[R] Bootstrap
Dear Any, Can someone please inform me, if they have a code to estimate the varaince using bootstrap resampling method under a two stage cluster design. Thanks for all your help and time. Murthy.M.N., PhD, Student, University of Canterbury, New Zealand. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] png problem
Hello, Thanks for the fast reply. Paul Murrell wrote: Hi Clément Calenge wrote: Dear R-users, I have a small problem with the function png(), when used with the argument colortype=pseudo.cube. png(toto.png, colortype=pseudo.cube) image(matrix(rnorm(1), 100, 100)) dev.off() R is blocked at the last command (R does not print any prompt after the last command). Nothing is written in the file (Gimp indicates that the file is corrupted). Did you wait long enough? This example took a little while to complete for me (may need someone more familiar with the code to tell us why it is so slow). You're right, it took 45 minutes for me. However, since I need to use this code to build a Sweave vignette, I cannot use it too often (I have about thirty files to create, this would take about 20 hours to build the vignette !). I also need the colortype argument (some graphics cards do not allow the compilation of the vignette without). Does anyone knows how to speed up the process ? Clément. Paul However, png(toto.png) image(matrix(rnorm(1), 100, 100)) dev.off() works fine. I tried: options(X11colortype = pseudo.cube) png(toto.png) image(matrix(rnorm(1), 100, 100)) dev.off() But, here again, R is blocked. I tried to replace dev.off() by graphics.off(), but this does not resolve the problem. The problem does not occurs when the function X11() is used instead of the function png(). I searched through the mail archive, the FAQ, on google, but I did not found any solution to this problem. On the help page on the function png(), it is indicated that The colour handling will be that of the 'X11' device in use. I never used these functions before, but maybe png() is not suitable with colortype=pseudo.cube ? Can you tell me where I have missed something ? Thanks in Advance, Clément Calenge. version _ platform sparc-sun-solaris2.9 arch sparc os solaris2.9 system sparc, solaris2.9 status major1 minor9.1 year 2004 month06 day 21 language R [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 [EMAIL PROTECTED] http://www.stat.auckland.ac.nz/~paul/ == UMR CNRS 5558 - Equipe Ecologie Statistique Laboratoire de Biométrie et Biologie Evolutive Université Claude Bernard Lyon 1 43, Boulevard du 11 novembre 1918 69622 Villeurbanne Cedex FRANCE tel. (+33) 04.72.43.27.57 fax. (+33) 04.72.43.13.88 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] t test problem?
Hello, I got two sets of data x=(124738, 128233, 85901, 33806, ...) y=(25292, 21877, 45498, 63973, ) When I did a t test, I got two tail p-value = 0.117, which is not significantly different. If I changed x, y to log scale, and re-do the t test, I got two tail p-value = 0.042, which is significantly different. Now I got confused which one is correct. Any help would be very appreciated. Thanks, Liu __ [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] t test problem?
Hi Liu, before applying a t-test (or any test) you should first check if the assumptions of the test are supported by your data, i.e., in a t-test x and y must be normally distributed. 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: +32/16/396887 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat/ http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: kan Liu [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Wednesday, September 22, 2004 9:52 AM Subject: [R] t test problem? Hello, I got two sets of data x=(124738, 128233, 85901, 33806, ...) y=(25292, 21877, 45498, 63973, ) When I did a t test, I got two tail p-value = 0.117, which is not significantly different. If I changed x, y to log scale, and re-do the t test, I got two tail p-value = 0.042, which is significantly different. Now I got confused which one is correct. Any help would be very appreciated. Thanks, Liu __ [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] t test problem?
Hi, maybe your data are distributed according a log-normal distribution, so logs are normally distributed. But remerber the significancy of t test can applied only on log transformated data and not on original data. See basic hypothesis for t testing, in alternative use non-parametric methods to compare results. Best Vito You wrote: Hello, I got two sets of data x=(124738, 128233, 85901, 33806, ...) y=(25292, 21877, 45498, 63973, ) When I did a t test, I got two tail p-value = 0.117, which is not significantly different. If I changed x, y to log scale, and re-do the t test, I got two tail p-value = 0.042, which is significantly different. Now I got confused which one is correct. Any help would be very appreciated. Thanks, Liu = Diventare costruttori di soluzioni Visitate il portale http://www.modugno.it/ e in particolare la sezione su Palese http://www.modugno.it/archivio/cat_palese.shtml ___ http://it.seriea.fantasysports.yahoo.com/ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] t test problem?
Hi Dimitris, you are describing a more stringent requirement than the t-test actually requires. It's the sampling distribution of the mean that should be normal, and this condition is addressed by the Central Limit Theorem. Whether or not the CLT can be invoked depends on numerous factors, including the distribution of the sample, and the size of the sample, neither of which we have any information about. Liu, the problem you describe is associated with the application of the test rather than the test itself. The difference between log- and natural- scaled data can often profitably be thought about by asking whether you would naturally assume that the variation is additive (natural scale) or multiplicative (log scale). Given the information that you've presented there's no way we can tell which version of the test is more reliable. I hope that this helps. Andrew On Wed, Sep 22, 2004 at 10:00:16AM +0200, Dimitris Rizopoulos wrote: Hi Liu, before applying a t-test (or any test) you should first check if the assumptions of the test are supported by your data, i.e., in a t-test x and y must be normally distributed. 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: +32/16/396887 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat/ http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: kan Liu [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Wednesday, September 22, 2004 9:52 AM Subject: [R] t test problem? Hello, I got two sets of data x=(124738, 128233, 85901, 33806, ...) y=(25292, 21877, 45498, 63973, ) When I did a t test, I got two tail p-value = 0.117, which is not significantly different. If I changed x, y to log scale, and re-do the t test, I got two tail p-value = 0.042, which is significantly different. Now I got confused which one is correct. Any help would be very appreciated. Thanks, Liu __ [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Andrew Robinson Ph: 208 885 7115 Department of Forest Resources Fa: 208 885 6226 University of Idaho E : [EMAIL PROTECTED] PO Box 441133W : http://www.uidaho.edu/~andrewr Moscow ID 83843 Or: http://www.biometrics.uidaho.edu No statement above necessarily represents my employer's opinion. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R]
Dear Any, Is there a fonction in R to change a string to uppercase ? Thanks for all your help __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] ordered probit and cauchit
On Wednesday, Sep 22, 2004, at 03:42 Europe/London, roger koenker wrote: What is the current state of the R-art for ordered probit models, and more esoterically is there any available R strategy for ordered cauchit models, i.e. ordered multinomial alternatives with a cauchy link function. MCMC is an option, obviously, but for a univariate latent variable model this seems to be overkill... standard mle methods should be preferable. (??) A quick look at polr (in the MASS package) suggests to me that it wouldn't be all that hard to extend it to link functions other than logistic. Is MLE known to be well behaved in these models if the latent variable is Cauchy? David Googling reveals that spss provides such functions... just to wave a red flag. url:www.econ.uiuc.edu/~rogerRoger Koenker email [EMAIL PROTECTED] Department of Economics vox:217-333-4558University of Illinois fax:217-244-6678Champaign, IL 61820 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Ever see a stata import problem like this?
I've had a similar problem once. What may have caused the problem then was a variate for which value lables had been defined for the highest and lowest values. What complicates things is that the file had been originally converted from SPSS to Stata. A workaround was to set convert.factor=FALSE and that seems to work here too (using R 1.91 and the latest update for foreign): m2-read.dta(morgen.dta,convert.factors=FALSE) summary(m2) CASEID yearidhrs1 Min. : 19721 Min. :1972 Min. : 1 Min. :0.00 1st Qu.: 1983475 1st Qu.:1978 1st Qu.: 445 1st Qu.: 37.00 Median : 1996808 Median :1987 Median : 905 Median : 40.00 Mean : 9963040 Mean :1986 Mean : 990 Mean : 41.05 3rd Qu.:19872187 3rd Qu.:1994 3rd Qu.:1358 3rd Qu.: 48.00 Max. :20002817 Max. :2000 Max. :3247 Max. : 89.00 NA's :17654.00 hrs2 prestigeagewed age Min. :0.00 Min. : 12.00 Min. : 12.00 Min. : 18.00 1st Qu.: 38.00 1st Qu.: 30.00 1st Qu.: 19.00 1st Qu.: 30.00 Median : 40.00 Median : 39.00 Median : 21.00 Median : 42.00 Mean : 39.79 Mean : 39.36 Mean : 22.10 Mean : 45.15 3rd Qu.: 45.00 3rd Qu.: 48.00 3rd Qu.: 24.00 3rd Qu.: 58.00 Max. : 89.00 Max. : 82.00 Max. : 73.00 Max. : 89.00 NA's :40159.00 NA's :1.00 NA's :15551.00 NA's :143.00 educpaeduc maeducspeduc Min. : 0.00 Min. :0.00 Min. : 0.00 Min. : 0.00 1st Qu.: 11.00 1st Qu.:8.00 1st Qu.: 8.00 1st Qu.: 12.00 Median : 12.00 Median : 11.00 Median : 12.00 Median : 12.00 Mean : 12.48 Mean : 10.21 Mean : 10.41 Mean : 12.53 3rd Qu.: 14.00 3rd Qu.: 12.00 3rd Qu.: 12.00 3rd Qu.: 14.00 Max. : 20.00 Max. : 20.00 Max. : 20.00 Max. : 20.00 NA's :127.00 NA's :11586.00 NA's :6782.00 NA's :18153.00 income Min. : 1.000 1st Qu.: 9.000 Median : 11.000 Mean : 9.756 3rd Qu.: 12.000 Max. : 13.000 NA's :3453.000 --- Paul Johnson [EMAIL PROTECTED] wrote: Greetings Everybody: I generated a 1.2MB dta file based on the general social survey with Stata8 for linux. The file can be re-opened with Stata, but when I bring it into R, it says all the values are missing for most of the variables. This dataset is called morgen.dta and I dropped a copy online in case you are interested http://www.ku.edu/~pauljohn/R/morgen.dta [snip] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R]
?toupper p.s: By default, generally everything on this list *is* regarding R; hence it would be nice to see a more imaginative subject line. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED] Sent: Wednesday, September 22, 2004 5:43 PM To: [EMAIL PROTECTED] Subject: [R] Dear Any, Is there a fonction in R to change a string to uppercase ? Thanks for all your help __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R]
The following function will do it. But be warned it will only work if all
the input strings are lower case.
Make.To.Upper.Case-function(my.string){
paste(LETTERS[match(strsplit(my.string,)[[1]],letters)],collapse=)
}
Make.To.Upper.Case(jhjhaskjdakdsj)
[1] JHJHASKJDAKDSJ
-Original Message-
From: [EMAIL PROTECTED]
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Sent: 22 September 2004 09:43
To: [EMAIL PROTECTED]
Subject: [R]
Dear Any,
Is there a fonction in R to change a string to uppercase ?
Thanks for all your help
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Re: [R] convert string to uppercase; was: nothing
[EMAIL PROTECTED] wrote: Dear Any, Is there a fonction in R to change a string to uppercase ? Thanks for all your help __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Please read the posting guide and learn a) to use a sensible subject line b) to use R's help facilities You are looking for ?toupper Uwe Ligges __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R]
help.search(upper) chartr(base)Character Translation and Casefolding ?chartr refers to the function toupper() Best wishes, Arne On Wednesday 22 September 2004 10:43, [EMAIL PROTECTED] wrote: Dear Any, Is there a fonction in R to change a string to uppercase ? Thanks for all your help __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Arne Henningsen Department of Agricultural Economics University of Kiel Olshausenstr. 40 D-24098 Kiel (Germany) Tel: +49-431-880 4445 Fax: +49-431-880 1397 [EMAIL PROTECTED] http://www.uni-kiel.de/agrarpol/ahenningsen/ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] t test problem?
Hi, Many thanks for your helpful comments and suggestions. The attached are the data in both log10 scale and original scale. It would be very grateful if you could suggest which version of test should be used. By the way, how to check whether the variation is additive (natural scale) or multiplicative (log scale) in R? How to check whether the distribution of the data is normal? PS, Can I confirm that do your suggestions mean that in order to check whether there is a difference between x and y in terms of mean I need check the distribution of x and that of y in both natual and log scales and to see which present normal distribution? and then perform a t test using the data scale which presents normal distribution? If both scales present normal distribution, then the t tests with both scales should give the similar results? Thanks again. Liu Andrew Robinson [EMAIL PROTECTED] wrote: Hi Dimitris, you are describing a more stringent requirement than the t-test actually requires. It's the sampling distribution of the mean that should be normal, and this condition is addressed by the Central Limit Theorem. Whether or not the CLT can be invoked depends on numerous factors, including the distribution of the sample, and the size of the sample, neither of which we have any information about. Liu, the problem you describe is associated with the application of the test rather than the test itself. The difference between log- and natural- scaled data can often profitably be thought about by asking whether you would naturally assume that the variation is additive (natural scale) or multiplicative (log scale). Given the information that you've presented there's no way we can tell which version of the test is more reliable. I hope that this helps. Andrew On Wed, Sep 22, 2004 at 10:00:16AM +0200, Dimitris Rizopoulos wrote: Hi Liu, before applying a t-test (or any test) you should first check if the assumptions of the test are supported by your data, i.e., in a t-test x and y must be normally distributed. 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: +32/16/396887 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat/ http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: kan Liu To: Sent: Wednesday, September 22, 2004 9:52 AM Subject: [R] t test problem? Hello, I got two sets of data x=(124738, 128233, 85901, 33806, ...) y=(25292, 21877, 45498, 63973, ) When I did a t test, I got two tail p-value = 0.117, which is not significantly different. If I changed x, y to log scale, and re-do the t test, I got two tail p-value = 0.042, which is significantly different. Now I got confused which one is correct. Any help would be very appreciated. Thanks, Liu __ [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Andrew Robinson Ph: 208 885 7115 Department of Forest Resources Fa: 208 885 6226 University of Idaho E : [EMAIL PROTECTED] PO Box 441133 W : http://www.uidaho.edu/~andrewr Moscow ID 83843 Or: http://www.biometrics.uidaho.edu No statement above necessarily represents my employer's opinion. - x y 37154 32211 114815 124738 10 128233 10 96383 10 85901 371535 338065 10 151008 56234 48978 34674 62087 758578 542001 14125 25645 26915 31696 10 119950 72444 56105 63096 39084 10 131522 33113 68077 37154 30409 26915 31842 70795 24322 93325 74989 10 101859 43652 50119 120226 86497 10 159956 10 44668 10 52602 10 82794 57544 24774 30200 19055 10 56624 10 39719 53703 51286 70795 17258 66069 52000 87096 140605 58884 36141 63096 74645 44668 32359 10 84140 15136 26915 43652 35075 794328 901571 20417 16218 147911 115345 57544 87498 10 73621 14454 19953 10 59429 72444 37670 199526 210378 38905 41020 79433 111944 10 141254 10 92045 23442 22751 18197 20606 316228 345144 83176 154170 48978 33806 10 84723 10 158855 20893 13552 141254 127350 67608 24774 10965 9290 17378 17742 120226 105925 23442 16943 56234 53211 66069 36392 38019 49774 75858 84140 42658 50466 56234 49204 12303 12474 120226 110154 131826 208449 104713 11508 70795 106905
[R] Create bitmap graphic in C
Hello, How to create a R graphic bitmap with C or C++ using R api ? Thanks, samples, ... ? [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] t test problem?
Hi, Many thanks for your helpful comments and suggestions. You're welcome. The attached are the data in both log10 scale and original scale. It would be very grateful if you could suggest which version of test should be used. I feel that it would be inappropriate. It depends on the origin of the data. You, as the analyst, must make that decision. If you're analyzing the data for someone else, then you should make that decision with their input, or have them make it. By the way, how to check whether the variation is additive (natural scale) or multiplicative (log scale) in R? Unfortuantely you can't do that. It depends on the context of the data, which is not amenable to testing. How to check whether the distribution of the data is normal? Tests exist for this, but there are problems with their usage. Try searching the help or your preferred search engine for more information. PS, Can I confirm that do your suggestions mean that in order to check whether there is a difference between x and y in terms of mean I need check the distribution of x and that of y in both natual and log scales and to see which present normal distribution? No, I don't agree with that statement, and I don't think that it reflects my earlier message. and then perform a t test using the data scale which presents normal distribution? Again, I don't agree. I would advise you to apply the appropriate test (maybe a t test, maybe not) to the data on the scale that is suggested by the sampling scheme, the origin of the data, and the hypothesis that interests you. I hope that this helps. Andrew -- Andrew Robinson Ph: 208 885 7115 Department of Forest Resources Fa: 208 885 6226 University of Idaho E : [EMAIL PROTECTED] PO Box 441133W : http://www.uidaho.edu/~andrewr Moscow ID 83843 Or: http://www.biometrics.uidaho.edu No statement above necessarily represents my employer's opinion. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] t test problem?
Hi Liu, I'd suggest you to use non-parametric tests (see http://www.cas.lancs.ac.uk/glossary_v1.1/nonparam.html) such as: wilcox.test() in stats package pairwise.wilcox.test() in stats package and see the result tou got (significancy/non significancy) and compare it with t test result; Parametric Test Analogous Non-Parametric test Student T-test Wilcoxon Rank Sum Test Paired t-testWilcoxon Signed Rank Test or the Sign Test to test normality you can use: shapiro.test() in stats package if you decide to use log scale, you must use this for both samples. bye Vito You wrote: Hi, Many thanks for your helpful comments and suggestions. The attached are the data in both log10 scale and original scale. It would be very grateful if you could suggest which version of test should be used. By the way, how to check whether the variation is additive (natural scale) or multiplicative (log scale) in R? How to check whether the distribution of the data is normal? PS, Can I confirm that do your suggestions mean that in order to check whether there is a difference between x and y in terms of mean I need check the distribution of x and that of y in both natual and log scales and to see which present normal distribution? and then perform a t test using the data scale which presents normal distribution? If both scales present normal distribution, then the t tests with both scales should give the similar results? Thanks again. Liu = Diventare costruttori di soluzioni Visitate il portale http://www.modugno.it/ e in particolare la sezione su Palese http://www.modugno.it/archivio/cat_palese.shtml ___ http://it.seriea.fantasysports.yahoo.com/ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Bootstrap
did you look at library(boot) ?boot On 9/22/04 9:19 AM, nmi13 [EMAIL PROTECTED] wrote: Dear Any, Can someone please inform me, if they have a code to estimate the varaince using bootstrap resampling method under a two stage cluster design. Thanks for all your help and time. Murthy.M.N., PhD, Student, University of Canterbury, New Zealand. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] t test problem?
Hi Kan Lui, I've had a quick look at the data. The logged data seems reasonably nicely distributed (roughly symmetrical + equal variance). Indeed the y variable passed the (very strict) shapiro.test for normality. However the main problem is that I do not get the same results as you for the significance of the t.test. The only significant test I see is the paired t.test on the logged data. Is your data paired data? To see what I mean check out: http://www.texasoft.com/winkpair.html The non-parametric tests show no significance (paired data or not) (logged and natural data). Although in general they do tend to be less strict than parametric tests. Unless the data is paired then the means of these samples most certainly do not significantly differ from one another. Here are my workings: Temp.Dat-read.table(data_natural.txt,header=T) hist(log(Temp.Dat$x,10)) hist(log(Temp.Dat$y,10)) shapiro.test(log(Temp.Dat$x,10)) shapiro.test(log(Temp.Dat$y,10)) t.test(log(Temp.Dat$x,10), log(Temp.Dat$y,10)) Welch Two Sample t-test data: log(Temp.Dat$x, 10) and log(Temp.Dat$y, 10) t = 0.9126, df = 195.806, p-value = 0.3626 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0.0599837 0.1633168 sample estimates: mean of x mean of y 4.891313 4.839647 t.test(log(Temp.Dat$x,10), log(Temp.Dat$y,10),paired=T) Paired t-test data: log(Temp.Dat$x, 10) and log(Temp.Dat$y, 10) t = 2.3535, df = 98, p-value = 0.02060 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.008101002 0.095232132 sample estimates: mean of the differences 0.05166657 wilcox.test(log(Temp.Dat$x,10), log(Temp.Dat$y,10),paired=T) Wilcoxon signed rank test with continuity correction data: log(Temp.Dat$x, 10) and log(Temp.Dat$y, 10) V = 2972.5, p-value = 0.0828 alternative hypothesis: true mu is not equal to 0 wilcox.test(log(Temp.Dat$x,10), log(Temp.Dat$y,10),paired=F) Wilcoxon rank sum test with continuity correction data: log(Temp.Dat$x, 10) and log(Temp.Dat$y, 10) W = 5206, p-value = 0.4491 alternative hypothesis: true mu is not equal to 0 wilcox.test(Temp.Dat$x, Temp.Dat$y,paired=F) Wilcoxon rank sum test with continuity correction data: Temp.Dat$x and Temp.Dat$y W = 5206, p-value = 0.4491 alternative hypothesis: true mu is not equal to 0 wilcox.test(Temp.Dat$x, Temp.Dat$y,paired=T) Wilcoxon signed rank test with continuity correction data: Temp.Dat$x and Temp.Dat$y V = 2896.5, p-value = 0.1417 alternative hypothesis: true mu is not equal to 0 t.test(Temp.Dat$x, Temp.Dat$y,paired=T) Paired t-test data: Temp.Dat$x and Temp.Dat$y t = 1.6731, df = 98, p-value = 0.0975 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -2351.81 27623.53 sample estimates: mean of the differences 12635.86 t.test(Temp.Dat$x, Temp.Dat$y,paired=F) Welch Two Sample t-test data: Temp.Dat$x and Temp.Dat$y t = 0.6432, df = 191.177, p-value = 0.5209 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -26116.18 51387.89 sample estimates: mean of x mean of y 120544.9 107909.0 -Original Message- From: kan Liu [mailto:[EMAIL PROTECTED] Sent: 22 September 2004 10:22 To: Andrew Robinson; Dimitris Rizopoulos Cc: [EMAIL PROTECTED] Subject: Re: [R] t test problem? Hi, Many thanks for your helpful comments and suggestions. The attached are the data in both log10 scale and original scale. It would be very grateful if you could suggest which version of test should be used. By the way, how to check whether the variation is additive (natural scale) or multiplicative (log scale) in R? How to check whether the distribution of the data is normal? PS, Can I confirm that do your suggestions mean that in order to check whether there is a difference between x and y in terms of mean I need check the distribution of x and that of y in both natual and log scales and to see which present normal distribution? and then perform a t test using the data scale which presents normal distribution? If both scales present normal distribution, then the t tests with both scales should give the similar results? Thanks again. Liu Andrew Robinson [EMAIL PROTECTED] wrote: Hi Dimitris, you are describing a more stringent requirement than the t-test actually requires. It's the sampling distribution of the mean that should be normal, and this condition is addressed by the Central Limit Theorem. Whether or not the CLT can be invoked depends on numerous factors, including the distribution of the sample, and the size of the sample, neither of which we have any information about. Liu, the problem you describe is associated with the application of the test rather than the test itself. The difference between log- and natural- scaled data
RE: [R] is.constant
Christian Hoffmann
x - c(1, 2, NA)
is.constant(x)
[1] TRUE
For data such as c(1, 1, 1, NA), I should think the safest
answer should be
NA, because one really doesn't know whether that last
number is 1 or not.
Andy
My version is
is.constant - function(x) {
if (is.numeric(x) !any(is.na(x))) identical(min(x),
max(x)) else FALSE
}
rendering
is.constant(c(1,1,NA))
[1] FALSE
As I said, the `safest' thing is to return NA, as the NA could be 1, or it
could be something else. We just don't know. It's probably a cleaner style
to have is.constant() return NA in the case that the input contains contants
and some NAs, and TRUE or FALSE otherwise; e.g., is.constant(c(1,2,NA))
should clearly be FALSE. Then the output of is.constant() should be checked
for possible NA.
Just my $0.02...
Andy
is.constant(c(1,1,NaN))
[1] FALSE
is.constant(rep(c(sin(pi/2),1),10)) # TRUE
[1] TRUE
--
Dr.sc.math.Christian W. Hoffmann,
http://www.wsl.ch/staff/christian.hoffmann
Mathematics + Statistical Computing e-mail:
[EMAIL PROTECTED]
Swiss Federal Research Institute WSL Tel: ++41-44-73922-
-77 (office)
CH-8903 Birmensdorf, Switzerland -11(exchange), -15 (fax)
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[R] asypow.noncent. Thanks!!
Thank y'all for a) pointing out what I was doing wrong b) being so patient with what, in retrospect, was such an obvious blunder by me. I'm afraid I was confused by the accompanying pdf file, which is not entirely consistent with the help files. ___ Most doctors use http://www.Doctors.net.uk e-mail. Move to a free professional address with spam and virus protection. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Re: R-help Digest, Vol 19, Issue 22
I will be out of the office 9/22/04 - 9/27/04. For immediate help, please call the JPSM main number, 301-314-7911. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] function to change a string to uppercase
On Wed, Sep 22, 2004 at 10:43:18AM +0200 - a wonderful day - [EMAIL PROTECTED] wrote: Is there a fonction in R to change a string to uppercase ? toupper() hint: basic functionality is in package base ;) which is easy to investigate in html-help - packages in my installation second entry from top of list ... All the best, Stefan. -- .o. e-mail: [EMAIL PROTECTED], web: www.sdrees.org, +49 700 SDREESDE ..o fingerprint = 516C C4EF 712A B26F 15C9 C7B7 5651 6964 D508 1B56 ooo stefan drees - consulting and lecturing - problems to tasks __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] impenetrable warning
Dear R-help, Can anyone explain the meaning of the warning, Singular precision matrix in level -1, block 1 ? Or how to track down where it comes from? More precisely, using the nlme package, I'm issued with the warning itt2 - lme(lrna~rx.nrti+lbrna, random=~1|patid, cor=corExp(form=~days|patid,nugget=T), weights=varPower( form=~lbrna),data=rna3) Warning messages: 1: Singular precision matrix in level -1, block 1 2: Singular precision matrix in level -1, block 1 the output is: Linear mixed-effects model fit by REML Data: rna3 Log-restricted-likelihood: -4990.142 Fixed: lrna ~ rx.nrti + lbrna (Intercept) rx.nrtiZDV+3TC rx.nrtiZDV+ABC lbrna 2.6597552 0.8589514 0.4259504 0.2929222 Random effects: Formula: ~1 | patid (Intercept) Residual StdDev:1.251113 0.2105329 Correlation Structure: Exponential spatial correlation Formula: ~days | patid Parameter estimate(s): range nugget 204.6422150 0.3349336 Variance function: Structure: Power of variance covariate Formula: ~lbrna Parameter estimates: power 0.9733358 Number of Observations: 2390 Number of Groups: 124 Thanks, Simon Bond. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] capitalize etc string
Dear Any,
Is there a fonction in R to change a string to uppercase ?
Thanks for all your help
Use the following:
capply - function(str, ff) {
sapply(lapply(strsplit(str, NULL), ff), paste, collapse=)
}
cap - function(char) {
# change lower letters to upper, others leave unchanged
if (any(ind - letters==char)) LETTERS[ind]
else char
}
capitalize - function(str) { # vector of words
ff - function(x) paste(lapply(unlist(strsplit(x,
NULL)),cap),collapse=)
capply(str,ff)
}
lower - function(char) {
# change upper letters to lower, others leave unchanged
if (any(ind - LETTERS==char)) letters[ind]
else char
}
lowerize - function(str) {
ff - function(x) paste(lapply(unlist(strsplit(x,
NULL)),lower),collapse=)
capply(str,ff)
}
CapLeading - function(str) {
ff - function(x) {r - x; r[1]-cap(x[1]); r}
capply(str,ff)
}
#cap(f)
#cap(R)
#capitalize(c(TruE,faLSe))
#capitalize(c(faLSe,TruE))
#lower(f)
#lower(R)
#lowerize(TruE)
#lowerize(faLSe)
--
Dr.sc.math.Christian W. Hoffmann,
http://www.wsl.ch/staff/christian.hoffmann
Mathematics + Statistical Computing e-mail: [EMAIL PROTECTED]
Swiss Federal Research Institute WSL Tel: ++41-44-73922- -77 (office)
CH-8903 Birmensdorf, Switzerland -11(exchange), -15 (fax)
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Re: [R] t test problem?
On 22-Sep-04 kan Liu wrote: Hi, Many thanks for your helpful comments and suggestions. The attached are the data in both log10 scale and original scale. It would be very grateful if you could suggest which version of test should be used. By the way, how to check whether the variation is additive (natural scale) or multiplicative (log scale) in R? How to check whether the distribution of the data is normal? As for additive vs multiplicative, this can only be judged in terms of the process by which the values are created in the real world. As for normality vs non-normality, an appraisal can often be made simply by looking at a histogram of the data. In your case, the commands hist(x,breaks=1*(0:100)) hist(y,breaks=1*(0:100)) indicate that the distributions of x and y do not look at all normal, since they both have considerable positive skewness (i.e. long upper tails relative to the main mass of the distribution). This does strongly suggest that a logarithmic transformation would give data which are more nearly normally distributed, as indeed is confirmed by the commands hist(log(x)) hist(log(y)) though in both cases the histograms show some irregularity compared with what you would expect from a sample from a normal distribution: the commands hist(log(x),breaks=0.2*(40:80)) hist(log(y),breaks=0.2*(40:80)) show that log(x) has an excessive peak at around 11.7, while log(y) has holes at around 11.1 and 12.1. Nevertheless, this inspection of the data shows that the use of log(x) and log(y) will come much closer to fulfilling the conditions of validity of the t test than using the raw data x and y. However, it is not merely the *normality* of each which is needed: the conditions for the usual t test also require that the two populations sampled for log(x) and log(y) should have the same standard deviations. In your case, this also turns out to be nearly enough true: sd(log(x)) [1] 0.902579 sd(log(y)) [1] 0.9314807 PS, Can I confirm that do your suggestions mean that in order to check whether there is a difference between x and y in terms of mean I need check the distribution of x and that of y in both natual and log scales and to see which present normal distribution? See above for an approach to this: the answer to your question is, in effect, yes. It could of course have happened that neither the raw nor the log scale would be satisfactory, in which case you would need to consider other possibilities. And, if the SDs had turned out to be very different, you should not use the standard t test but a variant which is adpated to the situation (e.g. the Welch test). You can, of course, also perform formal tests for skewness, for normality, and for equality of variances. Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 [NB: New number!] Date: 22-Sep-04 Time: 12:07:07 -- XFMail -- __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] is.constant 2
x - c(1, 2, NA)
is.constant(x)
[1] TRUE
For data such as c(1, 1, 1, NA), I should think the safest answer
should be
NA, because one really doesn't know whether that last number is 1 or
not.
Andy
My version is
is.constant - function(x) {
if (is.numeric(x) !any(is.na(x))) identical(min(x), max(x)) else
FALSE
}
Since the issue of factors surfaced, I improved my function:
is.constant - function(x) {
if (is.factor(x)) (length(attributes(x)$levels)==1)
(!any(is.na(as.character(x
else (is.numeric(x) !any(is.na(x)) identical(min(x), max(x)))
}
is.constant(rep(c(sin(pi/2),1),10)) # TRUE
x - factor(c(1,1,NA))
is.constant(x)# FALSE because of NA
is.constant(x[1:2]) # TRUE
is.constant(c(1,1,NA))# FALSE because of NA
is.constant(c(1,1,2)) # FALSE
is.constant(c(1,1,1)) # TRUE
--
Dr.sc.math.Christian W. Hoffmann,
http://www.wsl.ch/staff/christian.hoffmann
Mathematics + Statistical Computing e-mail: [EMAIL PROTECTED]
Swiss Federal Research Institute WSL Tel: ++41-44-73922- -77 (office)
CH-8903 Birmensdorf, Switzerland -11(exchange), -15 (fax)
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[R] Residuals, smoothers and estimates of noise
We have a clinical measurement on patients over time. Each patient has about 5 of these measurements over a period of two years, but the measurement are not necessarily taken at equal space in time. We want to use this data to establish test-retest variability. My first thought was to look at the residuals from fitting simple linear regression (the measurement may deteriorate over time and we wish to remove this trend -assuming of course that it is linear). But then what residual do I choose to represent the variability for each patient? Would it be good to use the mean of these? Has anyone got any better solutions with some known functions in R? Do running medians or other smoothers require the data to be a time series with equal spaces between time points? Any help/references/suggestions in R would be gratefully received. --- Dr. David Crabb School of Science, The Nottingham Trent University, Clifton Campus, Nottingham. NG11 8NS Tel: 0115 848 3275 Fax: 0115 848 6690 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] loops: pasting indexes in variables names
I cannot figure out how, using R, I can paste indexes or characters to the
variable
names which are used within loops. I will explain this with a simple
example:
Immagine I have a huge series of variables, each one taken two times, say
x1 x2 y1 y2 z1 z2.
Now, immagine that I want to compute a variable from the difference of
each couple, say dx=x1-x2, dy=y1-y2, dz=z1-z2...
In Stata, for example, this wold be straightforward:
foreach i in x y z {
gen d`i'= `i'1-`i'2
}
With R I tried to use paste( ) but I found that it applies to objects,
not to variable names.
best regards,
Umberto
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Re: [R] is.constant 2
Christian Hoffmann christian.hoffmann at wsl.ch writes:
x - c(1, 2, NA)
is.constant(x)
[1] TRUE
For data such as c(1, 1, 1, NA), I should think the safest answer
should be
NA, because one really doesn't know whether that last number is 1 or
not.
Andy
My version is
is.constant - function(x) {
if (is.numeric(x) !any(is.na(x))) identical(min(x), max(x)) else
FALSE
}
Since the issue of factors surfaced, I improved my function:
is.constant - function(x) {
if (is.factor(x)) (length(attributes(x)$levels)==1)
(!any(is.na(as.character(x
else (is.numeric(x) !any(is.na(x)) identical(min(x), max(x)))
}
Suggest you use an S3 generic and a separate methods for factor, and
in the future, other classes. Also to make it more consistent with
other R functions have an na.rm= argument which defaults to TRUE.
If na.rm = FALSE then it should return NA there are any NAs in
the same way that sum(c(1,2,NA), na.rm = FALSE) returns NA. There is
some question of how to handle zero length arguments (or ones
that become zero length after removing NAs).
is.constant - function(x, ...) UseMethod(is.constant)
is.constant.factor - function(x, na.rm = TRUE) ... code for factor ...
is.constant.default - function(x, na.rm = TRUE) ... code for default ...
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RE: [R] impenetrable warning
Generally you can set options(warn=2), run the code, then do traceback(). Andy From: Simon.Bond Dear R-help, Can anyone explain the meaning of the warning, Singular precision matrix in level -1, block 1 ? Or how to track down where it comes from? More precisely, using the nlme package, I'm issued with the warning itt2 - lme(lrna~rx.nrti+lbrna, random=~1|patid, cor=corExp(form=~days|patid,nugget=T), weights=varPower( form=~lbrna),data=rna3) Warning messages: 1: Singular precision matrix in level -1, block 1 2: Singular precision matrix in level -1, block 1 the output is: Linear mixed-effects model fit by REML Data: rna3 Log-restricted-likelihood: -4990.142 Fixed: lrna ~ rx.nrti + lbrna (Intercept) rx.nrtiZDV+3TC rx.nrtiZDV+ABC lbrna 2.6597552 0.8589514 0.4259504 0.2929222 Random effects: Formula: ~1 | patid (Intercept) Residual StdDev:1.251113 0.2105329 Correlation Structure: Exponential spatial correlation Formula: ~days | patid Parameter estimate(s): range nugget 204.6422150 0.3349336 Variance function: Structure: Power of variance covariate Formula: ~lbrna Parameter estimates: power 0.9733358 Number of Observations: 2390 Number of Groups: 124 Thanks, Simon Bond. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] loops: pasting indexes in variables names
Hi Umberto,
look at ?get, ?assign, ?paste and try the following:
x1 - rnorm(10)
x2 - rnorm(10)
y1 - rnorm(10)
y2 - rnorm(10)
z1 - rnorm(10)
z2 - rnorm(10)
##
names. - c(x, y, z)
for(i in names.){
res1 - get(paste(i,1,sep=))
res2 - get(paste(i,2,sep=))
assign(paste(d,i,sep=), res1-res2)
}
###
all.equal(dx, x1-x2)
all.equal(dy, y1-y2)
all.equal(dz, z1-z2)
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: +32/16/396887
Fax: +32/16/337015
Web: http://www.med.kuleuven.ac.be/biostat/
http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm
- Original Message -
From: Umberto Maggiore [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, September 22, 2004 2:12 PM
Subject: [R] loops: pasting indexes in variables names
I cannot figure out how, using R, I can paste indexes or characters
to the variable
names which are used within loops. I will explain this with a simple
example:
Immagine I have a huge series of variables, each one taken two
times, say
x1 x2 y1 y2 z1 z2.
Now, immagine that I want to compute a variable from the difference
of
each couple, say dx=x1-x2, dy=y1-y2, dz=z1-z2...
In Stata, for example, this wold be straightforward:
foreach i in x y z {
gen d`i'= `i'1-`i'2
}
With R I tried to use paste( ) but I found that it applies to
objects,
not to variable names.
best regards,
Umberto
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Re: [R] loops: pasting indexes in variables names
Umberto Maggiore umberto_maggiore at hotmail.com writes:
I cannot figure out how, using R, I can paste indexes or characters to the
variable
names which are used within loops. I will explain this with a simple
example:
Immagine I have a huge series of variables, each one taken two times, say
x1 x2 y1 y2 z1 z2.
Now, immagine that I want to compute a variable from the difference of
each couple, say dx=x1-x2, dy=y1-y2, dz=z1-z2...
In Stata, for example, this wold be straightforward:
foreach i in x y z {
gen d`i'= `i'1-`i'2
}
With R I tried to use paste( ) but I found that it applies to objects,
not to variable names.
best regards,
Umberto
You can try this:
paste. - function(...) paste(..., sep = )
for(i in c(x, y, z))
assign(paste.(d, i), get(paste.(i, 1) - paste.(i, 2)))
You also might review why you need this since you may prefer
to use vectors, matrices, arrays, lists or data frames for
this sort of processing. For example, if they were in a data
frame, DF, then you could write:
DF[,1] - DF[,2]
and perform all the subtractions at once.
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Re: [R] is.constant 2
On Wed, 22 Sep 2004 12:16:10 + (UTC), Gabor Grothendieck
[EMAIL PROTECTED] wrote :
Suggest you use an S3 generic and a separate methods for factor, and
in the future, other classes.
That's not a bad idea, but is it really worth the trouble? Why not
piggyback on the unique() generic, and define it as something like
is.constant - function(x, na.rm = FALSE, ...) {
vals - unique(x, ...)
if (na.rm) vals - vals[!is.na(vals)]
# What should the value be for c(1, NA)? If FALSE is wanted,
length(vals) == 1
# but if NA is desired
# ifelse (any(is.na(vals)), NA, length(vals) == 1)
}
Also to make it more consistent with
other R functions have an na.rm= argument which defaults to TRUE.
The more common default is FALSE.
Duncan Murdoch
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[R] dot density maps
Dear All, In the moment i'm using the map and maptools package to read shapefiles and display the maps. I'm looking for the possibility to draw points (randomly positioned or positioned according to a grid) into the polygons instead of filling the polygons with colors. For example: a map (shapefile) with 10 countries, 15 points in the polygon of country A, 20 points in the polygon of country B. Thank you in advance for your help Johannes [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] loops: pasting indexes in variables names
Umberto Maggiore wrote:
I cannot figure out how, using R, I can paste indexes or characters to
the variable
names which are used within loops. I will explain this with a simple
example:
Immagine I have a huge series of variables, each one taken two times, say
x1 x2 y1 y2 z1 z2.
Now, immagine that I want to compute a variable from the difference of
each couple, say dx=x1-x2, dy=y1-y2, dz=z1-z2...
In Stata, for example, this wold be straightforward:
foreach i in x y z {
gen d`i'= `i'1-`i'2
}
With R I tried to use paste( ) but I found that it applies to objects,
not to variable names.
best regards,
Umberto
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Try:
x1-10
x2-20
y1-5
y2-22
z1-4
z2-7
for(i in c(x,y,z)){
eval(parse(text=paste(d,i,-,i,1 - ,i,2,sep=)))
}
ls(pattern=d)
output-start
Wed Sep 22 14:38:28 2004
[1] dx dy dz
output-end
but why don't you store x1,y1,z1 and x2,y2,z2 in a list:
a-list(x=1:4, y=1:7, z=1:5)
b-list(x=(1:4)*10, y=1:7, z=(1:5)-20)
d-sapply(1:3, function(i) a[[i]]-b[[i]] )
@
output-start
Wed Sep 22 14:43:09 2004
[[1]]
[1] -9 -18 -27 -36
[[2]]
[1] 0 0 0 0 0 0 0
[[3]]
[1] 20 20 20 20 20
output-end
Peter Wolf
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[R] DCOM server high quality graphics ?
I'm using c# and I can produce some graphics with the DCOM Server on R but the graphics's quality is very bad, fonts problems, anti-aliasing ? How to produce high quality graphics with the DCOM server [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Multinomial Response Variable: Estimating the parameters
Hello! In a dyslexia-spelling-study we have a multinomial response variable with four categories. There are 12 subjects. Each subject was tested on 30 words, i.e. there are 30 responses for each subject. We are interested in estimating the parameters pi1, pi2, pi3, pi4, if possible point estimates and confidence intervals. Is there a possibility in R to conduct this analysis? -- Dr. Martin Plöderl Universität Salzburg Fachbereich Psychologie Abteilung Sozialpsychologie Hellbrunnerstr. 34 5020 Salzburg Phone: +43 662 8044 5130 [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R course
Apologies for cross-posting We would like to announce a 3 day course: R programming for beginners. When: 10-12 January 2005 (Monday-Wednesday). Location: The Ythan hotel in Newburgh (UK). Newburgh is a small coastal village 10 miles north of Aberdeen airport. Host: Organised by Highland Statistics Ltd. Price: 500 Euro for 3 days, excluding 17.5% VAT. The course fee includes a copy of Introductory Statistics with R by P. Dalgaard. You will need to bring you own laptop. Accommodation is available at the Ythan hotel at a special rate of 30 UK pounds per night (including breakfast). Early booking is essential! In this course, we teach how to use and program in the software package R. It is also relevant for S-Plus users who wish to learn script programming. The book Introductory Statistics with R from Peter Dalgaard is used as course material. We discuss how to import data into R, define vectors and nominal variables, make exploratory graphs, and apply ANOVA and linear regression. We assume that course attendants are familiar with the basic aspects of regression and ANOVA. Course attendants are encouraged to bring their own data. There is only place for 10 people on this course. Registration: http://www.brodgar.com/statscourse.htm Kind regards, Alain Zuur Dr. Alain F. Zuur Highland Statistics Ltd. 6 Laverock road UK - AB41 6FN Newburgh Tel: 0044 1358 788177 Email: [EMAIL PROTECTED] URL: www.highstat.com URL: www.brodgar.com (Brodgar complies with R GNU license) Our 5-day statistics course: Analysing Biological and Environmental Field data Brodgar: Software for multivariate analysis and multivariate time series analysis Statistical consultancy, courses, data analysis and software __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] RMySQL and Blob
Hi Jonathan, Currently RMySQL doesn't handle blob objects. The mechanics of inserting and extracting blob objects by itself is not too hard, but issues such as how should blobs be made available to R, how to prevent buffers overflows, how to prevent huge blobs from exhausting the available memory, should R callback functions be invoked as chunks of the blob are brought in, etc., need more consideration. And these issues are not R/MySQL specific, but also relevant to other databases and other non-dbms interfaces. BTW there are R facilities (e.g., external pointers, finalizers) that seems quite important for this type of implementation. What type and how big are the blobs that want to import? -- David [EMAIL PROTECTED] wrote: Dear R experts, Does RMySQL package handle Blob datatype in a MySQL database? Blob can represent an image, a sound or some other large and complex binary objects. In an article published by R-database special interest group, named A common database interface (DBI) (updated June 2003), it's mentioned in open issues and limitations that We need to carefully plan how to deal with binary objects. Before I invest time to try, I would appreciate any experts' opinions. Thanks, Jonathan __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] R 1.9.1 Fails to Start on WinXP SP2
One other suggestion: Run the msconfig System configuration utility to turn off most of the software that loads on your machine at startup, and see if that allows R to start. Then gradually add it back until you find the culprit, if there is one. Bingo! You rock Duncan. I had Rage3D (a video card overclocking utility) installed on the machine that wouldn't run R and I was able to isolate that as the problem. Once I uninstalled it, everything worked perfectly. Thanks a million! Scott __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] layout for xyplot
Dear all, I tried to use layout argument in xyplot to get one panel per page. I have a dataframe named 'data' with the following variables: x, y = coords, sub, bloc = 2-level factors, etat = 5-level factor, I did : lset(theme = col.whitebg()) xyplot(y ~ x | bloc*sub , data=data, groups=etat, + layout=c(0,1,4), + main=Etat des plantes dans chaque bloc, + auto.key=list(columns=5, cex=.8), + scales=list(relation=free, draw=FALSE), + xlab=, ylab=, + ylim=list(c(52, 69), c(16, 33), c(35, 51), c(-1, 15))) and received this error message : Error in if (!any(cond.max.level - cond.current.level 0) (row - 1) * : missing value where TRUE/FALSE needed I tried some changes in arguments - notably layout=c(0,1), but anything works. Thanks for helping... Jacques VESLOT CIRAD Réunion __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R 1.9.1 Fails to Start on WinXP SP2
On Wed, 22 Sep 2004 09:28:03 -0500, Scott Higginbotham [EMAIL PROTECTED] wrote : One other suggestion: Run the msconfig System configuration utility to turn off most of the software that loads on your machine at startup, and see if that allows R to start. Then gradually add it back until you find the culprit, if there is one. Bingo! You rock Duncan. I had Rage3D (a video card overclocking utility) installed on the machine that wouldn't run R and I was able to isolate that as the problem. Once I uninstalled it, everything worked perfectly. Thanks a million! You're welcome. You might want to let the Rage3D people know about the problem; they might be in a position to see what's going wrong with the R load process, and if it's their bug they might fix it. I'd be happy to correspond with them if they need info about R. Duncan Murdoch __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] t test problem?
Hi, thanks for all your suggestions. It is realy helpful. The data was not paired. Sorry for a wrong set number of sample, compared to the sample I used for tests. Best Liu Wayne Jones [EMAIL PROTECTED] wrote: Hi Kan Lui, I've had a quick look at the data. The logged data seems reasonably nicely distributed (roughly symmetrical + equal variance). Indeed the y variable passed the (very strict) shapiro.test for normality. However the main problem is that I do not get the same results as you for the significance of the t.test. The only significant test I see is the paired t.test on the logged data. Is your data paired data? To see what I mean check out: http://www.texasoft.com/winkpair.html The non-parametric tests show no significance (paired data or not) (logged and natural data). Although in general they do tend to be less strict than parametric tests. Unless the data is paired then the means of these samples most certainly do not significantly differ from one another. Here are my workings: Temp.Dat-read.table(data_natural.txt,header=T) hist(log(Temp.Dat$x,10)) hist(log(Temp.Dat$y,10)) shapiro.test(log(Temp.Dat$x,10)) shapiro.test(log(Temp.Dat$y,10)) t.test(log(Temp.Dat$x,10), log(Temp.Dat$y,10)) Welch Two Sample t-test data: log(Temp.Dat$x, 10) and log(Temp.Dat$y, 10) t = 0.9126, df = 195.806, p-value = 0.3626 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0.0599837 0.1633168 sample estimates: mean of x mean of y 4.891313 4.839647 t.test(log(Temp.Dat$x,10), log(Temp.Dat$y,10),paired=T) Paired t-test data: log(Temp.Dat$x, 10) and log(Temp.Dat$y, 10) t = 2.3535, df = 98, p-value = 0.02060 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.008101002 0.095232132 sample estimates: mean of the differences 0.05166657 wilcox.test(log(Temp.Dat$x,10), log(Temp.Dat$y,10),paired=T) Wilcoxon signed rank test with continuity correction data: log(Temp.Dat$x, 10) and log(Temp.Dat$y, 10) V = 2972.5, p-value = 0.0828 alternative hypothesis: true mu is not equal to 0 wilcox.test(log(Temp.Dat$x,10), log(Temp.Dat$y,10),paired=F) Wilcoxon rank sum test with continuity correction data: log(Temp.Dat$x, 10) and log(Temp.Dat$y, 10) W = 5206, p-value = 0.4491 alternative hypothesis: true mu is not equal to 0 wilcox.test(Temp.Dat$x, Temp.Dat$y,paired=F) Wilcoxon rank sum test with continuity correction data: Temp.Dat$x and Temp.Dat$y W = 5206, p-value = 0.4491 alternative hypothesis: true mu is not equal to 0 wilcox.test(Temp.Dat$x, Temp.Dat$y,paired=T) Wilcoxon signed rank test with continuity correction data: Temp.Dat$x and Temp.Dat$y V = 2896.5, p-value = 0.1417 alternative hypothesis: true mu is not equal to 0 t.test(Temp.Dat$x, Temp.Dat$y,paired=T) Paired t-test data: Temp.Dat$x and Temp.Dat$y t = 1.6731, df = 98, p-value = 0.0975 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -2351.81 27623.53 sample estimates: mean of the differences 12635.86 t.test(Temp.Dat$x, Temp.Dat$y,paired=F) Welch Two Sample t-test data: Temp.Dat$x and Temp.Dat$y t = 0.6432, df = 191.177, p-value = 0.5209 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -26116.18 51387.89 sample estimates: mean of x mean of y 120544.9 107909.0 -Original Message- From: kan Liu [mailto:[EMAIL PROTECTED] Sent: 22 September 2004 10:22 To: Andrew Robinson; Dimitris Rizopoulos Cc: [EMAIL PROTECTED] Subject: Re: [R] t test problem? Hi, Many thanks for your helpful comments and suggestions. The attached are the data in both log10 scale and original scale. It would be very grateful if you could suggest which version of test should be used. By the way, how to check whether the variation is additive (natural scale) or multiplicative (log scale) in R? How to check whether the distribution of the data is normal? PS, Can I confirm that do your suggestions mean that in order to check whether there is a difference between x and y in terms of mean I need check the distribution of x and that of y in both natual and log scales and to see which present normal distribution? and then perform a t test using the data scale which presents normal distribution? If both scales present normal distribution, then the t tests with both scales should give the similar results? Thanks again. Liu Andrew Robinson [EMAIL PROTECTED] wrote: Hi Dimitris, you are describing a more stringent requirement than the t-test actually requires. It's the sampling distribution of the mean that should be normal, and this condition is addressed by the Central Limit Theorem. Whether or not the CLT can be invoked depends on
Re: [R] Multinomial Response Variable: Estimating the parameters
hmm, the design sounds for me similiar to choice-data from choice-based-conjoint. IMHO check library(MNP) ? regrads,christian Am Mittwoch, 22. September 2004 15:04 schrieb Martin Plöderl: Hello! In a dyslexia-spelling-study we have a multinomial response variable with four categories. There are 12 subjects. Each subject was tested on 30 words, i.e. there are 30 responses for each subject. We are interested in estimating the parameters pi1, pi2, pi3, pi4, if possible point estimates and confidence intervals. Is there a possibility in R to conduct this analysis? __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] glmmPQL correlation structure
Running Mac OS 10.3.5 and R 2.0 Does glmmPQL use the same spatial dependence models as gls in nmle? It does not seem to - I get the following, for example: m2 - glmmPQL(S.Early ~ fertilized*watered, data=geodat, family=poisson, random=~1|col) iteration 1 iteration 2 plot(Variogram(m2, form=~col+row)) m3 - update(m2,correlation=corSpher(c(5,.5), form=~col+row, nugget=T)) iteration 1 Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid variable type Any leads on how to specify exponential or spherical models would be appreciated (e.g., pageg or chapter in Venables and Ripley 2002). Cheers, Hank Dr. Martin Henry H. Stevens, Assistant Professor 338 Pearson Hall Botany Department Miami University Oxford, OH 45056 Office: (513) 529-4206 Lab: (513) 529-4262 FAX: (513) 529-4243 http://www.cas.muohio.edu/botany/bot/henry.html http://www.muohio.edu/ecology/ http://www.muohio.edu/botany/ E Pluribus Unum __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] layout for xyplot
Have you read the posting guide, which says: quote For questions about unexpected behavior or a possible bug provide details about your platform (Windows2000, Linux, OS X) and R version (type version at the R prompt). State the full version number, e.g., `1.8.1', not just `1.8'. State whether you installed a pre-compiled binary version of R or compiled it yourself. If the function is in a package other than `base', include the header output from library(help=thatPackage). If you are using an old version of R and think it does not work properly, upgrade. /quote Further, we don't have access to your data, so there's no way we can reproduce what you have done. My guess is that you are using an old version of R and lattice, and this bug has already been fixed. I have no idea if it would help, but have you tried layout = c(1,1,4)? Deepayan On Wednesday 22 September 2004 09:32, Jacques VESLOT wrote: Dear all, I tried to use layout argument in xyplot to get one panel per page. I have a dataframe named 'data' with the following variables: x, y = coords, sub, bloc = 2-level factors, etat = 5-level factor, I did : lset(theme = col.whitebg()) xyplot(y ~ x | bloc*sub , data=data, groups=etat, + layout=c(0,1,4), + main=Etat des plantes dans chaque bloc, + auto.key=list(columns=5, cex=.8), + scales=list(relation=free, draw=FALSE), + xlab=, ylab=, + ylim=list(c(52, 69), c(16, 33), c(35, 51), c(-1, 15))) and received this error message : Error in if (!any(cond.max.level - cond.current.level 0) (row - 1) * : missing value where TRUE/FALSE needed I tried some changes in arguments - notably layout=c(0,1), but anything works. Thanks for helping... Jacques VESLOT CIRAD Réunion __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] block statistics with POSIX classes
I have a monthly price index series x, the related return series y = diff(log(x)) and a POSIXlt date-time variable dp. I would like to apply annual blocks to compute for example annual block maxima and mean of y. When studying the POSIX classes, in the first stage of the learning curve, I computed the maximum drawdown of x: mdd - maxdrawdown(x) max.dd - mdd$maxdrawdown from - as.character(dp[mdd$from]) to - as.character(dp[mdd$to]) from; to [1] 2000-08-31 [1] 2003-03-31 that gives me the POSIX dates of the start and end of the period and suggests that I have done something correctly. Two questions: (1) how to implement annual blocks and compute e.g. annual max, min and mean of y (each year's max, min, mean)? (2) how to apply POSIX variables with the 'block' argument in gev in the evir package? The S+FinMetrics function aggregateSeries does the job in that module; but I do not know, how handle it in R. My guess is that (1) is done by using the function aggregate, but how to define the 'by' argument with POSIX variables? Thanks! Hannu Kahra Progetti Speciali Monte Paschi Asset Management SGR S.p.A. Via San Vittore, 37 IT-20123 Milano, Italia Tel.: +39 02 43828 754 Mobile: +39 333 876 1558 Fax: +39 02 43828 247 E-mail: [EMAIL PROTECTED] Web: www.mpsam.it __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Sample without replacement
Hello, I have a simple problem (checked the archives and the appropriate
help pages, to no avail). I want to creat a vector that is length(2000).
The vector is to consist of two strings( std and dev) with 1760 stds
and 240 devs. Furthermore, for each element of the vector, i want the
selection of one of the strings to be approx. random. The end result will
be a vector with the following proportions: .12 dev and .88 std.
My solution, below, almost works, but i don't get the exact desired
proportions:
sample.from.vector - c(rep(std,1760),rep(dev,240))
make.vector - rep(99,2000)
for (i in 1:2000){
make.vector[i] - sample(sample.from.vector,1,replace=F)
}
As I understand the above code (which is not very well, obviously), each
iteration assigns one element from the sample.from.vector to the current
make.vector element; the element choosen from sample.from.vector is
tagged so that i is not selected again. However, after the loop,
make.vector contains, for example, 1758 stds and 242 devs. Each
iteration results in a different proportion of stds and devs (although
close to the desired, not right on).
Any help would be greatly apprecitated.
-Mark Orr
Mark G. Orr
Postdoctoral Research Fellow
Dept. of Neuroscience
RM 825 Kennedy Center
Albert Einstein College of Medicine
Bronx, NY 10461
718-430-2610
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Re: [R] Sample without replacement
Mark G Orr [EMAIL PROTECTED] writes:
Hello, I have a simple problem (checked the archives and the
appropriate help pages, to no avail). I want to creat a vector that
is length(2000). The vector is to consist of two strings( std and
dev) with 1760 stds and 240 devs. Furthermore, for each element
of the vector, i want the selection of one of the strings to be
approx. random. The end result will be a vector with the following
proportions: .12 dev and .88 std. My solution, below, almost
works, but i don't get the exact desired proportions:
sample.from.vector - c(rep(std,1760),rep(dev,240))
make.vector - rep(99,2000)
for (i in 1:2000){
make.vector[i] - sample(sample.from.vector,1,replace=F)
}
As I understand the above code (which is not very well, obviously),
each iteration assigns one element from the sample.from.vector to the
current make.vector element; the element choosen from
sample.from.vector is tagged so that i is not selected again.
However, after the loop, make.vector contains, for example, 1758 stds
and 242 devs. Each iteration results in a different proportion of
stds and devs (although close to the desired, not right on).
Any help would be greatly apprecitated.
It's staring you right in the face:
make.vector - sample(sample.from.vector)
--
O__ Peter Dalgaard Blegdamsvej 3
c/ /'_ --- Dept. of Biostatistics 2200 Cph. N
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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[R] pairs, panel.functions, xlim and ylim
Hi,
I have the following problem.
I wanted to get a matrix of scatterplots and I used pairs.
I wanted to add the line y=x in each plot and I created a panel
function for this scope.
I used points and abline in the following way:
## put y=x in each plot
panel.lin- function(x, y)
{
points(x,y, pch=21, bg=par(bg), col = black,cex=2)
abline(0,1,lwd=2, col=red)
}
and it works.
Now, I want that each plot has the same scale on the axis and I try
with this modification:
## put y=x in each plot - same scale for all the plots
panel.lincor- function(x, y)
{
points(x,y, pch=21, bg=par(bg), col =
black,cex=2,xlim=c(-1,1),ylim=c(-1,1))
abline(0,1,lwd=2, col=red)
}
but R tells me that xlim and ylim couldn't be set in high level plot
functions.
I try to use plot() instead of points, but I realized immediately that
I can't use plot as a panel function.
I know I could have added xlim and ylim directly in pairs() function,
if I hadn't had a panel function, but I need the line y=x in every plot
and I don't know other ways to get it.
Any idea?
Thank you very much
Valeria Edefonti
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RE: [R] ordered probit and cauchit
The Political Science Computational Laboratory at Stanford has R code for ordered probit (courtesy of Simon Jackman): http://pscl.stanford.edu/oprobit. David Reinke -Original Message- From: David Firth [mailto:[EMAIL PROTECTED] Sent: Wednesday, September 22, 2004 02:04 To: roger koenker Cc: [EMAIL PROTECTED] Subject: Re: [R] ordered probit and cauchit On Wednesday, Sep 22, 2004, at 03:42 Europe/London, roger koenker wrote: What is the current state of the R-art for ordered probit models, and more esoterically is there any available R strategy for ordered cauchit models, i.e. ordered multinomial alternatives with a cauchy link function. MCMC is an option, obviously, but for a univariate latent variable model this seems to be overkill... standard mle methods should be preferable. (??) A quick look at polr (in the MASS package) suggests to me that it wouldn't be all that hard to extend it to link functions other than logistic. Is MLE known to be well behaved in these models if the latent variable is Cauchy? David Googling reveals that spss provides such functions... just to wave a red flag. url:www.econ.uiuc.edu/~rogerRoger Koenker email [EMAIL PROTECTED] Department of Economics vox:217-333-4558University of Illinois fax:217-244-6678Champaign, IL 61820 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] block statistics with POSIX classes
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RE: [R] RMySQL and Blob
Hi David, The application I have in mind is for images. In my case, size of images is known and they are not big. As an example, a 64*32 image will have 2048 pixels. If they are 8-bit grey-level pixels, the image occupies 2KB memory. I may venture to guess that the unknown size and type of a blob object in MySQL prevent it from being very usable in R since R doesn't have a datatype for a binary blob? Thanks! Jonathan -Original Message- From: David James [mailto:[EMAIL PROTECTED] Sent: Wednesday, September 22, 2004 7:05 AM To: LI,JONATHAN (A-Labs,ex1) Cc: [EMAIL PROTECTED] Subject: Re: [R] RMySQL and Blob Hi Jonathan, Currently RMySQL doesn't handle blob objects. The mechanics of inserting and extracting blob objects by itself is not too hard, but issues such as how should blobs be made available to R, how to prevent buffers overflows, how to prevent huge blobs from exhausting the available memory, should R callback functions be invoked as chunks of the blob are brought in, etc., need more consideration. And these issues are not R/MySQL specific, but also relevant to other databases and other non-dbms interfaces. BTW there are R facilities (e.g., external pointers, finalizers) that seems quite important for this type of implementation. What type and how big are the blobs that want to import? -- David [EMAIL PROTECTED] wrote: Dear R experts, Does RMySQL package handle Blob datatype in a MySQL database? Blob can represent an image, a sound or some other large and complex binary objects. In an article published by R-database special interest group, named A common database interface (DBI) (updated June 2003), it's mentioned in open issues and limitations that We need to carefully plan how to deal with binary objects. Before I invest time to try, I would appreciate any experts' opinions. Thanks, Jonathan __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Issue with predict() for glm models
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Re: [R] glmmPQL correlation structure
Have you tried GLMM in lme4? Doug Bates is the primary architect of both nlme and lme4. Therefore, I would think that a spatial dependence model that works in nlme might also work in GLMM. hope this helps. spencer graves Martin Henry H. Stevens wrote: Running Mac OS 10.3.5 and R 2.0 Does glmmPQL use the same spatial dependence models as gls in nmle? It does not seem to - I get the following, for example: m2 - glmmPQL(S.Early ~ fertilized*watered, data=geodat, family=poisson, random=~1|col) iteration 1 iteration 2 plot(Variogram(m2, form=~col+row)) m3 - update(m2,correlation=corSpher(c(5,.5), form=~col+row, nugget=T)) iteration 1 Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid variable type Any leads on how to specify exponential or spherical models would be appreciated (e.g., pageg or chapter in Venables and Ripley 2002). Cheers, Hank Dr. Martin Henry H. Stevens, Assistant Professor 338 Pearson Hall Botany Department Miami University Oxford, OH 45056 Office: (513) 529-4206 Lab: (513) 529-4262 FAX: (513) 529-4243 http://www.cas.muohio.edu/botany/bot/henry.html http://www.muohio.edu/ecology/ http://www.muohio.edu/botany/ E Pluribus Unum __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Spencer Graves, PhD, Senior Development Engineer O: (408)938-4420; mobile: (408)655-4567 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] THanks, random/repl. prob. SOLVED.
Thanks for the help, it worked. Mark G. Orr Postdoctoral Research Fellow Dept. of Neuroscience RM 825 Kennedy Center Albert Einstein College of Medicine Bronx, NY 10461 718-430-2610 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] impenetrable warning
In this case, this trick will identify the specify command in the lme code that produce the error. If that does not lead you to an answer, have you tried simplifying your lme call to identify more clearly which part (or combination) seems to generate the problem? hope this helps. spencer graves Liaw, Andy wrote: Generally you can set options(warn=2), run the code, then do traceback(). Andy From: Simon.Bond Dear R-help, Can anyone explain the meaning of the warning, Singular precision matrix in level -1, block 1 ? Or how to track down where it comes from? More precisely, using the nlme package, I'm issued with the warning itt2 - lme(lrna~rx.nrti+lbrna, random=~1|patid, cor=corExp(form=~days|patid,nugget=T), weights=varPower( form=~lbrna),data=rna3) Warning messages: 1: Singular precision matrix in level -1, block 1 2: Singular precision matrix in level -1, block 1 the output is: Linear mixed-effects model fit by REML Data: rna3 Log-restricted-likelihood: -4990.142 Fixed: lrna ~ rx.nrti + lbrna (Intercept) rx.nrtiZDV+3TC rx.nrtiZDV+ABC lbrna 2.6597552 0.8589514 0.4259504 0.2929222 Random effects: Formula: ~1 | patid (Intercept) Residual StdDev:1.251113 0.2105329 Correlation Structure: Exponential spatial correlation Formula: ~days | patid Parameter estimate(s): range nugget 204.6422150 0.3349336 Variance function: Structure: Power of variance covariate Formula: ~lbrna Parameter estimates: power 0.9733358 Number of Observations: 2390 Number of Groups: 124 Thanks, Simon Bond. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Spencer Graves, PhD, Senior Development Engineer O: (408)938-4420; mobile: (408)655-4567 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Issue with predict() for glm models
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[R] pairs, panel.functions, xlim and ylim
Valerie,
A bit ugly, because you must ignore some warnings,
but for me works the code below.
Best
Jens Oehlschlägel
x - rnorm(100, sd=0.2)
x - cbind(x=x-0.1, y=x+0.1)
pairs(x
, panel=function(x,y){
function to be called with xlim=, ylim= parameters
points(x, y)
abline(0,1)
}
)
pairs(x
, panel=function(x,y, ...)
{ # adding the three points here allows your panel function to accept the
additional xlim=, ylim= parameters
points(x, y)
abline(0,1)
}
, xlim=c(-1, 1)
, ylim=c(-1, 1)
)
I have the following problem.
I wanted to get a matrix of scatterplots and I used pairs.
I wanted to add the line y=x in each plot and I created a panel
function for this scope.
I used points and abline in the following way:
## put y=x in each plot
panel.lin- function(x, y)
{
points(x,y, pch=21, bg=par(bg), col = black,cex=2)
abline(0,1,lwd=2, col=red)
}
and it works.
Now, I want that each plot has the same scale on the axis and I try
with this modification:
## put y=x in each plot - same scale for all the plots
panel.lincor- function(x, y)
{
points(x,y, pch=21, bg=par(bg), col =
black,cex=2,xlim=c(-1,1),ylim=c(-1,1))
abline(0,1,lwd=2, col=red)
}
but R tells me that xlim and ylim couldn't be set in high level plot
functions.
I try to use plot() instead of points, but I realized immediately that
I can't use plot as a panel function.
I know I could have added xlim and ylim directly in pairs() function,
if I hadn't had a panel function, but I need the line y=x in every plot
and I don't know other ways to get it.
Any idea?
Thank you very much
Valeria Edefonti
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[R] aparchFit()$fitted.value
Dear R people, I'm not able to have the component residuals, fitted.value from an aparchFit() estimation as explain in the Value of aparchFit Help, package fSeries. Could someone help me? Thanks in advance. Lisa __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Issue with predict() for glm models
Hello everyone, I am having a problem using the predict (or the predict.glm) function in R. Basically, I run the glm model on a training data set and try to obtain predictions for a set of new predictors from a test data set (i.e., not the predictors that were utilized to obtain the glm parameter estimates). Unfortunately, every time that I attempt this, I obtain the predictions for the predictors that were used to fit the glm model. I have looked at the R mailing list archives and don't believe I am making the same mistakes that have been made in the past and also have tried to closely follow the predict.glm example in the help file. Here is an example of what I am trying to do: set.seed(545345) # Necessary Variables # p - 2 train.n - 20 test.n - 25 mean.vec.1 - c(1,1) mean.vec.2 - c(0,0) Sigma.1 - matrix(c(1,.5,.5,1),p,p) Sigma.2 - matrix(c(1,.5,.5,1),p,p) ### # Load MASS Library # ### library(MASS) ### # Data to Parameters for Logistic Regression Model # ### train.data.1 - mvrnorm(train.n,mu=mean.vec.1,Sigma=Sigma.1) train.data.2 - mvrnorm(train.n,mu=mean.vec.2,Sigma=Sigma.2) train.class.var - as.factor(c(rep(1,train.n),rep(2,train.n))) predictors.train - rbind(train.data.1,train.data.2) ## # Test Data Where Predictions for Probabilities Using Logistic Reg. # # From Training Data are of Interest # ## test.data.1 - mvrnorm(test.n,mu=mean.vec.1,Sigma=Sigma.1) test.data.2 - mvrnorm(test.n,mu=mean.vec.2,Sigma=Sigma.2) predictors.test - rbind(test.data.1,test.data.2) ## # Run Logistic Regression on Training Data # ## log.reg - glm(train.class.var~predictors.train, family=binomial(link=logit)) log.reg # log.reg #Call: glm(formula = train.class.var ~ predictors.train, family = #binomial(link = logit)) # #Coefficients: # (Intercept) predictors.train1 predictors.train2 # 0.5105-0.2945-1.0811 # #Degrees of Freedom: 39 Total (i.e. Null); 37 Residual #Null Deviance: 55.45 #Residual Deviance: 41.67AIC: 47.67 ### # Predicted Probabilities for Test Data # ### New.Data - data.frame(predictors.train1=predictors.test[,1], predictors.train2=predictors.test[,2]) logreg.pred.prob.test - predict.glm(log.reg,New.Data,type=response) logreg.pred.prob.test #logreg.pred.prob.test # [1] 0.51106406 0.15597423 0.04948404 0.03863875 0.35587589 0.71331091 # [7] 0.17320087 0.14176632 0.30966718 0.61878952 0.12525988 0.21271139 #[13] 0.70068113 0.18340723 0.10295501 0.44591568 0.72285161 0.31499339 #[19] 0.65789420 0.42750139 0.14435889 0.93008117 0.70798465 0.80109005 #[25] 0.89161472 0.47480625 0.56520952 0.63981834 0.57595189 0.60075882 #[31] 0.96493393 0.77015507 0.87643986 0.62973986 0.63043351 0.45398955 #[37] 0.80855782 0.90835588 0.54809117 0.11568637 Of course, notice that the vector for the predicted probabilities has only 40 elements, while the New.Data has 50 elements (since n.test has 25 per group for 2 groups) and thus should have 50 predicted probabilities. As it turns out, the output is for the training data predictors and not for the New.Data as I would like it to be. I should also note that I have made sure that the names for the predictors in the New.Data are the same as the names for the predictors within the glm object (i.e., within log.reg) as this is what is done in the example for predict.glm() within the help files. Could some one help me understand either what I am doing incorrectly or what problems there might be within the predict() function? I should mention that I tried the same program using predict.glm() and obtained the same problematic results. Thanks and take care, Joe Joe Rausch, M.A. Psychology Liaison Lab for Social Research 917 Flanner Hall University of Notre Dame Notre Dame, IN 46556 (574) 631-3910 If we knew what it was we were doing, it would not be called research, would it? - Albert Einstein __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Issue with predict() for glm models
Dear Mark and Joe, Actually, the problem here appears to be caused by the use of a matrix on the RHS of the model formula. I'm not sure why this doesn't work (I must be missing something -- perhaps someone else can say what), but Joe can get the output he expects by specifying the columns of his matrix as individual predictors in the model formula. BTW, it's better form to call the generic predict() rather than the method predict.glm() directly, though the latter will work here. Editing the original input: x1 - predictors.train[,1] x2 - predictors.train[,2] log.reg - glm(train.class.var ~ x1 + x2, + family=binomial(link=logit)) log.reg Call: glm(formula = train.class.var ~ x1 + x2, family = binomial(link = logit)) Coefficients: (Intercept) x1 x2 0.5102 -0.6118 -0.3192 Degrees of Freedom: 39 Total (i.e. Null); 37 Residual Null Deviance: 55.45 Residual Deviance: 46.49AIC: 52.49 New.Data - data.frame(x1=predictors.test[,1], x2=predictors.test[,2]) logreg.pred.prob.test - predict(log.reg,New.Data, type=response) logreg.pred.prob.test [1] 0.2160246 0.2706139 0.3536572 0.6206490 0.5218391 0.2363767 0.1072153 [8] 0.6405459 0.443 0.6680043 0.3377492 0.5892127 0.3230353 0.7540425 [15] 0.2889855 0.5163141 0.6187335 0.1447511 0.5066670 0.4424428 0.4141701 [22] 0.3947212 0.4065674 0.6226195 0.5053101 0.4311552 0.4261810 0.4784102 [29] 0.5126050 0.6756437 0.6147516 0.7659146 0.5219031 0.3938457 0.6495470 [36] 0.5178400 0.8185613 0.7167129 0.5414552 0.8687371 0.5415976 0.8048741 [43] 0.7796451 0.5565636 0.6058371 0.7053130 0.1521769 0.7120320 0.4073465 [50] 0.6801101 I hope this helps, John On Wed, 22 Sep 2004 15:17:23 -0300 Fowler, Mark [EMAIL PROTECTED] wrote: Perhaps your approach reflects a method of producing a prediction dataframe that is just unfamiliar to me, but it looks to me like you have created two predictor variables based on the names of the levels of the original predictor (predictors.train1, predictors.train2). I don't know how the glm function would know that predictors.train1 and predictors.train2 are two subs for predictors.train. Maybe try just using one prediction variable, and give it the original variable name (predictors.train). If this works, just repeat for your second set of values. Mark Fowler Marine Fish Division Bedford Inst of Oceanography Dept Fisheries Oceans Dartmouth NS Canada [EMAIL PROTECTED] -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: September 22, 2004 2:53 PM To: [EMAIL PROTECTED] Subject: [R] Issue with predict() for glm models Hello everyone, I am having a problem using the predict (or the predict.glm) function in R. Basically, I run the glm model on a training data set and try to obtain predictions for a set of new predictors from a test data set (i.e., not the predictors that were utilized to obtain the glm parameter estimates). Unfortunately, every time that I attempt this, I obtain the predictions for the predictors that were used to fit the glm model. I have looked at the R mailing list archives and don't believe I am making the same mistakes that have been made in the past and also have tried to closely follow the predict.glm example in the help file. Here is an example of what I am trying to do: set.seed(545345) # Necessary Variables # p - 2 train.n - 20 test.n - 25 mean.vec.1 - c(1,1) mean.vec.2 - c(0,0) Sigma.1 - matrix(c(1,.5,.5,1),p,p) Sigma.2 - matrix(c(1,.5,.5,1),p,p) ### # Load MASS Library # ### library(MASS) ### # Data to Parameters for Logistic Regression Model # ### train.data.1 - mvrnorm(train.n,mu=mean.vec.1,Sigma=Sigma.1) train.data.2 - mvrnorm(train.n,mu=mean.vec.2,Sigma=Sigma.2) train.class.var - as.factor(c(rep(1,train.n),rep(2,train.n))) predictors.train - rbind(train.data.1,train.data.2) ## # Test Data Where Predictions for Probabilities Using Logistic Reg. # # From Training Data are of Interest # ## test.data.1 - mvrnorm(test.n,mu=mean.vec.1,Sigma=Sigma.1) test.data.2 - mvrnorm(test.n,mu=mean.vec.2,Sigma=Sigma.2) predictors.test - rbind(test.data.1,test.data.2) ## # Run Logistic Regression on Training Data # ## log.reg - glm(train.class.var~predictors.train, family=binomial(link=logit)) log.reg # log.reg #Call: glm(formula = train.class.var ~ predictors.train, family = #binomial(link = logit)) # #Coefficients: # (Intercept) predictors.train1 predictors.train2 # 0.5105-0.2945-1.0811 #
Re: [R] t-test problem
Kan Liu wrote: Hello, I got two sets of data x=(124738, 128233, 85901, 33806, ...) y=(25292, 21877, 45498, 63973, ) When I did a t test, I got two tail p-value = 0.117, which is not significantly different. If I changed x, y to log scale, and re-do the t test, I got two tail p-value = 0.042, which is significantly different. Now I got confused which one is correct. Any help would be very appreciated. If you are unsure about the metric of the attribute being measured, it would be preferable to use a rank test, the Wilcoxon-Mann-Whitney, rather than the parametric t-test. ** Cliff Lunneborg, Professor Emeritus, Statistics Psychology, University of Washington, Seattle [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Facing problems with C code compilation - Please help.
Hello,
I started using R a month ago - so I am a novice in this area. I am stuck with a
problem and need some help urgently.
I am using windows version of R 1.9.1. I am trying to compile C code in it. I have my
C code - hello.c is lying in C:\Program Files\R\rw1091
This code is -
#include R.h
void hello(int *n)
{
int i;
for(i=0;i *n; i++)
{
Rprintf(Hello World ! \n);
}
}
===
Code hello1.R is also lying in the same directory.
This code is -
hello2 - function(n)
{
.C(hello, as.integer))
}
===
From the command prompt, I go into the directory C:\Program Files\R\rw1091\bin
and I do
C:\Program Files\R\rw1091\binR CMD SHLIB hello.c
'make' is not recognized as an internal or external command,
operable program or batch file.
Perl is installed on my machine. I was wondering why am I getting this error. Could
someone please provide me with some pointers on this? Your help will be greatly
appreciated.
Thanks,
Neha
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Re: [R] png problem
Hi Clément Calenge wrote: Hello, Thanks for the fast reply. Paul Murrell wrote: Hi Clément Calenge wrote: Dear R-users, I have a small problem with the function png(), when used with the argument colortype=pseudo.cube. png(toto.png, colortype=pseudo.cube) image(matrix(rnorm(1), 100, 100)) dev.off() R is blocked at the last command (R does not print any prompt after the last command). Nothing is written in the file (Gimp indicates that the file is corrupted). Did you wait long enough? This example took a little while to complete for me (may need someone more familiar with the code to tell us why it is so slow). You're right, it took 45 minutes for me. However, since I need to use this code to build a Sweave vignette, I cannot use it too often (I have about thirty files to create, this would take about 20 hours to build the vignette !). I also need the colortype argument (some graphics cards do not allow the compilation of the vignette without). Does anyone knows how to speed up the process ? Wow! It didn't take that long for me (maybe a minute or two). A possibly useful observation in case anyone else takes a look at this before me: when I debugged it and ctrl-C'ed to interrupt every now and then, the interrupt always happened inside a call to XQueryColor (in bitgp(), in devX11.c), so that may be the slow part. Paul Paul However, png(toto.png) image(matrix(rnorm(1), 100, 100)) dev.off() works fine. I tried: options(X11colortype = pseudo.cube) png(toto.png) image(matrix(rnorm(1), 100, 100)) dev.off() But, here again, R is blocked. I tried to replace dev.off() by graphics.off(), but this does not resolve the problem. The problem does not occurs when the function X11() is used instead of the function png(). I searched through the mail archive, the FAQ, on google, but I did not found any solution to this problem. On the help page on the function png(), it is indicated that The colour handling will be that of the 'X11' device in use. I never used these functions before, but maybe png() is not suitable with colortype=pseudo.cube ? Can you tell me where I have missed something ? Thanks in Advance, Clément Calenge. version _ platform sparc-sun-solaris2.9 arch sparc os solaris2.9 system sparc, solaris2.9 status major1 minor9.1 year 2004 month06 day 21 language R [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 [EMAIL PROTECTED] http://www.stat.auckland.ac.nz/~paul/ == UMR CNRS 5558 - Equipe Ecologie Statistique Laboratoire de Biométrie et Biologie Evolutive Université Claude Bernard Lyon 1 43, Boulevard du 11 novembre 1918 69622 Villeurbanne Cedex FRANCE tel. (+33) 04.72.43.27.57 fax. (+33) 04.72.43.13.88 -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 [EMAIL PROTECTED] http://www.stat.auckland.ac.nz/~paul/ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Cox proportional hazards model
Good afternoon, I am currently trying to do some work on survival analysis. - I hope to seek your advice re: 2 questions (1 general and 1 specific) (1) I'm trying to do a stratified Cox analysis and subsequently plot(survfit(object)). It seems to work for some strata, but not for others. I have tumor grade, which is a range of 1 - 4. When I divide this range of 1:4 into 2 groups, it works fine for strata(grade2) and strata(grade 3). However, if I do a strata(grade1), there is an error when I do a survfit( coxph object ) Call: survfit.coxph(object = s) Error in print.survfit(structure(list(n = as.integer(46), time = c(22, : length of dimnames [1] not equal to array extent (2) As a general question, is it possible to distinguish between confounding and interaction in the Cox proportional hazards model? Thanks! Min-Han __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Re: Your Amazon.com Inquiry
Greetings from Amazon.com. We're sorry. You replied to a notification-only address that cannot accept incoming e-mail. But that's OK--this automated response will direct you to the right place at Amazon.com to answer your question or help you make changes to your order. To change any unshipped orders, make other changes to your account, or view your order history, visit: http://www.amazon.com/your-account For answers to questions about how to order, our shipping rates, and how to use any of our services, visit: http://www.amazon.com/help We hope our online resources meet all your needs. If you've explored the above links but find you still need to get in touch with us, please use the e-mail form available in our online Help department. Thanks for shopping at Amazon.com! Sincerely, Amazon.com Customer Service http://www.amazon.com P.S. You received this message because Amazon.com received the following message: From [EMAIL PROTECTED] Wed Sep 22 17:03:56 2004 Received: from mail-border-1001.vdc.amazon.com (mail-border-1001.vdc.amazon.com [10.139.9.251]) by mail-admin-1.amazon.com (8.12.7/) with ESMTP id i8N03jsQ016597 for [EMAIL PROTECTED]; Wed, 22 Sep 2004 17:03:46 -0700 Received: from service-5-internal.amazon.com by mail-border-1001.vdc.amazon.com with SMTP (crosscheck: service-5-internal.amazon.com [10.16.42.51]) id i8N03gKb025643 for [EMAIL PROTECTED]; Thu, 23 Sep 2004 00:03:43 GMT Message-Id: [EMAIL PROTECTED] X-Amazon-External-Source: yes X-Amazon-External-Envelope-Sender: [EMAIL PROTECTED] From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Received: from ga-cmng-cuda2-c2b-182.atlaga.adelphia.net ([68.71.182.182]) by service-5-internal.amazon.com via smtpd (for mail-border-1001.vdc.amazon.com [10.139.9.251]) with SMTP; 23 Sep 2004 00:03:42 UT Subject: %]Re: Hi Date: Wed, 22 Sep 2004 20:03:39 -0400 MIME-Version: 1.0 Content-Type: multipart/mixed; boundary==_NextPart_000_0016=_NextPart_000_0016 X-Priority: 3 X-MSMail-Priority: Normal X-PMX-Version: 4.7.0.111621, Antispam-Engine: 2.0.1.0, Antispam-Data: 2004.9.22.3 X-AMAZON-GAUGE: X X-PMX-REPORT: The following antispam rules were triggered by this message: Rule Score Description RELAY_IN_CBL 8.000 Composite Blocking List, see http://cbl.abuseat.org/: 182.182.71.68.cbl.abuseat.org MIME_BOUND_NEXTPART 2.100 Spam tool pattern in MIME boundary PRIORITY_NO_NAME 0.716 Message has priority setting, but no X-Mailer NO_REAL_NAME 0.000 From: does not include a real name __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Facing problems with C code compilation - Please help.
Read and follow the instructions in c:\Program
Files\R\rw1091\README.packages _very_, _very_ carefully. Stray from it even
a bit and you get what you deserve.
Andy
From: neha chaudhry
Hello,
I started using R a month ago - so I am a novice in this
area. I am stuck with a problem and need some help urgently.
I am using windows version of R 1.9.1. I am trying to compile
C code in it. I have my C code - hello.c is lying in
C:\Program Files\R\rw1091
This code is -
#include R.h
void hello(int *n)
{
int i;
for(i=0;i *n; i++)
{
Rprintf(Hello World ! \n);
}
}
===
Code hello1.R is also lying in the same directory.
This code is -
hello2 - function(n)
{
.C(hello, as.integer))
}
===
From the command prompt, I go into the directory C:\Program
Files\R\rw1091\bin
and I do
C:\Program Files\R\rw1091\binR CMD SHLIB hello.c
'make' is not recognized as an internal or external command,
operable program or batch file.
Perl is installed on my machine. I was wondering why am I
getting this error. Could someone please provide me with some
pointers on this? Your help will be greatly appreciated.
Thanks,
Neha
__
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[R] multinomial logistic regression
Hi, how can I do multinomial logistic regression in R? I think glm() can only handle binary response variable, and polr() can only handle ordinal response variable. how to do logistic regression with multinomial response variable? Thanks __ Y! Messenger - Communicate in real time. Download now. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] multinomial logistic regression
array chip wrote: I think glm() can only handle binary response variable, That's not true, have you looked at ?glm and ?family ? HTH, Kevin -- Ko-Kang Kevin Wang PhD Student Centre for Mathematics and its Applications Building 27, Room 1004 Mathematical Sciences Institute (MSI) Australian National University Canberra, ACT 0200 Australia Homepage: http://wwwmaths.anu.edu.au/~wangk/ Ph (W): +61-2-6125-2431 Ph (H): +61-2-6125-7407 Ph (M): +61-40-451-8301 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: RE: [R] Facing problems with C code compilation - Please help.
Hi Guys,
Thanks a ton for all the help. My code is finally compiling and running.
-Neha
- Original Message -
From: Liaw, Andy [EMAIL PROTECTED]
Date: Wednesday, September 22, 2004 5:12 pm
Subject: RE: [R] Facing problems with C code compilation - Please help.
Read and follow the instructions in c:\Program
Files\R\rw1091\README.packages _very_, _very_ carefully. Stray from it even
a bit and you get what you deserve.
Andy
From: neha chaudhry
Hello,
I started using R a month ago - so I am a novice in this
area. I am stuck with a problem and need some help urgently.
I am using windows version of R 1.9.1. I am trying to compile
C code in it. I have my C code - hello.c is lying in
C:\Program Files\R\rw1091
This code is -
#include R.h
void hello(int *n)
{
int i;
for(i=0;i *n; i++)
{
Rprintf(Hello World ! \n);
}
}
===
Code hello1.R is also lying in the same directory.
This code is -
hello2 - function(n)
{
.C(hello, as.integer))
}
===
From the command prompt, I go into the directory C:\Program
Files\R\rw1091\bin
and I do
C:\Program Files\R\rw1091\binR CMD SHLIB hello.c
'make' is not recognized as an internal or external command,
operable program or batch file.
Perl is installed on my machine. I was wondering why am I
getting this error. Could someone please provide me with some
pointers on this? Your help will be greatly appreciated.
Thanks,
Neha
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RE: [R] layout for xyplot
thanks a lot ! My version is 1.9.1. : _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major1 minor9.0 year 2004 month04 day 12 language R and I'm working on a Windows2000 platform. Besides I tried layout = c(1,1,4), but it doesn't work anymore (same error message). Please find my dataframe as attached file, and the first 6 columns summary just below: x yblocsub inoc etat Min. : 0.00 Min. : 0.0 1:240 C:240 2:135 : 83 1st Qu.: 3.25 1st Qu.:16.3 2:240 S:240 8:105 F : 55 Median : 6.00 Median :34.0 T:240 Fi : 38 Mean : 6.02 Mean :34.0 NF :302 3rd Qu.: 8.75 3rd Qu.:50.8 NFi: 2 Max. :12.00 Max. :68.0 Is there another means to operate better than with index.cond, by dropping all combination of conditioning variables but one and ploting it with layout=c(1,1). Thanks for helping... jacques -Message d'origine- De : Deepayan Sarkar [mailto:[EMAIL PROTECTED] Envoyé : mercredi 22 septembre 2004 19:18 À : [EMAIL PROTECTED]; [EMAIL PROTECTED] Objet : Re: [R] layout for xyplot Have you read the posting guide, which says: quote For questions about unexpected behavior or a possible bug provide details about your platform (Windows2000, Linux, OS X) and R version (type version at the R prompt). State the full version number, e.g., `1.8.1', not just `1.8'. State whether you installed a pre-compiled binary version of R or compiled it yourself. If the function is in a package other than `base', include the header output from library(help=thatPackage). If you are using an old version of R and think it does not work properly, upgrade. /quote Further, we don't have access to your data, so there's no way we can reproduce what you have done. My guess is that you are using an old version of R and lattice, and this bug has already been fixed. I have no idea if it would help, but have you tried layout = c(1,1,4)? Deepayan On Wednesday 22 September 2004 09:32, Jacques VESLOT wrote: Dear all, I tried to use layout argument in xyplot to get one panel per page. I have a dataframe named 'data' with the following variables: x, y = coords, sub, bloc = 2-level factors, etat = 5-level factor, I did : lset(theme = col.whitebg()) xyplot(y ~ x | bloc*sub , data=data, groups=etat, + layout=c(0,1,4), + main=Etat des plantes dans chaque bloc, + auto.key=list(columns=5, cex=.8), + scales=list(relation=free, draw=FALSE), + xlab=, ylab=, + ylim=list(c(52, 69), c(16, 33), c(35, 51), c(-1, 15))) and received this error message : Error in if (!any(cond.max.level - cond.current.level 0) (row - 1) * : missing value where TRUE/FALSE needed I tried some changes in arguments - notably layout=c(0,1), but anything works. Thanks for helping... Jacques VESLOT CIRAD Réunion __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html - CIRAD Reunion - MailScanner - NO VIRUS found - __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
