-- Forwarded message --
From: Jie TANG
Date: 2011/10/25
Subject: how can I install the latest version of r in linux?
To: r-help@r-project.org
HI R-users
I downloaded the latest version of R with the name R-latest.tar.gz from
the website
But when I tar the package and configure
HI R-users
I downloaded the latest version of R with the name R-latest.tar.gz from
the website
But when I tar the package and configure the package,
some error message showed that
configure: error: --with-readline=yes (default) and headers/libs are not
available
and I run make command:
error
>
> Thank you Petr!
>
> I have read on the merge help page, but I cant figure out how to write
this
> function.
> When I use your function it includes all data from "data3", but all
columns
> in "data" has "NA"(without "name" and "date". I hoped to keep these
values
Because name and date comm
Hi Bert
I am aware of factor features and frankly speaking I consider them quite
usefull despite of prevalent preference to character vectors. For the OP
question seems to me that ifelse construction is appropriate, based on his
statement he has 2 strings which shall be converted to another two
Dear Marc,
I would also like to request the R code for doing this nominal measure of
association analyses, thanks in advance. Your kindly helps are really
appreciated.
My email is hon...@mun.ca
Have a great day,
Hong
--
View this message in context:
http://r.789695.n4.nabble.com/lambda-uncert
Thanks all, for your great help!
--Peter
Op 24-10-2011 18:23, R. Michael Weylandt schreef:
You might also need the assign() function which is sort of the opposite of get()
Michael
On Mon, Oct 24, 2011 at 12:15 PM, jim holtman wrote:
Write a function that encapsulates the following three lin
Md Desa, Zairul Nor Deana Binti ku.edu> writes:
>
> Hello,
> Does anyone knows how to deal with zero subscript in R.
Try the "Oarray" package (that's a capital letter "O").
__
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Hi Kyle,
blups=ranef(modelrandom)[[1]]
should give you a data frame, the rownames of which are your strain (I
think). If that doesn't work, it would be great if you could post
some data so we see what you see.
Cheers,
Josh
On Mon, Oct 24, 2011 at 6:53 PM, kickout wrote:
> dfhfs...@ghghgr.com
dfhfs...@ghghgr.com
I ran a simple lme model:
modelrandom=lmer(y~ (1|Test) + (1|strain), data=tempsub)
Extracted the BLUPs:
blups=ranef(modelrandom)[1]
Even wrote myself a nice .csv file:
write.csv(ranef(modelrandom)[1],paste(x,"BLUPs.CSV"))
This all works great. I end up with a .csv fil
not sure about gplots, but the one in base R should work well if you
specify the scaling method to be 'none':
heatmap(a4[1:40, ], Rowv=NA, Colv=NA, col=c("grey", "blue", "purple",
"red"), scale='none')
Regards,
Yihui
--
Yihui Xie
Phone: 515-294-2465 Web: http://yihui.name
Department of Statisti
Note that this issue was raised on StackOverflow recently.
http://stackoverflow.com/questions/7678090/xts-merge-odd-behaviour
Here's the solution:
index(a) <- index(a)
index(b) <- index(b)
merge(a,b)
ZWD.UGX SCHB.Close
2010-03-31 NA 28.02
2010-04-01 7.6343 NA
2010-04
Hi All,
I'm having trouble with the automatized model generation (dredge) function
in the MuMIn package. I'm trying to use it to automatically generate subsets
of models from a global cox proportional hazards model, and rank them based
on AICc. These seems like it's possible, and the Mumin documen
Hello useRs
I am trying to build a package for personal use and for making easier working
with other people but I keep getting the same error message about the
DESCRIPTION file not existing.
when trying to install from a source tar.gz file:
Error in .read_description(dfile) :
file
'C:
On Oct 24, 2011, at 7:23 PM, Md Desa, Zairul Nor Deana Binti wrote:
Hello,
Does anyone knows how to deal with zero subscript in R. I have this
code:
for (i in 1:nitems){
+ for (j in 1:ncat-1) {
+ draw<-matrix(rnorm(nitems*(ncat-1),seed1,seed2),nitems,
(ncat-1))
+
Hi Deana,
No, R does not deal with zero subscripts. Could it be done? Of
course. The simplest approach is: 0 + 1 = 1, which is the R
equivalent. You need to adjust your code to go, for example, from 1
to 30 instead of 0 to 29.
Cheers,
Josh
On Mon, Oct 24, 2011 at 4:23 PM, Md Desa, Zairul No
Hello,
Does anyone knows how to deal with zero subscript in R. I have this code:
for (i in 1:nitems){
+ for (j in 1:ncat-1) {
+ draw<-matrix(rnorm(nitems*(ncat-1),seed1,seed2),nitems,(ncat-1))
+ d<-( sigma_d*draw ) + mu_d
+ draw<-matrix(rtnorm((nitems*(ncat-1)
Michael:
Thanks a lot. It worked.
We have to extract the core part of the time-series data.
Then do a xy-plot or linear regression.
-Arka
On Mon, Oct 24, 2011 at 10:34 AM, Michael Weylandt [via R] <
ml-node+s789695n3933306...@n4.nabble.com> wrote:
> Comments inline.
>
> On Oct 24, 2011, at 1:4
Hmm, I will see if I can see what Rcmdr is doing to possibly
replicate. I am experimenting with processing that ignores NULL
output from a function, BUT, this has to be done cautiously as NULL
may be a legitimate return so I do not want to simply not print any
NULL output, but I have not figured o
On Mon, 24 Oct 2011, B77S wrote:
The following might not be exactly the way to do this, but see the package
"reshape" and the line of code following your data:
Thank you, B77S. Bert pointed me to reshape and reshape2. I'll read the
help pages for them before responding. I'll be out of the of
On Mon, Oct 24, 2011 at 5:39 AM, Duncan Murdoch
wrote:
> Suppose I have data like this:
>
> A <- sample(letters[1:3], 1000, replace=TRUE)
> B <- sample(LETTERS[1:2], 1000, replace=TRUE)
> x <- rnorm(1000)
>
> I can get a table of means via
>
> tapply(x, list(A, B), mean)
>
> and I can add the marg
Could you dput() the structure of x and y: I'm having trouble
visualizing how your data is set up.
Michael
On Mon, Oct 24, 2011 at 12:07 PM, Julie wrote:
> The variable y is made of four columns, each paired to 20, 200, 2000 or 20
> 000.
>> y <- c(rdiktator20, rDiktator200, rDikt2000, rDikt2
Hi there,
I have a matrix like this:
> a4[1:20, 1:5]
194 211 294 314 315
GO:003 1 1 1 1 1
GO:072 0 0 0 0 0
GO:076 1 0 0 0 0
GO:082 1 3 1 1 1
GO:083 1 0 0 0 1
GO:086 0 1 0 1 1
GO:114 0 0 0 0
The following might not be exactly the way to do this, but see the package
"reshape" and the line of code following your data:
df <- structure(list(site = structure(c(1L, 1L, 2L, 4L, 2L, 2L, 2L,
3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("BC-0.5", "BC-2", "BC-3",
"BC-4"), class = "factor"), sampda
Hey Martin
Thats exactly what I intended but sent it to general R list. Will post it
again. Thanks for checking.
-A
On Mon, Oct 24, 2011 at 2:32 PM, Martin Morgan wrote:
> On 10/24/2011 02:23 PM, Abhishek Pratap wrote:
>
>> Hi All
>>
>> I am wondering if people based on their experience could
Glad to hear it worked for you.
There does seem to be some confusion on your end as to what the with()
command does, however. The following are all equivalent.
data(mtcars)
layout(matrix(1:4,2))
plot(mtcars$cyl, mtcars$mpg)
plot(mtcars[["cyl"]], mtcars[["mpg"]])
plot(mtcars[,"cyl"], mtcars[,"mpg
Just discovered It doesn't do this when run in the latest release Rcmdr, yet
still does in JGR, RKward and R-Gui.
Confusing.
Thanks,
L.A.
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View this message in context:
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Sent from the R help mailing list archive
Hello Michael,
You were absolutely right, it was not the database mtcars I wanted to
use. but the example you sent me helped me a lot.
I took the code that I adapted to this and what I wanted and resulted in
perfection.
Thank you for your help.
dataset
*with (dataset, plot (dataset
On Mon, 24 Oct 2011, B77S wrote:
Why not format the data like this:
site sampledate SO4 TDS NA Mg Cond Cl Ca
Because I don't know how to reformat the base data frame (chemdata) to
achieve this.
It seems to me that you summary doesn't make any sense. Those quantiles
are meaningless as they
On Mon, 24 Oct 2011, Daniel Nordlund wrote:
you need to reorganize your data so that you have a single record (i.e.
row) for each site/sampdate combination with the quant value for each of
your params save in a column named by the param value (I don't remember
all the param names so I used pn).
On Oct 24, 2011, at 5:25 PM, Dennis Murphy wrote:
Hi David:
When I try your code, I get the wireframe with the x, y, z axes sans
bounding cube and points, along with the error message
Error using packet 1
object 'pts' not found
Hi Dennis;
My code was run with 2 objects, one named 'pts1' an
Hi Michael:
Try this:
show.beta <- function(model, x = 'x',
x1, x2, label, col="black", ...) {
abline(model, col=col, lwd=2)
xs <- data.frame(c(x1, x2, x2))
names(xs) <- attr(model$coefficients, 'names')[2]
ys <- predict(model, xs)
lines(cbind(xs,ys[c(1,1,2)])
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Rich Shepard
> Sent: Monday, October 24, 2011 2:17 PM
> To: r-help@r-project.org
> Subject: Re: [R] Syntax Help for xyplot()
>
> On Mon, 24 Oct 2011, Rich Shepard wrote:
>
> >
Why not format the data like this:
site sampledate SO4 TDS NA Mg Cond Cl Ca
i.e. with a column for each parameter?
It seems to me that you summary doesn't make any sense. Those quantiles are
meaningless as they encompass all the parameters. Am I missing something?
Rich Shepard wrote:
>
>
On Mon, 24 Oct 2011, Dennis Murphy wrote:
xyplot('TDS'$quant ~ 'Cond'$quant | burns.tds.anal$site )
Dennis,
xyplot('TDS'$quant ~ 'Cond'$quant | burns.tds.anal$site)
Error in "TDS"$quant : $ operator is invalid for atomic vectors
I tried that earlier today.
Rich
__
On Mon, 24 Oct 2011, Bert Gunter wrote:
Where's TDS? -- 'cause it's sure not in your data frame.
Bert,
summary(burns.tds.anal)
sitesampdate param quant
BC-3 :460 Min. :1992-03-27 Ca : 65 Min. : 1.00
BC-2 :107 1st Qu.:1994-09-21 Cl :148
Hi:
You should perhaps do the following in the xyplot() call (untested
because the TDS and Cond data frames are missing):
xyplot('TDS'$quant ~ 'Cond'$quant | burns.tds.anal$site )
assuming that all three atomic objects have the same length. Caveat emptor.
Dennis
On Mon, Oct 24, 2011 at 2:17 PM
Where's TDS? -- 'cause it's sure not in your data frame.
-- Bert
On Mon, Oct 24, 2011 at 2:17 PM, Rich Shepard wrote:
> On Mon, 24 Oct 2011, Rich Shepard wrote:
>
> Perhaps because it's Monday I'm not successfully writing the xyplot()
>> command to show the quant, for example, for TDS by site.
On 10/24/2011 02:23 PM, Abhishek Pratap wrote:
Hi All
I am wondering if people based on their experience could share what methods
one could use to compare two gff/gtf files. The reason why I want to do so
is that we have constructed a RNA-Seq based transcriptome and would like to
compare it wit
Hi,
I got the same result with David's version, but defining the *lim values
prior to the function actually made it work. I also renamed the points pts
so it wouldn't matter whether it was pts1 or pts. The new code is:
data.frame(x = seq(-4, 0, 0.5), y = seq(0, 40, 5))-> df
expand.grid(x = df$x
I've written a simple function to draw a regression line in a plot and
annotate the line showing the slope
with a label. It works, as I'm using it, when the horizontal variable
is 'x', but gives incorrect results otherwise.
What's wrong?
# simple function to show the slope of a line
show.beta
Hi David:
When I try your code, I get the wireframe with the x, y, z axes sans
bounding cube and points, along with the error message
Error using packet 1
object 'pts' not found
> sessionInfo()
R version 2.13.1 (2011-07-08)
Platform: x86_64-pc-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=English
Hi All
I am wondering if people based on their experience could share what methods
one could use to compare two gff/gtf files. The reason why I want to do so
is that we have constructed a RNA-Seq based transcriptome and would like to
compare it with reference transcriptome we had from in-silico a
On Mon, 24 Oct 2011, Rich Shepard wrote:
Perhaps because it's Monday I'm not successfully writing the xyplot()
command to show the quant, for example, for TDS by site. What I need to do
is plot the quant values for TDS vs. Cond, TDS vs. SO4, etc.
I should have provided some of the attempts:
Thanks Uwe. This works perfectly.
###
owd <- setwd(pth)
fls <- list.files(pattern="^chr")
ufls <- unique(sapply(strsplit(fls, "_"), "[", 1))
for(i in ufls){
of <- strsplit(i, "\\.")[[1]]
of <- paste(of[1], tail(of, 1), sep=".")
impute2databel(genofile = i,
On Mon, Oct 24, 2011 at 4:52 PM, lauren mcdonagh
wrote:
> Dear R users,
>
> I hope you can help me as you have done before.
>
> I was wondering if I can make some of the points on my graph a different
> colour and symbol? Some of my soil samples are enriched and I wanted them to
> be shown with
Dear R users,
I hope you can help me as you have done before.
I was wondering if I can make some of the points on my graph a different colour
and symbol? Some of my soil samples are enriched and I wanted them to be shown
with a red triangle and the samples which are not enriched I wanted to s
Thanks to David's help I subset my large data set and produced a smaller
one for a single stream and 7 factors of interest. The structure of this
data frame is:
str(burns.tds.anal)
'data.frame': 718 obs. of 4 variables:
$ site: Factor w/ 143 levels "BC-0.5","BC-1",..: 1 1 4 6 4 4 4 5 5
On Oct 24, 2011, at 4:12 PM, Megan Bartlett wrote:
> Hi David,
>
> Thanks for the suggestion - I changed pts1 to pts, but I still got
> the same error as before. Do you know what else I'm doing wrong?
I do not. The code that works for me is:
wireframe(z ~ y*x, gridd, drape = TRUE, colorkey= T
Works beautifully.
Thanks!
--
Noah Silverman
UCLA Department of Statistics
8117 Math Sciences Building
Los Angeles, CA 90095
On Oct 24, 2011, at 1:32 PM, (Ted Harding) wrote:
> ix <- which(Seg[1013:1046] > 0)
> lines((1013:1046)[ix], Seg[1013:1046][ix], col=2)
[[alternative HTML ver
On 24-Oct-11 19:48:16, Noah Silverman wrote:
> Hi,
> I have a function that approximates some data and indicates "segments".
>
> I'd like to plot the original data, and then the linear approximations
> on top of it. (Ideally, just a subset of N rows at a time, as the data
> set is large.)
>
> I
What kind of tree do you want? The sos() package can help you find R
functions associated with a particular topic:
# install.packages('sos')
library('sos')
> findFn('tree')
found 2798 matches; retrieving 20 pages, 400 matches.
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
> findFn('classifica
Hi David,
Thanks for the suggestion - I changed pts1 to pts, but I still got the same
error as before. Do you know what else I'm doing wrong?
Thanks,
Megan
PS. New code is:
wireframe(z ~ y*x, gridd, drape = TRUE, colorkey= TRUE,
scales = list(arrows = FALSE),
pts = pts1,
Hi,
I have a function that approximates some data and indicates "segments".
I'd like to plot the original data, and then the linear approximations on top
of it. (Ideally, just a subset of N rows at a time, as the data set is large.)
I can't figure out a clean way to do this. Suggestions?
her
I believe Dr. Winsemius addressed this in your other thread, but I
would hesitate to do any sort of outlier identification based on
repeated application of a filter (for that matter, I'm not much of an
outlier removal guy generally, but let's suppose I were). You can
easily get into a situation whe
I'm not particularly familiar with EVT but it seems that here the term
exists to scale down the time base to that wherein extreme values were
identified. Just a hunch though...
If you want more accurate clarification you may wish to contact the
package maintainer, but to be honest I wouldn't worry
Dear all,
Does anyone know where can I find a tutorial on using 'fields' package
and related packages dealing with regular grids?
Thank you
--
Kind regards,
Antonio Rodriges
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On Oct 24, 2011, at 2:45 PM, flokke wrote:
Dear all,
I am a R user since about 3 weeks now and still struggeling with
things that
must be very
easy for you...
This week I am struggling with the function predict()
I want to use this function to get a 95% interval.
I understand that you have
Hello users.
I need a help again, please?
I runned my Arima model using difference order = 7
*
Y2=diff(Yst[,2],differences = 7)*
Yfit=arima(Yst2,order=c(3,0,0),seasonal=list(order=c(0,0,0),period=7),include.mean=TRUE)
and forecast
Yfor=forecast.Arima(Yfit, h=31)
Now, I would like to compare
On Oct 24, 2011, at 1:23 PM, Megan Bartlett wrote:
Hi,
I'm trying to follow the suggestions given by Deepayan Sarkar in this
message:
http://tolstoy.newcastle.edu.au/R/help/05/11/16135.html
to plot 3-D points on a wireframe plot. The problem is that I keep
getting a
partly formed plot- wit
This is a correction to a post from 3/10/09. I just wanted to get this in the
archive. It is the same thread as
http://www.mail-archive.com/r-help@r-project.org/msg48762.html
Thanks to Matt Oliver for bringing this to my attention.
The correct code for my.interp.surface() follows.
# A fun
I have the data set like this:
A B C D E
A2 235 327 278 2569
B244 5 122 248 259
C324 223513 98 69
D22232 7 6569
E232 2278 9
This i
Dear all,
I am a R user since about 3 weeks now and still struggeling with things that
must be very
easy for you...
This week I am struggling with the function predict()
I want to use this function to get a 95% interval.
I understand that you have to use it in such a way as:
lm_examplemodel<-
Are you sure you are using the data set mtcars? That data set doesn't
have variables Na or K nor does it look like what you provided...
For the real mtcars, you could try this which I think does something
like what you want.
with(mtcars, plot(cyl, mpg, pch = carb, col = gear))
with(mtcars, legend
I believe I already showed you how to do this (though your code
doesn't seem to do what your words ask, so I'm not sure which one to
answer), but didn't draw attention to it:
x[-1]/x[-length(x)]
This creates two vectors, one consisting of everything but the first
element of x and the other of eve
Hi Joe,
What about this? Btw, the real key to a reproducible example is one
that we can actually work withoutput of the summary of the data is
not the same as the data itself. That makes it tricky to know what
exactly you are working with (although sometimes the original form of
the data can
I see the problem, I fixed this bug for version 2.8 of TeachingDemos, but have
not submitted the new version to CRAN yet (I thought that I had fixed this
earlier, but apparently it is still only in the development version). An easy
fix is to install version 2.8 from R-forge (install.packages("T
Hi:
Aren't V1 and V2 factors in this data frame? If so, you should be
plotting bar charts rather than histograms (they're not the same). Do
you want separate graphs for V1 and V2, do you want them stacked, or
do you want them dodged (side-by side for each level 001, 002, 003)?
In the absence of su
On 24/10/2011 1:47 PM, Jim Bouldin wrote:
OK, what is the trick to extracting the overall p value from an lm object?
It shows up in the summary(lm(model)) output but I can't seem to extract it:
It's not part of the object, it is computed when the object is printed.
To see the print method, do
It's not directly extractable since it's calculated on the fly in the
printing method. If you type stats:::print.summary.lm, you can see
the code the leads to the calculation: It's basically (I'm leaving out
some formatting stuff):
pf(x$fstatistic[1L], x$fstatistic[2L], x$fstatistic[3L], lower.ta
Hi Jim,
Its a bit of a trick question. There isn't actually any overall p
value stored in the lm object, or even in the summary.lm object. It
is calculated from the f statistic by the print methods for
summary.lm. Of course, none of that helps, per se. Try this:
summary(lm(mpg ~ hp, data = mt
OK, what is the trick to extracting the overall p value from an lm object?
It shows up in the summary(lm(model)) output but I can't seem to extract it:
> test2 = apply(aa, 1, function(x) summary(lm(x[,1] ~ 0 + x[,3] + x[,6])))
> test2[[1]]
Call:
lm(formula = x[, 1] ~ 0 + x[, 3] + x[, 6])
[omitte
Hi,
I'm trying to follow the suggestions given by Deepayan Sarkar in this
message:
http://tolstoy.newcastle.edu.au/R/help/05/11/16135.html
to plot 3-D points on a wireframe plot. The problem is that I keep getting a
partly formed plot- with the colored lattice visible but no axis labels or
addit
It surely can be done. One way is to keep track of selected variables
in a set. If a new variable is selected, you expand the selected set
and set the frequency to be one, otherwise just increase the freqency
of the selected variable (if... else).
Also, you might want to have a look at glmulti pac
What about?
hist(Data$V1)
hist(Data$V2)
Cheers,
Josh
On Mon, Oct 24, 2011 at 9:38 AM, Joe Carl wrote:
>
> Hello all,
>
> I'm trying to make a histogram of the data contained in my dataframe.
> The summary of the data gives me exactly what I want
>
> summary (Data)
> V1 V2
> first001: 3 l
Hello all,
I have the following. Two sets of p x 3 matrices where p is say relatively
large.
My data looks like :
mat1
2 3 4
2 3 4
1 2 3
mat2
2 3 4
12 12 4
10 12 3
when p = 3.
I form a 2 x 3 contingency table using the i^th row from each matrix.
Its Fisher's exact te
Hello all,
I'm trying to make a histogram of the data contained in my dataframe.
The summary of the data gives me exactly what I want
summary (Data)
V1 V2
first001: 3 last001: 9
first002: 3 last002: 7
first003: 2 last003: 6
(Other) :52(Other): 27
But how do I capture the names and
The count() function in the plyr package works beautifully. Thanks to Jim,
Rainer and Dennis for your help.
Best.
-Original Message-
From: Dennis Murphy [mailto:djmu...@gmail.com]
Sent: Monday, October 24, 2011 12:05 PM
To: asindc
Cc: r-help@r-project.org
Subject: Re: [R] How to selecti
I am new to R coding and I am trying to model the returns on the ftse100
since 1990. I have got a vector with all the closing values on each trading
day. however, instead of using the difference in the closing values of two
consecutive days, (ie dx=diff(x) where x is the vector containing the
clos
The variable y is made of four columns, each paired to 20, 200, 2000 or 20
000.
> y <- c(rdiktator20, rDiktator200, rDikt2000, rDikt2)
So I guess the problem is in the fact that I did not specify it correctly,
is it so? How can I tell R properly that one part of y matches to one part
of x?
Th
Sorry, I attempted to paste the sample data but it must have been stripped
out when I posted. It is hopefully now listed below.
tapply looks useful. I will check it out further.
Here's the sample data:
> flights[1:10,]
PASSENGERS DISTANCE ORIGIN ORIGIN_CITY_NAME ORIGIN_WAC DEST
DEST_CITY_NA
Hi All,
Its a bit of a beginners question I'm afraid.
I have a looped stepwise regression (using MASS and StepAIC) to take random
predictors out of the total number. For this example a random sample of 5
out of a total of 20. The loop will continue until all combinations of
variables have been
Try it yourself:
x = seq(1, 11, by = 2)
diff(log(x))
log(x[-1]/x[-length(x)])
all.equal(diff(log(x)), log(x[-1]/x[-length(x)]))
It seems like you don't really understand logs / log returns and why
they are used by some in quant finance: might I suggest you read this:
http://quantivity.wordpress.
On Mon, 24 Oct 2011, David Winsemius wrote:
You could also have saved the subsetted data, applied `factor` to the
subsetted column and then used `xtabs`.
temp <- subset(chemdata, , )
temp$param <- factor(temp$param)
(Now only levels that exist are in the temp version.)
David,
Thank you. I
tom_pienkowski blueyonder.co.uk> writes:
> I'm trying to use the 'by' function to extract the co-efficients from a
> mixed model which is performed on multiple individuals. I basically have a
> group of individuals and for each individual I want the co-efficient for
> there change in 'pots_hauled
On Oct 24, 2011, at 12:10 PM, Rich Shepard wrote:
On Mon, 24 Oct 2011, David Winsemius wrote:
The appearance of levels with all zeroes is probably because I
didn't include drop.unused.levels = FALSE in the xtabs specification.
OK. Adding 'drop.unused.levels' does make a huge difference.
You might also need the assign() function which is sort of the opposite of get()
Michael
On Mon, Oct 24, 2011 at 12:15 PM, jim holtman wrote:
> Write a function that encapsulates the following three lines:
>
>
> city1997<- dataCleaning(read.csv2("C:\\city\\year1997.txt"))
> city1997<- wasteCalcu
On Mon, Oct 24, 2011 at 06:10, Prof Brian Ripley wrote:
> On Mon, 24 Oct 2011, Uwe Ligges wrote:
>
>>
>>
>> On 23.10.2011 22:33, Ali Tofigh wrote:
>>>
>>> Hi,
>>>
>>> When I plot text and use cex to change the text size, I notice that the
>>> cex
>>> multiplier is not exact. It looks as if the rea
X and y must have the same number of elements, and NA values must be removed
(?na.omit)
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Hi:
bestglmtest is your input data frame, is it not? From the names() line,
you can see that it has no variable named BestModel that corresponds
to a list containing a component named coefficients.
Were you perhaps looking for
output$BestModel$coefficients ??
Dennis
On Mon, Oct 24, 2011 at 8:3
Write a function that encapsulates the following three lines:
city1997<- dataCleaning(read.csv2("C:\\city\\year1997.txt"))
city1997<- wasteCalculations(city1997, year = 1997)
if (city1997[1,1] == "Time") {city1997<- timeCalculations(city1997)}
and then pass in the appropriate parameters.
On Mon
The merge function seems perfect for your problem. If you can't get it to work,
the problem may be in the data you are working with, which you have not
supplied (read the posting guide, and use the head and dput functions to make
your example small and reproducible).
One common mistake by beginn
Assuming that d(x) is equal to x, (I don't know a d() function in R)
these should be the same.
log(a/b) = log(a) - log(b) = diff(log(c(a,b))
If you mean simple returns instead of continuous/log returns, perhaps try this:
x[-1]/x[-length(x)] - 1
Michael
On Mon, Oct 24, 2011 at 11:44 AM, tynashy
On Mon, 24 Oct 2011, David Winsemius wrote:
The appearance of levels with all zeroes is probably because I didn't include
drop.unused.levels = FALSE in the xtabs specification.
OK. Adding 'drop.unused.levels' does make a huge difference.
Thanks,
Rich
__
Thanks so much, this is very very helpful.
I do have one remaining question here. I definitely see the value of
making a list of the datasets, an advise I will definitely follow.
However, for educational purposes, I would still like to know how to
automate the following without using a list:
See the count() function in the plyr package; it does fast summation.
Something like
library('plyr')
count(passengerData, c('ORIGIN_WAC', 'DEST_WAC'), 'npassengers')
HTH,
Dennis
On Mon, Oct 24, 2011 at 8:27 AM, asindc wrote:
> Hi, I am new to R so I would appreciate any help. I have some data t
... Well, this works in this simple case, but is too clumsy for a general
formulation of this problem: given a "dictionary" consisting of two
character vectors of unique "names" (or two columns in a data frame), x and
y, how does one convert a factor z with levels in x into the corresponding
equi
how do I code the following in R. I want to produce a vector where dx=log(
(d(x))/(d(x-1)) ). I can do it for dx=diff(log(x)). I am learning/trying to
model log returns of a stock market index. But instead of using the
difference of the closing values of two consecutive days, i want to use the
log
Hi all
I'm trying to use the 'by' function to extract the co-efficients from a
mixed model which is performed on multiple individuals. I basically have a
group of individuals and for each individual I want the co-efficient for
there change in 'pots_hauled' in response to a change in 'vpue' with m
It would be good to follow the posting guide and at least supply a
sample of the data.
Most likely 'tapply' is one way of doing it:
tapply(df$passenger, list(df$orig, df$dest), sum)
On Mon, Oct 24, 2011 at 11:27 AM, asindc wrote:
> Hi, I am new to R so I would appreciate any help. I have some d
Hi all,
I have been trying to run a bestglm in R for a while now and am struggling
to get it to run. When I thought I had succeeded, the "output" it gave me
was "NULL" and that's it. Below is my code:
bestglmtest<-read.table("C:\\Documents and
Settings\\clyons\\Desktop\\bestglmtest.txt",header=T,
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