >>> >> > A csv file is imported: > testdot category values 1a 10 2b 44 3c 51 4d 65 5a 27 6b 64 7c 71 8d 49 A dotplot is produced with the command: dotplot(category~values,data=testdot) The csv file is adjusted to move the values of 'a': > testdot category values 1b 44 2c 51 3d 65 4a 10 5b 64 6c 71 7d 49 8a 27 Why does the order of the categories not change from d,c,b,a to a,d,c,b (descending from top)? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] dotchart for matrix data
Readers,
The following commands were applied, to create a dot chart with black
dots and blue squares for data:
> library(lattice)
> testdot
category values
1b 44
2c 51
3d 65
4a 10
5b 64
6c 71
7d 49
8a 27
dotplot(category~values,col=c("black","black","black","black","blue","blue","blue","blue"),bg=c("black","black","black","black","blue","blue","blue","blue"),pch=c(21,21,21,21,22,22,22,22),xlab=NULL,
data=testdot)
The resultant graph shows correctly coloured points, but not filled,
only the border is coloured. The documentation for the command 'pch'
(?pch) indicates that the commands shown above should show
appropriately coloured solid symbols. What is causing this error
please?
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[R] percentage sign in expression
Readers, According to the documentation for the function 'plotmath' there is no apparent possibility to add the percent sign (%) to a plot function, e.g. plot(a[,1]~b[,2],ylab=expression(x~%),xlab=expression(z)) How to achieve this please? yours, rh...@conference.jabber.org r251 mandriva2009 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] percentage sign in expression
On 24/08/2010, Henrique Dallazuanna wrote: > You've tried: > > plot(a[,1]~b[,2],ylab=expression(x~'%'),xlab=expression(z)) ? > This is successful for me, thank you. This instruction should be added to the documentation for 'plotmath'. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] percentage sign in expression
On 24/08/2010, David Winsemius wrote:
>
> On Aug 24, 2010, at 9:37 AM, e-letter wrote:
>
>> Readers,
>>
>> According to the documentation for the function 'plotmath' there is no
>> apparent possibility to add the percent sign (%) to a plot function,
>
> Where did you see an assertion made???
>
Within R I entered the command:
?plotmath
Also accessed using:
help.start(browser="opera")
Navigating the web browser page:
packages
packages in /usr/lib/R/library
grdevices
plotmath
In the list headed 'syntax' and 'meaning' within the section 'details'.
>> e.g.
>>
>> plot(a[,1]~b[,2],ylab=expression(x~%),xlab=expression(z))
>
>
>>
>> How to achieve this please?
>
> Read the plotmath helo page more carefully. The section immediatedly
> below the plotmath expressions points you to the use of the symbol()
> expression-function and to the points help page where generation of
> the available glyphs proceeds according to the advice on help(plotmath):
>
In my system the paragraph immediately after the list of features
(i.e. 'syntax','meaning') describes a note to TeX users. I cannot see
reference to 'symbol()'.
> > TestChars <- function(sign=1, font=1, ...)
> + {
> +if(font == 5) { sign <- 1; r <- c(32:126, 160:254)
> +} else if (l10n_info()$MBCS) r <- 32:126 else r <- 32:255
> +if (sign == -1) r <- c(32:126, 160:255)
> +par(pty="s")
> +plot(c(-1,16), c(-1,16), type="n", xlab="", ylab="",
> + xaxs="i", yaxs="i")
> +grid(17, 17, lty=1)
> +for(i in r) try(points(i%%16, i%/%16, pch=sign*i, font=font,...))
> + }
> > TestChars(font=5)
>
> Notice that the "%" sign is three characters to the right (i.e.
> higher) of the "forall" symbol that is produced by the example code
I can't see 'forall' in the code above.
> they offer. (The numbering proceeds from bottom up which confused me
> at first.)
>
What numbering?
The documentation makes reference to the command:
demo(plotmath)
I applied this command and could not see an instruction to produce the
percent (%) symbol.
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Re: [R] percentage sign in expression
>On Wed, 2010-08-25 at 09:32 +0100, e-letter wrote:
>> On 24/08/2010, David Winsemius wrote:
>> >
>> > On Aug 24, 2010, at 9:37 AM, e-letter wrote:
>> >
>> >> Readers,
>> >>
>> >> According to the documentation for the function 'plotmath' there is no
>> >> apparent possibility to add the percent sign (%) to a plot function,
>> >
>> > Where did you see an assertion made???
>> >
>> Within R I entered the command:
>>
>> ?plotmath
>
>Surely you don't expect the R help pages to document everything that
>*isn't* possible with a given function? They'd be infinitely long! ;-)
>
No, following previous advice it would be preferable to add the
following (or similar) to the documentation:
'syntax' 'meaning'
...
'%' %
...
>What you meant to say was, "?plotmath doesn't show me how to add a %
>symbol to a plot. How do I do it?"
>
>> Also accessed using:
>>
>> help.start(browser="opera")
>>
>> Navigating the web browser page:
>>
>> packages
>> packages in /usr/lib/R/library
>> grdevices
>> plotmath
>>
>> In the list headed 'syntax' and 'meaning' within the section 'details'.
>>
>
>> > > TestChars <- function(sign=1, font=1, ...)
>> > + {
>> > +if(font == 5) { sign <- 1; r <- c(32:126, 160:254)
>> > +} else if (l10n_info()$MBCS) r <- 32:126 else r <- 32:255
>> > +if (sign == -1) r <- c(32:126, 160:255)
>> > +par(pty="s")
>> > +plot(c(-1,16), c(-1,16), type="n", xlab="", ylab="",
>> > + xaxs="i", yaxs="i")
>> > +grid(17, 17, lty=1)
>> > +for(i in r) try(points(i%%16, i%/%16, pch=sign*i, font=font,...))
>> > + }
>> > > TestChars(font=5)
>> >
>> > Notice that the "%" sign is three characters to the right (i.e.
>> > higher) of the "forall" symbol that is produced by the example code
>>
>> I can't see 'forall' in the code above.
>
>You need to run it in R. Then you will see a plot of the glyphs
>available in the symbol font. 'forall' is a mathematical symbol:
>
>http://en.wikipedia.org/wiki/Table_of_mathematical_symbols
>
>> > they offer. (The numbering proceeds from bottom up which confused me
>> > at first.)
>> >
>> What numbering?
>
>The numbering of the glyphs so you can use the number to draw the symbol
>you want. They are on the plot. You did run the code provided by David?
>
I did not read any instruction that it was code to apply! Anyway,
below is the result:
> TestChars <- function(sign=1, font=1, ...)
+ + {
+ +if(font == 5) { sign <- 1; r <- c(32:126, 160:254)
+ +} else if (l10n_info()$MBCS) r <- 32:126 else r <- 32:255
Error: syntax error, unexpected '}' in:
"
"
> +if (sign == -1) r <- c(32:126, 160:255)
Error in sign == -1 : comparison (1) is possible only for atomic and list types
> +par(pty="s")
Error in +par(pty = "s") : invalid argument to unary operator
> +plot(c(-1,16), c(-1,16), type="n", xlab="", ylab="",
+ + xaxs="i", yaxs="i")
Error: syntax error, unexpected EQ_ASSIGN, expecting ',' in "+
plot(c(-1,16), c(-1,16), type="n", xlab="", ylab="","
> +grid(17, 17, lty=1)
Error in int_abline(a = a, b = b, h = h, v = v, untf = untf, ...) :
plot.new has not been called yet
> +for(i in r) try(points(i%%16, i%/%16, pch=sign*i, font=font,...))
Error: object "r" not found
> + }
Error: syntax error, unexpected '}' in "+ }"
> TestChars(font=5)
Error: could not find function "TestChars"
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Re: [R] percentage sign in expression
>It's possible that my help page is different than yours. Right after
>the syntax/meaning description on mine (which is a Mac OSX system) is
>a paragraph:
>
>"The symbol font uses Adobe Symbol encoding so, for example, a lower
>case mu can be obtained either by the special symbol mu or by
>symbol("m"). This provides access to symbols that have no special
>symbol name, for example, the universal, or forall, symbol is
>symbol("\042"). To see what symbols are available in this way
>useTestChars(font=5) as given in the examples for points: some are
>only available on some devices."
>
>(In this case I would be surprised if the help pages were different
>because this makes a cross-reference to the examples in points. I am
>not surprised about cross-platform differences in descriptions of
>graphical devices and would have included a caveat if I were
>corresponding on rhelp about such. I suppose the font issues could be
>platform specific so if you want to correct me on this point, I will
>try to file it away. I did, however, give you the code needed to to
>display Symbols and it sounds further on that it succeeded)
>
I agree and seems that in this example that the documentation for
graphical devices is platform dependent. This should be stated in the
manual if not already. Please see previous post for the errors trying
to apply the commands provided.
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[R] accessing a data frame with row names
Readers,
I have entered a file into r:
,column1,column2
row1,0.1,0.2
row2,0.3,0.4
using the command:
dataframe<-read.table("/path/to/file.csv",header=T,row.names=1)
When I try the command:
dataframe[,2]
I receive the response:
NULL
I was expecting:
row1 0.2
row2 0.4
What is my error with the syntax please?
Yours,
r251
mandriva2009
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Re: [R] accessing a data frame with row names
On 31/05/2010, Gabor Grothendieck wrote: > Use read.csv or read.table(..., sep = ","). Also note that if you > delete the first comma of the header (as in the second example below) > you won't have to specify row.names since it can figure it out from > the fact that there is one fewer column name than data fields. > >> Lines <- ",column1,column2 > + row1,0.1,0.2 > + row2,0.3,0.4" >> >> read.csv(textConnection(Lines), row.names = 1) > column1 column2 > row1 0.1 0.2 > row2 0.3 0.4 Thank you. When I enter the command: max(dataframe[,2]) The response is: [1] 0.4 But I want to receive the row name, i.e.: [1] row2 0.4 Is this possible? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] data extrapolation function
Readers,
Data was imported using the read csv command:
dataimport<-read.csv("/path/to/dataimport.csv")
10,2000
12,2001
13,2002
15,2003
17,2004
Using the help contents for 'predict.lm' (i.e. ?predict.lm) a new data
frame was created
dataimportextra<-data.frame(x=seq(1990,2010,1))
predict(lm(dataimport),dataimportextra[,2],se.fit=TRUE)
write.csv<-(dataimportextraout,"/path/to/dataimportextra.csv")
I was expecting to see in the file dataimportextra.csv something like:
1,1990
2,1991
3,1993
...
to previously known data
10,2000
...
final extrapolated value, e.g.
20,2010
I didn't ; this suggests that I chose the wrong function! Can someone
please advise me of the correct function to use for this extrapolation
task?
yours,
r251
mandriva2009
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[R] create data set from selection of rows
Readers, For a data set: text1,23,text2,45 text1,23,text3,78 text1,23,text3,56 text1,23,text2,45 The following command was entered: datasubset<-data.frame(dataset[,3]=="text3") The result of datasubset is TRUE TRUE The required result is text1,23,text3,78 text1,23,text3,56 What is the correct command to use please? Thanks in advance. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] create data set from selection of rows
On 15/03/2011, Francisco Gochez wrote: > Hi, > > What you are after is: > > datasubset <- dataset[ dataset[,3] == "text3", ] Thank you. For the set text1,23,text2,45 text1,23,text3,78 text1,23,text3,56 text1,23,text2,45 Is it possible to write a function that selects rows containing 'text3' and applies the function 'sum' to values '78' and '56'? The control statements described in the document 'an introduction to r' (Venables and Smith, 2010) suggest that the if statement would return 'true', to prevent a sum function being applied to 'true' results. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] histogram of time-stamp data
Readers,
A data set consists of time-stamp values:
00:00:00
23:11:00
06:22:00
The data set was imported:
timestamps<-read.table("path/to/timestampsvalues")
hist(timestamps)
...error... x must be numeric
Then tried:
plot(timestamps).
How to adjust the graph to create a histogram where the intervals
ranges can be specified, e.g. intervals of 60 minutes?
Thanks in advance.
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Re: [R] histogram of time-stamp data
On 16/07/2012, r-help-requ...@r-project.org
wrote:
> --
>
> Message: 77
> Date: Mon, 16 Jul 2012 10:48:39 +0100
> From: Rui Barradas
> To: e-letter
> Cc: r-help@r-project.org
> Subject: Re: [R] histogram of time-stamp data
> Message-ID: <5003e377.3000...@sapo.pt>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
>
> timestamps <- as.POSIXct(Sys.Date()) + sample(24*60*60, 1e3, TRUE)
>
Why is it necessary to apply the sample to the current date? Looking
at the dataframe, I noticed that values have been changed:
file 'test.txt':
12:19:00
09:30:00
16:56:00
01:56:00
10:44:00
10:31:00
02:14:00
05:05:00
12:52:00
21:50:00
R command terminal input:
> timestamps<-read.table("test.txt")
> timestamps
V1
1 12:19:00
2 09:30:00
3 16:56:00
4 01:56:00
5 10:44:00
6 10:31:00
7 02:14:00
8 05:05:00
9 12:52:00
10 21:50:00
> timestamps <- as.POSIXct(Sys.Date()) + sample(24*60*60, 1e3, TRUE)
> write.csv(timestamps,file="test1.txt")
test1.txt:
"","x"
"1",2012-07-16 04:52:48
"2",2012-07-16 21:21:28
"3",2012-07-16 18:58:27
"4",2012-07-16 22:17:25
"5",2012-07-16 11:13:52
"6",2012-07-16 03:17:35
"7",2012-07-16 02:14:17
"8",2012-07-16 14:18:27
"9",2012-07-16 14:39:16
Why is this happening?
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Re: [R] R-help Digest, Vol 113, Issue 19
On 16/07/2012, r-help-requ...@r-project.org
wrote:
> --
>
> Message: 77
> Date: Mon, 16 Jul 2012 10:48:39 +0100
> From: Rui Barradas
> To: e-letter
> Cc: r-help@r-project.org
> Subject: Re: [R] histogram of time-stamp data
> Message-ID: <5003e377.3000...@sapo.pt>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
>
> timestamps <- as.POSIXct(Sys.Date()) + sample(24*60*60, 1e3, TRUE)
>
Why is it necessary to apply the sample to the current date? Looking
at the dataframe, I noticed that values have been changed:
file 'test.txt':
12:19:00
09:30:00
16:56:00
01:56:00
10:44:00
10:31:00
02:14:00
05:05:00
12:52:00
21:50:00
R command terminal input:
> timestamps<-read.table("test.txt")
> timestamps
V1
1 12:19:00
2 09:30:00
3 16:56:00
4 01:56:00
5 10:44:00
6 10:31:00
7 02:14:00
8 05:05:00
9 12:52:00
10 21:50:00
> timestamps <- as.POSIXct(Sys.Date()) + sample(24*60*60, 1e3, TRUE)
> write.csv(timestamps,file="test1.txt")
test1.txt:
"","x"
"1",2012-07-16 04:52:48
"2",2012-07-16 21:21:28
"3",2012-07-16 18:58:27
"4",2012-07-16 22:17:25
"5",2012-07-16 11:13:52
"6",2012-07-16 03:17:35
"7",2012-07-16 02:14:17
"8",2012-07-16 14:18:27
"9",2012-07-16 14:39:16
Why is this happening?
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Re: [R] histogram of time-stamp data
On 16/07/2012, Rui Barradas wrote: > Hello, > > Em 16-07-2012 22:45, e-letter escreveu: >> On 16/07/2012, r-help-requ...@r-project.org >>wrote: >> > -- >> > >> > Message: 77 >> > Date: Mon, 16 Jul 2012 10:48:39 +0100 >> > From: Rui Barradas >> > To: e-letter >> > Cc: r-help@r-project.org >> > Subject: Re: [R] histogram of time-stamp data >> > Message-ID: <5003e377.3000...@sapo.pt> >> > Content-Type: text/plain; charset=ISO-8859-1; format=flowed >> > >> > >> > timestamps <- as.POSIXct(Sys.Date()) + sample(24*60*60, 1e3, TRUE) >> > >> >> Why is it necessary to apply the sample to the current date? Looking >> at the dataframe, I noticed that values have been changed: > > No! That instruction is just to create a data example with more > date/time values, in this case with a total of 1e3 different values. > What's important is the way to plot the histogram, namely, the cut() > with two example time periods, and that hist() needs numbers, not cut's > levels. > With the original data provided, R reports an error, that "'x' must be numeric'. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] histogram of time-stamp data
On 17/07/2012, Rui Barradas wrote: > Hello, > > That's not the error I've had. You must be aware that read.table creates > a data.frame and therefore the object 'timestamps' is NOT holding time > stamps, it's holding a vector, 'V1', of time stamps. > Was not aware of the significance of the data frame of vector values. > ..., stringsAsFactors=FALSE) > Can confirm the success of adding this parameter to the command 'read.table' > timestamps <- as.POSIXct(timestamps$V1, format="%H:%M:%S") # here > > h1 <- cut(timestamps, breaks="hour") > > And the rest works. > Confirmed, thanks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] knapsack problem limitation
Readers, Attempting to solve the knapsack problem (e.g. see: http://rosettacode.org/wiki/Knapsack_problem/Unbounded), the following error occurred: ... result would be too long a vector Is this indicative that R is not suitable to solve this problem when combinations are large? Is there a known limit? Is there an alternative, or better to use a different program? Thanks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to solve a power series linear coefficient equation
Readers,
I am having difficulty understanding how to enter commands into r.
I have data arranged as:
100, 200, 300
5.6, 6.7, 7.8
8.9, 9.0, 0.1
1.2, 2.3, 3.4
The data is saved in csv format and I use the command 'read.table' to
import into r.
The values 5.6...3.4 are a function of values 100,...300, i.e.
100,...300 are independent variables (x). The function is a power
series:
y=a+bx+cx^2
How do I obtain the values of a,b,c by minimisation of the sum of
squares of deviations?
So far looking at the documentation 'r-intro' section 10, I tried the following:
functionname<-function(a,b,c){
+ a<-sum(a)
+ b<-sum(b)
+ c<-sum(c)
+ y<-(a+bx+cx^2)
+ x<-(x)
+ }
The documentation states: "...using a call such as...". What does this mean?
Yours,
[EMAIL PROTECTED]
gnu/linux mandriva 2008
r 251 (27-06-07)
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[R] Inconsistent linear model calculations
Readers, Using version 251 I tried the following command: lm(y~a+b,data=datafile) Resulting in, inter alia: ... coefficients (intercept) a 1.2 3.4 Packages installed: acepack ace() and avas() for selecting regression transformations adlift An adaptive lifting scheme algorithm akima Interpolation of irregularly spaced data alr3Methods and data to accompany Applied Linear Regression 3rd edition car Companion to Applied Regression codaOutput analysis and diagnostics for MCMC drc Analysis of dose-response curves EbayesThreshEmpirical Bayes thresholding and related methods emplik Empirical likelihood ratio for censored/truncated data gllmGeneralised log-linear model glmcFitting Generalized Linear Models Subject to Constraints glmmML Generalized linear models with clustering glmpath L1 Regularization Path for Generalized Linear Models and Cox Proportional Hazards Model HydroMe Estimation of Soil Hydraulic Parameters from Experimental Data JGR JGR - Java Gui for R lmtest Testing Linear Regression Models locfit Local Regression, Likelihood and Density Estimation. mvtnorm Multivariate Normal and t Distributions NISTnls Nonlinear least squares examples from NIST nlmeLinear and Nonlinear Mixed Effects Models nlstoolsTools for nonlinear regression diagnostics nlt A nondecimated lifting transform for signal denoising nlts(non)linear time series analysis nnlsThe Lawson-Hanson algorithm for non-negative least squares (NNLS) plotrix Various plotting functions quantregQuantile Regression Rcmdr ** No title available (pre-2.0.0 install?) ** sandwichRobust Covariance Matrix Estimators SparseM Sparse Linear Algebra sspline Smoothing Splines on the Sphere zoo Z's ordered observations Packages in library '/usr/lib/R/library': baseThe R Base Package bootBootstrap R (S-Plus) Functions (Canty) class Functions for Classification cluster Cluster Analysis Extended Rousseeuw et al. codetools Code Analysis Tools for R datasetsThe R Datasets Package foreign Read Data Stored by Minitab, S, SAS, SPSS, Stata, Systat, dBase, ... graphicsThe R Graphics Package grDevices The R Graphics Devices and Support for Colours and Fonts gridThe Grid Graphics Package KernSmooth Functions for kernel smoothing for Wand & Jones (1995) lattice Lattice Graphics MASSMain Package of Venables and Ripley's MASS methods Formal Methods and Classes mgcvGAMs with GCV smoothness estimation and GAMMs by REML/PQL nlmeLinear and Nonlinear Mixed Effects Models nnetFeed-forward Neural Networks and Multinomial Log-Linear Models quadprogFunctions to solve Quadratic Programming lattice Lattice Graphics MASSMain Package of Venables and Ripley's MASS methods Formal Methods and Classes mgcvGAMs with GCV smoothness estimation and GAMMs by REML/PQL nlmeLinear and Nonlinear Mixed Effects Models nnetFeed-forward Neural Networks and Multinomial Log-Linear Models quadprogFunctions to solve Quadratic Programming Problems. rcompgenCompletion generator for R rpart Recursive Partitioning spatial Functions for Kriging and Point Pattern Analysis splines Regression Spline Functions and Classes stats The R Stats Package stats4 Statistical Functions using S4 Classes survivalSurvival analysis, including penalised likelihood. tcltk Tcl/Tk Interface tools Tools for Package Development utils The R Utils Package When using version 171 I entered the same command: lm(y~a+b,data=datafile
Re: [R] Inconsistent linear model calculations
2008/5/15 Douglas Bates <[EMAIL PROTECTED]>: > Did you happen to notice the part at the bottom of every message about > "provide commented, minimal, self-contained, reproducible code"? > Sorry, don't understand what that statement means in my context. > Considering that the result you quote from "251" has 2 coefficients > and the result from "171" has 3 coefficients one might contemplate the > possibility that you are fitting different models or perhaps using > different data. However we can't verify anything about the causes > because we have no data regarding the problem. data below: yabcd e f 1 300 39.87 39.85 39.90 39.87333 9 2 400 45.16 45.23 45.17 45.18667 16 3 500 50.72 51.03 50.90 50.88333 25 4 600 56.85 56.80 57.02 56.89000 36 5 700 63.01 63.09 63.14 63.08000 49 6 800 69.52 59.68 69.63 66.27667 64 so substitute a in my generic equation below for data column e (e.g. 39.87...) and b for f (e.g. 9e4). > > Perhaps you are Brian Ripley's evil twin trying to provoke him. I > would say that references to a nonexistent version "171" are > deliberately provocative. No idea and couldn't care less who br is: R : Copyright 2003, The R Development Core Team Version 1.7.1 (2003-06-16) Get your facts correct before spouting further rubbish. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Inconsistent linear model calculations
Professor Kubovy As you instructed, below is the command terminal output: sessionInfo() R version 2.5.1 (2007-06-27) i586-mandriva-linux-gnu locale: LC_CTYPE=en_GB.UTF-8;LC_NUMERIC=C;LC_TIME=en_GB.UTF-8;LC_COLLATE=en_GB.UTF-8;LC_MONETARY=en_GB.UTF-8;LC_MESSAGES=en_GB.UTF-8;LC_PAPER=en_GB.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_GB.UTF-8;LC_IDENTIFICATION=C attached base packages: [1] "stats" "graphics" "grDevices" "utils" "datasets" "methods" [7] "base" Yours, __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
