[R] Generating random integers
Dear R users, I need to generate random integer(s) in a range (say that beetween 1 to 100) in R. Any help is deeply appreciated. Kind Regards Seyit Ali Dr. Seyit Ali KAYIS Selcuk University, Faculty of Agriculture Kampus/Konya, Turkey s_a_ka...@yahoo.com, s_a_ka...@hotmail.com Tel: +90 332 223 2830 Mobile: +90 535 587 1139 Greetings from Konya, Turkey http://www.ziraat.selcuk.edu.tr/skayis/ __ R-help@r-project.org 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] Generating random integers
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of skayis selcuk Sent: Saturday, April 11, 2009 11:24 PM To: r-help@r-project.org Subject: [R] Generating random integers Dear R users, I need to generate random integer(s) in a range (say that beetween 1 to 100) in R. Any help is deeply appreciated. Kind Regards Seyit Ali Look at ?sample Hope this is helpful, Dan Daniel Nordlund Bothell, WA USA __ R-help@r-project.org 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 may I add is.outer method
Hello I have a problem like this: Glass$RI[is.outlier(Glass$RI)] Error: could not find function is.outlier Which commend I may add is.outer to my application? -- View this message in context: http://www.nabble.com/How-may-I-add-%22is.outer%22-method-tp23006216p23006216.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] Labeling points on plot on relative warp scores?
stephalope wrote: Hi there, I am plotting relative warp scores (equivalent to pca scores) and I want to label (color code and shape) the points by group. I can't figure out how to do this beyond simple plotting. plot(RW1, RW2); Do I need to make vectors of each group and then plot them separately onto the same plot? How do I go about this? Hi stephalope, I've done this with boxed.labels in the plotrix package for factor analyses or PCAs when there aren't many variables. I use strong background colors for the box and white text for the variable names. You may have to shift some labels apart if they overlap, but it gives an easy to understand illustration of simple factor structures or PCAs. Jim __ R-help@r-project.org 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] Fedora 10 KDE plasma font rendering issue
I checked with my KDE 4.2 (Mandriva 2009 system, kde.org binaries)
with no font rendering issues in running demo(graphics).
It is quite possible that pango was not installed with Fedora (as it
is unnecessary for KDE 4.2 systems). Installing the relevant rpm may
fix things.
The packaging of the Fedora 2.8.1 rpm may need to take this into
account - if compiling, you'd have discovered at configure stage.
Paul Bivand
Paul Bivand
2009/4/1 Martyn Plummer plum...@iarc.fr:
On Tue, 2009-03-31 at 18:36 -0700, dfermin wrote:
Nope. I checked this. Both those fonts are installed.
Well it is some kind of font rendering problem. The default device is
the Cairo X11 device, which uses the Pango layout engine for font
rendering.
If you set the environment variable FC_DEBUG to 1 before launching your
R session, you will get some debugging information. It is very verbose,
but we only need to see this bit:
First font Pattern has 15 elts (size 15)
family: Nimbus Sans L(w)
style: Regular(w)
slant: 0(i)(w)
weight: 80(i)(w)
width: 100(i)(w)
foundry: urw(w)
file: /usr/share/fonts/default/Type1/n019003l.pfb(w)
index: 0(i)(w)
outline: FcTrue(w)
scalable: FcTrue(w)
Martyn Plummer-2 wrote:
Quoting dfermin dfer...@umich.edu:
Hello.
I've got a new workstation running Fedora 10 linux and I use the KDE 4.2
desktop which uses some kind of new desktop environment called 'plasma'.
If I start up R and generate a plot (for example: hist(rnorm(1,
mean=0,
sd=1), breaks=100) ). The plot appears but all text (the x/y axes, title,
etc..) is replaced by a square box. No font is rendered at all.
Has anyone else got this problem? If so do you have a work around or a
solution?
I'm using R version 2.8.1 installed from the Fedora 10 repositories if
that
helps.
Thanks in advance.
It sounds like you are missing some fonts. Check that the urw-fonts and
liberation-fonts RPMs are installed.
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Re: [R] Generating random integers
floor(runif(1000, 1,101)) On Sun, Apr 12, 2009 at 2:24 AM, skayis selcuk ska...@selcuk.edu.tr wrote: Dear R users, I need to generate random integer(s) in a range (say that beetween 1 to 100) in R. Any help is deeply appreciated. Kind Regards Seyit Ali Dr. Seyit Ali KAYIS Selcuk University, Faculty of Agriculture Kampus/Konya, Turkey s_a_ka...@yahoo.com, s_a_ka...@hotmail.com Tel: +90 332 223 2830 Mobile: +90 535 587 1139 Greetings from Konya, Turkey http://www.ziraat.selcuk.edu.tr/skayis/ __ R-help@r-project.org 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org 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 may I add is.outer method
I am not quite sure how you define your outlier, but the definition
that I am familiar with is that an outlier is greater than 1.5*Inter
quartile range above the 3rd quartile or below the 1st quartile (this
is the default for the whiskers in boxplot). This can be easily found
by
x[(x quantile(x)[2] - 1.5*IQR(x)) | (x quantile(x)[4] + 1.5*IQR
(x))]
or if you prefer
is.outlier - function(x) {(x quantile(x)[2] - 1.5*IQR(x)) | (x
quantile(x)[4] + 1.5*IQR(x))}
x[is.outlier(x)]
HTH.
Andrew.
On Apr 12, 8:51 am, Grześ gregori...@gmail.com wrote:
Hello
I have a problem like this:
Glass$RI[is.outlier(Glass$RI)]
Error: could not find function is.outlier
Which commend I may add is.outer to my application?
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context:http://www.nabble.com/How-may-I-add-%22is.outer%22-method-tp23006216p...
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Re: [R] Help with postscript (huge file size)
Sorry about some mistakes in the code. The correct one is: library(rimage) image size: 458 x 372 type: rgb laplacian_result - normalize(laplacian(image)) postscript(laplacian_result.eps) plot.imagematrix(laplacian_result) dev.off() Talita 2009/4/11 Talita Perciano talitaperci...@gmail.com Ok... I'm using the rimage package to manipulate an image. So, the image I have in R is of the type imagematrix, which is a matrix with the pixel values of the R, G anf B bands. What I'm doing is applying some operation (like laplacian filter for example) and plotting the result as an image: library(rimage) image size: 458 x 372 type: rgb laplacian_result - normalize(laplacian(image)) postscript(laplacian_result) plot.imagematrix(laplacian) dev.off() Talita 2009/4/11 Ben Bolker bol...@ufl.edu Do you mean you're importing jpegs or other bitmaps into R and writing them out (possibly with annotation etc.) as PostScript? Can you give a small example of some sort? It would help for giving advice. Talita Perciano wrote: Thank you for the answer. Just to clear things out, I'm generating plots of rgb images. Talita 2009/4/11 Ben Bolker bol...@ufl.edumailto:bol...@ufl.edu Talita Perciano wrote: Dear users, I'm generating some images in R to put into a document that I'm producing using Latex. This document in Latex is following a predefined model, which does not accept compilation with pdflatex, so I have to compile with latex - dvi - pdf. Because of that, I have to generate the images in R with postscript (I want a vector format to keep the quality). The problem is that the files of the images are very huge (10MB) and I have many images to put into the pdf document. I want to know if there is a way to reduce the size of those images generated by R using postscript. Thank you in advance, Talita Not in any extremely easy way. The fundamental problem is that if you have a whole lot of points in your graph, it's hard to make them take less file space even if they're overplotted (and hence not visible in the actual image). This has been discussed in various forms on the R list in the past, but I can't locate those posts easily. It's a little hard without knowing what kind of plot you're generating, but I'm assuming that you have many, many points or lines in the graphic (or a very high-resolution image plot), and that the details don't all show up in the figure anyway. A few general strategies: * thin the points down to a random subset * use a 2D density plot or hexagonal binning * create a bitmap (PNG) plot, then use image manipulation tools (ImageMagick etc.) to convert that back to a PostScript file * there was some discussion earlier about whether one could embed a bitmap of just the internals of the plot, leaving the axes, labels etc. in vector format, but I don't think that came to anything good luck Ben Bolker -- View this message in context: http://www.nabble.com/Help-with-postscript-%28huge-file-size%29-tp23003428p23004309.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.orgmailto:R-help@r-project.org 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. -- Talita Perciano Instituto de Matemática e Estatísitca Universidade de São Paulo - USP PhD Student in Computer Science São Paulo, SP, Brazil Tel: +55 11 8826 7092 Success is not final, failure is not fatal: it is the courage to continue that counts. (Winston Churchill) -- Ben Bolker Associate professor, Biology Dep't, Univ. of Florida bol...@ufl.edu / www.zoology.ufl.edu/bolker GPG key: www.zoology.ufl.edu/bolker/benbolker-publickey.asc -- Talita Perciano Instituto de Matemática e Estatísitca Universidade de São Paulo - USP PhD Student in Computer Science São Paulo, SP, Brazil Tel: +55 11 8826 7092 Success is not final, failure is not fatal: it is the courage to continue that counts. (Winston Churchill) -- Talita Perciano Instituto de Matemática e Estatísitca Universidade de São Paulo - USP PhD Student in Computer Science São Paulo, SP, Brazil Tel: +55 11 8826 7092 Success is not final, failure is not fatal: it is the courage to continue that counts. (Winston Churchill) [[alternative HTML version deleted]] __ R-help@r-project.org 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] Help with postscript (huge file size)
I'm generating some images in R to put into a document that I'm producing using Latex. This document in Latex is following a predefined model, which does not accept compilation with pdflatex, so I have to compile with latex - dvi - pdf. Because of that, I have to generate the images in R with postscript (I want a vector format to keep the quality). The problem is that the files of the images are very huge (10MB) and I have many images to put into the pdf document. I want to know if there is a way to reduce the size of those images generated by R using postscript. Just use a high-resolution png or tiff. At 300 dpi you won't be able to tell the difference when it's printed. Hadley -- http://had.co.nz/ __ R-help@r-project.org 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] lmer overdispersion
I got a similar problem when I used family=quasibinomial with my data. But, the problem disappeared when I used family=binomial. I assumed that Douglas Bates et al. had amended the lmer program to detect over-dispersion, so that it is no longer necessary to specify its possible presence with family=quasi... But, I may be wrong. If you get more information about this from the great man, then would you please let me know? Thanks, Jonathan Williams __ R-help@r-project.org 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] Bug in col2rgb?
On 31/03/2009 12:53 PM, Duncan Murdoch wrote: On 3/31/2009 12:29 PM, hadley wickham wrote: col2rgb(#0079, TRUE) [,1] red 0 green0 blue 0 alpha 121 col2rgb(#0080, TRUE) [,1] red255 green 255 blue 255 alpha0 col2rgb(#0081, TRUE) [,1] red 0 green0 blue 0 alpha 129 Any ideas? The #0080 string converts into the hex integer 0x8000, which, by an unfortunate coincidence, is the NA_integer value. Since NA_integer becomes the background colour, you get white instead of black with alpha=0x80. This can probably be fixed (if you have a string, there's a different way to know you have an NA). Want to work out a patch? The file to look at is src/main/colors.c. I've added a fix for this to R-devel. After 2.9.0 is released, I'll move it into R-patched. Duncan Murdoch __ R-help@r-project.org 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] Running random forest using different training and testing schemes
Hi, I would like to run random Forest classification algorithm and check the accuracy of the prediction according to different training and testing schemes. For example, extracting 70% of the samples for training and the rest for testing, or using 10-fold cross validation scheme. How can I do that? Is there a function? Thanks a lot, Chrysanthi. [[alternative HTML version deleted]] __ R-help@r-project.org 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] Re : Running random forest using different training and testing schemes
Hi Chysanthi, check out the randomForest package, with the function randomForest. It has a CV option. Sorry for not providing you with a lengthier response at the moment but I'm rather busy on a project. Let me know if you need more help. Also, to split your data into two parts- the training and the test set you can do (n the number of data points): n-length(data[,1]) indices-sample(rep(c(TRUE,FALSE),each=n/2),round(n/2),replace=TRUE) training_indices-(1:n)[indices] test_indices-(1:n)[!indices] Then, data[train,] is the training set and data[test,] is the test set. Best, Pierre De : Chrysanthi A. chrys...@gmail.com À : r-help@r-project.org Envoyé le : Dimanche, 12 Avril 2009, 17h26mn 59s Objet : [R] Running random forest using different training and testing schemes Hi, I would like to run random Forest classification algorithm and check the accuracy of the prediction according to different training and testing schemes. For example, extracting 70% of the samples for training and the rest for testing, or using 10-fold cross validation scheme. How can I do that? Is there a function? Thanks a lot, Chrysanthi. [[alternative HTML version deleted]] __ R-help@r-project.org 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. [[alternative HTML version deleted]] __ R-help@r-project.org 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] Re : Running random forest using different training and testing schemes
you need to include in your code something like: tree-rpart(result~., data, control=rpart.control(xval=10)). this xval=10 is 10-fold CV. Best, Pierre De : Chrysanthi A. chrys...@gmail.com À : r-help@r-project.org Envoyé le : Dimanche, 12 Avril 2009, 17h26mn 59s Objet : [R] Running random forest using different training and testing schemes Hi, I would like to run random Forest classification algorithm and check the accuracy of the prediction according to different training and testing schemes. For example, extracting 70% of the samples for training and the rest for testing, or using 10-fold cross validation scheme. How can I do that? Is there a function? Thanks a lot, Chrysanthi. [[alternative HTML version deleted]] __ R-help@r-project.org 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. [[alternative HTML version deleted]] __ R-help@r-project.org 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] Generating random integers
On Sun, Apr 12, 2009 at 1:21 PM, jim holtman jholt...@gmail.com wrote: floor(runif(1000, 1,101)) I need to generate random integer(s) in a range (say that beetween 1 to 100) in R. Another way: sample(1:100,1000,replace=T) Paul __ R-help@r-project.org 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] Convert string to time
One variable contains values (1.30 - one hour and thirty minutes, 1.2 (which is supposed to be 1.20 - one hour and twenty minutes)). I would like to convert to a minute variable so 1.2 is converted to 80 minutes. How? __ R-help@r-project.org 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] goodness of fit between two samples of size N (discrete variable)
Hello list: I generate by simulation (using different procedures) two sample vectors of size N, each corresponding to a discrete variable and I want to text if these samples can be considered as having the same probability distribution (which is unknown). What is the best test for that? I've read that Kolmogorov-Smirnov and Anderson-Darling tests are restricted to continuous data (http://cran.r-project.org/doc/contrib/Ricci-distributions-en.pdf), while chi-square can handle discrete data, but how do i test (in R) equivalence of ditribution in 2 samples using it? Are there better tests than those i mentioned? Thanks and regards, jlrp __ R-help@r-project.org 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] Convert string to time
On 12 April 2009 at 21:00, Peter Kraglund Jacobsen wrote:
| One variable contains values (1.30 - one hour and thirty minutes, 1.2
| (which is supposed to be 1.20 - one hour and twenty minutes)). I would
| like to convert to a minute variable so 1.2 is converted to 80
| minutes. How?
The littler sources have a script pace.r (see below) that does some simple
transformations so that it can come up with min/mile and min/km expression
given a distance (in miles) and a time (in fractional minutes) as 'read' --
eg in the example below 76.02 stands for 1 hour 16 minutes and 2 seconds.
In a nutshell, you divide and keep score of remainders.
There may be more compact ways to do this---pace.r is a pretty linear
translation of an earlier Octave script pace.m.
Hth, Dirk
e...@ron:~ pace.r 10.20 76.02
Miles: 10.2
Time : 76.02
Pace/m : 7 min 27.25 sec
Pace/km : 4 min 37.91 sec
e...@ron:~
e...@ron:~ cat bin/pace.r
#!/usr/bin/env r
#
# a simple example to convert miles and times into a pace
# where the convention is that we write e.g. 37 min 15 secs
# as 37.15 -- so a call 'pace.r 4.5 37.15' yields a pace of
# 8.1667, ie 8 mins 16.67 secs per mile
if (is.null(argv) | length(argv)!=2) {
cat(Usage: pace.r miles time\n)
q()
}
dig - 6
rundist - as.numeric(argv[1])
runtime - as.numeric(argv[2])
cat(Miles\t :, format(rundist, digits=dig), \n)
cat(Time\t :, format(runtime, digits=dig), \n)
totalseconds - floor(runtime)*60 + (runtime-floor(runtime))*100
totalsecondspermile - totalseconds / rundist
minutespermile - floor(totalsecondspermile/60)
secondspermile - totalsecondspermile - minutespermile*60
totalsecondsperkm - totalseconds / (rundist * 1.609344)
minutesperkm - floor(totalsecondsperkm/60)
secondsperkm - totalsecondsperkm - minutesperkm*60
pace - minutespermile + secondspermile/100
pacekm - minutesperkm + secondsperkm/100
cat(sprintf(Pace/m\t : % 2.0f min %05.2f sec\n,
minutespermile, secondspermile))
cat(sprintf(Pace/km\t : % 2.0f min %05.2f sec\n,
minutesperkm, secondsperkm))
e...@ron:~
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[R] p-values from bootstrap - what am I not understanding?
Dear stats experts: Me and my little brain must be missing something regarding bootstrapping. I understand how to get a 95%CI and how to hypothesis test using bootstrapping (e.g., reject or not the null). However, I'd also like to get a p-value from it, and to me this seems simple, but it seems no-one does what I would like to do to get a p-value, which suggests I'm not understanding something. Rather, it seems that when people want a p-value using resampling methods, they immediately jump to permutation testing (e.g., destroying dependencies so as to create a null distribution). SO - here's my thought on getting a p-value by bootstrapping. Could someone tell me what is wrong with my approach? Thanks: STEPS TO GETTING P-VALUES FROM BOOTSTRAPPING - PROBABLY WRONG: 1) sample B times with replacement, figure out theta* (your statistic of interest). B is large ( 1000) 2) get the distribution of theta* 3) the mean of theta* is generally near your observed theta. In the same way that we use non-centrality parameters in other situations, move the distribution of theta* such that the distribution is centered around the value corresponding to your null hypothesis (e.g., make the distribution have a mean theta = 0) 4) Two methods for finding 2-tailed p-values (assuming here that your observed theta is above the null value): Method 1: find the percent of recentered theta*'s that are above your observed theta. p-value = 2 * this percent Method 2: find the percent of recentered theta*'s that are above the absolute value of your observed value. This is your p-value. So this seems simple. But I can't find people discussing this. So I'm thinking I'm wrong. Could someone explain where I've gone wrong? J Jackson [[alternative HTML version deleted]] __ R-help@r-project.org 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] Running random forest using different training and testing schemes
There is also the train function in the caret package. The trainControl function can be used to try different resampling schemes. There is also a package vignette with details. Max On Apr 12, 2009, at 12:26 PM, Chrysanthi A. chrys...@gmail.com wrote: Hi, I would like to run random Forest classification algorithm and check the accuracy of the prediction according to different training and testing schemes. For example, extracting 70% of the samples for training and the rest for testing, or using 10-fold cross validation scheme. How can I do that? Is there a function? Thanks a lot, Chrysanthi. [[alternative HTML version deleted]] __ R-help@r-project.org 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@r-project.org 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 Loop and overwritten results
Dear all, I am a newbie to R and practising at the moment. Here is my problem: I have a programme with 2 loops involved. The inner loop get me matrices as output and safes all values for me. Now once I wrote a 2nd loop around the other loop in order to repeat the inner loop a couple of times, the results are overwritten and i found no way how to actually put the output results in a vector. I can receive single results only, like this [1] number [1] number2 but it want it rather like this [1] number number2 number3 ... any advice? Thanks unbekannter weise -- View this message in context: http://www.nabble.com/Problem-with-Loop-and-overwritten-results-tp23013391p23013391.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] First Derivative of Data Matrix
I am really new to R and ran across a need to take a data matrix and calculate an approximation of the first derivative of the data. I am more than happy to do an Excel kind of calculation (deltaY/deltaX) for each pair of rows down the matrix, but I don't know how to get R to do that kind of calculation. I'd like to store it as a 3rd column in the matrix as well. My data looks like this: acflong 1 1.000 2 0.9875858 3 0.9871751 4 0.9867585 5 0.9863358 6 0.9859070 7 0.9854721 8 0.9850316 9 0.9817161 10 0.9812650 and I'd like to generate a table like this: acflong dacflong/dx 1 1.000 2 0.9875858-0.01241 #delta(acflong)/delta(index) 3 0.9871751-0.00041 4 0.9867585-0.00042 5 0.9863358-0.00042 6 0.9859070-0.00043 7 0.9854721-0.00043 8 0.9850316-0.00044 9 0.9817161-0.00033 10 0.9812650 -0.00045 Is there a way to do this in R and how do I eliminate the first line of the data? Thanks, -Chris -- View this message in context: http://www.nabble.com/First-Derivative-of-Data-Matrix-tp23012026p23012026.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] Generalised Rejection Sampling
Hi,
I am trying to figure out the observed acceptance rate and M, using generalised
rejection sampling to generate a sample from the posterior distribution for p.
I have been told my code doesn't work because I need to take the log of the
expression for M, evaluate it and then exponentiate the result. This is
because R is unable to calculate high powers such as 545.501.
As you can see in my code I have tried to taking the log of M and then the
exponential of the result, but I clearly must be doing something wrong.
I keep getting the error message:
Error in if (U = ratio/exp(M)) { : missing value where TRUE/FALSE needed
Any ideas how I go about correctly taking the log and then the exponential?
rvonmises.norm - function(n,alpha,beta) {
out - rep(0,n)
counter - 0
total.sim - 0
p-alpha/(alpha+beta)
M
-logalpha-1)^(alpha-1))*((beta-1)^(beta-1)))/((beta+alpha-2)^(alpha+beta-2)))
while(counter n) {
total.sim - total.sim+1
proposal - runif(1)
if(proposal = 0 proposal = 1) {
U - runif(1)
ratio- (p^(alpha-1))*((1-p)^(beta-1))
if(U =ratio/exp(M)) {
counter - counter+1
out[counter] - proposal
}
}
}
obs.acc.rate - n/total.sim
return(out,obs.acc.rate,M)
}
set.seed(220)
temp - rvonmises.norm(1,439.544,545.501)
print(temp$obs.acc.rate)
Louisa
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and provide commented, minimal, self-contained, reproducible code.
[R] taking the log then later exponentiate the result query
Hi,
I am trying to figure out the observed acceptance rate and M, using generalised
rejection sampling to generate a sample from the posterior distribution for p.
I have been told my code doesn't work because I need to take the log of the
expression for M, evaluate it and then exponentiate the result. This is
because R is unable to calculate high powers such as 545.501.
As you can see in my code I have tried to taking the log of M and then the
exponential of the result, but I clearly must be doing something wrong.
I keep getting the error message:
Error in if (U = ratio/exp(M)) { : missing value where TRUE/FALSE needed
Any ideas how I go about correctly taking the log and then the exponential?
rvonmises.norm - function(n,alpha,beta) {
out - rep(0,n)
counter - 0
total.sim - 0
p-alpha/(alpha+beta)
M
-logalpha-1)^(alpha-1))*((beta-1)^(beta-1)))/((beta+alpha-2)^(alpha+beta-2)))
while(counter n) {
total.sim - total.sim+1
proposal - runif(1)
if(proposal = 0 proposal = 1) {
U - runif(1)
ratio- (p^(alpha-1))*((1-p)^(beta-1))
if(U =ratio/exp(M)) {
counter - counter+1
out[counter] - proposal
}
}
}
obs.acc.rate - n/total.sim
return(out,obs.acc.rate,M)
}
set.seed(220)
temp - rvonmises.norm(1,439.544,545.501)
print(temp$obs.acc.rate)
Louisa
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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] Problem with Loop and overwritten results
Dear unbekannt; The construction that would append a number to a numeric vector would be: vec - c(vec , number) You can create an empty vector with vec - c() or vec - NULL -- David Winsemius On Apr 12, 2009, at 2:10 PM, unbekannt wrote: Dear all, I am a newbie to R and practising at the moment. Here is my problem: I have a programme with 2 loops involved. The inner loop get me matrices as output and safes all values for me. Now once I wrote a 2nd loop around the other loop in order to repeat the inner loop a couple of times, the results are overwritten and i found no way how to actually put the output results in a vector. I can receive single results only, like this [1] number [1] number2 but it want it rather like this [1] number number2 number3 ... any advice? Thanks unbekannter weise -- View this message in context: http://www.nabble.com/Problem-with-Loop-and-overwritten-results-tp23013391p23013391.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org 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] taking the log then later exponentiate the result query
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Mary Winter
Sent: Sunday, April 12, 2009 1:39 PM
To: r-help@r-project.org
Subject: [R] taking the log then later exponentiate the result query
Hi,
I am trying to figure out the observed acceptance rate and M,
using generalised rejection sampling to generate a sample
from the posterior distribution for p.
I have been told my code doesn't work because I need to
take the log of the expression for M, evaluate it and then
exponentiate the result. This is because R is unable to
calculate high powers such as 545.501.
As you can see in my code I have tried to taking the log of M
and then the exponential of the result, but I clearly must be
doing something wrong.
I keep getting the error message:
Error in if (U = ratio/exp(M)) { : missing value where
TRUE/FALSE needed
Any ideas how I go about correctly taking the log and then
the exponential?
rvonmises.norm - function(n,alpha,beta) {
out - rep(0,n)
counter - 0
total.sim - 0
p-alpha/(alpha+beta)
M
-logalpha-1)^(alpha-1))*((beta-1)^(beta-1)))/((beta+alpha
-2)^(alpha+beta-2)))
while(counter n) {
total.sim - total.sim+1
proposal - runif(1)
if(proposal = 0 proposal = 1) {
U - runif(1)
ratio- (p^(alpha-1))*((1-p)^(beta-1))
if(U =ratio/exp(M)) {
counter - counter+1
out[counter] - proposal
}
}
}
obs.acc.rate - n/total.sim
return(out,obs.acc.rate,M)
}
set.seed(220)
temp - rvonmises.norm(1,439.544,545.501)
print(temp$obs.acc.rate)
Louisa
I think when someone told you to take the log of the calculation, they
meant for you to simplify the logarithmic calculation algebraically so that
you are not exponentiating large numbers. Try changing your calculation of
M to (I think this right)
M - (alpha-1)*log(alpha-1) + (beta-1)*log(beta-1) -
(alpha+beta-2)*log(alpha+beta-2)
Hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
__
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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] p-values from bootstrap - what am I not understanding?
Johan Jackson wrote: Dear stats experts: Me and my little brain must be missing something regarding bootstrapping. I understand how to get a 95%CI and how to hypothesis test using bootstrapping (e.g., reject or not the null). However, I'd also like to get a p-value from it, and to me this seems simple, but it seems no-one does what I would like to do to get a p-value, which suggests I'm not understanding something. Rather, it seems that when people want a p-value using resampling methods, they immediately jump to permutation testing (e.g., destroying dependencies so as to create a null distribution). SO - here's my thought on getting a p-value by bootstrapping. Could someone tell me what is wrong with my approach? Thanks: STEPS TO GETTING P-VALUES FROM BOOTSTRAPPING - PROBABLY WRONG: 1) sample B times with replacement, figure out theta* (your statistic of interest). B is large ( 1000) 2) get the distribution of theta* 3) the mean of theta* is generally near your observed theta. In the same way that we use non-centrality parameters in other situations, move the distribution of theta* such that the distribution is centered around the value corresponding to your null hypothesis (e.g., make the distribution have a mean theta = 0) 4) Two methods for finding 2-tailed p-values (assuming here that your observed theta is above the null value): Method 1: find the percent of recentered theta*'s that are above your observed theta. p-value = 2 * this percent Method 2: find the percent of recentered theta*'s that are above the absolute value of your observed value. This is your p-value. So this seems simple. But I can't find people discussing this. So I'm thinking I'm wrong. Could someone explain where I've gone wrong? There's nothing particularly wrong about this line of reasoning, or at least not (much) worse than the calculation of CI. After all, one definition of a CI at level 1-alpha is that it contains values of theta0 for which the hypothesis theta=theta0 is accepted at level alpha. (Not the only possible definition, though.) The crucial bit in both cases is the assumption of approximate translation invariance, which holds asymptotically, but maybe not well enough in small samples. There are some braintwisters connected with the bootstrap; e.g., if the bootstrap distribution is skewed to the right, should the CI be skewed to the right or to the left? The answer is that it cannot be decided based on the distribution of theta* alone since that depends only on the true theta, and we need to know what the distribution would have been had a different theta been the true one. The point is that these things get tricky, so most people head for the safe haven of permutation testing, where it is rather more easy to feel that you know what you are doing. For a rather different approach, you might want to look into the theory of empirical likelihood (book by Art Owen, or just Google it). -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907 __ R-help@r-project.org 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] First Derivative of Data Matrix
delta(index) is identically 1, so taking first differences is all that is needed. If the dtatframe's name is df then: df$dacflong_dx - c(NA, diff(acflong)) # the slash would not be a legal character in a variable name unless you jumped through some hoops that appear entirely without value If you want to get rid of the first line of df then df[-1] -- David Winsemius On Apr 12, 2009, at 11:55 AM, thaumaturgy wrote: I am really new to R and ran across a need to take a data matrix and calculate an approximation of the first derivative of the data. I am more than happy to do an Excel kind of calculation (deltaY/deltaX) for each pair of rows down the matrix, but I don't know how to get R to do that kind of calculation. I'd like to store it as a 3rd column in the matrix as well. My data looks like this: acflong 1 1.000 2 0.9875858 3 0.9871751 4 0.9867585 5 0.9863358 6 0.9859070 7 0.9854721 8 0.9850316 9 0.9817161 10 0.9812650 and I'd like to generate a table like this: acflong dacflong/dx 1 1.000 2 0.9875858-0.01241 #delta(acflong)/delta(index) 3 0.9871751-0.00041 4 0.9867585-0.00042 5 0.9863358-0.00042 6 0.9859070-0.00043 7 0.9854721-0.00043 8 0.9850316-0.00044 9 0.9817161-0.00033 10 0.9812650 -0.00045 Is there a way to do this in R and how do I eliminate the first line of the data? Thanks, -Chris -- View this message in context: http://www.nabble.com/First-Derivative-of-Data-Matrix-tp23012026p23012026.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org 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] taking the log then later exponentiate the result query
Your problem is that with the alpha beta you've specified
(((alpha-1)^(alpha-1))*((beta-1)^(beta-1)))/((beta+alpha-2)^(alpha+beta-2))
is
Inf/Inf
which is NaN.
On Sun, Apr 12, 2009 at 5:39 PM, Mary Winter statsstud...@hotmail.com wrote:
Hi,
I am trying to figure out the observed acceptance rate and M, using
generalised rejection sampling to generate a sample from the posterior
distribution for p.
I have been told my code doesn't work because I need to take the log of the
expression for M, evaluate it and then exponentiate the result. This is
because R is unable to calculate high powers such as 545.501.
As you can see in my code I have tried to taking the log of M and then the
exponential of the result, but I clearly must be doing something wrong.
I keep getting the error message:
Error in if (U = ratio/exp(M)) { : missing value where TRUE/FALSE needed
Any ideas how I go about correctly taking the log and then the exponential?
rvonmises.norm - function(n,alpha,beta) {
out - rep(0,n)
counter - 0
total.sim - 0
p-alpha/(alpha+beta)
M
-logalpha-1)^(alpha-1))*((beta-1)^(beta-1)))/((beta+alpha-2)^(alpha+beta-2)))
while(counter n) {
total.sim - total.sim+1
proposal - runif(1)
if(proposal = 0 proposal = 1) {
U - runif(1)
ratio- (p^(alpha-1))*((1-p)^(beta-1))
if(U =ratio/exp(M)) {
counter - counter+1
out[counter] - proposal
}
}
}
obs.acc.rate - n/total.sim
return(out,obs.acc.rate,M)
}
set.seed(220)
temp - rvonmises.norm(1,439.544,545.501)
print(temp$obs.acc.rate)
Louisa
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_
[[alternative HTML version deleted]]
__
R-help@r-project.org 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.
--
Mike Lawrence
Graduate Student
Department of Psychology
Dalhousie University
Looking to arrange a meeting? Check my public calendar:
http://tinyurl.com/mikes-public-calendar
~ Certainty is folly... I think. ~
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Re: [R] p-values from bootstrap - what am I not understanding?
Hi Johan, Interesting question. I'm (trying) to write a lecture on this as we speak. I'm no expert, but here are my two cents. I think that your method works fine WHEN the sampling distribution doesn't change its variance or shape depending on where it's centered. Of course, for normally, t-, or chi-square distributed statistics, this is the case, which is why it's fine to do this using traditional statistical methods. However, there are situations where this might not be the case (e.g., there may be a mean-variance relationship), and since we would like a general method of getting valid p-values that doesn't depend on strong assumptions, this probably isn't the way to go. Permutation would seem to work better because you are simulating the null process. However, figuring out how to permute the data in a way that creates the null you want while retaining all the dependencies, missingness patterns, etc in your data can be difficult/impossible. Hope that helps... Matt On Sun, Apr 12, 2009 at 4:38 PM, Peter Dalgaard p.dalga...@biostat.ku.dk wrote: Johan Jackson wrote: Dear stats experts: Me and my little brain must be missing something regarding bootstrapping. I understand how to get a 95%CI and how to hypothesis test using bootstrapping (e.g., reject or not the null). However, I'd also like to get a p-value from it, and to me this seems simple, but it seems no-one does what I would like to do to get a p-value, which suggests I'm not understanding something. Rather, it seems that when people want a p-value using resampling methods, they immediately jump to permutation testing (e.g., destroying dependencies so as to create a null distribution). SO - here's my thought on getting a p-value by bootstrapping. Could someone tell me what is wrong with my approach? Thanks: STEPS TO GETTING P-VALUES FROM BOOTSTRAPPING - PROBABLY WRONG: 1) sample B times with replacement, figure out theta* (your statistic of interest). B is large ( 1000) 2) get the distribution of theta* 3) the mean of theta* is generally near your observed theta. In the same way that we use non-centrality parameters in other situations, move the distribution of theta* such that the distribution is centered around the value corresponding to your null hypothesis (e.g., make the distribution have a mean theta = 0) 4) Two methods for finding 2-tailed p-values (assuming here that your observed theta is above the null value): Method 1: find the percent of recentered theta*'s that are above your observed theta. p-value = 2 * this percent Method 2: find the percent of recentered theta*'s that are above the absolute value of your observed value. This is your p-value. So this seems simple. But I can't find people discussing this. So I'm thinking I'm wrong. Could someone explain where I've gone wrong? There's nothing particularly wrong about this line of reasoning, or at least not (much) worse than the calculation of CI. After all, one definition of a CI at level 1-alpha is that it contains values of theta0 for which the hypothesis theta=theta0 is accepted at level alpha. (Not the only possible definition, though.) The crucial bit in both cases is the assumption of approximate translation invariance, which holds asymptotically, but maybe not well enough in small samples. There are some braintwisters connected with the bootstrap; e.g., if the bootstrap distribution is skewed to the right, should the CI be skewed to the right or to the left? The answer is that it cannot be decided based on the distribution of theta* alone since that depends only on the true theta, and we need to know what the distribution would have been had a different theta been the true one. The point is that these things get tricky, so most people head for the safe haven of permutation testing, where it is rather more easy to feel that you know what you are doing. For a rather different approach, you might want to look into the theory of empirical likelihood (book by Art Owen, or just Google it). -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907 __ R-help@r-project.org 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. -- Matthew C Keller Asst. Professor of Psychology University of Colorado at Boulder www.matthewckeller.com __ R-help@r-project.org 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
Re: [R] First Derivative of Data Matrix
However, estimating derivatives from differencing data amplifies minor errors. Less noisy estimates can be obtained by first smoothing and then differentiating the smooth. The fda package provides substantial facilities for this. Hope this helps. Spencer Graves David Winsemius wrote: delta(index) is identically 1, so taking first differences is all that is needed. If the dtatframe's name is df then: df$dacflong_dx - c(NA, diff(acflong)) # the slash would not be a legal character in a variable name unless you jumped through some hoops that appear entirely without value If you want to get rid of the first line of df then df[-1] __ R-help@r-project.org 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] goodness of fit between two samples of size N (discrete variable)
On Apr 12, 2009, at 3:09 PM, jose romero wrote: Hello list: I generate by simulation (using different procedures) two sample vectors of size N, each corresponding to a discrete variable and I want to text if these samples can be considered as having the same probability distribution (which is unknown). What is the best test for that? I've read that Kolmogorov-Smirnov and Anderson-Darling tests are restricted to continuous data (http://cran.r-project.org/doc/contrib/Ricci-distributions-en.pdf ), while chi-square can handle discrete data, but how do i test (in R) equivalence of ditribution in 2 samples using it? Are there better tests than those i mentioned? The question of whether two discrete samples are independent, conditional on their joint marginals is generally handled with a chi- square test. The theoretical distribution is only approximately chi- square, but is seems close enough that most people will accept it. This is not a test of equivalence. Ricci deals with the cases where one sample is fitted to a theoretical distribution. You do not seem to have that situation. ?chisq.test I find myself wondering to what purpose you are seeking these answers. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org 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] Quantative procedure assessing if data is normal
Hi, As part of an R code assingment I have been asked to find a quantitative procedure for assessing whether or not the data are normal? I have previously used the graphical procedure using the qqnorm command. Any help/tips would be greatly appreciated as to how I should start going about this! Henry _ [[alternative HTML version deleted]] __ R-help@r-project.org 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] Physical Units in Calculations
Back in 2005 I had been doing most of my work in Mathcad for over 10 years. For a number of reasons I decided to switch over to R. After much effort and help from you folks I am finally thinking in R rather than thinking in Mathcad and trying to translating to R. Anyway, the only task I still use Mathcad for is calculations that involve physical quantities and units. For example, in Mathcad I can add 1 kilometer to 1 mile and get the right answer in the units of length I choose. Likewise, if I try to add 1 kilometer to 1 kilogram I get properly chastised. Is there a way in R to assign quantities and units to numbers and have R keep track of them like Mathcad does? Tom -- View this message in context: http://www.nabble.com/Physical-Units-in-Calculations-tp23016092p23016092.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] First Derivative of Data Matrix
David, Thank you! -Chris -- View this message in context: http://www.nabble.com/First-Derivative-of-Data-Matrix-tp23012026p23015941.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] p-values from bootstrap - what am I not understanding?
There is really nothing wrong with this approach, which differs primarily from the permutation test in that sampling is with replacement instead of without replacement (multinomial vs. multiple hypergeometric). One of the issues that permutation tests don't have is bias in the statistic. In order for bootstrap p-values to be reasonably accurate, you need a reasonable dataset size, so that sampling with replacement isn't a big effect, and so that enough patterns arise in resampling. It also helps if the data is continuous instead of categorical or binary. The same issues affect permutation tests, but untroubled by bias. The usual methods for p-values (e.g., see Fisher's test in Agresti's Categorical Analysis) work here. Typically there is some ambiguity on how to treat the values equal to the observed statistic. If you include it, the p-value is conservative for rejection. If you don't, it's liberal for rejection. If you include 1/2 weight, it averages correctly in the long run. Ditto for 2-tailed p-values vs. single tails. Several different methods (some of which you listed) are used. As a general rule, if you have data from which you wish a p-value, a permutation (i.e., without replacement) test is used, but for confidence intervals, bootstrapping (i.e., with replacement) is used. For reasonably large datasets, both methods will agree closely. But permutation tests are typically used for smaller size datasets. (Think binomial vs. hypgeometric distributions for p-values, and when they agree.) At 05:47 PM 4/12/2009, Johan Jackson wrote: Dear stats experts: Me and my little brain must be missing something regarding bootstrapping. I understand how to get a 95%CI and how to hypothesis test using bootstrapping (e.g., reject or not the null). However, I'd also like to get a p-value from it, and to me this seems simple, but it seems no-one does what I would like to do to get a p-value, which suggests I'm not understanding something. Rather, it seems that when people want a p-value using resampling methods, they immediately jump to permutation testing (e.g., destroying dependencies so as to create a null distribution). SO - here's my thought on getting a p-value by bootstrapping. Could someone tell me what is wrong with my approach? Thanks: STEPS TO GETTING P-VALUES FROM BOOTSTRAPPING - PROBABLY WRONG: 1) sample B times with replacement, figure out theta* (your statistic of interest). B is large ( 1000) 2) get the distribution of theta* 3) the mean of theta* is generally near your observed theta. In the same way that we use non-centrality parameters in other situations, move the distribution of theta* such that the distribution is centered around the value corresponding to your null hypothesis (e.g., make the distribution have a mean theta = 0) 4) Two methods for finding 2-tailed p-values (assuming here that your observed theta is above the null value): Method 1: find the percent of recentered theta*'s that are above your observed theta. p-value = 2 * this percent Method 2: find the percent of recentered theta*'s that are above the absolute value of your observed value. This is your p-value. So this seems simple. But I can't find people discussing this. So I'm thinking I'm wrong. Could someone explain where I've gone wrong? J Jackson [[alternative HTML version deleted]] __ R-help@r-project.org 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. Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: r...@lcfltd.com Least Cost Formulations, Ltd.URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239Fax: 757-467-2947 Vere scire est per causas scire __ R-help@r-project.org 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] Clustered data with Design package--bootcov() vs. robcov()
Hi, I am trying to figure out exactly what the bootcov() function in the Design package is doing within the context of clustered data. From reading the documentation/source code it appears that using bootcov() with the cluster argument constructs standard errors by resampling whole clusters of observations with replacement rather than resampling individual observations. Is that right, and is there any more detailed documentation on the math behind this? Also, what is the difference between these two functions: bootcov(my.model, cluster.id) robcov(my.model, cluster.id) Thank you. -- View this message in context: http://www.nabble.com/Clustered-data-with-Design-package--bootcov%28%29-vs.-robcov%28%29-tp23016400p23016400.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] Quantative procedure assessing if data is normal
www.rseek.org normality test On Sun, Apr 12, 2009 at 8:45 PM, Henry Cooper henry.1...@hotmail.co.uk wrote: Hi, As part of an R code assingment I have been asked to find a quantitative procedure for assessing whether or not the data are normal? I have previously used the graphical procedure using the qqnorm command. Any help/tips would be greatly appreciated as to how I should start going about this! Henry _ [[alternative HTML version deleted]] __ R-help@r-project.org 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 Lawrence Graduate Student Department of Psychology Dalhousie University Looking to arrange a meeting? Check my public calendar: http://tinyurl.com/mikes-public-calendar ~ Certainty is folly... I think. ~ __ R-help@r-project.org 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] Cross-platforms solution to export R graphs
Philippe Grosjean wrote: ..I would be happy to receive your comments and suggestions to improve this document. All the best, PhG LaTeX is my personal tool of choice and the vector format I use most often is http://sourceforge.net/projects/pgf/ PGF (Portable Graphics Format), implemented via a LaTeX package written by Till Tantu. There exists a very nice converter called http://sourceforge.net/projects/eps2pgf/ eps2pgf which is written in java and does an excellent job of translating R eps output. The primary advantage of PGF is that figure text gets typeset by the LaTeX engine instead of by R which unifies font choices and gives the final document a very consistent, professional look. LaTeX commands, such as mathematical typesetting, can also be embedded in the figure. Along with a friend of mine, I have been working on a R package that extends Sweave to include pgf graphics output. Currently http://www.rforge.net/pgfSweave pgfSweave uses eps2pgf to perform the conversions but a native R graphics device is planned to help speed up the process. The package is currently very much a beta and has been developed and tested on Mac OS X and runs quite well. Limited testing has been conducted on Linux and Windows and we have produced documents on those systems. Heavy development is expected to take place this summer. PGF is a human-readable format and can be be easily annotated by adding additional commands to the resulting file. However, editing the original content is possible but difficult due to the lack of structure in the eps2pgf output. The LaTeX environment can even be switched from pgfpicture to tikzpicture which allows the use of TiKZ- a high level graphics language built on top of PGF. TiKZ/PGF is easy to learn and the manual is one of the best pieces of software documentation I have seen. Since I came across PGF a couple of years ago, Adobe Illustrator has languished unused on my hard drive except for the occasional application of Live Trace. An excellent showcase of PGF/TiKZ examples along with additional tools is hosted at http://www.texample.net Texample . The end result of the PGF/TiKZ build process is a PDF which makes it very portable. All the best! -Charlie - Charlie Sharpsteen Undergraduate Environmental Resources Engineering Humboldt State University -- View this message in context: http://www.nabble.com/Cross-platforms-solution-to-export-R-graphs-tp22970668p23016682.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] Physical Units in Calculations
At 08:00 PM 4/12/2009, Tom La Bone wrote: Back in 2005 I had been doing most of my work in Mathcad for over 10 years. For a number of reasons I decided to switch over to R. After much effort and help from you folks I am finally thinking in R rather than thinking in Mathcad and trying to translating to R. Anyway, the only task I still use Mathcad for is calculations that involve physical quantities and units. For example, in Mathcad I can add 1 kilometer to 1 mile and get the right answer in the units of length I choose. Likewise, if I try to add 1 kilometer to 1 kilogram I get properly chastised. Is there a way in R to assign quantities and units to numbers and have R keep track of them like Mathcad does? Yes, but it's a lot of work: Create objects with units as an attribute. Perhaps someone else can tell you if such a set of definitions already exists. Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: r...@lcfltd.com Least Cost Formulations, Ltd.URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239Fax: 757-467-2947 Vere scire est per causas scire __ R-help@r-project.org 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] value of strptime in R 1.8.0
Dear R friends, I have a data frame, I need to get a time interval between the two columns. The times are recorded in 24 hour clock. My data frame is called version.one. my commands are: t.s.one-paste(version.one[,9]) t.s.two-paste(version.one[,61]) x-strptime(t.s.one,format=%H:%M) x y-strptime(t.s.two ,format=%H:%M) y z-difftime(y,x, units = mins) z But now in my z object i have negative numbers. Would you please let me know why? And how can I get rid of this problem? Thanks Mitra [[alternative HTML version deleted]] __ R-help@r-project.org 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: modelling a nested student-school-district model in lmer
Dear all, I have a dataset with students nested in schools and also schools belong to each district. The data was explicitly nested as previous examples. In my case, I don't care the variance between schools or district,and I just want to assess the effect of different teaching methods,traditionally,the model can be specified in lmer like : lmer(score~method+(1|district/school),data) notes: the method was a factor(m1...m10) in my study ,I want to know the variance between methods,and I also use some covariates at method level to explain the variance between methods. I construct the unconditional model and conditional models like these: unconditional model :lmer(score~1+(1|method)+(1/district/school),data) conditional model :lmer(score~1+CK+(1|method)+(1/district/school),data) the CK indicates the evaluation score for each method. What I want to confirm is whether the specification of all the models was reasonable and correct in lmer. Any help will be appreciated. yours, Lei Chen [[alternative HTML version deleted]] __ R-help@r-project.org 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] Physical Units in Calculations
Is there anything available off the shelf in R for this? I don't think so. It is, however, an interesting problem and there are the tools there to handle it. Basically you need to create a class for each kind of measure you want to handle (length, area, volume, weight, and so on) and then overload the arithmetic operators so that they can handle arguments of the appropriate class. This may be a case where S4 classes do have a distinct advantage, as they can more easily dispatch methods on combinations of classes rather than the class of a single argument, as in the case of S3. It looks like an interesting problem, in fact, but with the potential to get well out of hand if you set your sights too general. Limited versions would be simple enough, though. Something like it exists for times in the POSIXt and Date classes, of course, but with many limitations. For example you cannot divide one time difference by another to get a pure number, but you can divide one by a pure number to get another time difference. This could be remedied, of course. I was about the say that this is another case where the USA is mainly to blame, *again*, because of the dogged clinging to an outdated system of weights and measures, (not to mention the peverse practice putting the month *first* in their date format), but it's not entirely true. The UK uses metres for most lengths but miles for road distances - the worst of all worlds. They even measure fuel performance in litres per 100 *miles*, if you can believe it. Bill Venables http://www.cmis.csiro.au/bill.venables/ -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Robert A LaBudde Sent: Monday, 13 April 2009 12:20 PM To: Tom La Bone Cc: r-help@r-project.org Subject: Re: [R] Physical Units in Calculations At 08:00 PM 4/12/2009, Tom La Bone wrote: Back in 2005 I had been doing most of my work in Mathcad for over 10 years. For a number of reasons I decided to switch over to R. After much effort and help from you folks I am finally thinking in R rather than thinking in Mathcad and trying to translating to R. Anyway, the only task I still use Mathcad for is calculations that involve physical quantities and units. For example, in Mathcad I can add 1 kilometer to 1 mile and get the right answer in the units of length I choose. Likewise, if I try to add 1 kilometer to 1 kilogram I get properly chastised. Is there a way in R to assign quantities and units to numbers and have R keep track of them like Mathcad does? Yes, but it's a lot of work: Create objects with units as an attribute. Perhaps someone else can tell you if such a set of definitions already exists. Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: r...@lcfltd.com Least Cost Formulations, Ltd.URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239Fax: 757-467-2947 Vere scire est per causas scire __ R-help@r-project.org 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@r-project.org 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] Running random forest using different training and testing schemes
Thanks a lot for your help.. But, using this function, how can identify the size of the training set? and how will I identify my data? There is not any example and I am a bit confused. Many thanks, Chrysanthi 2009/4/12 Max Kuhn mxk...@gmail.com There is also the train function in the caret package. The trainControl function can be used to try different resampling schemes. There is also a package vignette with details. Max On Apr 12, 2009, at 12:26 PM, Chrysanthi A. chrys...@gmail.com wrote: Hi, I would like to run random Forest classification algorithm and check the accuracy of the prediction according to different training and testing schemes. For example, extracting 70% of the samples for training and the rest for testing, or using 10-fold cross validation scheme. How can I do that? Is there a function? Thanks a lot, Chrysanthi. [[alternative HTML version deleted]] __ R-help@r-project.org 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. [[alternative HTML version deleted]] __ R-help@r-project.org 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] Re : Running random forest using different training and testing schemes
Hi Pierre, Thanks a lot for your help.. So, using that script, I just separate my data in two parts, right? For using as training set the 70 % of the data and the rest as test, should I multiply the n with the 0.70 (for this case)? Many thanks, Chrysanthi 2009/4/12 Pierre Moffard pier.m...@yahoo.fr Hi Chysanthi, check out the randomForest package, with the function randomForest. It has a CV option. Sorry for not providing you with a lengthier response at the moment but I'm rather busy on a project. Let me know if you need more help. Also, to split your data into two parts- the training and the test set you can do (n the number of data points): n-length(data[,1]) indices-sample(rep(c(TRUE,FALSE),each=n/2),round(n/2),replace=TRUE) training_indices-(1:n)[indices] test_indices-(1:n)[!indices] Then, data[train,] is the training set and data[test,] is the test set. Best, Pierre -- *De :* Chrysanthi A. chrys...@gmail.com *À :* r-h...@r-project..org *Envoyé le :* Dimanche, 12 Avril 2009, 17h26mn 59s *Objet :* [R] Running random forest using different training and testing schemes Hi, I would like to run random Forest classification algorithm and check the accuracy of the prediction according to different training and testing schemes. For example, extracting 70% of the samples for training and the rest for testing, or using 10-fold cross validation scheme. How can I do that? Is there a function? Thanks a lot, Chrysanthi. [[alternative HTML version deleted]] __ R-help@r-project.org 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. [[alternative HTML version deleted]] __ R-help@r-project.org 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] value of strptime in R 1.8.0
Mitra; n Apr 12, 2009, at 8:09 PM, Mitra Jazayeri wrote: Dear R friends, I have a data frame, I need to get a time interval between the two columns. The times are recorded in 24 hour clock. My data frame is called version.one. my commands are: t.s.one-paste(version.one[,9]) t.s.two-paste(version.one[,61]) x-strptime(t.s.one,format=%H:%M) x y-strptime(t.s.two ,format=%H:%M) y z-difftime(y,x, units = mins) z But now in my z object i have negative numbers. Would you please let me know why? Given that neither version.one[,9] nor version[,61] is available to us, how can readers of this be expected to answer other than with the trivial hypothesis that t.s.one is after t.s.two? And how can I get rid of this problem? Change the order of the calculation? (You cannot use abs, since that function is not defined for difftime objects.) -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org 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] Using trace
I would like to trace functions, displaying their arguments and return value, but I haven't been able to figure out how to do this with the 'trace' function. After some thrashing, I got as far as this: fact - function(x) if(x1) 1 else x*fact(x-1) tracefnc - function() dput(as.list(parent.frame()), # parent.frame() holds arg list control=NULL) trace(fact,tracer=tracefnc,print=FALSE) but I couldn't figure out how to access the return value of the function in the 'exit' parameter. The above also doesn't work for ... arguments. (More subtly, it forces the evaluation of promises even if they are otherwise unused -- but that is, I suppose, a weird and obscure case.) Surely someone has solved this already? What I'm looking for is something very simple, along the lines of old-fashioned Lisp trace: defun fact (i) (if ( i 1) 1 (* i (fact (+ i -1) FACT (trace fact) (FACT) (fact 3) 1 (FACT 3) 2 (FACT 2) 3 (FACT 1) 4 (FACT 0) 4 (FACT 1) 3 (FACT 1) 2 (FACT 2) 1 (FACT 6) 6 Can someone help? Thanks, -s __ R-help@r-project.org 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.
