Bonjour,
je travaille actuellement sur le logiciel R et mets en place des procédures
d'analyse de données :
sur des procédures de classification employant la commande dist(), je rencontre
des problèmes de memoire : avec des data.frame d'environ 100 000 lignes
j'obtiens le message d'erreur suivan
dear list,
i have a problem using the par function.
in one graphic device i want to have two plots so i tried to use
par(mfror=c(1,2)).
of course it worked out, but the height now is twice the length of the width
for each single plot.
what i actually wanted is something like par(mfrow=c(2,2)) whe
Le 28.04.2006 09:51, anthony BOND a écrit :
> Bonjour,
> je travaille actuellement sur le logiciel R et mets en place des procédures
> d'analyse de données :
> sur des procédures de classification employant la commande dist(), je
> rencontre des problèmes de memoire : avec des data.frame d'enviro
hello
I need to know more about logistic regression with
interactive parameters.
Could you provide document, link etc..?
kind regards
--
Ahmet Temiz
Jeoloji Müh.
Afet İşleri Genel Müdürlüğü
Deprem Araştırma Dairesi
Tel: (312) 287 89 51 veya (312) 287 26 80/1547
Faks: (312) 287 89 51
E. Posta: [
On Fri, 2006-04-28 at 10:07 +0200, Stefan Semmeling wrote:
> dear list,
>
> i have a problem using the par function.
>
> in one graphic device i want to have two plots so i tried to use
> par(mfror=c(1,2)).
> of course it worked out, but the height now is twice the length of the width
> for each
Another approach is to use the framework provided by package grid.
If you want to use R base graphics, you might want to look at package
gridBase as well.
BTW: There is the book on R Graphics by Paul Murrell ...
Best,
Uwe Ligges
Gavin Simpson wrote:
> On Fri, 2006-04-28 at 10:07 +0200, Stefan
Hi all,
I am trying to filter the element of a df that start with "TF", like
this:
alfa =
c(123221,"TF13124",41243,"TF1234",32414,"TF13124","TF14333",2134123,"TF1234")
beta =
c("type_a","type_b","type_a","type_g","type_d","type_a","type_g","type_a","type_g")
mydf = data.frame(alfa,beta)
mydf
tf =
Albert Vilella wrote:
> Hi all,
>
> I am trying to filter the element of a df that start with "TF", like
> this:
>
> alfa =
> c(123221,"TF13124",41243,"TF1234",32414,"TF13124","TF14333",2134123,"TF1234")
> beta =
> c("type_a","type_b","type_a","type_g","type_d","type_a","type_g","type_a","type_g"
Hi,
I'm using the power.law.fit function from the igraph package to fit a
power law distribution to some data. This function returns the power
law exponent as it's only result. I would like to have some sort of
goodness-of-fit and/or error estimate of the exponent returned. This
paper:
ht
> "Jennifer" == Jennifer Lai <[EMAIL PROTECTED]>
> on Thu, 27 Apr 2006 15:36:04 -0400 writes:
Jennifer> Hi,
Jennifer> How do I document setReplaceMethod, such as this,
Jennifer> setGeneric("x<-", function(.Object, value)
Jennifer> standardGeneric("myM
I've written a series of functions that evaluates an integral from -inf to a or
b to +inf using equally spaced quadrature points along a normal distribution
from -10 to +10 moving in increments of .01. These functions are working and
give very good approximations, but I think they are computatio
Dear R help
I have a question on the variance of the binomial probit model.
I have fitted the following model :
> lmer1<-lmer(mp ~ l + op + l*op+ us_lev + bw_lev +(1|tatu) ,
+family = binomial(link="probit"),
+method = 'Laplace',
+data = matings,
I think that you want to use 'grep' after converting the factors to
characters:
> tf = mydf[grep("TF", as.character(mydf$alfa)),]
> tf
alfa beta
2 TF13124 type_b
4 TF1234 type_g
6 TF13124 type_a
7 TF14333 type_g
9 TF1234 type_g
>
On 4/28/06, Albert Vilella <[EMAIL PROTECTED]> wrote:
>
Dear Jinsong Zhao,
In proc reg in SAS, selection=stepwise does (modified) forward selection. In
step() in R, the default method is "backward" when the scope argument is
absent. To do (modified) forward selection, you can specify an initial model
with only a constant, and use the scope argument to
Hello,
I have a question about margins when plotting an unrooted tree
(type="unrooted") with the 'ape' package ver. 1.7. When I plot an unrooted
tree with:
no.margin=TRUE
it seems that the margins are still there. It appears to be only when
type="unrooted".
I'm plotting multiple plots using l
Hi
This could be near to what you want. You need to play around with
width and height to get the exact result you want.
> pdf("test.pdf", width=4, height=8)
> par(mfrow=c(2,1))
> plot(1:10,1:10)
> plot(10:1,1:10)
> dev.off()
HTH
Petr
On 28 Apr 2006 at 10:07, Stefan Semmeling wrote:
From:
Hi R-Experts,
I have a vector of length 72. I want to break it into 12 parts and want to take
standerd deviation of each group. Please help me in this regard.
Thanks,
Sumanta.
-
[[alternative HTML version deleted]]
_
sumanta basak wrote:
> Hi R-Experts,
>
> I have a vector of length 72. I want to break it into 12 parts and want to
> take standerd deviation of each group. Please help me in this regard.
x <- 1:72
apply(matrix(x, ncol=12), 2, sd)
Uwe Ligges
> Thanks,
> Sumanta.
>
>
> --
Jinsong Zhao wrote:
> Dear all,
>
> I have encountered a problem when perform stepwise regression.
You have more problems than you know.
> The dataset have more 9 independent variables, but 7 observation.
Why collect any data? You can get great fits using random numbers using
this procedure.
Try this:
tapply(x, cut(x, 12), sd)
On 4/28/06, sumanta basak <[EMAIL PROTECTED]> wrote:
> Hi R-Experts,
>
> I have a vector of length 72. I want to break it into 12 parts and want to
> take standerd deviation of each group. Please help me in this regard.
>
> Thanks,
> Sumanta.
>
>
> --
Brian:
> Thanks for the suggestion, but tomorrow I am teaching a little seminar
> for my department trying to convince people about how wonderful R is.
> These people are all Stata users, and they really like the idea that
> they only have to type ", robust" to get het. consistent std. errors.
Wi
You didn't say _how_ you want the vector to be broken up,
so you get two different answers from Uwe and Gabor. Uwe's
answer group every six elements into one group, in the order
they appear in the vector (which, BTW, can be simplified to
just sd(matrix(x, ncol=12)). Gabor's answer put the smalles
Good point.
Following Andy's comment sd(matrix(sort(x), nc=12))
could also be used if you want them broken up
by 6 smallest, next 6 smallest, etc. although
there might be differences in the case of ties.
Using tapply here are a number of ways of breaking
it up (the first three give the same answe
Thanks, Martin! It worked. With other methods I didn't have to quote the
function name to export them in the NAMESPACE file. But with
replaceMethod, I have to quote the function name without getting any
error during package build time. Thanks!
Regards,
Jennifer
Martin Maechler wrote:
>>"
Hello r-help,
I have a couple of time-series of different length and I would like to
produce a simple overview plot showing the autocorrelation functions of
the series. The time-series are stored in a dataframe like this:
> test.data
item year value
1 xxx 1961
On 4/28/2006 1:48 AM, Saptarshi Guha wrote:
> Hi,
> I just installed R-2.3 for Tiger 10.4.6 on a PPC. However, though i
> successfully installed RGL on R.2.2, this time it doesn't compile.
> I get this error when trying the R INSTALL command
>
> In file included from Texture.hp
Try this:
lapply(names(tslist), function(nm) acf(tslist[[nm]], main = nm))
On 4/28/06, Ulf Mehlig <[EMAIL PROTECTED]> wrote:
> Hello r-help,
>
> I have a couple of time-series of different length and I would like to
> produce a simple overview plot showing the autocorrelation functions of
> the
Hi,
this function uses maximum likelihood estimation to fit the exponent, and
returns an mle object. See ?"mle-class" for the details.
Just to give you an example for getting confidence intervals:
library(igraph)
data <- sample (1:10, 10, rep=TRUE, prob=(1:10)^-2.4)
res <- power.law.
On Fri, 28 Apr 2006, Jinsong Zhao wrote:
> Dear all,
>
> I have encountered a problem when perform stepwise regression.
> The dataset have more 9 independent variables, but 7 observation.
>
The functions in the "leaps" package can do subset selection for data sets
with more variables than observ
On Thu, 27 Apr 2006, Brian Quinif wrote:
> John,
>
> Thanks for the suggestion, but tomorrow I am teaching a little seminar
> for my department trying to convince people about how wonderful R is.
> These people are all Stata users, and they really like the idea that
> they only have to type ", rob
copydir.bat wont work for libraries such as clim.pact, haplo.stats,
hier.part, pls.pcr, R.matlab, R.oo. It will truncate new directories as
clim, haplo, hier, pls, R.
On 4/26/06, Thomas Harte <[EMAIL PROTECTED]> wrote:
>
> hi all,
>
> is there a new mechanism in R 2.3.0 for copying libraries from,
Nope. I've never even used tcltk (to my knowledge!).
Oddly, I can't find any error messages anywhere. Even when I try to
launch R GUI with the OS X Console utility open, nothing is being
reported (that I can find).
On Apr 27, 2006, at 9:04 PM, Rob J Goedman wrote:
> Hi Joran
>
> Do you pos
Thomas Lumley wrote:
> On Thu, 27 Apr 2006, Brian Quinif wrote:
>
>
>>John,
>>
>>Thanks for the suggestion, but tomorrow I am teaching a little seminar
>>for my department trying to convince people about how wonderful R is.
>>These people are all Stata users, and they really like the idea that
>>
I figured out my problem. I just used locator() to find where I wanted my
x.lim and y.lim, and set them manually.
Jerry
>Hello,
> I have a question about margins when plotting an unrooted tree
>(type="unrooted") with >the 'ape' package ver. 1.7. When I plot an unrooted
>tree with: no.margin
I have a very basic question about limma.
Assume I have experiments from 3 or more RNA sources in a reference
design. It is easy to define individual contrasts but I want to specify a
contrast matrix that tests for significant differences among ALL the
different RNA sources (i.e. the analogous
Hi Max,
Max Kauer wrote:
> I have a very basic question about limma.
> Assume I have experiments from 3 or more RNA sources in a reference
> design. It is easy to define individual contrasts but I want to specify a
> contrast matrix that tests for significant differences among ALL the
> differ
A corrected version is now in batchfiles_0.2-8.zip in:
http://cran.r-project.org/contrib/extra/batchfiles/
and will propogate to the mirrors shortly.
On 4/28/06, Xiaohua Dai <[EMAIL PROTECTED]> wrote:
> copydir.bat wont work for libraries such as clim.pact, haplo.stats,
> hier.part, pls.pcr,
I keep my windows XP R updated as soon as a new version is available in
www.r-project.org. After re-installing the new version of R I reinstall all
the extra packages my programs need by manually selecting them in the list
popping up from the "install packages" menu (from an Italian mirror), aft
Hi,
I am trying to use rm.outlier method but encountering following error:
> y <- rnorm(100)
> rm.outlier(y)
Error:
Error in if (nrow(x) != ncol(x)) stop("x must be a square matrix") :
argument is of length zero
Whats wrong here?
TIA
Sachin
__
I am stuck in the installation of "R" on ALPHA runing OSF1/V5.1 for
individual use.
When I ran "./configure CC=cc F77=f77 Make=gmake ", the following
message came out,
506426:/usr/users/1/mwang2/R/R-2.1.1/bin/exec/R: /sbin/loader: Error:
libreadline.so.4: symbol "tgetflag" unresolved
506426:/u
On Fri, 2006-04-28 at 11:17 -0700, Sachin J wrote:
> Hi,
>
> I am trying to use rm.outlier method but encountering following error:
>
> > y <- rnorm(100)
> > rm.outlier(y)
>
> Error:
> Error in if (nrow(x) != ncol(x)) stop("x must be a square matrix") :
> argument is
Hi Marc:
I am using rm.outlier() function from outliers package (reference: CRAN
package help).
You are right. I too couldn't find this error message in rm.outlier function.
Thats why I am unable to understand the cause of error. Any further thoughts? I
will take a look at the robust ana
Sachin,
I don't have a definitive thought, but some possibilities might be a
conflict somewhere in your environment with a local function or with one
in the searchpath.
Use ls() to review the current objects in your environment to see if
something looks suspicious. It did not look like 'outliers'
Salut
pour y remedier tu peux employer les mécanismes de base de données
si tu bosses avec Linux essaye RMySQL ou RODBC sous windows. Tu pourras ainsi
faire ce que tu veux si ça ne marche pas lits les archives on y a longuement
discuté de ce sujet
- Message d'origine
De
Thank you Marc. That was of great help. There was some problem with the
environment. I closed and reopened the workspace. Works fine now.
Sachin
"Marc Schwartz (via MN)" <[EMAIL PROTECTED]> wrote: Sachin,
I don't have a definitive thought, but some possibilities might be a
conflict somew
Hello,
I recently installed R 2.3 for Tiger 10.4.6. Though the commandline
R works (i.e R at the Terminal), the R GUI hangs while trying to load R.
Once it did run, and gave many errors regarding Quicktime and DivX
libraries.
Is there anything I can do?
as.character.factor contains this line (where cx=levels(x)[x]):
if ("NA" %in% levels(x)) cx[is.na(x)] <- ""
Is it possible that this is no longer the desired behavior? These
two results don't seem very consistent:
> as.character(as.factor(c("AB", "CD", NA)))
[1] "AB" "CD" NA
> is.na(.Last.va
I would like to use aggregate() to combine statistics
for several days in a data frame. My data frame looks
similar to this:
datetype count value
1 2006-04-01 A 10 99.6
2 2006-04-01 B 4 33.2
3 2006-04-02 A 22 43.2
4 2006-04-02 B 8 44.9
5
Does this do what you want?
> x
date type count value
1 2006-04-01A10 99.6
2 2006-04-01B 4 33.2
3 2006-04-02A22 43.2
4 2006-04-02B 8 44.9
5 2006-04-03A12 12.4
6 2006-04-03B14 18.5
> y <- lapply(split(1:nrow(x), x$type), function(.ind){
Here are three possibilities:
1. aggregate on the columns that you want to sum and aggregate on
the columns that you want to average and then merge them:
By <- A[, 2, drop = FALSE]
merge(aggregate(A[, 3, drop = FALSE], By, sum),
aggregate(A[, 4, drop = FALSE], By, mean))
2. use by:
f <- fu
I am using R 2.2.1 on a Windows 2000 PC.
When I do a grid() after the boxplot
it overprints the boxplot:
> boxplot(count ~ spray, data = InsectSprays, col = "lightgray")
> grid(nx=NA, ny=NULL)
>
if I try the panel.first
> boxplot(count ~ spray, data = InsectSprays, col = "lightgray",
+
I have to be making a riddiculously silly ommission.
when I run the fillowing i get the cloud plot ok. But I cant figure out what
I am missing out when I call wireframe.
Any help would be appreciated.
x<-runif(100)
y<-rnorm(100)
z<-runif(100)
temp <-data.frame(x,y,z)
wireframe(x~y*z,temp)
cloud
> RSiteSearch("imagemagick")
reveals a few functions whose help pages reference ImageMagick and R-help
postings that pertain to image manipulation.
However, I did notice anything that amounts to an 'interface'.
Certainly, you can use 'shell' to run command line tools like 'convert' et
cetera
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