Hi all,
I am trying to use plm to estimate coefficients in a model consisting of a
system of equations. So far I used mle2 from the package "bbmle", but now I
need to test for autocorrelation and mle2 does not provide for the necessary
tests. mle2 needs a 'function' as input that might as well con
I have a set of data that is basically 3+ years of data. It is daily sales for
this year and then back 3 years so there are 3*365 + 231 days or 1326 days of
data. Since this is a time series I have constructed it as:
Start = c(1, 1)
End = c(4, 231)
Frequency = 365
In trying to ayalyze this d
On Fri, 12 Sep 2008, sandsky wrote:
I use "while" loop but it produces an errro. I have no idea about this.
Error in "[<-"(`*tmp*`, i, value = numeric(0)) :
nothing to replace with
What version of R is this? I suspect not a recent one. I get
Error in verpi[i] <- c(5^alpha[i - 1], 1
On Fri, 12 Sep 2008, Bert Gunter wrote:
Looks like a bug to me. Compare:
A bug fixed in the R-devel version of R (because of a related problem with
duplicated).
match (1:3,1:3,incom=3)
[1] 1 2 3
match (1:3,1:3,incom=2:3)
[1] 1 2 3
match (1:3,1:3,incom=1:2)
[1] NA NA 3
match (1:3,1:3
The function symbols() is in the package 'graphics', so you don't need
any add-on packages.
For example:
# similar code to the example you mentioned
set.seed(1)
symbols(x = rnorm(50), y = rnorm(50), circles = abs(rnorm(50)),
main = "Circles Plot")
Yihui
On Sat, Sep 13, 2008 at 11:48 AM, Wen
Hi:
On Sep 12, 2008, at 8:35 PM, Wen Gu wrote:
I need help creating a bubbleplot, like a simple pseudo three
dimensional scatterplot of circles whose sizes index a 3rd
variable. I initially came across this at http://addictedtor.free.fr/graphiques/graphcode.php?graph=73
but the circle
?symbols
use the argument "circles"
On Sat, Sep 13, 2008 at 11:35 AM, Wen Gu <[EMAIL PROTECTED]> wrote:
>
> I need help creating a bubbleplot, like a simple pseudo three dimensional
> scatterplot of circles whose sizes index a 3rd variable.I initially
> came across this at
> http://add
I need help creating a bubbleplot, like a simple pseudo three dimensional
scatterplot of circles whose sizes index a 3rd variable.I initially
came across this at
http://addictedtor.free.fr/graphiques/graphcode.php?graph=73 but the circleplot
function does not exist in fbasic as listed
Kevin- this is a simple rescaling of the axes so that the "area under the
curve" remains constant (and is half of the variance since you only look at
the positive frequencies). In this case, freq(x) = 1/dx, where dx is the
time between points. It is basically a graphic device so that you get
p
Hi,
I am working with QCC and would like help with the following:
a) Change the color of the background. I tried bg="#FF" in par and also
in qcc and it did not work.
b) I want to remove the X Axis Text (i.e. 1, 2, 3 ... n) -- but xaxt="n"
does not work in qcc.
Thanks
--
View this message
I don't have a solution but I receive the same error when trying to fit a GARCH
model using the garchFit in the fGarch package.
Kevin
Desislava Kavrakova <[EMAIL PROTECTED]> wrote:
> Hello everyone,
> I'm trying to estimate the parameters of the returns series attached using
> the GARCH c
Hi, I'm a beginner with R, but I'm getting excellent results with it.
Well I've got an capscale object (vegan package) and I want to made a biplot
with symbols representing six groups of areas.
The plot.cca function and some par attributes (like 'labels') I was able to
substitute the samples name
I use "while" loop but it produces an errro. I have no idea about this.
Error in "[<-"(`*tmp*`, i, value = numeric(0)) :
nothing to replace with
The problem description is
The likelihood includes two parameters to be estimated: lambda
(=beta0+beta1*x) and alpha. The algorithm for the e
Hello everyone,
I'm trying to estimate the parameters of the returns series attached using the
GARCH code below, but I get the following error message:
Error in solve.default(Hessian) :
system is computationally singular: reciprocal condition number = 0
Error in diag(solve(Hessian)) :
erro
On Fri, Sep 12, 2008 at 5:29 PM, Amanda Young <[EMAIL PROTECTED]> wrote:
> Hi,
> I am a complete R rookie so this question is probably really simple
> but I haven't found an answer on the web that I can understand.
>
> My data frame has 3 columns, A, B and C. A and B have numbers (about
> 8000 rows
Looks like a bug to me. Compare:
> match (1:3,1:3,incom=3)
[1] 1 2 3
> match (1:3,1:3,incom=2:3)
[1] 1 2 3
> match (1:3,1:3,incom=1:2)
[1] NA NA 3
> match (1:3,1:3,incom=1)
[1] NA 2 3
Cheers,
Bert
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Adam
Dear Amanda,
Try this:
# Data set
set.seed(123)
DF=data.frame(A=rnorm(10),B=rpois(10,10),C=sample(1:3,10,replace=T))
attach(DF)
# Plot
plot(A,B,pch=16,col=C)
legend('topleft',paste('Group',1:3,sep=" "),col=1:3,pch=16)
HTH,
Jorge
On Fri, Sep 12, 2008 at 6:29 PM, Amanda Young <[EMAIL PROTECTED
See argument 'col' to plot()/points(). Setup a 'col' vector of length
equal to the number of data points and have the 'C' variable specify
the colors of the individual elements. Then call plot()/points() with
argument 'col'.
My $.02
/HB
On Fri, Sep 12, 2008 at 3:29 PM, Amanda Young <[EMAIL PRO
Hi El,
As does my PowerBook G4 with 2GB of RAM. The iBook shipped with 256MB,
however, which Leopard does not run well on. My point was more that Linux
was superior to MacOS in the case of very low memory...not that the chip was
somehow the problem.
...but now I am solidly off-topic!
Cheers,
Ad
Hi,
I am a complete R rookie so this question is probably really simple
but I haven't found an answer on the web that I can understand.
My data frame has 3 columns, A, B and C. A and B have numbers (about
8000 rows), C is a factor which is either true or false. So I can
plot A vs B with plot
Adam,
I (or rather the kids and the wife) got two iMinis and one iBook
with G4. They all run 10.5.4 happily on their 1GB of RAM...
greetings, el
On 12 Sep 2008, at 23:43 , Adam D. I. Kramer wrote:
To Stephen's credit, Apple will no longer support PowerPC chips
(such as his
G4) in the next op
I can replicate this and also do not understand it.
match(1:3,1:3,incomparables=5)
[1] NA 2 3
match(1:3,1:3,incomparables=4)
[1] 1 2 3
match(1:3,1:3,incomparables=3)
[1] 1 2 3
match(1:3,1:3,incomparables=2)
[1] 1 2 3
match(1:3,1:3,incomparables=1)
[1] NA 2 3
...every other integer
To Stephen's credit, Apple will no longer support PowerPC chips (such as his
G4) in the next operating system (Snow Leopard), so some sort of switchover
will be necessary in order to maintain SOME sort of state-of-the-art
software packaging for PPC-based Mac users in the near future.
Also, it is
On 12 September 2008 at 16:30, stephen sefick wrote:
| This is an operating system question, but it is with the intent of
| using R on that operating system. I have an ibook G4 Power PC that I
| am going to install linux on. Is there a better, worse, or perhaps
| easier (I am a linux newby migra
Yes, of course.
That's why the :-)-O was there
el
On 12 Sep 2008, at 23:01 , Doran, Harold wrote:
Are you aware that a BSD unix OS runs on the mac already?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
yes, but the os now is 10.3.9 and I am starting to have a couple of
troubles with some things, and the os boxed with my mac mini are intel
specific, so they won't install on my poor old booger ppc. So, it has
precipitated this move. I can always go back if I don't like it.
it's just time right?
Are you aware that a BSD unix OS runs on the mac already?
-Original Message-
From: [EMAIL PROTECTED] on behalf of Dr Eberhard W Lisse
Sent: Fri 9/12/2008 4:38 PM
To: stephen sefick
Cc: R-help Mailing List; Dr Eberhard W Lisse
Subject: Re: [R] Power PC with a linux distribution and R
Yes
Yes,
I'd install the OS X Leopard (10.5.4) distribution on it :-)-O
el
On 12 Sep 2008, at 22:30 , stephen sefick wrote:
This is an operating system question, but it is with the intent of
using R on that operating system. I have an ibook G4 Power PC that I
am going to install linux on. Is th
This is an operating system question, but it is with the intent of
using R on that operating system. I have an ibook G4 Power PC that I
am going to install linux on. Is there a better, worse, or perhaps
easier (I am a linux newby migrating from mac) distribution that I
should look at. I apprecia
Hello,
I was playing around with the newly implemented 'incomparables' argument
in 'match' and realized the argument does not behave anything like I
expected. Can someone explain what is going on here? Sorry if I'm
misreading the documentation.
> match(1:3, 1:3, incomparables=1)
[1] NA 2 3 # Th
On Fri, Sep 12, 2008 at 6:02 AM, Antje <[EMAIL PROTECTED]> wrote:
> Hi there,
>
> I have a nested call of lapply in which I do a tryCatch statement like this
>
> lapply(1:length(p_names), function(w_idx) {
>r <- as.numeric(pos_r[[w_idx]][1])
>c <- as.numeric(pos_c[[w_idx]][1])
>
Hi Monica:
On Sep 12, 2008, at 11:59 AM, Monica Pisica wrote:
I am wondering if there is a function which will do a join between 2
data.frames by minimum distance, as it is done in ArcGIS for
example. For people who are not familiar with ArcGIS here it is an
explanation:
Suppose you ha
I am wondering if there is a function which will do a join between 2
data.frames by minimum distance, as it is done in ArcGIS for example. For
people who are not familiar with ArcGIS here it is an explanation:
Suppose you have a data.frame with x, y, coordinates called track, and a second
data
I've tracked this down to the clipping bug fix reported in the CHANGES
file. There is another bug fix in R-devel that does not affect printing,
so your first option is to use one of the R-devel snapshots on CRAN.
I'll move the R-devel fix to R-patched shortly, so tonight's R-patched
snapshot
Why not use
con <- pipe(COMMAND)
foo <- read.delim(con, colClasses="numeric")
close(con)
? See the 'R Data Input/Output Manual'.
On Fri, 12 Sep 2008, Michael A. Gilchrist wrote:
Hello,
I am currently using R to run an external program and then read the results
the external program sends to
Hi Adai,
It works!! Thanks a lot!!
Hui-Yi
On Fri, Sep 12, 2008 at 11:26 AM, Adaikalavan Ramasamy <
[EMAIL PROTECTED]> wrote:
> help(rowttests) says that fac needs to be a factor. So how about ?
>
> m <- matrix( rnorm(30), nc=6 )
> genotype <- c("a", "a", "b", "b", "c", "c")
>
> w1 <- which(
Hi,
a few comments below.
On Fri, Sep 12, 2008 at 9:34 AM, Michael A. Gilchrist <[EMAIL PROTECTED]> wrote:
> Hello,
>
> I am currently using R to run an external program and then read the results
> the external program sends to the stdout which are tsv data.
>
> When R reads the results in it con
The conversion to a data frame and the transpose might be time consuming. In
addition to other comments you have received, try this:
matrix(rawinput, ncol = 6, byrow = TRUE)
and if you don't really need a data frame eliminate the conversion from matrix
to data.frame.
You might time this against
on 09/12/2008 11:34 AM Michael A. Gilchrist wrote:
> Hello,
>
> I am currently using R to run an external program and then read the
> results the external program sends to the stdout which are tsv data.
>
> When R reads the results in it converts it to to a list of strings which
> I then have to
Hello,
I am currently using R to run an external program and then read the results
the external program sends to the stdout which are tsv data.
When R reads the results in it converts it to to a list of strings which I
then have to maniuplate with a whole slew of commands (which, figuring out
Not really the ceiling and floor, though I haven't tried your code
(and my brain isn't at 100% capacity today to simply parse the code in
my head :) )
Though I think ?firstof and ?lastof as well as startof/endof might be
something along those lines. ?to.period has the ability to change
periodici
On Fri, Sep 12, 2008 at 10:53 AM, Jeff Ryan <[EMAIL PROTECTED]> wrote:
>
> I'm still not entirely sure I follow the desired usage, as the original post
> made no reference to ggplot2, but as Gabor mentioned the yearmon etc stuff
> is quite useful.
Well, I said "I need to be able to correct draw da
One last item that could be of use:
?timeBasedSeq
Lets you create ranges in any standard format, with simple ISO style
formatting. The level of detail you specify is interpreted as the resolution
that is desired.
e.g.
> timeBasedSeq(2000/2008)
[1] "2000-01-01" "2001-01-01" "2002-01-01" "2003-01
I'm still not entirely sure I follow the desired usage, as the original post
made no reference to ggplot2, but as Gabor mentioned the yearmon etc stuff
is quite useful.
If you are formatting arbitrary precision dates, take a look at
axTicksByTime in xts. Both xts and quantmod use it for plotting
I am more used to getting an error if you try to take the log of 0, like
this (in Perl):
perl -le 'for my $num (1, 0, -1, -2) { print log $num; }'
0
Can't take log of 0 at -e line 1.
R is different. With R, you do not even get a *warning* about log(0). Only
log() of negative number produces a war
on 09/12/2008 10:36 AM Marc Schwartz wrote:
> on 09/12/2008 10:07 AM cathelf wrote:
>> Thank you for your guys reply for my previous question. But I got one more
>> question about the boxplot. With the code in the R-help:
>>
>> boxplot(len ~ dose, data = ToothGrowth,
>> boxwex = 0.25, at =
?factor
On Fri, Sep 12, 2008 at 11:33 AM, Adaikalavan Ramasamy
<[EMAIL PROTECTED]> wrote:
> I just changed the 'at' argument and added an 'xlim' option.
>
> boxplot(len ~ dose, data = ToothGrowth,
> boxwex = 0.25, at = 1:3,
> subset = which(supp == "VC"), col = "yellow",
>
I just changed the 'at' argument and added an 'xlim' option.
boxplot(len ~ dose, data = ToothGrowth,
boxwex = 0.25, at = 1:3,
subset = which(supp == "VC"), col = "yellow",
main = "Guinea Pigs' Tooth Growth",
xlab = "Vitamin C dose mg",
ylab = "t
on 09/12/2008 10:07 AM cathelf wrote:
> Thank you for your guys reply for my previous question. But I got one more
> question about the boxplot. With the code in the R-help:
>
> boxplot(len ~ dose, data = ToothGrowth,
> boxwex = 0.25, at = 1:3 - 0.2,
> subset = supp == "VC", col =
help(rowttests) says that fac needs to be a factor. So how about ?
m <- matrix( rnorm(30), nc=6 )
genotype <- c("a", "a", "b", "b", "c", "c")
w1 <- which( genotype %in% c("a", "b") )
w2 <- which( genotype %in% c("a", "c") )
w3 <- which( genotype %in% c("b", "c") )
list( ab = rowttes
> As per my attached script I am ploting monthly mean data which has
missing
> months.
> But in the plot missing months are not shown (plot attached).
> Kindly help how to show complete plot (Jan-Dec) with missing months.
I don't accept scripts from strangers, but I guess the problem looks like
Thank you for your guys reply for my previous question. But I got one more
question about the boxplot. With the code in the R-help:
boxplot(len ~ dose, data = ToothGrowth,
boxwex = 0.25, at = 1:3 - 0.2,
subset = supp == "VC", col = "yellow",
main = "Guinea Pigs' Tooth Grow
Just out of curiosity, why do you want the spearman rank correlation to error
in this case?
One of the advantages of the spearman correlation is that it is invariant to
monotone transformations, so most people that use it see the fact that
corr(x,y, method='spearman') gives the same answer as c
1/(x-axis value) = frequency in time
(the x-axis is in cycles per time)
2008/9/12 Petr PIKAL <[EMAIL PROTECTED]>:
> Hi
>
> [EMAIL PROTECTED] napsal dne 09.09.2008 18:44:34:
>
>> For the command 'spectrum' I read:
>>
>> The spectrum here is defined with scaling 1/frequency(x), following
> S-PLUS.
>
FAQ 7.31
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED]
> Sent: Friday, September 12, 2008 7:36 AM
> To: r-help@r-
Hi Tobias,
Thanks for your answer but it doesn't change the error message
clustot <- svydesign(id=~num+Id_An, fpc=~fpc1+Totanim, data=tab1)
Erreur dans as.fpc(fpc, strata, ids) :
FPC implies >100% sampling in some strata
Maybe the probleme is that in some farms all the animals were examined
AFAIK, tapply() only works for one variable (apart from the grouping
variable). It might be perhaps better to use split() here:
df <- data.frame(ID = c(111, 111, 111, 178, 178, 138, 138, 138, 138),
value = c(5, 6, 2, 7, 3, 3, 8, 7, 6),
Seg = c(2, 2, 2,
Hi Yihui,
That's good, I like it! Very nice site.
Regards, Mark.
Yihui Xie wrote:
>
> Well, his talk seems to have attracted a lot of people... You may
> simply use gray text in your plot. Here is an example:
>
> ##
> x = runif(10)
> y = runif(
An alternative is to use semi-transparency, which can be helpful if you
want to do more than one such annoation. E.g. add
text(1, 1, "More", col=rgb(0,0,0,0.2), cex=3)
to this example. (I'd say it was closer to what "ghosted" might mean.)
On Fri, 12 Sep 2008, Mark Difford wrote:
Hi Agust
Thats nice thanks =) .. I can trick R to do multivariate armax with
lagged inputs as well and I bet R people didnt designed it that way
(but the idea is the same when doing MLE, it must work)..
anyway.. I wrote a small code (you can change it if you want) that
does armax with multiple inputs in ma
Well, his talk seems to have attracted a lot of people... You may
simply use gray text in your plot. Here is an example:
##
x = runif(10)
y = runif(10)
z = runif(10, 0.1, 0.3)
cl = rgb(runif(10), runif(10), runif(10), 0.5) # transparent colors!
par(m
Hi Agustin,
>> Is there any way of having a greyed ("ghosted") text...
Yes!
##
plot(1, type="n")
text(1, "Old Grey Whistle Test", col="slategray4", cex=2)
text(1, y=1.2, "OH!", col="grey95", cex=4)
Then plot what you want on top. If you export or plot to a PDF/ps device the
last-plotted items
--
From: "Ahoussou Sylvie" <[EMAIL PROTECTED]>
Sent: Friday, September 12, 2008 9:48 AM
To: "Thomas Lumley" <[EMAIL PROTECTED]>
Subject: Re: [R] Complex sampling survey _ Use of survey package
Thanks for your answer
I think I made a mistake when
Adai,
Thank you so much for your help. I like your code the best. :) So simple. I
have another question though, if you don't mind. I'd like to include another
variable in "res". This variable defines the segmentation of each person
(ranges, say, from 1 to 4).
ID value Seg
111 5 2
Dear R users,
As per my attached script I am ploting monthly mean data which has missing
months.
But in the plot missing months are not shown (plot attached).
Kindly help how to show complete plot (Jan-Dec) with missing months.
Regards,
Yogesh
file<-read.csv("CRI-run109-run110.csv")
names(file)
a
Hi!
Is there any way of having a greyed ("ghosted") text
(i.e, 2006) in the background of a plot?
I'm making a dynamic plot and would like to show the
year of each time step as a big greyed text in the background.
(the idea comes from Hans Rosling video:
http://video.google.com/videoplay?docid=4
Hello,
Below is a simple loop that should place a 3 in the first row of each
column in the array "List".
iterations<-3
tweaksize<-0.00
ii <-1
List=array(-1000, dim=c(iterations,11))
colnames(List) <- c("orig", "0.05", "0.10",
"0.15","0.20","0.25","0.30","0.35","0.40","0.45","0.50")
Entry=3
Dear all,
There are some R package that provide posterior
distribution of logit, probit models. Does anyone know any R package that can
provide posterior distribution for Gamma-GLM ?
Thank you.
Alex
__
R-help@r-project.org mailing list
https://stat.
Hi,
I have looked at packages like SNOW and FF , for distributed computing
but got overwhelmed a bit. I need to know how to put a R GUI (like
Rattle) on say a Hadoop backed platform on remote computing (like
Amazon Web Services). Please let me know regarding any help in this.
--
Regards,
Ajay O
Thank you all for the help! I had a friend look at it and basically what he
did was do an initial plot() of the line (which was all the mean values) and
then use ylower and yupper to plot 2 addition line() which he set lyt=3 to
so they would look like points. I know its a dirty hack but it worke
On Fri, 12 Sep 2008, Oliver Bandel wrote:
Hello,
I have strings with date-/time-data from a logfile.
For a sort-by-time (and other time-related analysis)
I need something to convert the
times to the Unix-epoche-time (better also with parts
of seconds).
Is there something that does that job (
If you are using POSIXct, just 'unclass' the value:
> x <- as.POSIXct('2008-09-12 09:00')
> str(x)
POSIXct[1:1], format: "2008-09-12 09:00:00"
> unclass(x)
[1] 122121
attr(,"tzone")
[1] ""
On Fri, Sep 12, 2008 at 9:15 AM, Oliver Bandel
<[EMAIL PROTECTED]> wrote:
> Hello,
>
>
> I have string
d secondly(n) -> 4000 + n. If 'n' divides evenly
into 3600 for 'secondly(n)', the return code will be the same as
'hourly(n/3600)'. For 'secondly(n)' and 'minutely(n)', if 'n'
divides evenly into 60, the return code
[EMAIL PROTECTED] napsal dne 12.09.2008 15:24:17:
> Dear Ramon,
>
> I don´t know anything about time series, but if you are looking for to
> identify thresholds (i.e. breaking points ; broken stick ; piece-wise
> regression) segmented package may help you.
Or maybe strucchange.
Regards
Petr
>
Hi
[EMAIL PROTECTED] napsal dne 09.09.2008 18:44:34:
> For the command 'spectrum' I read:
>
> The spectrum here is defined with scaling 1/frequency(x), following
S-PLUS.
> This makes the spectral density a density over the range
(-frequency(x)/2,
> +frequency(x)/2], whereas a more common sca
Dear Ramon,
I don´t know anything about time series, but if you are looking for to
identify thresholds (i.e. breaking points ; broken stick ; piece-wise
regression) segmented package may help you.
Cheers,
miltinho astronauta
brazil
On Fri, Sep 12, 2008 at 9:39 AM, Ramon Hidalgo <[EMAIL PROTECTE
Hi
[EMAIL PROTECTED] napsal dne 11.09.2008 19:11:55:
> I'm working on Human Exon Array 1.0 ST. I'm getting normalized data
> fine but I'm running into problems with QC. QCReport gives me the
> following error:
>
> > load(file= "huex10stv2cdf.rda")
> > [EMAIL PROTECTED] <- "huex10stv2cdf"
Hello,
I have strings with date-/time-data from a logfile.
For a sort-by-time (and other time-related analysis)
I need something to convert the
times to the Unix-epoche-time (better also with parts
of seconds).
Is there something that does that job (or a similar conversion)
in R?
TIA,
Olive
Hi there,
I have a nested call of lapply in which I do a tryCatch statement like this
lapply(1:length(p_names), function(w_idx) {
r <- as.numeric(pos_r[[w_idx]][1])
c <- as.numeric(pos_c[[w_idx]][1])
pos <- c(r,c)
lapply(1:length(p_names[[w_idx]]), function(p_idx,
I've been trying to install R on a PS3 for several weeks now without success
and have read various posts here, and elsewhere, on this and installing R on
other PPC64 architecture machines which have given me some pointers.
Yesterday, after much frustration, I finally succeeded and it runs well. A
Try this:
t(sapply(strsplit(gsub("a", "", ln), "(,)|(,0+)"), as.numeric))
On Fri, Sep 12, 2008 at 9:55 AM, Ajay ohri <[EMAIL PROTECTED]> wrote:
> Hi Henrique ,
>
> Thanks for this. An additional problem is that each row begins and ends
> with a "
>
> So when i try and copy and paste (94 rows) ,
Hi Henrique ,
Thanks for this. An additional problem is that each row begins and ends with a "
So when i try and copy and paste (94 rows) , it does not work with the
last value within a row.
Each row has some 50 x1,y1,0 values.
How to modify the code for this?
Regards,
Ajay
_
> I have a bunch of lines I want to plot using plotCI()
>
> What Id like to know is, how can I connect the points with a line and how
> can I print multiple lines on the same graph?
You might want to have a look at ggplot2, e.g.
http://had.co.nz/ggplot2/geom_errorbar.html, for an alternative way t
Hi,
As I can obtain, of a temporary series, if it has some point where a
significant change of tendency takes place.
Tank's
Ramón Hidalgo
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Thanks very much! options(useFancyQuotes=FALSE) did the trick!
Joanne
P.S.: For reference I'm running R on Linux Fedora 8:
> Sys.getlocale()
[1]
"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
Try this:
lines <- "77.56787,22.09909,0 73.26787,21.09909,0
76.53487,23.09909,0 75.56787,22.09909,0
76.54587,22.09909,0 76.56787,22.03509,0
75.56457,22.05609,0 76.56787,22.05609,0"
ln <- readLines(textConnection(lines))
t(sapply(strsplit(ln, "(,)|(,0+)"), as.numeric))
On Fri, Sep 12, 2008 at 9
Hi All,
I have the following data in one column for each row , (each value is
like x1,y1,z1 space x2,y2,z2 space)
77.56787,22.09909,0 73.26787,21.09909,0
76.53487,23.09909,0 75.56787,22.09909,0
76.54587,22.09909,0 76.56787,22.03509,0
75.56457,22.05609,0 76.56787,22.05609,0
Note all z valu
On Fri, 12 Sep 2008, Joanne Demmler wrote:
Does anyone know what file type sink is outputting to?
To be pedantic, whatever the connection you diverted to is using. For a
default file() connection, this is a plain text file in the encoding of
the current locale.
I've been reverting output
Thanks, that's very helpful! Also, many thanks to Dimitris Rizopoulos and
Robin Williams who both came up with good solutions to my problem.
Pall
-Original Message-
From: Henrique Dallazuanna [mailto:[EMAIL PROTECTED]
Sent: 12 September 2008 12:26
To: Jonsson, Pall
Cc: r-help@r-project
I would like to know about the skewness function bundled in fBasics package.
Is it population measure of skewness or sample estimate?
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Try this:
order(row_median)[names(row_median) == "R3"]
On Fri, Sep 12, 2008 at 5:22 AM, Jonsson, Pall <[EMAIL PROTECTED]
> wrote:
> Dear R gurus,
>
> I have been struggling with this for a while and thought you might be able
> to give me some guidance.
>
> I have a data frame, on which I apply a
Does anyone know what file type sink is outputting to? I've been
reverting output from R using sink and wanted to add it to my Appendix
in LaTex (Lyx) using "Verbatiminput". However, I get an error message as
LaTex is expecting an ASCII file.
Can I set sink somehow to ASCII or is there another
bioinformatics_guy wrote:
I have a bunch of lines I want to plot using plotCI()
What Id like to know is, how can I connect the points with a line and how
can I print multiple lines on the same graph?
At the moment, plotCI doesn't accept a type="b" argument. I'll see if I
can fix it. You migh
Muhammad Azam wrote:
Thanks for the effort but still we are far from the desired result. May be this
example will help you to understand the situation. Example
a1=c(1:12); a1=array(a1,dim=c(3,4)); a2=c(1:12); a2=array(a2,dim=c(3,4));
a3=c(1:16)
a3=array(a3,dim=c(4,4));
a=list(a1,a2,a3);
I am finding real trouble in installing evd library in R for linux
--
View this message in context:
http://www.nabble.com/library-instal-tp19453453p19453453.html
Sent from the R help mailing list archive at Nabble.com.
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try this:
dframe <- data.frame(
Col1 = c(10,20,30),
Col2 = c(2,4,6),
Col3 = c(5,10,7),
row.names = c("R1","R2","R3")
)
mat <- data.matrix(dframe)
rowMedians <- apply(mat, 1, median)
rowMedians[names(rowMedians) %in% "R3"]
I hope it helps.
Best,
Dimitris
> Dear
Thanks Yihui and to others who replied privately.
Very helpful information.
Regards,
Paul
pgseye wrote:
>
> Hi,
>
> Am wanting to save the summary of a PCA to file.
>
> Have tried:
>
>> write.table(summary(PCA), file="PCAvar.txt", sep="\t")
>
> but receive:
>
> Error in as.data.frame.de
Dear R gurus,
I have been struggling with this for a while and thought you might be able to
give me some guidance.
I have a data frame, on which I apply a row function. The result looks to me
like a vector that retains the old row names. I then sort the vector and
subsequently need to be abl
Ah, homework!
Uwe Ligges
Talina Ruiz wrote:
Hi,
I have this problem:
X is hazardous variable N(mean 2, sd=3)
question 1) Find the c value, so that P(X>c)=0.10. using R
question 2) Graph the function N(2,3) and with this graph, explain what
you do in question number 1.
I just found que
well, first you could have a look at
?"[.data.frame"
?subset
and then check the following:
dat <- data.frame(year = sample(seq(2000, 2008, 2), 100, TRUE), y =
rnorm(100))
subset(dat, year == 2002)
dat[dat$year == 2002, ]
# or
subset(dat, year > 2002)
dat[dat$year > 2002, ]
I hope it helps.
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