[R] Problems with passing ... to a function
Dear useRs!
I have written a function that should pass argument m to the next
function, however it does not! Please have a look at the function below that
shows a problem and tell me what I am missing. As you can see, the blocks
argument is passed corectly, while m is not.
Best,
Ales Ziberna
opt.par.new-function(
#function for optimizig partition in blockmodeling
M, #matrix
clu, #initial partition
maxiter=50, #maximum number of iterations
trace.iter=FALSE, #save a result of each iteration or only the best
(minimal error)
switch.names=is.null(BLOCKS), #should partitions that only differ in group
names be considert equal (is c(1,1,2)==c(2,2,1))
save.initial.param=TRUE, #should the initial parameters be saved
approach,
... #other arguments to called functions - to 'crit.fun'
){
if(save.initial.param)initial.param-tryCatch(lapply(as.list(sys.frame(sys.nframe())),eval),error=function(...)return(error))#saves
the inital parameters f-function(blocks,...)(print(blocks)) f(...)
f-function(m,...)(print(m))
f(...)}opt.par.new(M=M,m=1,clu=rep(1:2,times=c(7,8)),blocks=c(null,reg),approach=val)
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Re: [R] Huber location estimate
On Thu, 22 Dec 2005, Murray Jorgensen wrote:
We have a choice when calculating the Huber location estimate:
set.seed(221205)
y - 7 + 3*rt(30,1)
That's Cauchy, BTW, a very extreme case.
library(MASS)
huber(y)$mu
[1] 5.9117
coefficients(rlm(y~1))
(Intercept)
5.9204
I was surprised to get two different results. The function huber() works
directly with the definition whereas rlm() uses iteratively reweighted
least squares.
My surprise is because I vaguely remember
@ARTICLE{hw77,
author = {Holland, P. W. and Welsch, R. E.},
title = {Robust Regression using Iteratively Reweighted Least-Squares},
journal = {Communications in Statistics: Theory and Methods},
volume = {A6(9)},
number = {},
pages = {813-827},
year= {1977}
}
as saying that the two methods were equivalent. Obviously they aren't
quite. Comments welcome.
Scale estimation differs. You have (unfairly to the uncredited author)
not included all the output:
huber(y)
$mu
[1] 5.911719
$s
[1] 4.096697
rlm(y~1)
Call:
rlm(formula = y ~ 1)
Converged in 5 iterations
Coefficients:
(Intercept)
5.920354
Degrees of freedom: 30 total; 29 residual
Scale estimate: 3.75
Note that the huber() scale estimate is the initial MAD, whereas rlm
iterates. Because during iteration the 'center' for MAD is known to be
zero, the results differ. For symmetric distributions there is little
difference, but your sample is not close to symmetric.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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[R] program work
Hi i'm enrico pavan and I work for TSW in Treviso, Italy and also I study Statistic at Treviso's University. TSW is an agency that work in search engine marketing and with R we would like to make some statistic about the site web work using logs file. Is it possible with R? Does R support and read logs file? I'll wait a reply, Best Regard Enrico Pavan Stat. Analytic [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Hiding a function from ls()
I need to define a small helper function which should not be listed by ls(). What is the best and cleanest way of achieving this? -- Erich Neuwirth, University of Vienna Faculty of Computer Science Computer Supported Didactics Working Group Visit our SunSITE at http://sunsite.univie.ac.at Phone: +43-1-4277-38624 Fax: +43-1-4277-9386 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Huber location estimate
Prof Brian Ripley wrote: On Thu, 22 Dec 2005, Murray Jorgensen wrote: We have a choice when calculating the Huber location estimate: set.seed(221205) y - 7 + 3*rt(30,1) That's Cauchy, BTW, a very extreme case. Sure, the sort of situation where one might want a robust estimator. [...] Note that the huber() scale estimate is the initial MAD, whereas rlm iterates. Because during iteration the 'center' for MAD is known to be zero, the results differ. For symmetric distributions there is little difference, but your sample is not close to symmetric. [Blush] yes I knew that and somehow I forgot. But leave rlm() alone for a while and do IRLS with fixed scale: th - median(y) s - mad(y) # paste this in a few times: w - ifelse((y-th 1.345*s y-th-1.345*s), 1, 1.345*s/abs(y-th)) th - weighted.mean(y,w) th We converge to th [1] 5.9203 close to the answer given by rlm() different from huber(y)$mu [1] 5.9117 So the variable scale does not account for the difference. Murray Jorgensen -- Dr Murray Jorgensen http://www.stats.waikato.ac.nz/Staff/maj.html Department of Statistics, University of Waikato, Hamilton, New Zealand Email: [EMAIL PROTECTED]Fax 7 838 4155 Phone +64 7 838 4773 wkHome +64 7 825 0441Mobile 021 1395 862 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Hiding a function from ls()
Erich == Erich Neuwirth [EMAIL PROTECTED]
on Thu, 22 Dec 2005 10:06:38 +0100 writes:
Erich I need to define a small helper function
Erich which should not be listed by ls().
Erich What is the best and cleanest way of achieving this?
it depends:
1) if it's part of a package, the best and cleanest is
- have the package use a namespace
- do nothing special at, so your helper function is not
exported from the namespace, and hence not much visible at all.
The encouraged use of much more such helper function
(i.e. modular programming) is one prominent reason
for using name spaces.
2) If it's just part of a small script {anything else should be
in a package anyway :-)},
just start the function name with a . (dot).
Then, ls() only would list it when used with the additional
argument setting ls(..., all.names = TRUE)
Martin Maechler, ETH Zurich
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Re: [R] bivariate kernel density estimates at point locations (rather than at grid locations)
On Wed, 21 Dec 2005 12:21:54 -0500, Strickland, Matthew wrote: SM Hello Dr. Adelchi Azzalini, SM Dr. Strickland, your message was directed to the whole r-help list with no CC to myself, and sometimes I do not have the chance to browse the r-help list for weeks.. SM Thank you for your quick response to my question that I posted on SM the r-help board about bivariate kernel density estimates. I have SM been using your sm package the past few days and have encountered SM a problem with estimating the density for only 1 point. I am using SM R 2.2.0, sm version 2.1-0 on a Windows machine. Example code SM below: SM SM #The code below creates 3 point locations SM SM x.locs = c(74, 75, 77) SM y.locs = c(64, 63, 61) SM points = cbind(x.locs, y.locs) SM SM #If I send this data into sm.density everything works fine. SM SM dens = sm.density(points, h=c(1, 1)) SM SM #However, if I only wish to send 1 point location to sm.density, SM #i.e., SM SM points.2 =points[1,] SM dens.2 = sm.density(points.2, h=c(1, 1)) SM SM R returns to me the error:length(h) != 1 formally, the error is due to this R is.vector(points.2) [1] TRUE sm.density receives a vector of length 2, and it works for that case: estimation of a one-dimensional density from which you supplied two data values. Then it finds a two-dimensional h and there it complains. On the statistical side, I cannot follow the logic of estimating nonparametrically a density function on the basis of only one (supposed bivariate) observation. SM SM It appears to me that sm.density thinks that my 1 point is a SM 1-dimensional location rather than a 2-dimensional location, and I SM am getting an error when I request a bivariate kernel. Do you SM have any suggestions? SM I am not sure to grasp what you have in mind; is it perhaps that you want the following? dens.2 = sm.density(points, h=c(1, 1), eval.points=points, eval.grid=FALSE) print(dens.2$estimate[1]) best regards, Adelchi SM Best, SM Matt SM SM SM -Original Message- SM From: Adelchi Azzalini [mailto:[EMAIL PROTECTED] SM Sent: Friday, December 16, 2005 2:42 AM SM To: Strickland, Matthew SM Cc: r-help@stat.math.ethz.ch; [EMAIL PROTECTED] SM Subject: Re: [R] bivariate kernel density estimates at point SM locations (rather than at grid locations) SM SM On Thu, 15 Dec 2005 14:21:17 -0500, Strickland, Matthew wrote: SM SM SM Hi, SM SM SM SM My data consists of a set of point locations (x,y). SM SM SM SM I would like to know if there is a procedure for bivariate SM SM kernel density estimation in R that returns the density SM SM estimates at the observed point locations rather than at grid SM SM locations. I have looked at a number of different routines SM SM and they all seem to return estimates at grid locations. SM SM SM SM One option is to use (from package sm), SM sm.density(xy, eval.points=xy, eval.grid=FALSE) where xy in a SM (n\times 2) matrix. SM SM Best wishes, SM Adelchi Azzalini SM SM -- SM Adelchi Azzalini [EMAIL PROTECTED] Dipart.Scienze SM Statistiche, Università di Padova, Italia tel. +39 049 8274147, SM http://azzalini.stat.unipd.it/ SM SM __ SM R-help@stat.math.ethz.ch mailing list SM https://stat.ethz.ch/mailman/listinfo/r-help SM PLEASE do read the posting guide! SM http://www.R-project.org/posting-guide.html SM __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Huber location estimate
Murray == Murray Jorgensen [EMAIL PROTECTED] on Thu, 22 Dec 2005 22:13:45 +1300 writes: Murray Prof Brian Ripley wrote: On Thu, 22 Dec 2005, Murray Jorgensen wrote: We have a choice when calculating the Huber location estimate: set.seed(221205) y - 7 + 3*rt(30,1) That's Cauchy, BTW, a very extreme case. Murray Sure, the sort of situation where one might want a robust estimator. Murray [...] Note that the huber() scale estimate is the initial MAD, whereas rlm iterates. Because during iteration the 'center' for MAD is known to be zero, the results differ. For symmetric distributions there is little difference, but your sample is not close to symmetric. Murray [Blush] yes I knew that and somehow I forgot. But leave rlm() alone for Murray a while and do IRLS with fixed scale: Murray th - median(y) Murray s - mad(y) Murray # paste this in a few times: Murray w - ifelse((y-th 1.345*s y-th-1.345*s), 1, 1.345*s/abs(y-th)) Murray th - weighted.mean(y,w) Murray th Murray We converge to th Murray [1] 5.9203 Murray close to the answer given by rlm() different from huber(y)$mu Murray [1] 5.9117 Murray So the variable scale does not account for the difference. No, the main difference is the different default: huber() has k=1.5 and rlm() has k=1.345 : Try this set.seed(221205) y - 7 + 3*rt(30,1) str(huber(y, k = 1.345), digits = 5) ## List of 2 ## $ mu: num 5.9203 ## $ s : num 4.0967 str(rlm(y ~ 1)[c(coefficients, s)], digits = 5) # ## (edited to) ## $ coefficients: num 5.9204 ## $ s : num 3.7463 which gives 'mu' very close, even for the iterated vs. non-iterated scales. Martin Maechler, ETH Zurich __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Problems with passing ... to a function
Thanks to the helpful R-hlep-er, the problem has been solved.
'm' was matching 'maxiter'.
Thanks again,
Ales Ziberna
- Original Message -
From: Ales Ziberna [EMAIL PROTECTED]
To: R-help r-help@stat.math.ethz.ch
Sent: Thursday, December 22, 2005 9:25 AM
Subject: Problems with passing ... to a function
Dear useRs!
I have written a function that should pass argument m to the next
function, however it does not! Please have a look at the function below that
shows a problem and tell me what I am missing. As you can see, the blocks
argument is passed corectly, while m is not.
Best,
Ales Ziberna
opt.par.new-function(
#function for optimizig partition in blockmodeling
M, #matrix
clu, #initial partition
maxiter=50, #maximum number of iterations
trace.iter=FALSE, #save a result of each iteration or only the best
(minimal error)
switch.names=is.null(BLOCKS), #should partitions that only differ in group
names be considert equal (is c(1,1,2)==c(2,2,1))
save.initial.param=TRUE, #should the initial parameters be saved
approach,
... #other arguments to called functions - to 'crit.fun'
){
if(save.initial.param)initial.param-tryCatch(lapply(as.list(sys.frame(sys.nframe())),eval),error=function(...)return(error))#savesthe
inital parameters f-function(blocks,...)(print(blocks))
f(...)f-function(m,...)(print(m))f(...)}opt.par.new(M=M,m=1,clu=rep(1:2,times=c(7,8)),blocks=c(null,reg),approach=val)
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Re: [R] Huber location estimate
D'oh! Apologies for wasting everybody's time! Murray Martin Maechler wrote: Murray == Murray Jorgensen [EMAIL PROTECTED] on Thu, 22 Dec 2005 22:13:45 +1300 writes: Murray Prof Brian Ripley wrote: On Thu, 22 Dec 2005, Murray Jorgensen wrote: We have a choice when calculating the Huber location estimate: set.seed(221205) y - 7 + 3*rt(30,1) That's Cauchy, BTW, a very extreme case. Murray Sure, the sort of situation where one might want a robust estimator. Murray [...] Note that the huber() scale estimate is the initial MAD, whereas rlm iterates. Because during iteration the 'center' for MAD is known to be zero, the results differ. For symmetric distributions there is little difference, but your sample is not close to symmetric. Murray [Blush] yes I knew that and somehow I forgot. But leave rlm() alone for Murray a while and do IRLS with fixed scale: Murray th - median(y) Murray s - mad(y) Murray # paste this in a few times: Murray w - ifelse((y-th 1.345*s y-th-1.345*s), 1, 1.345*s/abs(y-th)) Murray th - weighted.mean(y,w) Murray th Murray We converge to th Murray [1] 5.9203 Murray close to the answer given by rlm() different from huber(y)$mu Murray [1] 5.9117 Murray So the variable scale does not account for the difference. No, the main difference is the different default: huber() has k=1.5 and rlm() has k=1.345 : Try this set.seed(221205) y - 7 + 3*rt(30,1) str(huber(y, k = 1.345), digits = 5) ## List of 2 ## $ mu: num 5.9203 ## $ s : num 4.0967 str(rlm(y ~ 1)[c(coefficients, s)], digits = 5) # ## (edited to) ## $ coefficients: num 5.9204 ## $ s : num 3.7463 which gives 'mu' very close, even for the iterated vs. non-iterated scales. Martin Maechler, ETH Zurich -- Dr Murray Jorgensen http://www.stats.waikato.ac.nz/Staff/maj.html Department of Statistics, University of Waikato, Hamilton, New Zealand Email: [EMAIL PROTECTED]Fax 7 838 4155 Phone +64 7 838 4773 wkHome +64 7 825 0441Mobile 021 1395 862 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Logistic regression to select genes and estimate cutoff point?
You could take a look at www.bioconductor.org limma would be a good starting point. hth ido __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] No PNG support 2.2.1
Hi everybody, When trying to save my plots as PNG images, I get this error message? Can anybody give me a solution? png(my_plot.png) Error in X11(paste(png::, filename, sep = ), width, height, pointsize, : unable to start device PNG In addition: Warning message: no png support in this version of R Thank you very much, Sylvain -- Sylvain Brohée Service de Conformation des Macromolécules Biologiques et de Bioinformatique (SCMBB) Université Libre de Bruxelles Boulevard du Triomphe - CP263 1050 Bruxelles (Belgium) Tel : +32(0)2/650.54.34 Fax : +32(0)2/650.54.25 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] program work
Does R support and read logs file? R can read log files if they are text files. ?read.table hth ido __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] No PNG support 2.2.1
I am effectively running R under Suse linux 9.2... However libpng is compiled and works! Thanks , Sylvain On Thursday 22 December 2005 12:02, Landini Massimiliano wrote: On Thu, 22 Dec 2005 12:01:00 +0100, you wrote: |=[:o) Hi everybody, |=[:o) |=[:o) When trying to save my plots as PNG images, I get this error | message? =[:o) Can anybody give me a solution? |=[:o) |=[:o) png(my_plot.png) |=[:o) Error in X11(paste(png::, filename, sep = ), width, height, | pointsize, : =[:o) unable to start device PNG |=[:o) In addition: Warning message: |=[:o) no png support in this version of R |=[:o) |=[:o) Thank you very much, |=[:o) you are using kmail so i think that your OS is *nix: do you have libpng compiled??? ;o) |=[:o) Sylvain |=[:o) |=[:o) -- |=[:o) Sylvain Brohée |=[:o) |=[:o) Service de Conformation des Macromolécules |=[:o) Biologiques et de Bioinformatique (SCMBB) |=[:o) Université Libre de Bruxelles |=[:o) Boulevard du Triomphe - CP263 |=[:o) 1050 Bruxelles (Belgium) |=[:o) Tel : +32(0)2/650.54.34 |=[:o) Fax : +32(0)2/650.54.25 |=[:o) |=[:o) __ |=[:o) R-help@stat.math.ethz.ch mailing list |=[:o) https://stat.ethz.ch/mailman/listinfo/r-help |=[:o) PLEASE do read the posting guide! | http://www.R-project.org/posting-guide.html --- -- Landini dr. Massimiliano Tel. mob. (+39) 347 140 11 94 Tel./Fax. (+39) 051 762 196 e-mail: numero (dot) primo (at) tele2 (dot) it --- -- Legge di Hanggi: Più stupida è la tua ricerca, più verrà letta e approvata. Corollario alla Legge di Hanggi: Più importante è la tua ricerca, meno verrà capita. --- -- -- Sylvain Brohée Service de Conformation des Macromolécules Biologiques et de Bioinformatique (SCMBB) Université Libre de Bruxelles Boulevard du Triomphe - CP263 1050 Bruxelles (Belgium) Tel : +32(0)2/650.54.34 Fax : +32(0)2/650.54.25 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] strsplit with dataframes
Hello fellow R people, I can not figure out a pretty way to use strplit with vectors Imagine that I got the following data from someone with ID's representing several factors ID data A1-B1-t10 A1-B1-t21 A1-B2-t15 A1-B2-t210 A1-B10-t1 0 A1-B10-t2 1 A1-B20-t1 5 A1-B20-t2 10 ... I would like to turn this dataframe to station substation time data A1 B1 t1 0 A1 B1 t2 1 A1 B2 t1 5 A1 B2 t2 10 A1 B10 t1 0 A1 B10 t2 1 A1 B20 t1 5 A1 B20 t2 10 ... This must surely be done easily, but there are not an example like this in ?strsplit Cheers, - Henrik Andersson Netherlands Institute of Ecology - Centre for Estuarine and Marine Ecology P.O. Box 140 4400 AC Yerseke Phone: +31 113 577472 [EMAIL PROTECTED] http://www.nioo.knaw.nl/ppages/handersson __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] strsplit with dataframes
try: cbind.data.frame(do.call(rbind, strsplit(as.character(yourdf$ID), -)), yourdf$data) Henrik Andersson a écrit : Hello fellow R people, I can not figure out a pretty way to use strplit with vectors Imagine that I got the following data from someone with ID's representing several factors ID data A1-B1-t1 0 A1-B1-t2 1 A1-B2-t1 5 A1-B2-t2 10 A1-B10-t1 0 A1-B10-t2 1 A1-B20-t1 5 A1-B20-t2 10 ... I would like to turn this dataframe to station substation time data A1 B1 t1 0 A1 B1 t2 1 A1 B2 t1 5 A1 B2 t2 10 A1 B10 t1 0 A1 B10 t2 1 A1 B20 t1 5 A1 B20 t2 10 ... This must surely be done easily, but there are not an example like this in ?strsplit Cheers, - Henrik Andersson Netherlands Institute of Ecology - Centre for Estuarine and Marine Ecology P.O. Box 140 4400 AC Yerseke Phone: +31 113 577472 [EMAIL PROTECTED] http://www.nioo.knaw.nl/ppages/handersson __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Plot problems: xlim
Hi, Still fresh in R, tried to figure this out, now on my second day running with no luck (and a pile of hair on my desk) so I have thrown in the towel and would like to ask for some help. Here is what I am trying to do. I am trying to plot a distribution, I have 99 points, bound in the range xlim.min: -0.0173 xlim.max: 0.02103 However, I have a value outside this range (0.2454959) which I would like to add to the plot as a line and to do this I use abline(v=0.2454959) This is what I write xlim = c(-0.02, 0.3) denz - density(morp) plot.density(denz, xlim = xlim, ylim = c(0,70)) hist(morp, freq=F, add= T) abline(v=0.2454959) Without any options, plot.density spreads out nicely, however, naturally, the line I want to add is not plotted since it is well outside the range automatically determined by plot.density hence the need to add xlim however this produces something I dont find aesthetically appealing. The plot is squeezed out into a very lean bell shape. So (finally) my question, how can i widen the spread of my plot and yet also be able to add my xline. Many thanks Ronnie __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Plot problems: xlim
Please give an example of your data. set.seed(231) morp - rnorm(20) range(morp) [1] -2.311664 1.650254 You can plot 2 histograms, one of them with the extreme value: par(mfrow=c(2,1)) hist(morp, breaks=10, freq=F) lines(density(morp)) par(mfrow=c(1,2)) hist(morp, breaks=10, freq=F) lines(density(morp)) hist(morp, breaks=seq(min(morp), max(morp), length=10), xlim=c(-3, 13), freq=F) lines(density(morp)) abline(v=7.5, lty=3) Ronnie Babigumira a écrit : Hi, Still fresh in R, tried to figure this out, now on my second day running with no luck (and a pile of hair on my desk) so I have thrown in the towel and would like to ask for some help. Here is what I am trying to do. I am trying to plot a distribution, I have 99 points, bound in the range xlim.min: -0.0173 xlim.max: 0.02103 However, I have a value outside this range (0.2454959) which I would like to add to the plot as a line and to do this I use abline(v=0.2454959) This is what I write xlim = c(-0.02, 0.3) denz - density(morp) plot.density(denz, xlim = xlim, ylim = c(0,70)) hist(morp, freq=F, add= T) abline(v=0.2454959) Without any options, plot.density spreads out nicely, however, naturally, the line I want to add is not plotted since it is well outside the range automatically determined by plot.density hence the need to add xlim however this produces something I dont find aesthetically appealing. The plot is squeezed out into a very lean bell shape. So (finally) my question, how can i widen the spread of my plot and yet also be able to add my xline. Many thanks Ronnie __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Install Rmpi on Fedora with mpich2 installed.
I don't know Fedora but you should have installed the dev(-elopment) packages too (like mpich-dev or similar for instance). Usually under unixes those supplementary packages contain all the header files such as mpi.h needed to compile a C program. Vittorio Alle 20:45, martedì 20 dicembre 2005, Ye, Bin ha scritto: Thank you very much, Martin! I've tried that already, but it still can't find the mpi.h file. Any other suggestions? Bin -Original Message- From: Martin Morgan [mailto:[EMAIL PROTECTED] Sent: Tue 12/20/2005 2:58 PM To: Ye, Bin Subject: Re: [R] Install Rmpi on Fedora with mpich2 installed. Hi Bin, I don't have direct experience installing Rmpi on mpich2, but you can specify the location of the mpi.h files with commands like ./configure --with-mpi=/usr/local/mpich2 when in the unpacked Rmpi packate, or R CMD INSTALL Rmpi_... --configure-args=--with-mpi=/usr/local/mpich2 when installing the package from the command line. The ... are the results of tab completion to the Rmpi tarball, and the path /usr/local/mpich2 should lead to a direcotry hierarchy such that mpi.h will be found in something like /usr/local/mpich2/include/mpi.h (some insight into what is going on is in the configure.in file). Hope that helps! Martin Ye, Bin [EMAIL PROTECTED] writes: Hi, everyone, I want to install Rmpi on a cluster with Fedora linux. It already installed mpich2, but not lam-mpi. I installed R-2.2.0 on it already. And I got error as below: * Installing *source* package 'Rmpi' ... Try to find mpi.h ... checking for gcc... gcc checking for C compiler default output file name... a.out checking whether the C compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C compiler... yes checking whether gcc accepts -g... yes checking for gcc option to accept ANSI C... none needed checking how to run the C preprocessor... gcc -E checking for egrep... grep -E checking for ANSI C header files... yes checking for sys/types.h... yes checking for sys/stat.h... yes checking for stdlib.h... yes checking for string.h... yes checking for memory.h... yes checking for strings.h... yes checking for inttypes.h... yes checking for stdint.h... yes checking for unistd.h... yes checking mpi.h usability... no checking mpi.h presence... no checking for mpi.h... no Try to find mpi.h ... Cannot find mpi head file Please check if --with-mpi=/usr/local/mpich2/bin/ is right ERROR: configuration failed for package 'Rmpi' ** Removing '/usr/local/R-2.2.0/library/Rmpi' Somehow it can not find the mpi.h which is in usr/local/mpich2. Can anyone kindly give me some hint on what should be done? Will installing lam-mpi solve the problem? If so, should mpich2 be uninstalled first? Or just modify the path will do? Thanks a lot! Bin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] panel order in xyplot
Hi all,
I have a question concerning xyplot. My data is a data.frame looking like that:
In the first column I have numbers from 0 to 23 (hours of a day), the second
column contains the name of a weekday (Day as factor) and the third column
contains the number I am interested in. So as an example, the first five rows
look like that:
Hour Day Freq
1 0 Mo23
2 1 Mo20
3 2 Mo14
4 3 Mo27
5 4 Mo26
To read: On Monday between 0 and 1 o'clock 23 things happened.
The code of the xyplot looks like that:
trellis.device(new = FALSE, col = FALSE)
xyplot(Freq ~ Hour | Day, data = mydata,
xlab = hour, ylab = number, main = xxx, ylim = 1:30,
scales = list(x = list(at = seq(0,24,6), labels = c(0, 6, 12, 18, 24),
cex = 0.7, relation = free),
y = list(tick.number = 5, cex = 0.7)), between = list(x = 0.5, y = 1.5),
layout = c(4,2), aspect = 1,
panel = function(x,y){
panel.grid()
panel.barchart(x, y, horiz = F, aspect = 1, col = red)
}
)
Now my problem:
The plot is ordered by default, that in the first row I can see Friday,
Saturday and Sunday, in the second row there is Monday, Tuesday, Wednesday and
Thursday. I like to order the panels that way that in the first row there
should be Monday-Thursday, the last row to contain Friday - Sunday.
Is there any parameter in xyplot I can add to achieve this result? Any
parameter where I can tell how the panels have to be arranged?
Thanks a lot for any help.
Antje
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Re: [R] Logistic regression to select genes and estimate cutoff point?
Lingsheng Dong wrote: Hi, all, I am new to R or even to statistics. Not sure if the question has a answer. But I couldn't find a straight forward answer in the help mailing list. I need use MicroArray data to select several diagnostic genes between Normal samples and Tumor samples and use these genes to predict unknow samples. Since the sample size is so small and data doesn't follow normal distribution, I am thinking to use logistic regression instead of Student T test to select genes. To make the problem simpler, I assume each gene is independent to each other without interactions. My questions is how I should build up the model: one model for each gene or a multiple variable model to include all genes? Which is the test to compare the discrimination power of each gene? I am thinking it is Wald statistic for the multiple variable model and Maximum likelihood for the single gene models? Am I correct? To estimate the cutoff point, I guess the answer is the gene expression when p=0.5 in the model. Am I on the right direction? Any suggestion is appreciated! Thanks a lot. Lingsheng Just a comment: Do you not have a statistician to work with at your institution? You are new to statistics and are asking a question that would be very difficult to deal with for someone with a PhD in statistics and 20 years of experience. Some of the issues involved are multiple comparisons, false discovery rate, shrinkage, array geometry effects, nonparametric vs. parametric statistics, stability of selected genes, discovery validation, ... -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] data frame
Dear R users: s4 - seq(length=10, from=1, by=5) s-data.frame(s4,s4,s4) I would like to do some modification to s. And I want the form like the following,if it is possible, how should I do? The last column is the sum of previous three column. s4 s4.1 s4.2sum 1 11 26 6 3 11112 4 16622 5 21 111 33 6 26 166 48 7 3121 11 63 8 3626 16 78 9 4131 21 93 10 4636 26 108 Thanks for any help !! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] bVar slot of lmer objects and standard errors
Hello, I am looking for a way to obtain standard errors for emprirical Bayes estimates of a model fitted with lmer (like the ones plotted on page 14 of the document available at http://www.eric.ed.gov/ERICDocs/data/ericdocs2/content_storage_01/000b/80/2b/b3/94.pdf). Harold Doran mentioned (http://tolstoy.newcastle.edu.au/~rking/R/help/05/08/10638.html) that the posterior modes' variances can be found in the bVar slot of lmer objects. However, when I fit e.g. this model: lmertest1-lmer(mathtot~1+(m_escs_c|schoolid),hlmframe) then [EMAIL PROTECTED] is a three-dimensional array with dimensions (2,2,28). The factor schoolid has 28 levels, and there are random effects for the intercept and m_escs_c, but what does the third dimension correspond to? In other words, what are the contents of bVar, and how can I use them to get standard errors? Thanks in advance for your answers and Merry Christmas, Uli Keller __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] data frame
Here is one way. You can change depending on what you want the offsets to be: s4 - seq(length=10, from=1, by=5) s4 [1] 1 6 11 16 21 26 31 36 41 46 f.x - function(vec, n) c(rep(0,n), vec)[1:length(vec)] f.x(s4,2) [1] 0 0 1 6 11 16 21 26 31 36 df - data.frame(s4=s4, s4.1=f.x(s4,2), s4.2=f.x(s4,4)) df s4 s4.1 s4.2 1 100 2 600 3 1110 4 1660 5 21 111 6 26 166 7 31 21 11 8 36 26 16 9 41 31 21 10 46 36 26 df$sum - rowSums(df) df s4 s4.1 s4.2 sum 1 100 1 2 600 6 3 1110 12 4 1660 22 5 21 111 33 6 26 166 48 7 31 21 11 63 8 36 26 16 78 9 41 31 21 93 10 46 36 26 108 On 12/22/05, Rhett Eckstein [EMAIL PROTECTED] wrote: Dear R users: s4 - seq(length=10, from=1, by=5) s-data.frame(s4,s4,s4) I would like to do some modification to s. And I want the form like the following,if it is possible, how should I do? The last column is the sum of previous three column. s4 s4.1 s4.2sum 1 11 26 6 3 11112 4 16622 5 21 111 33 6 26 166 48 7 3121 11 63 8 3626 16 78 9 4131 21 93 10 4636 26 108 Thanks for any help !! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Jim Holtman Cincinnati, OH +1 513 247 0281 What the problem you are trying to solve? [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] data frame
On 12/22/05 9:11 AM, Rhett Eckstein [EMAIL PROTECTED] wrote: Dear R users: s4 - seq(length=10, from=1, by=5) s-data.frame(s4,s4,s4) I would like to do some modification to s. And I want the form like the following,if it is possible, how should I do? The last column is the sum of previous three column. s4 s4.1 s4.2sum 1 11 26 6 3 11112 4 16622 5 21 111 33 6 26 166 48 7 3121 11 63 8 3626 16 78 9 4131 21 93 10 4636 26 108 Rhett, You might want to look at ?embed. It would need some modification, but you can hopefully get the idea. embed(s4,3) Sean __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] panel order in xyplot
On 12/22/05, Antje Döring [EMAIL PROTECTED] wrote:
Hi all,
I have a question concerning xyplot. My data is a data.frame looking like
that:
In the first column I have numbers from 0 to 23 (hours of a day), the second
column contains the name of a weekday (Day as factor) and the third column
contains the number I am interested in. So as an example, the first five
rows look like that:
Hour Day Freq
1 0 Mo23
2 1 Mo20
3 2 Mo14
4 3 Mo27
5 4 Mo26
To read: On Monday between 0 and 1 o'clock 23 things happened.
The code of the xyplot looks like that:
trellis.device(new = FALSE, col = FALSE)
xyplot(Freq ~ Hour | Day, data = mydata,
xlab = hour, ylab = number, main = xxx, ylim = 1:30,
scales = list(x = list(at = seq(0,24,6), labels = c(0, 6, 12, 18,
24), cex = 0.7, relation = free),
y = list(tick.number = 5, cex = 0.7)), between = list(x = 0.5, y =
1.5),
layout = c(4,2), aspect = 1,
panel = function(x,y){
panel.grid()
panel.barchart(x, y, horiz = F, aspect = 1, col = red)
}
)
Now my problem:
The plot is ordered by default, that in the first row I can see Friday,
Saturday and Sunday, in the second row there is Monday, Tuesday, Wednesday
and Thursday. I like to order the panels that way that in the first row
there should be Monday-Thursday, the last row to contain Friday - Sunday.
Is there any parameter in xyplot I can add to achieve this result?
'as.table'.
Any parameter where I can tell how the panels have to be arranged?
Deepayan
--
http://www.stat.wisc.edu/~deepayan/
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[R] add factor to dataframe given ranges
Hi all, I would like to factorize the entries in a dataframe given some groupings. E.g: mydf = data.frame( a = rnorm(100,10), b = rnorm(100,10), c = rgamma(100, 1, scale=1)) group = hist(mydf$c, breaks=FD) group$breaks The idea is to create a factor mydf$d with levels corresponding to the ranges in group$breaks. There must be an easy way to do this that I haven't found out. Thanks in advance, Albert. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] from Colombia - help
At 13:50 20/12/05, andres felipe wrote: Hi, my name is Andres Felipe Barrientos, I'm a student of Statistic and don't speak in english. En mi trabajo de grado necesito implementar la funcion smooth.spline y necesito saber con que tipo de spline trabaja (b-splines o naturales). Since I have not yet seen a reply here goes. It is very honest of you to tell us it is your 'trabajo de grado' but should you not ask your tutor for help? Have you tried ?smooth.spline which tells you what the function does (admittedly fairly tersely)? Ademas me gustaria saber cual es la base que se usa para encontrar estos splines, por ejemplo, cosenos, senos, polinomios entre otros Otra pregunta que tengo consiste en saber cual es la relacion que sostiene la funcion smooth.spline y ns Agradeciendo la atencion prestada y esperando una respuesta desde la universidad del valle, quien le escribe. Andres Felipe - 1GB gratis, Antivirus y Antispam Correo Yahoo!, el mejor correo web del mundo Abrí tu cuenta aquí [[alternative HTML version deleted]] Michael Dewey http://www.aghmed.fsnet.co.uk __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] boot problem
What are you trying to do? boot does the resampling for you so you should
pass the statistic that you want calculated on the resamples. Read the
helpfile regarding the format of the statistic. It should take two
arguments, one of which is a vector of indices specifying the resample.
On Tue, 20 Dec 2005, david v wrote:
Hello,
This is the code that is giving me problems
library(boot)
data-read.table(test,header=FALSE,sep=\t,row.names=1)
data
V2 V3 V4
A 5 8 9
B 12 54 89
C 65 89 23
D 32 69 44
E 21 84 97
F 33 59 71
G 16 45 93
H2 46 55
I 22 33 88
resample - function(x,index) {
sample(data,replace=TRUE)
}
dist-boot(data,resample,R=1000)
Erreur : nombre d'indices incorrect sur la matrice (french)
Error: number of indices wrong in the matrix (moreless)
Can anybody help???
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| Mathematics and Statistics Phone: (905) 525-9140 x 27079 |
| McMaster UniversityFax : (905) 522-0935 |
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Re: [R] add factor to dataframe given ranges
On Thu, 2005-12-22 at 16:58 +0100, Albert Vilella wrote: Hi all, I would like to factorize the entries in a dataframe given some groupings. E.g: mydf = data.frame( a = rnorm(100,10), b = rnorm(100,10), c = rgamma(100, 1, scale=1)) group = hist(mydf$c, breaks=FD) group$breaks The idea is to create a factor mydf$d with levels corresponding to the ranges in group$breaks. There must be an easy way to do this that I haven't found out. Thanks in advance, Albert. See ?cut Then: myfac.c - cut(mydf$c, breaks = group$breaks) Take note of the additional arguments in cut() relative to the nature of the interval cutpoints (ie. 'right' and 'include.lowest'). HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] add factor to dataframe given ranges
On Thu, 2005-12-22 at 10:08 -0600, Marc Schwartz (via MN) wrote: On Thu, 2005-12-22 at 16:58 +0100, Albert Vilella wrote: Hi all, I would like to factorize the entries in a dataframe given some groupings. E.g: mydf = data.frame( a = rnorm(100,10), b = rnorm(100,10), c = rgamma(100, 1, scale=1)) group = hist(mydf$c, breaks=FD) group$breaks The idea is to create a factor mydf$d with levels corresponding to the ranges in group$breaks. There must be an easy way to do this that I haven't found out. Thanks in advance, Albert. See ?cut Then: myfac.c - cut(mydf$c, breaks = group$breaks) One quick update here which is more clearly based upon your comment above: mydf$d - cut(mydf$c, breaks = group$breaks) Which will add the resultant factor as a new column 'd' to your existing mydf. Marc __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] reading long matrix
Hi, I'm needing some help finding a function to read a large text file into an array in R. The data are essentially presence / absence / na data for many species and come as a grid with each species name (after two spaces) at the beginning of the matrix defining the map for that species. An excerpt could therefore be: SPECIES1 999001099 900110109 011101000 901100101 110100019 901110019 SPECIES2 99999 900110119 011101100 901010101 11019 90019 SPECIES3 999001099 900100109 011100010 901100100 110100019 901110019 where 9 is actually na, 0 is absence and 1 presence. The final array I want to create should have dimensions that are the x and y coordinates and the number of species (known in advance). (In this example dim = c(9,6,3)). It would be sort of neat if the code could also read the species name into the appropriate names attribute, but this is a refinement that I could probably do if someone can help me read the data into R and into an array in the first place. I'm currently thinking a line by line approach using readLines might be the best option, but I've got a very long file - well over 100 species, each a matrix of 70 x 100 datapoints. making this option rther time consuming, I expect - especially as the next dataset has 1300 species and a much larger grid... Any hints would be gratefully recieved. Colin Beale Macaulay Land Use Research Institute __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Plot problems: xlim
Ronnie Babigumira rb.glists at gmail.com writes: It sounds like you might want to break your axis. plotrix provides a function to draw the axis break, but you have to mess around with the data scaling and axis labels yourself. See RSiteSearch(axis break); most of these discussions are about breaking y axes but the same techniques apply to the x axis. good luck, Ben Bolker __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to put constraint in R?
What are you trying to do? Suppose, for example, that you have a categorical variable representing 3 different age groups, and you want to estimate a linear model. As long as R recognizes the variable as categorical (of class factor or ordered factor, R handles this automatically using contrasts. If this is your question, I suggest you read about factors, Formulae for Statistical Models and Contrasts in An Introduction to R, which is the upper-left choice after 'help.start()' in recent versions of R (or downloadable from www.r-project.org - Documentation - Manuals). If you would like more help from this group, I believe you will increase your chances of getting quickly the information you seek in you PLEASE do read the posting guide! www.R-project.org/posting-guide.html. hope this helps. spencer graves Oana Mocila wrote: Dear all, I have a problem when I was working on Age-Period-Cohort study in R. I tried to put constraint to two coefficients on age (so that to solve the identification problem due to linear dependency). But I don't know how to do this in R(put constraint). If you could give me some suggestion, it will be very helpful! Oana __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] reading long matrix
Here is a way of reading the data into a 'list'. You can convert the list
to any array of the proper dimensions.
input - scan('/tempxx.txt.r', what='')
Read 21 items
input
[1] SPECIES1 999001099 900110109 011101000 901100101 110100019
[7] 901110019 SPECIES2 99999 900110119 011101100 901010101
[13] 11019 90019 SPECIES3 999001099 900100109 011100010
[19] 901100100 110100019 901110019
# find the names
breaks - grep([[:alpha:]][[:alnum:]]+, input)
# determine the sizes
map - cbind(breaks, diff(c(breaks, length(input)+1)))
out - list()
# repeat for each data block
for (i in 1:nrow(map)){
+ .set - NULL
+ for (j in 1:(map[i, 2] - 1)){
+ .set - rbind(.set, strsplit(input[map[i, 1] + j], '')[[1]])
+ }
+ out[[input[map[i, 1 - .set
+ }
out
$SPECIES1
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] 9 9 9 0 0 1 0 9 9
[2,] 9 0 0 1 1 0 1 0 9
[3,] 0 1 1 1 0 1 0 0 0
[4,] 9 0 1 1 0 0 1 0 1
[5,] 1 1 0 1 0 0 0 1 9
[6,] 9 0 1 1 1 0 0 1 9
$SPECIES2
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] 9 9 9 0 0 0 0 9 9
[2,] 9 0 0 1 1 0 1 1 9
[3,] 0 1 1 1 0 1 1 0 0
[4,] 9 0 1 0 1 0 1 0 1
[5,] 1 1 0 0 0 0 0 1 9
[6,] 9 0 0 0 0 0 0 1 9
$SPECIES3
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] 9 9 9 0 0 1 0 9 9
[2,] 9 0 0 1 0 0 1 0 9
[3,] 0 1 1 1 0 0 0 1 0
[4,] 9 0 1 1 0 0 1 0 0
[5,] 1 1 0 1 0 0 0 1 9
[6,] 9 0 1 1 1 0 0 1 9
On 12/22/05, Colin Beale [EMAIL PROTECTED] wrote:
Hi,
I'm needing some help finding a function to read a large text file into an
array in R. The data are essentially presence / absence / na data for many
species and come as a grid with each species name (after two spaces) at the
beginning of the matrix defining the map for that species. An excerpt could
therefore be:
SPECIES1
999001099
900110109
011101000
901100101
110100019
901110019
SPECIES2
99999
900110119
011101100
901010101
11019
90019
SPECIES3
999001099
900100109
011100010
901100100
110100019
901110019
where 9 is actually na, 0 is absence and 1 presence. The final array I
want to create should have dimensions that are the x and y coordinates and
the number of species (known in advance). (In this example dim = c(9,6,3)).
It would be sort of neat if the code could also read the species name into
the appropriate names attribute, but this is a refinement that I could
probably do if someone can help me read the data into R and into an array in
the first place. I'm currently thinking a line by line approach using
readLines might be the best option, but I've got a very long file - well
over 100 species, each a matrix of 70 x 100 datapoints. making this option
rther time consuming, I expect - especially as the next dataset has 1300
species and a much larger grid...
Any hints would be gratefully recieved.
Colin Beale
Macaulay Land Use Research Institute
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Cincinnati, OH
+1 513 247 0281
What the problem you are trying to solve?
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Re: [R] reading long matrix
Here's one possibility, if you know the number of species and the numbers of
rows and columns before hand, and the dimension for all species are the
same.
readSpeciesMap - function(fname, nspecies, nr, nc) {
spcnames - character(nspecies)
spcdata - array(0, c(nc, nr, nspecies))
## open the file for reading, and close it upon exit.
f - file(fname, open=r)
on.exit(close(f))
for (i in seq(along=spcnames)) {
## read the name
spcnames[i] - readLines(f, 1)[[1]]
## read the grid
spcdata[, , i] - as.numeric(unlist(strsplit(readLines(f, nr), )))
## pick up the empty line
readLines(f, 1)
}
## replace the 9s with NAs
spcdata[spcdata == 9] - NA
dimnames(spcdata)[[3]] - spcnames
## transpose the array in each species
aperm(spcdata, c(2, 1, 3))
}
Using the example you supplied (saved in the file species.txt):
readSpeciesMap(species.txt, 3, 6, 9)
, , SPECIES1
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] NA NA NA0010 NA NA
[2,] NA0011010 NA
[3,]011101000
[4,] NA01100101
[5,]11010001 NA
[6,] NA0111001 NA
, , SPECIES2
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] NA NA NA0000 NA NA
[2,] NA0011011 NA
[3,]011101100
[4,] NA01010101
[5,]11000001 NA
[6,] NA0000001 NA
, , SPECIES3
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] NA NA NA0010 NA NA
[2,] NA0010010 NA
[3,]011100010
[4,] NA01100100
[5,]11010001 NA
[6,] NA0111001 NA
Andy
From: Colin Beale
Hi,
I'm needing some help finding a function to read a large text
file into an array in R. The data are essentially presence /
absence / na data for many species and come as a grid with
each species name (after two spaces) at the beginning of the
matrix defining the map for that species. An excerpt could
therefore be:
SPECIES1
999001099
900110109
011101000
901100101
110100019
901110019
SPECIES2
99999
900110119
011101100
901010101
11019
90019
SPECIES3
999001099
900100109
011100010
901100100
110100019
901110019
where 9 is actually na, 0 is absence and 1 presence. The
final array I want to create should have dimensions that are
the x and y coordinates and the number of species (known in
advance). (In this example dim = c(9,6,3)). It would be sort
of neat if the code could also read the species name into the
appropriate names attribute, but this is a refinement that I
could probably do if someone can help me read the data into R
and into an array in the first place. I'm currently thinking
a line by line approach using readLines might be the best
option, but I've got a very long file - well over 100
species, each a matrix of 70 x 100 datapoints. making this
option rther time consuming, I expect - especially as the
next dataset has 1300 species and a much larger grid...
Any hints would be gratefully recieved.
Colin Beale
Macaulay Land Use Research Institute
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[R] Testing a linear hypothesis after maximum likelihood
I'd like to be able to test linear hypotheses after setting up and running a model using optim or perhaps nlm. One hypothesis I need to test are that the average of several coefficients is less than zero, so I don't believe I can use the likelihood ratio test. I can't seem to find a provision anywhere for testing linear combinations of coefficients after max. likelihood. Cheers happy holidays, Peter __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Reading in large file in pieces
I have a large file (millions of lines) and would like to read it in pieces. The file is logically separated into little modules, but these modules do not have a common size, so I have to scan the file to know where they are. They are independent, so I don't have to read one at the end to interpret one at the beginning. Is there a way to read one line at a time and parse it on the fly and do so quickly, or do I need to read say 100k lines at a time and then work with those? Only a small piece of each module will remain in memory after parsing is completed on each module. My direct question is: Is there a fast way to parse one line at a time looking for breaks between modules, or am I better off taking large but manageable chunks from the file and parsing that chunk all at once? Thanks, Sean __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] (no subject)
Hi,
I am new in writing R extension.
I read the Writing R Extensions, and search online but just cannot
find the answer.
I am trying to add c-code, which depend on GSL C library, into my R
package.
I am using Max OS-X 10.4.3, powerpc-apple-darwin8-gcc-4.0.1
I begin wtih package.skeleton, and then add my c file into folder src
and then add file Makevars:
PKG_LIBS=-lgsl -lgslcblas -lm
In the zzz.R file, I add the function .First.lib:
.First.lib - function(lib, pkg) {
library.dynam(anovaGSL, pkg, lib)
}
I think these two things are enough. then I just run R CMD check
anovaGSL (my package name)
followed is the error message:
Error: .First.lib failed for 'anovaGSL'
Call sequence:
2: stop(gettextf(.First.lib failed for '%s', libraryPkgName(package)),
domain = NA)
1: library(package, lib.loc = lib.loc, character.only = TRUE, verbose
= FALSE)
Execution halted
See section 'Generic functions and methods' of the 'Writing R
Extensions'
manual.
* checking replacement functions ... WARNING
Error: .First.lib failed for 'anovaGSL'
Call sequence:
2: stop(gettextf(.First.lib failed for '%s', libraryPkgName(package)),
domain = NA)
1: library(package, lib.loc = lib.loc, character.only = TRUE, verbose
= FALSE)
Execution halted
In R, the argument of a replacement function which corresponds to the
right
hand side must be named 'value'.
* checking foreign function calls ... WARNING
Error: .First.lib failed for 'anovaGSL'
Call sequence:
2: stop(gettextf(.First.lib failed for '%s', libraryPkgName(package)),
domain = NA)
1: library(package, lib.loc = lib.loc, character.only = TRUE, verbose
= FALSE)
Execution halted
See section 'System and foreign language interfaces' of the 'Writing R
Extensions' manual.
* checking Rd files ... OK
* checking for missing documentation entries ... ERROR
Error: .First.lib failed for 'anovaGSL'
appreciate any help!!!
wei
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[R] convolution of the double exponential distribution
Is there any R function that computes the convolution of the double
exponential distribution?
If not, is there a good way to integrate ((q+x)^n)*exp(-2x) over x from
0 to Inf for any value of q and for any positive integer n? I need to
perform the integration within a function with q and n as arguments. The
function integrate() is giving me this message:
evaluation of function gave a result of wrong length
David
___
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Research Scientist
Pioneer Hi-Bred International (DuPont)
Bioinformatics and Exploratory Research
7200 NW 62nd Ave.; PO Box 184
Johnston, IA 50131-0184
515-334-4739 Tel
515-334-4473 Fax
[EMAIL PROTECTED], [EMAIL PROTECTED]
This communication is for use by the intended recipient and ...{{dropped}}
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Re: [R] convolution of the double exponential distribution
On 12/22/2005 7:56 PM, Bickel, David wrote: Is there any R function that computes the convolution of the double exponential distribution? If not, is there a good way to integrate ((q+x)^n)*exp(-2x) over x from 0 to Inf for any value of q and for any positive integer n? I need to perform the integration within a function with q and n as arguments. The function integrate() is giving me this message: evaluation of function gave a result of wrong length Under the substitution of y = q+x, that looks like a gamma integral. The x = 0 to Inf range translates into y = q to Inf, so you'll need an incomplete gamma function, such as pgamma. Be careful to get the constant multiplier right. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] how to specify dev.print target by a variable?
I want to do the following: DEVw=500 DEVh=350 fname=my_plot dev.print(file=fname, device=FOO, width=DEVw, height=DEVh, bg=transparent) How do I do this such that I can specify FOO to be one of several choices? (GDD, PNG, postscript, etc.) If I make FOO a character variable, then dev.print complains. I tried a simpled substitute but didn't get it to work... I'm thinking it's going to involve a do.call and substitute but I'm not sure. Using: $platform[1] i386-pc-mingw32 $arch[1] i386 $os[1] mingw32 $system[1] i386, mingw32 $status[1] $major[1] 2 $minor[1] 2.0 $year[1] 2005 $month[1] 10 $day[1] 06 $svn rev[1] 35749 $language[1] R and also running the same code on: $platform[1] i686-redhat-linux-gnu $arch[1] i686 $os[1] linux-gnu $system[1] i686, linux-gnu $status[1] $major[1] 2 $minor[1] 0.0 $year[1] 2004 $month[1] 10 $day[1] 04 $language[1] R -Leif S. Kirschenbaum, Ph.D. Yield Integration Engineer Reflectivity, Inc. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to specify dev.print target by a variable?
G'day Leif, LK == Leif Kirschenbaum [EMAIL PROTECTED] writes: LK How do I do this such that I can specify FOO to be one of LK several choices? (GDD, PNG, postscript, etc.) If I make FOO a LK character variable, then dev.print complains. Mmh, I am not sure what the complaint of R 2.2.0 on MS Windows is, but I guess it is the same as under linux: DEVw=500 DEVh=350 fname=my_plot plot(rnorm(300)) FOO - png dev.print(file=fname, device=FOO, width=DEVw, height=DEVh, bg=transparent) Error in dev.copy(device = png, file = fname, width = DEVw, height = DEVh, : 'device' should be a function Which is very informative. `device' is supposed to be a function, not a character variable, thus: FOO - png dev.print(file=fname, device=FOO, width=DEVw, height=DEVh, bg=transparent) X11 2 FOO - pdf dev.print(file=fname, device=FOO, width=DEVw, height=DEVh, bg=transparent) X11 2 all seem to work. HTH. Cheers, Berwin == Full address Berwin A Turlach Tel.: +61 (8) 6488 3338 (secr) School of Mathematics and Statistics+61 (8) 6488 3383 (self) The University of Western Australia FAX : +61 (8) 6488 1028 35 Stirling Highway Crawley WA 6009e-mail: [EMAIL PROTECTED] Australiahttp://www.maths.uwa.edu.au/~berwin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] reading long matrix
One way to do this is to use read.fwf. I have borrowed Jim's use of scan and use a similar calculation to get the indexes of the breaks, breaks. We then determine the common number of rows and columns in each species. The second group of statements replaces all 9's with spaces so that upon parsing them as numbers they will be NAs and then sets up a text connection to the resulting character vector. These are then read in by read.fwf, nr rows at a time and the result is unlist'ed to a numeric vector, nums. The last statement reshapes it into an array and adds the species names as the last dimension names. # read data in L - scan(clipboard, what = ) breaks - grep(^[[:alpha:]], L) nr - breaks[2] - breaks[1] - 1; nc - nchar(L[2]) # parse numbers n - length(L[-breaks]) / nr con - textConnection(gsub(9, , L[-breaks])) nums - unlist(replicate(n, read.fwf(con, widths = rep(1, nc), n = nr))) result - array(nums, c(6,9,3), c(NULL, NULL, L[breaks])) On 12/22/05, Colin Beale [EMAIL PROTECTED] wrote: Hi, I'm needing some help finding a function to read a large text file into an array in R. The data are essentially presence / absence / na data for many species and come as a grid with each species name (after two spaces) at the beginning of the matrix defining the map for that species. An excerpt could therefore be: SPECIES1 999001099 900110109 011101000 901100101 110100019 901110019 SPECIES2 99999 900110119 011101100 901010101 11019 90019 SPECIES3 999001099 900100109 011100010 901100100 110100019 901110019 where 9 is actually na, 0 is absence and 1 presence. The final array I want to create should have dimensions that are the x and y coordinates and the number of species (known in advance). (In this example dim = c(9,6,3)). It would be sort of neat if the code could also read the species name into the appropriate names attribute, but this is a refinement that I could probably do if someone can help me read the data into R and into an array in the first place. I'm currently thinking a line by line approach using readLines might be the best option, but I've got a very long file - well over 100 species, each a matrix of 70 x 100 datapoints. making this option rther time consuming, I expect - especially as the next dataset has 1300 species and a much larger grid... Any hints would be gratefully recieved. Colin Beale Macaulay Land Use Research Institute __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] reading long matrix
One correction. I had hard coded the last statement for testing with the data provided. Change it to this for generality: result - array(nums, c(nr, nc, n), c(NULL, NULL, L[breaks])) On 12/22/05, Gabor Grothendieck [EMAIL PROTECTED] wrote: One way to do this is to use read.fwf. I have borrowed Jim's use of scan and use a similar calculation to get the indexes of the breaks, breaks. We then determine the common number of rows and columns in each species. The second group of statements replaces all 9's with spaces so that upon parsing them as numbers they will be NAs and then sets up a text connection to the resulting character vector. These are then read in by read.fwf, nr rows at a time and the result is unlist'ed to a numeric vector, nums. The last statement reshapes it into an array and adds the species names as the last dimension names. # read data in L - scan(clipboard, what = ) breaks - grep(^[[:alpha:]], L) nr - breaks[2] - breaks[1] - 1; nc - nchar(L[2]) # parse numbers n - length(L[-breaks]) / nr con - textConnection(gsub(9, , L[-breaks])) nums - unlist(replicate(n, read.fwf(con, widths = rep(1, nc), n = nr))) result - array(nums, c(6,9,3), c(NULL, NULL, L[breaks])) On 12/22/05, Colin Beale [EMAIL PROTECTED] wrote: Hi, I'm needing some help finding a function to read a large text file into an array in R. The data are essentially presence / absence / na data for many species and come as a grid with each species name (after two spaces) at the beginning of the matrix defining the map for that species. An excerpt could therefore be: SPECIES1 999001099 900110109 011101000 901100101 110100019 901110019 SPECIES2 99999 900110119 011101100 901010101 11019 90019 SPECIES3 999001099 900100109 011100010 901100100 110100019 901110019 where 9 is actually na, 0 is absence and 1 presence. The final array I want to create should have dimensions that are the x and y coordinates and the number of species (known in advance). (In this example dim = c(9,6,3)). It would be sort of neat if the code could also read the species name into the appropriate names attribute, but this is a refinement that I could probably do if someone can help me read the data into R and into an array in the first place. I'm currently thinking a line by line approach using readLines might be the best option, but I've got a very long file - well over 100 species, each a matrix of 70 x 100 datapoints. making this option rther time consuming, I expect - especially as the next dataset has 1300 species and a much larger grid... Any hints would be gratefully recieved. Colin Beale Macaulay Land Use Research Institute __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Reading in large file in pieces
See ?scan or maybe ?readLines - Original Message - From: Sean Davis [EMAIL PROTECTED] To: r-help r-help@stat.math.ethz.ch Sent: Friday, December 23, 2005 12:08 AM Subject: [R] Reading in large file in pieces I have a large file (millions of lines) and would like to read it in pieces. The file is logically separated into little modules, but these modules do not have a common size, so I have to scan the file to know where they are. They are independent, so I don't have to read one at the end to interpret one at the beginning. Is there a way to read one line at a time and parse it on the fly and do so quickly, or do I need to read say 100k lines at a time and then work with those? Only a small piece of each module will remain in memory after parsing is completed on each module. My direct question is: Is there a fast way to parse one line at a time looking for breaks between modules, or am I better off taking large but manageable chunks from the file and parsing that chunk all at once? Thanks, Sean __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Reading in large file in pieces
On Thu, 22 Dec 2005, Sean Davis wrote: I have a large file (millions of lines) and would like to read it in pieces. The file is logically separated into little modules, but these modules do not have a common size, so I have to scan the file to know where they are. They are independent, so I don't have to read one at the end to interpret one at the beginning. Is there a way to read one line at a time and parse it on the fly and do so quickly, or do I need to read say 100k lines at a time and then work with those? Only a small piece of each module will remain in memory after parsing is completed on each module. My direct question is: Is there a fast way to parse one line at a time looking for breaks between modules, or am I better off taking large but manageable chunks from the file and parsing that chunk all at once? On any reasonable OS (you have not told us yours), it will make no difference as the file reads will be buffered. Assuming you are doing something like opening a connection and calling readLines(n=1), of course. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] (C programming problem on MacOS X)
Please do INSTALL the package and test loading it before R CMD check.
You will get better errror messages. This has been discussed on the
appropriate R list (R-devel, not here), many times.
Since you mention -lgslcblas and no BLAS, my guess is that your shared
object has unsatisfied entry points.
However, can you not obtain advice locally on R / MacOS X programming:
your address suggests that you have neighbours who are long-term
practitioners?
On Thu, 22 Dec 2005, wei sun wrote:
Hi,
I am new in writing R extension.
I read the Writing R Extensions, and search online but just cannot
find the answer.
You seeem to have failed to find the R posting guide and its request to
use a proper subject line, not to send HTML mail and to use an appropriate
list.
I am trying to add c-code, which depend on GSL C library, into my R
package.
I am using Max OS-X 10.4.3, powerpc-apple-darwin8-gcc-4.0.1
I begin wtih package.skeleton, and then add my c file into folder src
and then add file Makevars:
PKG_LIBS=-lgsl -lgslcblas -lm
In the zzz.R file, I add the function .First.lib:
.First.lib - function(lib, pkg) {
library.dynam(anovaGSL, pkg, lib)
}
I think these two things are enough. then I just run R CMD check
anovaGSL (my package name)
followed is the error message:
Error: .First.lib failed for 'anovaGSL'
Call sequence:
2: stop(gettextf(.First.lib failed for '%s', libraryPkgName(package)),
domain = NA)
1: library(package, lib.loc = lib.loc, character.only = TRUE, verbose
= FALSE)
Execution halted
See section 'Generic functions and methods' of the 'Writing R
Extensions'
manual.
* checking replacement functions ... WARNING
Error: .First.lib failed for 'anovaGSL'
Call sequence:
2: stop(gettextf(.First.lib failed for '%s', libraryPkgName(package)),
domain = NA)
1: library(package, lib.loc = lib.loc, character.only = TRUE, verbose
= FALSE)
Execution halted
In R, the argument of a replacement function which corresponds to the
right
hand side must be named 'value'.
* checking foreign function calls ... WARNING
Error: .First.lib failed for 'anovaGSL'
Call sequence:
2: stop(gettextf(.First.lib failed for '%s', libraryPkgName(package)),
domain = NA)
1: library(package, lib.loc = lib.loc, character.only = TRUE, verbose
= FALSE)
Execution halted
See section 'System and foreign language interfaces' of the 'Writing R
Extensions' manual.
* checking Rd files ... OK
* checking for missing documentation entries ... ERROR
Error: .First.lib failed for 'anovaGSL'
appreciate any help!!!
wei
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Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] how to specify dev.print target by a variable?
On Fri, 23 Dec 2005, Berwin A Turlach wrote: G'day Leif, LK == Leif Kirschenbaum [EMAIL PROTECTED] writes: LK How do I do this such that I can specify FOO to be one of LK several choices? (GDD, PNG, postscript, etc.) If I make FOO a LK character variable, then dev.print complains. Mmh, I am not sure what the complaint of R 2.2.0 on MS Windows is, but I guess it is the same as under linux: DEVw=500 DEVh=350 fname=my_plot plot(rnorm(300)) FOO - png dev.print(file=fname, device=FOO, width=DEVw, height=DEVh, bg=transparent) Error in dev.copy(device = png, file = fname, width = DEVw, height = DEVh, : 'device' should be a function Which is very informative. `device' is supposed to be a function, not a character variable, thus: FOO - png dev.print(file=fname, device=FOO, width=DEVw, height=DEVh, bg=transparent) X11 2 FOO - pdf dev.print(file=fname, device=FOO, width=DEVw, height=DEVh, bg=transparent) X11 2 all seem to work. And my guess is that Leif wants to be able to do mydevice - png dev.print(device = get(mydevice), ...) ^^^ -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
