Re: [R] biplot label size
cex works for me. On Tue, 5 Sep 2006, Nair, Murlidharan T wrote: Which is the parameter that is used to decrease the size of ylabs plotted in biplot? I tried playing with cex and cex.lab I am not getting it right pc - princomp(USArrests) biplot(pc, xlabs = rep(, nrow(USArrests)),ylabs=(colnames(USArrests))) Thanks../Murli -- 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 and provide commented, minimal, self-contained, reproducible code.
[R] Question on AffyBatch
Dear list, I'm trying to find out the following in an AffyBatch. To get the indices from a loction on a chip I use the function xy2i() for the hgu133plus2 by Affymetrix. But now I want to know the name of the probe located at this spot. How can this be done? And what about the locations used as QC? Gunther __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Quick question about lm()
Thanks a lot for your help. tong - Original Message - From: Christos Hatzis [EMAIL PROTECTED] Date: Monday, September 4, 2006 10:54 pm Subject: RE: [R] Quick question about lm() To: 'Tong Wang' [EMAIL PROTECTED], r-help@stat.math.ethz.ch Say, my.lm - lm(y ~ x, data=my.data) Then if you try: names(summary(my.lm)) you will see the components of the summary.lm object. The coefficients and t-statistics can be extracted by summary(my.lm)$coefficients and similarly for the r-squared and other statistics provided in the summary report. -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Tong Wang Sent: Tuesday, September 05, 2006 1:35 AM To: r-help@stat.math.ethz.ch Subject: [R] Quick question about lm() Hi, Feel awkward to ask , but really couldn't find a answer anywhere, How could I extract the R^2 and t-stat. from the result of lm()? Thanks a lot. best __ 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.htmland provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] A question about gc()
Yes, I am using R in windows, sorry for not being specific. You are right, I have stored too much stuff, and I need to trash some data in the process. Problem solved for me, Thank you very much. tong - Original Message - From: Prof Brian Ripley [EMAIL PROTECTED] Date: Tuesday, September 5, 2006 3:42 am Subject: Re: [R] A question about gc() To: Tong Wang [EMAIL PROTECTED] Cc: R help r-help@stat.math.ethz.ch On Tue, 5 Sep 2006, Tong Wang wrote: Hi everyone, I am doing some intensive computation: 5 regressions of the form Y~X with each y of size (1,1000) , even if I break invoke gc() for a few time in the loop, it still breaks down at some point with the error message: gc() does not reclaim memory for you beyond what R has already done. Error in .signalSimpleWarning(Reached total allocation of 1024Mb: see help(memory.size), : recursive default argument reference After getting this, even if I call gc() and resume the computation, it won't move at all. May I get some suggestions what should I do to get around this problem ? Consult the rw-FAQ (since you seem to be using Windows without telling us) and the help page the message mentions. It looks as if you are trying to store too many objects. If you have lots of RAM you can increase that limit (see the previous para), but you are getting uncomfortably close to the address space limit of your OS. Perhaps you can only save the part of the fit you need, or save() the objects to separate files and rm() them, and postprocess them in a later session? -- 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 and provide commented, minimal, self-contained, reproducible code.
[R] What is the matrix version of min()
Hi, Is there a function which operates on a matrix and return a vector of min/max of each rol/col ? say, X= 2, 1 3, 4 min.col(X)=c(2,1) thanks a lot. tong __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fitting generalized additive models with constraints?
On Tuesday 05 September 2006 20:32, David Reiss wrote: I am trying to fit a GAM for a simple model, a simple model, y ~ s(x0) + s(x1) ; with a constraint that the fitted smooth functions s(x0) and s(x1) have to each always be 0. From the library documentation and a search of the R-site and R-help archives I have not been able to decipher whether the following is possible using this, or other GAM libraries, or whether I will have to try to roll my own. I see from the mgcv docs that GAMs need to be constrained such that the smooth functions have zero mean. Is there a way around this? Is such a constraint possible? It is possible to estimate a GAM subject to this constraint, but be aware that the mean levels of your component smooths are not identifiable, so there is an unavoidable abitrariness in the estimate You have to have some sort of constraint on the smooths in a GAM to ensure identifiability, and a convenient way to set the model up is to write it as e.g. E(y) = a + f0(x0) + f1(x1) where `a' is the intercept and f0 and f1 are smooth functions which sum to zero over their respective covariate values. In this parameterization your constraint implies that a + f0(x0) + f1(x1) 0 for all x0, x1. If this constraint is met then you can find constants b and c such that b+c=a such that f0(x0)+b0 and f1(x1)+c0 for all x0,x1. i.e. you redefine f0 as f0+b and f1 as f1+c, and you have a fitted model meeting the constraints. To fit the GAM subject to the constraints you can use mgcv:::pcls... ?pcls has some examples, but it does involve moderately low level programming. It's hard to impose the constraint exactly, so the usual approach would be to impose the constraint over a fairly fine grid of x0, x1 values. Also, you'll need to figure out how to select smoothing parameters. For many problems it suffices to estimate smoothing parameters on the unconstrained fit, and then use these to fit subject to constraints, but it depends on the problem Hope that's some use. Simon Hi Simon, thanks very much for the advice. I will try to parse your response and the pcls docs and see if I can get this to work. In the meantime, I found a paper that tries to achieve a similar thing with the same constraints as I am working with, using quadprog: http://www.esajournals.org/esaonline/?request=get-abstractissn=0012-9658v olume=083issue=08page=2256 -David - The additional wrinkle in gradient matching is that you have the equality constraints that the functions pass through zero at zero (a zero population can't produce offspring or corpses! ), which you don't have do you? If you do then it removes the ambiguity in the model, and makes everything a bit easier. best, Simon -- Simon Wood, Mathematical Sciences, University of Bath, Bath, BA2 7AY UK +44 1225 386603 www.maths.bath.ac.uk/~sw283 __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] What is the matrix version of min()
Tong you need to use apply(). The second argument specifies whether you want to work with rows or columns. The point of this is that min() and max() operate on vectors and give a single value, and you want to apply this function to all rows or all columns: a - matrix(rnorm(30),5,6) apply(a,2,max) [1] 2.6413241 0.9842076 1.7989560 0.6999855 2.0542201 0.1162821 apply(a,1,max) [1] 1.1771370 0.9811693 2.6413241 0.9842076 2.0542201 HTH rksh On 6 Sep 2006, at 09:37, Tong Wang wrote: Hi, Is there a function which operates on a matrix and return a vector of min/max of each rol/col ? say, X= 2, 1 3, 4 min.col(X)=c(2,1) thanks a lot. tong __ 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 and provide commented, minimal, self-contained, reproducible code. -- Robin Hankin Uncertainty Analyst National Oceanography Centre, Southampton European Way, Southampton SO14 3ZH, UK tel 023-8059-7743 __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] What is the matrix version of min()
you could use something like: # for row min and max apply(X, 1, min) apply(X, 1, max) # for column min and max apply(X, 2, min) apply(X, 2, max) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: Tong Wang [EMAIL PROTECTED] To: R help r-help@stat.math.ethz.ch Sent: Wednesday, September 06, 2006 10:37 AM Subject: [R] What is the matrix version of min() Hi, Is there a function which operates on a matrix and return a vector of min/max of each rol/col ? say, X= 2, 1 3, 4 min.col(X)=c(2,1) thanks a lot. tong __ 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 and provide commented, minimal, self-contained, reproducible code. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] What is the matrix version of min()
Tong Wang a écrit : Hi, Is there a function which operates on a matrix and return a vector of min/max of each rol/col ? say, X= 2, 1 3, 4 min.col(X)=c(2,1) thanks a lot. tong see ?pmin, which.min, which.max, max.col hih __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] package ltm -- version 0.6-0
Dear R-users, I'd like to announce the release of the new version of package 'ltm' for analyzing multivariate dichotomous and polytomous data under the Item Response Theory approach. New features: * function tpm() (along with supporting methods, i.e., anova, plot, margins, factor.scores, etc.) has been added for fitting Birnbaum's Three Parameter Model. * grm() can now handle mix of dichotomous and polytomous items. * descript() returns more output, especially for dichotomous manifest variables. For more details check the CHANGES file. Future plans include development of functions for goodness-of-fit, and the Partial Credit Model. Any kind of feedback (questions, suggestions, bug-reports, etc.) is more than welcome. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm ___ R-packages mailing list R-packages@stat.math.ethz.ch https://stat.ethz.ch/mailman/listinfo/r-packages __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] What is the matrix version of min()
see ?apply min.row - apply(X, 1, min) min.col - apply(X, 2, min) JeeBee On Wed, 06 Sep 2006 01:37:22 -0700, Tong Wang wrote: Hi, Is there a function which operates on a matrix and return a vector of min/max of each rol/col ? say, X= 2, 1 3, 4 min.col(X)=c(2,1) thanks a lot. tong __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] What is the matrix version of min()
On Wed, 6 Sep 2006, Robin Hankin wrote: Tong you need to use apply(). The second argument specifies whether you want to work with rows or columns. The point of this is that min() and max() operate on vectors and give a single value, and you want to apply this function to all rows or all columns: a - matrix(rnorm(30),5,6) apply(a,2,max) [1] 2.6413241 0.9842076 1.7989560 0.6999855 2.0542201 0.1162821 apply(a,1,max) [1] 1.1771370 0.9811693 2.6413241 0.9842076 2.0542201 HTH rksh Or in some circumstances you can use pmin (or pmax): a-matrix(rnorm(15),5,3) a [,1] [,2][,3] [1,] 1.5175319 -0.4428964 -0.55473327 [2,] -0.2235937 1.0157411 0.08653748 [3,] 0.3240530 -0.4251498 -0.28565732 [4,] 0.4663556 1.1933213 0.60395935 [5,] 0.4078475 0.1739074 1.85645664 pmin(a[,1],a[,2],a[,3]) [1] -0.5547333 -0.2235937 -0.4251498 0.4663556 0.1739074 David Scott _ David Scott Visiting (July 06 to January 07) Department of Probability and Statistics The University of Sheffield The Hicks Building Hounsfield Road Sheffield S3 7RH United Kingdom Phone: +44 114 222 3908 Email: [EMAIL PROTECTED] __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] What is the matrix version of min()
Hi, THANK YOU ALL for the prompt reply. cheers. - Original Message - From: Robin Hankin [EMAIL PROTECTED] Date: Wednesday, September 6, 2006 1:42 am Subject: Re: [R] What is the matrix version of min() To: Tong Wang [EMAIL PROTECTED] Cc: R help r-help@stat.math.ethz.ch Tong you need to use apply(). The second argument specifies whether you want to work with rows or columns. The point of this is that min() and max() operate on vectors and give a single value, and you want to apply this function to all rows or all columns: a - matrix(rnorm(30),5,6) apply(a,2,max) [1] 2.6413241 0.9842076 1.7989560 0.6999855 2.0542201 0.1162821 apply(a,1,max) [1] 1.1771370 0.9811693 2.6413241 0.9842076 2.0542201 HTH rksh On 6 Sep 2006, at 09:37, Tong Wang wrote: Hi, Is there a function which operates on a matrix and return a vector of min/max of each rol/col ? say, X= 2, 1 3, 4 min.col(X)=c(2,1) thanks a lot. tong __ 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 and provide commented, minimal, self-contained, reproducible code. -- Robin Hankin Uncertainty Analyst National Oceanography Centre, Southampton European Way, Southampton SO14 3ZH, UK tel 023-8059-7743 __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] plot axises on both sides of a graph
Usually the y-axis is shown on the left-hand-side of a graph, is it possible to artifically creat one more y-axis on the right-hand-side in R? What is the main reference? Thank you in advance. [[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 and provide commented, minimal, self-contained, reproducible code.
[R] histogram in the background?
I intend to draw a plot of y against x. In the background of this graph I wish to creat a histogram of the horizontal variable x. Does any expert know how to produce such a plot? [[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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] winDialog UNIX equivalent?
Yes, there are many. To give one example, you could consider using tcltk. library(tcltk) tkmessageBox(title=This is terrible, message=What did you do?\nPromise not to do this again!, icon=error, type=ok) On Tue, 05 Sep 2006 23:10:41 +0200, Richard Müller wrote: Hi all, I'm using winDialog and winDialogString in scripts running on a XP-machine. Since we're using some Linux-machines (Suse 10.0 and 10.1 on x86) I'm interested in equivalents of the above functions usable under Linux-OS. Are there any? Thanks, Richard __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] histogram in the background?
How about this? http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=78 JeeBee On Wed, 06 Sep 2006 18:19:28 +0800, gallon li wrote: I intend to draw a plot of y against x. In the background of this graph I wish to creat a histogram of the horizontal variable x. Does any expert know how to produce such a plot? [[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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot axises on both sides of a graph
Look at: ?axis (try the examples) Further, a nice example I found on this mailing lists archive, from somebody who says this has been asked many times already :) x - 1:10 y1 - 1:10 y2 - rev(seq(1,1000, length=10)) plot(x,y1,ann=FALSE) axis(2, at=c(2,4,6,8), labels=as.character(c(2,4,6,8))) points(x,y2/100,col=red) axis(4, at=c(2,4,6,8), labels=as.character(c(200, 400, 600, 800))) On Wed, 06 Sep 2006 18:10:33 +0800, gallon li wrote: Usually the y-axis is shown on the left-hand-side of a graph, is it possible to artifically creat one more y-axis on the right-hand-side in R? What is the main reference? Thank you in advance. [[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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] histogram in the background?
gallon li wrote: I have found this one before. However, my intension is slightly differing from this plot: I wish to plot the histogram in the backgroun instead of in the margin. Thanks anyway! On Wed, 06 Sep 2006 12:29:51 +0200, JeeBee wrote: How about this? http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=78 JeeBee __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] Fwd: plot axises on both sides of a graph
-- Forwarded message -- From: gallon li [EMAIL PROTECTED] Date: Sep 6, 2006 7:48 PM Subject: Re: [R] plot axises on both sides of a graph To: Jim Lemon [EMAIL PROTECTED] Both of your suggestions are so helpful. By combining what you told me, now I am able to produce a second y-axis on the right-hand-side. Still one problem remains: how can I put a definition of this y-axis in the space left? Clearly there is enough room and I have to check some functions for defining the margin of a plot. Moreoever, it seems not straight forward to put some vertical text directly on the plot for this second ylab. On 9/7/06, Jim Lemon [EMAIL PROTECTED] wrote: gallon li wrote: Usually the y-axis is shown on the left-hand-side of a graph, is it possible to artifically creat one more y-axis on the right-hand-side in R? What is the main reference? Thank you in advance. Have a look at: http://cran.r-project.org/doc/contrib/Lemon-kickstart/kr_addat.html Jim [[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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] winDialog UNIX equivalent?
I'm using winDialog and winDialogString in scripts running on a XP-machine. Since we're using some Linux-machines (Suse 10.0 and 10.1 on x86) I'm Thanks to the respondents. Use of TCl/Tk is a fine idea, because you can use the same scripts on Win and Linux OS. Richard -- Richard Müller - Am Spring 9 - D-58802 Balve-Eisborn www.oeko-sorpe.de __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] problem with putting objects in list
Hi
I use the following code and it stores the results of density() in the
list dr:
dens - function(run) { density( positions$X[positions$run==run], bw=3,
cut=-2 ) }
dr - lapply(1:5, dens)
but the results are stored in dr[[i]] and not dr[i], i.e. plot(dr[[1]])
works, but plot([1]) doesn't.
Is there any way that I can store them in dr[i]?
Thanks a lot,
Rainer
--
Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation
Biology (UCT)
Department of Conservation Ecology and Entomology
University of Stellenbosch
Matieland 7602
South Africa
Tel:+27 - (0)72 808 2975 (w)
Fax:+27 - (0)21 808 3304
Cell: +27 - (0)83 9479 042
email: [EMAIL PROTECTED]
[EMAIL PROTECTED]
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fwd: plot axises on both sides of a graph
See these two examples. plot(1:2) axis(4) mtext(right y axis, side=4, line=-1.5) par(mar=c(5,4,4,5)+.1) plot(1:2) axis(4) mtext(right y axis, side=4, line=3) Good luck finding the right combination again ;) On Wed, 06 Sep 2006 20:08:43 +0800, gallon li wrote: -- Forwarded message -- From: gallon li [EMAIL PROTECTED] Date: Sep 6, 2006 7:48 PM Subject: Re: [R] plot axises on both sides of a graph To: Jim Lemon [EMAIL PROTECTED] Both of your suggestions are so helpful. By combining what you told me, now I am able to produce a second y-axis on the right-hand-side. Still one problem remains: how can I put a definition of this y-axis in the space left? Clearly there is enough room and I have to check some functions for defining the margin of a plot. Moreoever, it seems not straight forward to put some vertical text directly on the plot for this second ylab. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] histogram in the background?
--- JeeBee [EMAIL PROTECTED] wrote: gallon li wrote: I have found this one before. However, my intension is slightly differing from this plot: I wish to plot the histogram in the backgroun instead of in the margin. Thanks anyway! On Wed, 06 Sep 2006 12:29:51 +0200, JeeBee wrote: How about this? http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=78 JeeBee Can you not just plot the histogram and then the main plot? Something like this, although it is a barplot rather than a histogram. Hours - c(0,1,2,3,4,5) Nam - c( alf, bet, cet, det, NA , fet) En - c(5, 7, 8, 9, NA, 9) L1 - c(10, 9, 7, 5, 3 ,6) L2 - c(7, 4, 3, 2, 5, 7) MyLabels - seq(200, 500, length=13) mp -barplot(En) # adjust margins to accommodate titles and labels especially the 4 axis label. par(mar=c(5,5,5,5)) barplot(En, ylim=c(0, 12),axes =FALSE, ann=FALSE, xlab=Hours, ylab=Volume, col.lab=blue) points (mp,L1, type=p, pch=19, col = red) points (mp,L2, type = l, col=blue) axis ( 2, 0:12, las=1, font=2) axis (3, at=1:6, tick =F, labels =c( alf, bet, cet, det, NA , fet)) axis (4, at = 0:12, labels = MyLabels, font=2) mtext(Number of Units Detected, 4, line=3,col=blue) title ( main=A combined line and bar chart \n with different x-axis labels, line=3, font=3, col.main=blue) box() __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question on AffyBatch
Hi Gunther, Gunther Höning wrote: Dear list, I'm trying to find out the following in an AffyBatch. This question is related specifically to a Bioconductor package, so should be asked on the bioconductor listserv rather than R-help. Best, Jim To get the indices from a loction on a chip I use the function xy2i() for the hgu133plus2 by Affymetrix. But now I want to know the name of the probe located at this spot. How can this be done? And what about the locations used as QC? Gunther __ 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 and provide commented, minimal, self-contained, reproducible code. -- James W. MacDonald, M.S. Biostatistician Affymetrix and cDNA Microarray Core University of Michigan Cancer Center 1500 E. Medical Center Drive 7410 CCGC Ann Arbor MI 48109 734-647-5623 ** Electronic Mail is not secure, may not be read every day, and should not be used for urgent or sensitive issues. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] histogram in the background?
I intend to draw a plot of y against x. In the background of this graph I wish to creat a histogram of the horizontal variable x. Does any expert know how to produce such a plot? When constructing such a plot, you need to be careful that you don't end up constructing a pretty picture instead of a statistical graphic. In this case you need to ask yourself, what would the y-axis represent? What scale would it have? Hadley __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with putting objects in list
Use 'sapply' instead of 'lapply'. Type
?lapply
for details
Antonio, Fabio Di Narzo.
University of Bologna, Italy
2006/9/6, Rainer M Krug [EMAIL PROTECTED]:
Hi
I use the following code and it stores the results of density() in the
list dr:
dens - function(run) { density( positions$X[positions$run==run], bw=3,
cut=-2 ) }
dr - lapply(1:5, dens)
but the results are stored in dr[[i]] and not dr[i], i.e. plot(dr[[1]])
works, but plot([1]) doesn't.
Is there any way that I can store them in dr[i]?
Thanks a lot,
Rainer
--
Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation
Biology (UCT)
Department of Conservation Ecology and Entomology
University of Stellenbosch
Matieland 7602
South Africa
Tel:+27 - (0)72 808 2975 (w)
Fax:+27 - (0)21 808 3304
Cell: +27 - (0)83 9479 042
email: [EMAIL PROTECTED]
[EMAIL PROTECTED]
__
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Re: [R] problem with putting objects in list
Antonio, Fabio Di Narzo wrote:
Use 'sapply' instead of 'lapply'. Type
If I use sapply it seems to simplify / collapse to much.
?lapply
for details
Antonio, Fabio Di Narzo.
University of Bologna, Italy
2006/9/6, Rainer M Krug [EMAIL PROTECTED]:
Hi
I use the following code and it stores the results of density() in the
list dr:
dens - function(run) { density( positions$X[positions$run==run], bw=3,
cut=-2 ) }
dr - lapply(1:5, dens)
but the results are stored in dr[[i]] and not dr[i], i.e. plot(dr[[1]])
works, but plot([1]) doesn't.
Is there any way that I can store them in dr[i]?
Thanks a lot,
Rainer
--
Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation
Biology (UCT)
Department of Conservation Ecology and Entomology
University of Stellenbosch
Matieland 7602
South Africa
Tel:+27 - (0)72 808 2975 (w)
Fax:+27 - (0)21 808 3304
Cell: +27 - (0)83 9479 042
email: [EMAIL PROTECTED]
[EMAIL PROTECTED]
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation
Biology (UCT)
Department of Conservation Ecology and Entomology
University of Stellenbosch
Matieland 7602
South Africa
Tel:+27 - (0)72 808 2975 (w)
Fax:+27 - (0)21 808 3304
Cell: +27 - (0)83 9479 042
email: [EMAIL PROTECTED]
[EMAIL PROTECTED]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] how to loop through 2 lists with different indexes
Dear all, I am a newbie in R and need some help please. (I do apologise if my email is not as informative as it should be, I've tried to include the relevant details without overcrowding it with the rest of the code) I would like to sample (without replacement) Y objects based on the number of objects in X in 5 different bins. I'm having trouble because the list object in which the number of objects of X is stored in doesn't start its index from 0. # number of X objects in each of the 5 bins x.bin.size - lapply(x.by.bins, nrow) x.bin.size $`3` [1] 1 $`4` [1] 3 $`5` [1] 10 # no. of objects in each of the 5 bins of Y y.bin.size - lapply(y.by.bins, nrow) y.bin.size $`2` [1] 4 $`3` [1] 42 $`4` [1] 253 $`5` [1] 945 how do I loop through Y and sample from X when the index of Y starts from 2 and that of X starts from 3? in X, the missing index $`2` is assumed to have 0 objects in it. hence, I would only sample from Y and X starting from index 3. sample (y.bin.size$`3`, x.bin.size$`3`, replace=FALSE) but how should I do this in an R command without knowing which indexes (of X ) are empty? Any pointers would be greatly appreciated. Many thanks in advance, tania __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] how to loop through 2 lists with different indexes
Very sorry if this mail is sent out twice to the list, I wasn't sure if the email address I used in the first go was correct. From: [EMAIL PROTECTED] Subject:how to loop through 2 lists with different indexes Date: 6 September 2006 15:16:21 BDT To: [EMAIL PROTECTED] Dear all, I am a newbie in R and need some help please. (I do apologise if my email is not as informative as it should be, I've tried to include the relevant details without overcrowding it with the rest of the code) I would like to sample (without replacement) Y objects based on the number of objects in X in 5 different bins. I'm having trouble because the list object in which the number of objects of X is stored in doesn't start its index from 0. # number of X objects in each of the 5 bins x.bin.size - lapply(x.by.bins, nrow) x.bin.size $`3` [1] 1 $`4` [1] 3 $`5` [1] 10 # no. of objects in each of the 5 bins of Y y.bin.size - lapply(y.by.bins, nrow) y.bin.size $`2` [1] 4 $`3` [1] 42 $`4` [1] 253 $`5` [1] 945 how do I loop through Y and sample from X when the index of Y starts from 2 and that of X starts from 3? in X, the missing index $`2` is assumed to have 0 objects in it. hence, I would only sample from Y and X starting from index 3. sample (y.bin.size$`3`, x.bin.size$`3`, replace=FALSE) but how should I do this in an R command without knowing which indexes (of X ) are empty? Any pointers would be greatly appreciated. Many thanks in advance, tania --- Tania Oh D.Phil student Department of Physiology, Anatomy and Genetics University of Oxford OX1 3TU __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to loop through 2 lists with different indexes
To find what names are common to both, use 'intersect'
x.bin.size - list('3'=1, '4'=4, '5'=10)
y.bin.size - list('2'=4, '3'=42, '4'=253, '5'=954)
sameNames - intersect(names(x.bin.size), names(y.bin.size))
sameNames
[1] 3 4 5
lapply(sameNames, function(x) sample(seq(y.bin.size[[x]]), x.bin.size[[x]]))
[[1]]
[1] 12
[[2]]
[1] 95 145 228 51
[[3]]
[1] 858 901 630 599 59 196 168 651 364 728
On 9/6/06, Tania Oh [EMAIL PROTECTED] wrote:
Dear all,
I am a newbie in R and need some help please. (I do apologise if my
email is not as informative as it should be, I've tried to include
the relevant details without overcrowding it with the rest of the code)
I would like to sample (without replacement) Y objects based on the
number of objects in X in 5 different bins. I'm having trouble
because the list object in which the number of objects of X is stored
in doesn't start its index from 0.
# number of X objects in each of the 5 bins
x.bin.size - lapply(x.by.bins, nrow)
x.bin.size
$`3`
[1] 1
$`4`
[1] 3
$`5`
[1] 10
# no. of objects in each of the 5 bins of Y
y.bin.size - lapply(y.by.bins, nrow)
y.bin.size
$`2`
[1] 4
$`3`
[1] 42
$`4`
[1] 253
$`5`
[1] 945
how do I loop through Y and sample from X when the index of Y starts
from 2 and that of X starts from 3? in X, the missing index $`2` is
assumed to have 0 objects in it. hence, I would only sample from Y
and X starting from index 3.
sample (y.bin.size$`3`, x.bin.size$`3`, replace=FALSE)
but how should I do this in an R command without knowing which
indexes (of X ) are empty? Any pointers would be greatly appreciated.
Many thanks in advance,
tania
__
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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[R] About the Skew Student distribution
Hello everybody, I need your help about the package SN and the skew student distribution. Il will be very grateful if I have the solution. I construct a stochastic model with a white noise not gaussian but following a skew student distribution. I fit the noise on monthly data to obtain the four parameters. The question is : how to annualize the parameters to use my model for simulate daily data for example ? If the volatility is estimated to 3 for example, I need to multiply this by sqrt(12) to have for the parameter of volatility of the skew student : 3*sqrt(12)*sqrt(dt) with dt the time increment parameter (1/12 for monthly data, 1/261 for daily data, and so on). Do I do the same thing (and what is the multiplicative factor ?) for the parameters of asymmetry and the degree of freedom ? Thanks a lot ! Best regards Pierre. [[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 and provide commented, minimal, self-contained, reproducible code.
[R] Matrix multiplication using apply() or lappy() ?
I am trying to divide the columns of a matrix by the first row in the
matrix.
I have tried to get this using apply and I seem to be missing a concept
regarding the apply w/o calling a function but rather command args %*% /
etc. Would using apply be more efficient than this approach?
I have observed examples in the archives using this type of approach. Does
anybody have a snippet of a call to apply() that would accomplish this as
well?
Thanks!
seed=50
$a = array(rnorm(20),dim=c(4,5))
$b = matrix(a[1,],dim(a)[1],dim(a)[2],byrow=T)
$a
[,1] [,2] [,3][,4] [,5]
[1,] -1.3682810 -0.4314462 1.57572752 0.67928882 -0.3672346
[2,] 0.4328180 0.6556479 0.64289931 0.08983289 0.1852306
[3,] -0.8113932 0.3219253 0.08976065 -2.99309008 0.5818237
[4,] 1.4441013 -0.7838389 0.27655075 0.28488295 1.3997368
$a/b
[,1] [,2] [,3] [,4] [,5]
[1,] 1.000 1.000 1. 1.000 1.00
[2,] -0.3163225 -1.5196515 0.40800157 0.1322455 -0.504393
[3,] 0.5930018 -0.7461539 0.05696457 -4.4062113 -1.584338
[4,] -1.0554128 1.8167710 0.17550671 0.4193841 -3.811560
CONFIDENTIALITY NOTICE: This electronic mail transmission (i...{{dropped}}
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and provide commented, minimal, self-contained, reproducible code.
[R] density plots????
Dear all, I arrive to do density plots using the function kde2d , and from this do a countour plot. My problem is that I do not really understand what the labels for the different levels mean??? What I would like to obtain is a surface encompassing the 95 percentile of my values. In other words I would like the levels to represent, for example, the 90th, 95th and 99th percentiles of my values. I hope I have been clear. Do you think you can help me??? I would be VERY grateful. Thanks in advance Luis Barreiro __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix multiplication using apply() or lappy() ?
See ?sweep
sweep(a, 2, a[1,],/)
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
[EMAIL PROTECTED]
Sent: Wednesday, September 06, 2006 11:49 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Matrix multiplication using apply() or lappy() ?
I am trying to divide the columns of a matrix by the first row in the
matrix.
I have tried to get this using apply and I seem to be missing a concept
regarding the apply w/o calling a function but rather command args %*% /
etc. Would using apply be more efficient than this approach?
I have observed examples in the archives using this type of approach. Does
anybody have a snippet of a call to apply() that would accomplish this as
well?
Thanks!
seed=50
$a = array(rnorm(20),dim=c(4,5))
$b = matrix(a[1,],dim(a)[1],dim(a)[2],byrow=T)
$a
[,1] [,2] [,3][,4] [,5]
[1,] -1.3682810 -0.4314462 1.57572752 0.67928882 -0.3672346 [2,] 0.4328180
0.6556479 0.64289931 0.08983289 0.1852306 [3,] -0.8113932 0.3219253
0.08976065 -2.99309008 0.5818237 [4,] 1.4441013 -0.7838389 0.27655075
0.28488295 1.3997368
$a/b
[,1] [,2] [,3] [,4] [,5]
[1,] 1.000 1.000 1. 1.000 1.00 [2,] -0.3163225
-1.5196515 0.40800157 0.1322455 -0.504393 [3,] 0.5930018 -0.7461539
0.05696457 -4.4062113 -1.584338 [4,] -1.0554128 1.8167710 0.17550671
0.4193841 -3.811560
CONFIDENTIALITY NOTICE: This electronic mail transmission (i...{{dropped}}
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[R] Covariance/Correlation matrix for repeated measures data frame
All, I have a repeated measures data frame and was wondering if the covariance matrix can be calculated via some created indexing or built-in R function. Specifically, say there are 3 variables, where potassium concentration is measured 6 times on each patient. Patient number (discrete) Time (1 to 6, discrete) Potassium (continuous variable) I want the covariance/correlation matrix for the cov/corr between Potassium at time i and time j. Is this possible in the current dataframe format? Or do I have to define new varialbes, say Time i and Time j, and then compute the cov/corr between Time i and Time j for all combinations? Cheers, Dave David Afshartous, PhD University of Miami School of Business Rm KE-408 Coral Gables, FL 33124 [[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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix multiplication using apply() or lappy() ?
Here are a few possibilities:
a - matrix(1:24, 4) # test data
a / rep(a[1,], each = 4)
a / outer(rep(1, nrow(a)), a[1,])
a %*% diag(1/a[1,])
sweep(a, 2, a[1,], /)
On 9/6/06, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
I am trying to divide the columns of a matrix by the first row in the
matrix.
I have tried to get this using apply and I seem to be missing a concept
regarding the apply w/o calling a function but rather command args %*% /
etc. Would using apply be more efficient than this approach?
I have observed examples in the archives using this type of approach. Does
anybody have a snippet of a call to apply() that would accomplish this as
well?
Thanks!
seed=50
$a = array(rnorm(20),dim=c(4,5))
$b = matrix(a[1,],dim(a)[1],dim(a)[2],byrow=T)
$a
[,1] [,2] [,3][,4] [,5]
[1,] -1.3682810 -0.4314462 1.57572752 0.67928882 -0.3672346
[2,] 0.4328180 0.6556479 0.64289931 0.08983289 0.1852306
[3,] -0.8113932 0.3219253 0.08976065 -2.99309008 0.5818237
[4,] 1.4441013 -0.7838389 0.27655075 0.28488295 1.3997368
$a/b
[,1] [,2] [,3] [,4] [,5]
[1,] 1.000 1.000 1. 1.000 1.00
[2,] -0.3163225 -1.5196515 0.40800157 0.1322455 -0.504393
[3,] 0.5930018 -0.7461539 0.05696457 -4.4062113 -1.584338
[4,] -1.0554128 1.8167710 0.17550671 0.4193841 -3.811560
CONFIDENTIALITY NOTICE: This electronic mail transmission (i...{{dropped}}
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Covariance/Correlation matrix for repeated measures data frame
try the following:
dat - data.frame(id = rep(1:100, each = 6), time = rep(1:6, 100), pot
= rnorm(600))
# for a balanced data-set
mat - matrix(dat$pot, ncol = 6, byrow = TRUE)
cor(mat)
# for a unbalanced data-set
dat - dat[-sample(600, 100), ]
mat - t(sapply(split(dat, dat$id), function(x){
out - rep(NA, 6)
out[x$time] - x$pot
out
}))
cor(mat, use = pairwise.complete.obs)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: Afshartous, David [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Wednesday, September 06, 2006 5:59 PM
Subject: [R] Covariance/Correlation matrix for repeated measures data
frame
All,
I have a repeated measures data frame and was wondering if the
covariance matrix can be
calculated via some created indexing or built-in R function.
Specifically, say there are 3 variables, where potassium
concentration
is measured 6 times on each patient.
Patient number (discrete)
Time (1 to 6, discrete)
Potassium (continuous variable)
I want the covariance/correlation matrix for the cov/corr between
Potassium at time i and time j.
Is this possible in the current dataframe format? Or do I have to
define new varialbes, say Time i and Time j,
and then compute the cov/corr between Time i and Time j for all
combinations?
Cheers,
Dave
David Afshartous, PhD
University of Miami
School of Business
Rm KE-408
Coral Gables, FL 33124
[[alternative HTML version deleted]]
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Matrix multiplication using apply() or lappy() ?
And here is one more:
t(apply(a, 1, function(x) x/a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here are a few possibilities:
a - matrix(1:24, 4) # test data
a / rep(a[1,], each = 4)
a / outer(rep(1, nrow(a)), a[1,])
a %*% diag(1/a[1,])
sweep(a, 2, a[1,], /)
On 9/6/06, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
I am trying to divide the columns of a matrix by the first row in the
matrix.
I have tried to get this using apply and I seem to be missing a concept
regarding the apply w/o calling a function but rather command args %*% /
etc. Would using apply be more efficient than this approach?
I have observed examples in the archives using this type of approach. Does
anybody have a snippet of a call to apply() that would accomplish this as
well?
Thanks!
seed=50
$a = array(rnorm(20),dim=c(4,5))
$b = matrix(a[1,],dim(a)[1],dim(a)[2],byrow=T)
$a
[,1] [,2] [,3][,4] [,5]
[1,] -1.3682810 -0.4314462 1.57572752 0.67928882 -0.3672346
[2,] 0.4328180 0.6556479 0.64289931 0.08983289 0.1852306
[3,] -0.8113932 0.3219253 0.08976065 -2.99309008 0.5818237
[4,] 1.4441013 -0.7838389 0.27655075 0.28488295 1.3997368
$a/b
[,1] [,2] [,3] [,4] [,5]
[1,] 1.000 1.000 1. 1.000 1.00
[2,] -0.3163225 -1.5196515 0.40800157 0.1322455 -0.504393
[3,] 0.5930018 -0.7461539 0.05696457 -4.4062113 -1.584338
[4,] -1.0554128 1.8167710 0.17550671 0.4193841 -3.811560
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Re: [R] Matrix multiplication using apply() or lappy() ?
On Wed, 6 Sep 2006, Christos Hatzis wrote:
See ?sweep
sweep(a, 2, a[1,],/)
That is less efficient than
a/rep(a[1,], each=nrow(a))
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
[EMAIL PROTECTED]
Sent: Wednesday, September 06, 2006 11:49 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Matrix multiplication using apply() or lappy() ?
I am trying to divide the columns of a matrix by the first row in the
matrix.
I have tried to get this using apply and I seem to be missing a concept
regarding the apply w/o calling a function but rather command args %*% /
etc. Would using apply be more efficient than this approach?
I have observed examples in the archives using this type of approach. Does
anybody have a snippet of a call to apply() that would accomplish this as
well?
Thanks!
seed=50
$a = array(rnorm(20),dim=c(4,5))
$b = matrix(a[1,],dim(a)[1],dim(a)[2],byrow=T)
$a
[,1] [,2] [,3][,4] [,5]
[1,] -1.3682810 -0.4314462 1.57572752 0.67928882 -0.3672346 [2,] 0.4328180
0.6556479 0.64289931 0.08983289 0.1852306 [3,] -0.8113932 0.3219253
0.08976065 -2.99309008 0.5818237 [4,] 1.4441013 -0.7838389 0.27655075
0.28488295 1.3997368
$a/b
[,1] [,2] [,3] [,4] [,5]
[1,] 1.000 1.000 1. 1.000 1.00 [2,] -0.3163225
-1.5196515 0.40800157 0.1322455 -0.504393 [3,] 0.5930018 -0.7461539
0.05696457 -4.4062113 -1.584338 [4,] -1.0554128 1.8167710 0.17550671
0.4193841 -3.811560
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and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix multiplication using apply() or lappy() ?
This last one could also be written slightly shorter as:
t(apply(a, 1, /, a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
And here is one more:
t(apply(a, 1, function(x) x/a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here are a few possibilities:
a - matrix(1:24, 4) # test data
a / rep(a[1,], each = 4)
a / outer(rep(1, nrow(a)), a[1,])
a %*% diag(1/a[1,])
sweep(a, 2, a[1,], /)
On 9/6/06, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
I am trying to divide the columns of a matrix by the first row in the
matrix.
I have tried to get this using apply and I seem to be missing a concept
regarding the apply w/o calling a function but rather command args %*% /
etc. Would using apply be more efficient than this approach?
I have observed examples in the archives using this type of approach. Does
anybody have a snippet of a call to apply() that would accomplish this as
well?
Thanks!
seed=50
$a = array(rnorm(20),dim=c(4,5))
$b = matrix(a[1,],dim(a)[1],dim(a)[2],byrow=T)
$a
[,1] [,2] [,3][,4] [,5]
[1,] -1.3682810 -0.4314462 1.57572752 0.67928882 -0.3672346
[2,] 0.4328180 0.6556479 0.64289931 0.08983289 0.1852306
[3,] -0.8113932 0.3219253 0.08976065 -2.99309008 0.5818237
[4,] 1.4441013 -0.7838389 0.27655075 0.28488295 1.3997368
$a/b
[,1] [,2] [,3] [,4] [,5]
[1,] 1.000 1.000 1. 1.000 1.00
[2,] -0.3163225 -1.5196515 0.40800157 0.1322455 -0.504393
[3,] 0.5930018 -0.7461539 0.05696457 -4.4062113 -1.584338
[4,] -1.0554128 1.8167710 0.17550671 0.4193841 -3.811560
CONFIDENTIALITY NOTICE: This electronic mail transmission (i...{{dropped}}
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix multiplication using apply() or lappy() ?
Yet another one using the idempotent apply in reshape package
that eliminates the transpose:
library(reshape)
iapply(a, 1, /, a[1,])
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
This last one could also be written slightly shorter as:
t(apply(a, 1, /, a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
And here is one more:
t(apply(a, 1, function(x) x/a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here are a few possibilities:
a - matrix(1:24, 4) # test data
a / rep(a[1,], each = 4)
a / outer(rep(1, nrow(a)), a[1,])
a %*% diag(1/a[1,])
sweep(a, 2, a[1,], /)
On 9/6/06, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
I am trying to divide the columns of a matrix by the first row in the
matrix.
I have tried to get this using apply and I seem to be missing a concept
regarding the apply w/o calling a function but rather command args %*% /
etc. Would using apply be more efficient than this approach?
I have observed examples in the archives using this type of approach.
Does
anybody have a snippet of a call to apply() that would accomplish this
as
well?
Thanks!
seed=50
$a = array(rnorm(20),dim=c(4,5))
$b = matrix(a[1,],dim(a)[1],dim(a)[2],byrow=T)
$a
[,1] [,2] [,3][,4] [,5]
[1,] -1.3682810 -0.4314462 1.57572752 0.67928882 -0.3672346
[2,] 0.4328180 0.6556479 0.64289931 0.08983289 0.1852306
[3,] -0.8113932 0.3219253 0.08976065 -2.99309008 0.5818237
[4,] 1.4441013 -0.7838389 0.27655075 0.28488295 1.3997368
$a/b
[,1] [,2] [,3] [,4] [,5]
[1,] 1.000 1.000 1. 1.000 1.00
[2,] -0.3163225 -1.5196515 0.40800157 0.1322455 -0.504393
[3,] 0.5930018 -0.7461539 0.05696457 -4.4062113 -1.584338
[4,] -1.0554128 1.8167710 0.17550671 0.4193841 -3.811560
CONFIDENTIALITY NOTICE: This electronic mail transmission
(i...{{dropped}}
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Help on estimated variance in lme4
Dear all, I get an error message when I run my model and I am not sure what to do about it. I try to determine what factors influence the survival of voles. I use a mixed-model because I have several voles per site (varying from 2 to 19 voles). Here is the model: ### fm5 -lmer(data=cdrgsaou2, alive~factor(pacut)+factor(agecamp)+factor(sex)+ResCondCorp+(1|factor(cdrgsa ou2$ids)), family=binomial, method=Laplace, ) ### Description of variables Alive: 0 or 1; dead or alive pacut: 0 or 1; presence of parasites agecamp: a or j; adult or juvenile sex: m or f; male or female ResCondCorp: body condition, continuous; cdrgsaou2$ids: name of the site. Here is the output: ### Generalized linear mixed model fit using Laplace Formula: alive ~ factor(pacut) + factor(agecamp) + factor(sex) + ResCondCorp + (1 | factor(cdrgsaou2$ids)) Data: cdrgsaou2 Family: binomial(logit link) AIC BIClogLik deviance 305.7418 328.7331 -146.8709 293.7418 Random effects: GroupsNameVariance Std.Dev. factor(cdrgsaou2$ids) (Intercept) 0.034382 0.18542 number of obs: 341, groups: factor(cdrgsaou2$ids), 36 Estimated scale (compare to 1) 2.174681 Fixed effects: Estimate Std. Error z value Pr(|z|) (Intercept) 0.971458 0.250951 3.8711 0.0001083 *** factor(pacut)1 -0.831888 0.358583 -2.3199 0.0203447 * factor(agecamp)j -1.294236 0.330638 -3.9144 9.065e-05 *** factor(sex)m 0.581713 0.296229 1.9637 0.0495616 * ResCondCorp -0.176251 0.020263 -8.6982 2.2e-16 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Correlation of Fixed Effects: (Intr) fct()1 fctr(g) fctr(s) factr(pct)1 -0.334 fctr(gcmp)j -0.417 0.066 factor(sx)m -0.505 -0.002 -0.173 ResCondCorp -0.309 -0.010 0.302 -0.032 ### Here is the error message: ### Warning message: Estimated variance for factor 'factor(cdrgsaou2$ids)' is effectively zero in: LMEopt(x = mer, value = cv) ### Thank you very much by advance for any help. Jérôme Lemaître Ph.D. student Silviculture-wildlife research chair in irregular boreal forests Départment of biology, Faculty of Sciences and Engineering Alexandre-Vachon building University Laval Quebec, QC G1K 7P4 Phone : (418) 656-2131 poste 2917 Office : VCH-2044 Email: [EMAIL PROTECTED] __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix multiplication using apply() or lappy() ?
Prof. Brian Ripley wrote: On Wed, 6 Sep 2006, Christos Hatzis wrote: See ?sweep sweep(a, 2, a[1,],/) That is less efficient than a/rep(a[1,], each=nrow(a)) *My* first instinct was to use t(t(a)/a[1,]) (which has not heretofore been suggested). This seems to be more efficient still (at least in respect of Prof. Grothendieck's toy example) by between 20 and 25 percent: a - matrix(1:24,4) system.time(for(i in 1:1000) junk - a / rep(a[1,], each = 4)) [1] 0.690 0.080 1.051 0.000 0.000 system.time(for(i in 1:1000) junk - t(t(a)/a[1,])) [1] 0.520 0.120 0.647 0.000 0.000 system.time(for(i in 1:1) junk - a / rep(a[1,], each = 4)) [1] 7.08 0.99 10.08 0.00 0.00 system.time(for(i in 1:1) junk - t(t(a)/a[1,])) [1] 5.530 0.940 7.856 0.000 0.000 cheers, Rolf Turner __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix multiplication using apply() or lappy() ?
The apply was exactly what I was after. And, I will check out the others
as well. great tips!
Gabor Grothendieck [EMAIL PROTECTED]
09/06/2006 11:11 AM
To
[EMAIL PROTECTED] [EMAIL PROTECTED]
cc
r-help@stat.math.ethz.ch
Subject
Re: [R] Matrix multiplication using apply() or lappy() ?
This last one could also be written slightly shorter as:
t(apply(a, 1, /, a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
And here is one more:
t(apply(a, 1, function(x) x/a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here are a few possibilities:
a - matrix(1:24, 4) # test data
a / rep(a[1,], each = 4)
a / outer(rep(1, nrow(a)), a[1,])
a %*% diag(1/a[1,])
sweep(a, 2, a[1,], /)
On 9/6/06, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
I am trying to divide the columns of a matrix by the first row in
the
matrix.
I have tried to get this using apply and I seem to be missing a
concept
regarding the apply w/o calling a function but rather command args
%*% /
etc. Would using apply be more efficient than this approach?
I have observed examples in the archives using this type of
approach. Does
anybody have a snippet of a call to apply() that would accomplish
this as
well?
Thanks!
seed=50
$a = array(rnorm(20),dim=c(4,5))
$b = matrix(a[1,],dim(a)[1],dim(a)[2],byrow=T)
$a
[,1] [,2] [,3][,4] [,5]
[1,] -1.3682810 -0.4314462 1.57572752 0.67928882 -0.3672346
[2,] 0.4328180 0.6556479 0.64289931 0.08983289 0.1852306
[3,] -0.8113932 0.3219253 0.08976065 -2.99309008 0.5818237
[4,] 1.4441013 -0.7838389 0.27655075 0.28488295 1.3997368
$a/b
[,1] [,2] [,3] [,4] [,5]
[1,] 1.000 1.000 1. 1.000 1.00
[2,] -0.3163225 -1.5196515 0.40800157 0.1322455 -0.504393
[3,] 0.5930018 -0.7461539 0.05696457 -4.4062113 -1.584338
[4,] -1.0554128 1.8167710 0.17550671 0.4193841 -3.811560
CONFIDENTIALITY NOTICE: This electronic mail transmission
(i...{{dropped}}
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and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix multiplication using apply() or lappy() ?
In terms of speed Toby's original idea was actually the fastest. Here they are decreasing order of the largest timing in each row of system.time. I also tried it with a 100x10 matrix and got almost the same order: library(reshape) system.time(for(i in 1:1000) iapply(a, 1, /, a[1,])) [1] 11.51 0.01 18.65NANA system.time(for(i in 1:1000) t(apply(a, 1, /, a[1,]))) [1] 0.83 0.00 1.36 NA NA system.time(for(i in 1:1000) sweep(a, 2, a[1,], /)) [1] 0.27 0.00 0.39 NA NA system.time(for(i in 1:1000) a/outer(rep(1, nrow(a)), a[1,])) [1] 0.23 0.00 0.39 NA NA system.time(for(i in 1:1000) a %*% diag(1/a[1,])) [1] 0.25 0.00 0.38 NA NA system.time(for(i in 1:1000) a/rep(a[1,], each = nrow(a))) [1] 0.09 0.00 0.16 NA NA system.time(for(i in 1:1000) t(t(a)/a[1,])) [1] 0.10 0.00 0.13 NA NA system.time(for(i in 1:1000) a/matrix(a[1,], nrow(a), ncol(a), byrow = TRUE)) [1] 0.05 0.00 0.12 NA NA On 9/6/06, Rolf Turner [EMAIL PROTECTED] wrote: Prof. Brian Ripley wrote: On Wed, 6 Sep 2006, Christos Hatzis wrote: See ?sweep sweep(a, 2, a[1,],/) That is less efficient than a/rep(a[1,], each=nrow(a)) *My* first instinct was to use t(t(a)/a[1,]) (which has not heretofore been suggested). This seems to be more efficient still (at least in respect of Prof. Grothendieck's toy example) by between 20 and 25 percent: a - matrix(1:24,4) system.time(for(i in 1:1000) junk - a / rep(a[1,], each = 4)) [1] 0.690 0.080 1.051 0.000 0.000 system.time(for(i in 1:1000) junk - t(t(a)/a[1,])) [1] 0.520 0.120 0.647 0.000 0.000 system.time(for(i in 1:1) junk - a / rep(a[1,], each = 4)) [1] 7.08 0.99 10.08 0.00 0.00 system.time(for(i in 1:1) junk - t(t(a)/a[1,])) [1] 5.530 0.940 7.856 0.000 0.000 cheers, Rolf Turner __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix multiplication using apply() or lappy() ?
What version of R was this? In 2.4.0 alpha a - matrix(1:24,4) system.time(for(i in 1:1000) junk - a / rep(a[1,], each = 4)) [1] 0.014 0.000 0.014 0.000 0.000 system.time(for(i in 1:1000) junk - t(t(a)/a[1,])) [1] 0.057 0.000 0.058 0.000 0.000 shows a large margin the other way, which increases with bigger matrices a - matrix(pi*1:100, 100, 1000) system.time(for(i in 1:1000) junk - t(t(a)/a[1,])) [1] 18.329 2.238 20.595 0.000 0.000 system.time(for(i in 1:1000) junk - a / rep(a[1,], each = 4)) [1] 2.589 1.021 3.610 0.000 0.000 On Wed, 6 Sep 2006, Rolf Turner wrote: Prof. Brian Ripley wrote: On Wed, 6 Sep 2006, Christos Hatzis wrote: See ?sweep sweep(a, 2, a[1,],/) That is less efficient than a/rep(a[1,], each=nrow(a)) *My* first instinct was to use t(t(a)/a[1,]) (which has not heretofore been suggested). This seems to be more efficient still (at least in respect of Prof. Grothendieck's toy example) by between 20 and 25 percent: a - matrix(1:24,4) system.time(for(i in 1:1000) junk - a / rep(a[1,], each = 4)) [1] 0.690 0.080 1.051 0.000 0.000 system.time(for(i in 1:1000) junk - t(t(a)/a[1,])) [1] 0.520 0.120 0.647 0.000 0.000 system.time(for(i in 1:1) junk - a / rep(a[1,], each = 4)) [1] 7.08 0.99 10.08 0.00 0.00 system.time(for(i in 1:1) junk - t(t(a)/a[1,])) [1] 5.530 0.940 7.856 0.000 0.000 cheers, Rolf Turner __ 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 and provide commented, minimal, self-contained, reproducible code. -- 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 and provide commented, minimal, self-contained, reproducible code.
[R] singular factor analysis
Are there any functions available to do a factor analysis with fewer observations than variables? As long as you have more than 3 observations, my computations suggest you have enough data to estimate a factor analysis covariance matrix, even though the sample covariance matrix is singular. I tried the naive thing and got an error: set.seed(1) X - array(rnorm(50), dim=c(5, 10)) factanal(X, factors=1) Error in solve.default(cv) : system is computationally singular: reciprocal condition number = 4.8982e-018 I can write a likelihood for a multivariate normal and solve it, but I wondered if there is anything else available that could do this? Thanks, Spencer Graves __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with putting objects in list
I suspect you are not thinking about the list and the
subsetting/extraction operators in the right way.
A list contains a number of components.
To get a subset of the list, use the '[' operator. The subset can
contain zero or more components of the list, and it is a list itself.
So, if x is a list, then x[2] is a list containing a single component.
To extract a component from the list, use the '[[' operator. You can
only extract one component at a time. If you supply a vector index with
more than one element, it will index recursively.
x - list(1,2:3,letters[1:3])
x
[[1]]
[1] 1
[[2]]
[1] 2 3
[[3]]
[1] a b c
# a subset of the list
x[2:3]
[[1]]
[1] 2 3
[[2]]
[1] a b c
# a list with one component:
x[2]
[[1]]
[1] 2 3
# the second component itself
x[[2]]
[1] 2 3
# recursive indexing
x[[c(2,1)]]
[1] 2
x[[c(3,2)]]
[1] b
Rainer M Krug wrote:
Hi
I use the following code and it stores the results of density() in the
list dr:
dens - function(run) { density( positions$X[positions$run==run], bw=3,
cut=-2 ) }
dr - lapply(1:5, dens)
but the results are stored in dr[[i]] and not dr[i], i.e. plot(dr[[1]])
works, but plot([1]) doesn't.
Is there any way that I can store them in dr[i]?
Thanks a lot,
Rainer
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and provide commented, minimal, self-contained, reproducible code.
[R] deleting an arow added to a graphic
I know this has got to be simple, but I have a added an arrow to a graph with: arrows(5,8,8, predict(lmfit,data.frame(x=8)), length=0.1) but its in the wrong position, correcting it and running again adds an new arrow (which is what you would expect) so how do I a) edit the existing arrow, and b) delete it all together As so often seems to be the case, some of the simplist things seem also to be the most difficult to find the answer to. Many thanks, Graham [[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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] deleting an arow added to a graphic
On 9/6/2006 3:04 PM, Graham Smith wrote: I know this has got to be simple, but I have a added an arrow to a graph with: arrows(5,8,8, predict(lmfit,data.frame(x=8)), length=0.1) but its in the wrong position, correcting it and running again adds an new arrow (which is what you would expect) so how do I a) edit the existing arrow, and b) delete it all together As so often seems to be the case, some of the simplist things seem also to be the most difficult to find the answer to. Generally classic graphics in R are like drawing in ink on paper: you can't remove items that you've drawn there. So the way to do what you want is to save the commands that produced the entire graph, and edit them until you get them right. Then you can run them and produce a new graph that's just right. grid allows items to be removed from an existing graph, so lattice and ggplot inherit this nice property. rgl (for 3d graphics) also allows items to be removed in a fairly inflexible way (only in the reverse order of the order drawn); the next release will make this more flexible. 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] deleting an arow added to a graphic
This does not actually remove it but you could overwrite it with an arrow the same color as the background and then plot a new arrow: x - 1:10 plot(x ~ x) arrows(1, 1, 2, 2) # revise it arrows(1, 1, 2, 2, col = white) arrows(2, 2, 3, 3) On 9/6/06, Graham Smith [EMAIL PROTECTED] wrote: I know this has got to be simple, but I have a added an arrow to a graph with: arrows(5,8,8, predict(lmfit,data.frame(x=8)), length=0.1) but its in the wrong position, correcting it and running again adds an new arrow (which is what you would expect) so how do I a) edit the existing arrow, and b) delete it all together As so often seems to be the case, some of the simplist things seem also to be the most difficult to find the answer to. Many thanks, Graham [[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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] continuation lines in R script files
I literally copy/pasted your second version (without the + signs) in a file and the sourced that file. It loaded without error and the a matrix was as I expected. On 9/6/06, Evan Cooch [EMAIL PROTECTED] wrote: When I have to enter a very large matrix into the R console, I can make use of the continuation feature in the console to enter the matrix in pieces (e.g., on a row by row basis). So, for example, the console would show the + sign for continuation lines - something like what I've written below: a=matrix(c(0,20,50, + 0.05,0,0, + 0,0.1,0), + 3,3,byrow=T) (obviously, for a matrix this small - 3x3 - I could enter it all on a single line, this is just to demonstrate) My question is - how do you accomplish the same thing in an R script file? I've tried literally copying the preceding - syntax error at the second line. I've also tried a=matrix(c(0,20,50, 0.05,0,0, 0,0.1,0), 3,3,byrow=T) Again, syntax error, at the second line... After multiple searches for 'continuation line', with no luck (everything I found refers to the R console, no a script file), I'll ask here. Basically, I want to know how to get an R script to handle a structure entered over multiple lines (e.g., a matrix). This is default behaviour in .m files in Matlab, and most other environments I've ever worked in (e.g., SAS looks for the ; to indicate end of a line). Thanks in advance... __ 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 and provide commented, minimal, self-contained, reproducible code. -- Mike Meyer, Seattle WA [[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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] continuation lines in R script files
Are your sure your second solution does not work? Try again...
Evan Cooch wrote:
When I have to enter a very large matrix into the R console, I can make
use of the continuation feature in the console to enter the matrix in
pieces (e.g., on a row by row basis). So, for example, the console
would show the + sign for continuation lines - something like what
I've written below:
a=matrix(c(0,20,50,
+ 0.05,0,0,
+ 0,0.1,0),
+ 3,3,byrow=T)
(obviously, for a matrix this small - 3x3 - I could enter it all on a
single line, this is just to demonstrate)
My question is - how do you accomplish the same thing in an R script
file? I've tried literally copying the preceding - syntax error at the
second line. I've also tried
a=matrix(c(0,20,50,
0.05,0,0,
0,0.1,0),
3,3,byrow=T)
Again, syntax error, at the second line...
After multiple searches for 'continuation line', with no luck
(everything I found refers to the R console, no a script file), I'll ask
here. Basically, I want to know how to get an R script to handle a
structure entered over multiple lines (e.g., a matrix). This is default
behaviour in .m files in Matlab, and most other environments I've ever
worked in (e.g., SAS looks for the ; to indicate end of a line).
Thanks in advance...
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The information contained in this e-mail is confidential and...{{dropped}}
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Re: [R] About the Skew Student distribution
pierre clauss pierreclauss at yahoo.fr writes: Hello everybody, I need your help about the package SN and the skew student distribution. Il will be very grateful if I have the solution. I construct a stochastic model with a white noise not gaussian but following a skew student distribution. I fit the noise on monthly data to obtain the four parameters. The question is : how to annualize the parameters to use my model for simulate daily data for example ? If the volatility is estimated to 3 for example, I need to multiply this by sqrt(12) to have for the parameter of volatility of the skew student : 3*sqrt(12)*sqrt(dt) with dt the time increment parameter (1/12 for monthly data, 1/261 for daily data, and so on). Do I do the same thing (and what is the multiplicative factor ?) for the parameters of asymmetry and the degree of freedom ? I'm not sure about your application, but I'll take a crack at it. It sounds like you've got a continuous-time stochastic process (you don't say explicitly). For a Brownian motion, you would just do what you suggest -- scale the variance by sqrt(dt) (I don't see exactly why daily data have dt=1/261 -- although 365*5/7 = 261, so I guess you're counting weekdays (trading days??) only). Unfortunately, it's not nearly as transparent (to me) what stochastic differential equation would lead to aggregated data that were skew-Student. The review paper on Azzalini (SN's author)'s web site ( http://azzalini.stat.unipd.it/SN/review-web.ps ) cites some papers in computational finance, which I'm guessing is your area -- your best bet is probably to go back to those papers and see if they deal with the effects of temporal aggregation. VERY crudely, you can just experiment with this yourself by simulating values from a particular skew-Student distribution, aggregating them, and then looking at the properties of the resulting distribution -- *do* the asymmetry and df change? (I bet they do -- in some sense temporal aggregation must lead to a distribution that is more normal, larger df and smaller skew -- but reversing this could be quite ugly). good luck Ben Bolker finance __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] How to get multiple partial matches?
Hi, I'm very new to R, and am not at all a software programmer of any sort.I appreciate any help you may have. I have figured out how to get my data into a dataframe and order it alphabetically according to a particular column. Now, I would like to seperate out certain rows based on partial character matches. Here is an (extremely) abreviated example of my data set Probe Ch1 Median - B Ch1 Mean - B 72 5S_F_1501 567 7700 5S_F_2338 611 7517 5S_F_3412 467 10687 5S_F_4380 428 4870 5S_F_5315 368 6035 5S_F_6300 359 3826 5S_F_7350 386 8754 5S_F_8450 473 6399 5S_F_9439 494 749 5S_F_10334 384 I would like to be able to select out all rows with, for example, 5S_F_ in the Probe column (there are non-5S_F_ containing values in the real, larger data set). I think pmatch does this for instances where there is only 1 match, but I would like to recover all the matches. I have tried to use charmatch, match, pmatch, agrep and grep for this purpose, but with no luck. When I grep for 5S_F_ with value = T, I get character(0) Adding wildcards (either * or .) does not change this outcome. I thought maybe the underscores were messing it up, so I tried to grep 5S* with value = T, and I get a long list of numbers back [1] 55 95 56 57 58 59 65 75 85 105 [11] 115 125 135 5555 555 These numbers make no sense to me. They don't seem to correlate with where the 5S's occur in the dataframe, and they don't look like any values in the Probe column (there are no numeric vaules in the Probe column, just strings of character digit combinations). How can I select out all the rows with the same partial character match? __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to get multiple partial matches?
Try using 'grep' and regular expressions:
x - 72 5S_F_1501 567
+ 7700 5S_F_2338 611
+ 7517 5S_F_3412 467
+ 10687 5S_F_4380 428
+ 4870 5S_F_5315 368
+ 6035 5S_F_6300 359
+ 3826 5S_F_7350 386
+ 8754 5S_F_8450 473
+ 6399 5S_F_9439 494
+ 749 5S_F_10334 384
+
df - read.table(textConnection(x))
df
V1 V2 V3 V4
1 72 5S_F_1 501 567
2 7700 5S_F_2 338 611
3 7517 5S_F_3 412 467
4 10687 5S_F_4 380 428
5 4870 5S_F_5 315 368
6 6035 5S_F_6 300 359
7 3826 5S_F_7 350 386
8 8754 5S_F_8 450 473
9 6399 5S_F_9 439 494
10 749 5S_F_10 334 384
# select only ones with '5S_F_1'
df[grep('5S_F_1', as.character(df$V2)),]
V1 V2 V3 V4
1 72 5S_F_1 501 567
10 749 5S_F_10 334 384
On 9/6/06, Sarah Tucker [EMAIL PROTECTED] wrote:
Hi,
I'm very new to R, and am not at all a software
programmer of any sort.I appreciate any help you
may have. I have figured out how to get my data into
a dataframe and order it alphabetically according to a
particular column. Now, I would like to seperate out
certain rows based on partial character matches. Here
is an (extremely) abreviated example of my data set
Probe Ch1 Median - B Ch1 Mean - B
72 5S_F_1501 567
7700 5S_F_2338 611
7517 5S_F_3412 467
10687 5S_F_4380 428
4870 5S_F_5315 368
6035 5S_F_6300 359
3826 5S_F_7350 386
8754 5S_F_8450 473
6399 5S_F_9439 494
749 5S_F_10334 384
I would like to be able to select out all rows with,
for example, 5S_F_ in the Probe column (there are
non-5S_F_ containing values in the real, larger data
set).
I think pmatch does this for instances where there is
only 1 match, but I would like to recover all the
matches. I have tried to use charmatch, match,
pmatch, agrep and grep for this purpose, but with no
luck.
When I grep for 5S_F_ with value = T, I get
character(0)
Adding wildcards (either * or .) does not change
this outcome.
I thought maybe the underscores were messing it up, so
I tried to grep 5S* with value = T, and I get a long
list of numbers back
[1] 55 95 56 57 58 59 65
75 85 105
[11] 115 125 135 5555
555
These numbers make no sense to me. They don't seem to
correlate with where the 5S's occur in the
dataframe, and they don't look like any values in the
Probe column (there are no numeric vaules in the Probe
column, just strings of character digit combinations).
How can I select out all the rows with the same
partial character match?
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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Re: [R] Help on estimated variance in lme4
Could you try this model fit again adding control = list(usePQL = FALSE, msVerbose=TRUE) to the argument list of the call to lmer? By default PQL iterations are used at the beginning of a generalized linear mixed model fit followed by optimization of the Laplace approximation to the log-likelihood when method = Laplace. Sometimes the PQL iterations do more harm than good and you do better going straight to the optimization of the Laplace approximation. On 9/6/06, jerome lemaitre [EMAIL PROTECTED] wrote: Dear all, I get an error message when I run my model and I am not sure what to do about it. I try to determine what factors influence the survival of voles. I use a mixed-model because I have several voles per site (varying from 2 to 19 voles). Here is the model: ### fm5 -lmer(data=cdrgsaou2, alive~factor(pacut)+factor(agecamp)+factor(sex)+ResCondCorp+(1|factor(cdrgsa ou2$ids)), family=binomial, method=Laplace, ) ### Description of variables Alive: 0 or 1; dead or alive pacut: 0 or 1; presence of parasites agecamp: a or j; adult or juvenile sex: m or f; male or female ResCondCorp: body condition, continuous; cdrgsaou2$ids: name of the site. Here is the output: ### Generalized linear mixed model fit using Laplace Formula: alive ~ factor(pacut) + factor(agecamp) + factor(sex) + ResCondCorp + (1 | factor(cdrgsaou2$ids)) Data: cdrgsaou2 Family: binomial(logit link) AIC BIClogLik deviance 305.7418 328.7331 -146.8709 293.7418 Random effects: GroupsNameVariance Std.Dev. factor(cdrgsaou2$ids) (Intercept) 0.034382 0.18542 number of obs: 341, groups: factor(cdrgsaou2$ids), 36 Estimated scale (compare to 1) 2.174681 Fixed effects: Estimate Std. Error z value Pr(|z|) (Intercept) 0.971458 0.250951 3.8711 0.0001083 *** factor(pacut)1 -0.831888 0.358583 -2.3199 0.0203447 * factor(agecamp)j -1.294236 0.330638 -3.9144 9.065e-05 *** factor(sex)m 0.581713 0.296229 1.9637 0.0495616 * ResCondCorp -0.176251 0.020263 -8.6982 2.2e-16 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Correlation of Fixed Effects: (Intr) fct()1 fctr(g) fctr(s) factr(pct)1 -0.334 fctr(gcmp)j -0.417 0.066 factor(sx)m -0.505 -0.002 -0.173 ResCondCorp -0.309 -0.010 0.302 -0.032 ### Here is the error message: ### Warning message: Estimated variance for factor 'factor(cdrgsaou2$ids)' is effectively zero in: LMEopt(x = mer, value = cv) ### Thank you very much by advance for any help. Jérôme Lemaître Ph.D. student Silviculture-wildlife research chair in irregular boreal forests Départment of biology, Faculty of Sciences and Engineering Alexandre-Vachon building University Laval Quebec, QC G1K 7P4 Phone : (418) 656-2131 poste 2917 Office : VCH-2044 Email: [EMAIL PROTECTED] __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with Variance Components (and general glmm confusion)
On 9/4/06, Toby Gardner [EMAIL PROTECTED] wrote: Dear list, I am having some problems with extracting Variance Components from a random-effects model: I am running a simple random-effects model using lme: model-lme(y~1,random=~1|groupA/groupB) which returns the output for the StdDev of the Random effects, and model AIC etc as expected. Until yesterday I was using R v. 2.0, and had no problem in calling the variance components of the above model using VarCorr(model), together with their 95% confidence intervals using intervals() - although for some response variables a call to intervals() returns the error: Cannot get confidence intervals on var-cov components: Non-positive definite approximate variance-covariance. I have now installed R v. 2.3.1 and am now experiencing odd behaviour with VarCorr(lme.object), with an error message typically being returned: Error in VarCorr(model) : no direct or inherited method for function 'VarCorr' for this call Is this known to happen? For instance could it be due to the subsequent loading of new packages? (lme4 for instance?). Yes. Avoid loading lme4 and nlme simultaneously. To get around this problem I have tried running the same model using lmer: model2-lmer(y~1 + (1|groupA) + (1|groupB)) In recent versions of lme4 you can use the specification model2 - lmer(y ~ 1 + (1|groupA/groupB)) Your version may be correct or not. It depends on what the distinct levels of groupB correspond to. The version with the / is more reliable. Should this not produce the same model? The variance components are very similar but not identical, making me think that I am doing something wrong. I am also correct in thinking that intervals() does not work with lmer? I get: Error in intervals(model2) : no applicable method for intervals That is correct. Currently there is no intervals method for an lmer model. You can use mcmcsamp to get a Markov chain Monte Carlo sample to which you can apply HPDinterval from the coda package. However, these are stochastic intervals so it is best to try on a couple of chains to check on the reproducibility or the intervals. GLMM I have a general application question - please excuse my ignorance, I am relatively new to this and trying to find a way through the maze. In short I need to compile generalized linear mixed models both for (a) Poisson data and (b) binonial data incorporating a two nested random factors, and I need to be able to extract AIC values as I am taking an information-theoretic approach to model selection. Prior to sending an email to the list I have spent quite a few days reading the background on a number of functions, all of which offer potential for this; glmmML, glmmPQL, lmer, and glmmADMB. I can understand that glmmPQL is unsuitable because there is no way of knowing the maximised likelihood, but is there much difference between the remaining three options? I have seen simulation comparisons published on this list between glmmADMB and glmmPQL and lmer, but it seems these are before the latest release of lmer, and also they do not evaluate glmmML. To a newcomer this myria! d ! of options is bewildering, can anyone offer advice as to the most robust approach? Goran can correct me if I am wrong but I don't believe that glmmML can be used with multiple levels of random effects. I'm not sure what the status of glmmADMB is these days. There was some controversy regarding the license applied to some of that code a while back. I don't know if it has been resolved to everyone's satisfaction. When using lmer I would suggest using method = Laplace and perhaps control = list(usePQL = FALSE, msVerbose = 1) as I mentioned in another reply to the list a few minutes ago. Let us know how it works out. Many thanks for your time and patience, Toby Gardner School of Environmental Sciences University of East Anglia Norwich, NR4 7TJ United Kingdom Email: [EMAIL PROTECTED] Website: www.uea.ac.uk/~e387495 [[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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] graphics - joining repeated measures with a line
I would like to join repeated measures for patients across two visits using a line. The program below uses symbols to represent each patient. Basically, I would like to join each pair of symbols. library(lattice) patient - c(1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9) var - c(826,119,168,90,572,323,122,10,42,900,250,180,120,650,400,130,12,33) visit - c(1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2) symbols - c(1,2,3,4,5,6,7,8,9) xyplot(var ~ visit, pch = symbols[patient], key = list(points = list(pch = symbols), space = list(right),text = list(c(1,2,3,4,5,6,7,8,9 # grid.lines(x = visit,y = var,draw = TRUE) ?? I am thinking I may need to use a function that joins coordinates (for example join (1,826) with (2,900)), but am hoping there may be a better way. Thanks for any help. Murray -- Murray Pung Statistician, Datapharm Australia Pty Ltd 0404 273 283 [[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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] graphics - joining repeated measures with a line
Make each pair of points a separate group using group= and specify that both points and lines be used via type = b. Also set the symbols in par.settings= so that they are accessed by both the main plot and the legend: xyplot(var ~ visit, group = symbols[patient], type = b, auto.key = list(space = right), par.settings = list(superpose.symbol = list(pch = symbols))) On 9/6/06, Murray Pung [EMAIL PROTECTED] wrote: I would like to join repeated measures for patients across two visits using a line. The program below uses symbols to represent each patient. Basically, I would like to join each pair of symbols. library(lattice) patient - c(1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9) var - c(826,119,168,90,572,323,122,10,42,900,250,180,120,650,400,130,12,33) visit - c(1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2) symbols - c(1,2,3,4,5,6,7,8,9) xyplot(var ~ visit, pch = symbols[patient], key = list(points = list(pch = symbols), space = list(right),text = list(c(1,2,3,4,5,6,7,8,9 # grid.lines(x = visit,y = var,draw = TRUE) ?? I am thinking I may need to use a function that joins coordinates (for example join (1,826) with (2,900)), but am hoping there may be a better way. Thanks for any help. Murray -- Murray Pung Statistician, Datapharm Australia Pty Ltd 0404 273 283 [[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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] graphics - joining repeated measures with a line
Just one correction (although in this case it does not change the output) -- use group = patient rather than group = symbol[patient]: xyplot(var ~ visit, group = patient, type = b, auto.key = list(space = right), par.settings = list(superpose.symbol = list(pch = symbols))) On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote: Make each pair of points a separate group using group= and specify that both points and lines be used via type = b. Also set the symbols in par.settings= so that they are accessed by both the main plot and the legend: xyplot(var ~ visit, group = symbols[patient], type = b, auto.key = list(space = right), par.settings = list(superpose.symbol = list(pch = symbols))) On 9/6/06, Murray Pung [EMAIL PROTECTED] wrote: I would like to join repeated measures for patients across two visits using a line. The program below uses symbols to represent each patient. Basically, I would like to join each pair of symbols. library(lattice) patient - c(1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9) var - c(826,119,168,90,572,323,122,10,42,900,250,180,120,650,400,130,12,33) visit - c(1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2) symbols - c(1,2,3,4,5,6,7,8,9) xyplot(var ~ visit, pch = symbols[patient], key = list(points = list(pch = symbols), space = list(right),text = list(c(1,2,3,4,5,6,7,8,9 # grid.lines(x = visit,y = var,draw = TRUE) ?? I am thinking I may need to use a function that joins coordinates (for example join (1,826) with (2,900)), but am hoping there may be a better way. Thanks for any help. Murray -- Murray Pung Statistician, Datapharm Australia Pty Ltd 0404 273 283 [[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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] stratified poisson regression
Hello, I'm fitting poisson regression to mortality data and wish to stratify by age. Is there any way to perform this stratification and use the glm function in R? Thanks, Hannah 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 and provide commented, minimal, self-contained, reproducible code.
