Re: [R] Installing 2.4.0
These are old compilers (2.95.3 is really old, much older than your OS). Are they really built for Solaris 10? The underlying problem in both cases seems to be a mismatch between your OS and your compiler, something discussed in the R-admin manual as a problem on Solaris. I would suggest getting a current version of gcc for Solaris 10 and using that. (www.freesunware.com has 3.4.6, and that works for us on Solaris 8.) On Tue, 21 Nov 2006, Denise Mauldin wrote: Hello, I'm trying to install R 2.4.0 on a SunOS 5.10 Generic_118833-24 sun4v sparc SUNW,Sun-Fire-T200 machine. The machine has gcc version 3.3.2 and gcc version 2.95.3 20010315 (release). When I try to compile using 3.3.2 I get an error with signal.h included from dcigettext.c. When I compile using gcc 2.95.3 I get an error with va_copy. Undefined symbol va_copy first referenced in file connections.o ld: fatal: Symbol referencing errors. No output written to R.bin collect2: ld returned 1 exit status A friend suggested this was an error with stdarg.h and so I searched through the gcc2.95.3 directories and found a version of it and tried to include it using the -I and -L arguments to gcc, but this results in the same error. I've tried using both the R-2.4.0 (R-latest) and R-patched gzip source files. Is there any additional information that you need in order to help me? What steps should I take to solve this problem? Thanks, Denise __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] saving graphics in jpeg format
On Tue, 21 Nov 2006, Paulo Barata wrote: Dear R users, I need to save a graph in jpeg format. After plotting the graph, when the graphics window is active, in the File menu, the Save as / Jpeg / 100% quality correctly saves the graph in jpeg format. But I would like to know, how could I control the resolution (in dpi) of the saved jpeg file? I need to produce a jpeg at 1200 dpi. jpeg files do not have a dpi: they are dimensioned in pixels. I think you mean you want a file at 1200 ppi (pixels per inch), as 'dpi' (dots per inch) is a printer parameter (sometimes also used of monochrome screens). I have tried also the jpeg function in the package grDevices. A simple command like this works correctly: jpeg(filename=test01.jpg, width = 800, height = 600, quality = 100, pointsize = 250) barplot(1:5,col=1:5) dev.off() But I can't figure out the relation between pointsize, pixels, points and dpi. For example, to be specific, to save a graph measuring width = 6 inches, height = 4 inches, at 1200 dpi, which parameters should I use in the jpeg function? width=6*1200, height=4*1200, res=1200 Note that 1200ppi is a very high resolution, and 200dpi is more usual. Please take local advice on this issue, as I believe the problem is a misunderstanding of resolution. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to merge these dataframes
Hi, Having 3 dataframes with different row numbers, but equal column names (see below) I want to merge them by Var1 so I've tried: merge(j1,j2,j3,by=Var1) merge(j,j1,j2,by=names(Var1)) But always got the same message: Erro en fix.by(by.x, x) : 'by' must specify column(s) as numbers, names or logical What I'm doing wrong? Thanks, Antonio j1 Var1 Freq 1 1988-01-131 2 1988-01-161 3 1988-01-203 4 1988-01-252 5 1988-01-301 6 1988-02-015 7 1988-02-084 8 1988-02-141 9 1988-02-161 10 1988-02-184 j2 Var1 Freq 1 1988-01-131 2 1988-01-161 3 1988-01-203 4 1988-01-252 5 1988-01-301 6 1988-02-014 7 1988-02-084 8 1988-02-161 9 1988-02-184 10 1988-02-242 11 1988-03-041 12 1988-03-071 j3 Var1 Freq 1 1988-01-131 2 1988-01-161 3 1988-01-203 4 1988-01-252 5 1988-01-301 6 1988-02-014 7 1988-02-084 8 1988-02-161 9 1988-02-184 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to merge these dataframes
merge() merges *two* data frames, as is very clearly stated on its help page. In your first version, j3 is matching by.x (and the second is wrong, as j does not exist and names() applies to an object). You can do j4 - merge(j1, j2, by=Var1, all=TRUE) j5 - merge(j4, j3, all=TRUE) j5 Var1 Freq.x Freq.y Freq 1 1988-01-13 1 11 2 1988-01-16 1 11 3 1988-01-20 3 33 4 1988-01-25 2 22 5 1988-01-30 1 11 6 1988-02-01 5 44 7 1988-02-08 4 44 8 1988-02-14 1 NA NA 9 1988-02-16 1 11 10 1988-02-18 4 44 11 1988-02-24 NA 2 NA 12 1988-03-04 NA 1 NA 13 1988-03-07 NA 1 NA (or omit all=TRUE if you only want columns which are in all 3 data frames). On Wed, 22 Nov 2006, antonio rodriguez wrote: Hi, Having 3 dataframes with different row numbers, but equal column names (see below) I want to merge them by Var1 so I've tried: merge(j1,j2,j3,by=Var1) merge(j,j1,j2,by=names(Var1)) But always got the same message: Erro en fix.by(by.x, x) : 'by' must specify column(s) as numbers, names or logical What I'm doing wrong? Thanks, Antonio j1 Var1 Freq 1 1988-01-131 2 1988-01-161 3 1988-01-203 4 1988-01-252 5 1988-01-301 6 1988-02-015 7 1988-02-084 8 1988-02-141 9 1988-02-161 10 1988-02-184 j2 Var1 Freq 1 1988-01-131 2 1988-01-161 3 1988-01-203 4 1988-01-252 5 1988-01-301 6 1988-02-014 7 1988-02-084 8 1988-02-161 9 1988-02-184 10 1988-02-242 11 1988-03-041 12 1988-03-071 j3 Var1 Freq 1 1988-01-131 2 1988-01-161 3 1988-01-203 4 1988-01-252 5 1988-01-301 6 1988-02-014 7 1988-02-084 8 1988-02-161 9 1988-02-184 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to merge these dataframes [Solved]
Thanks to everybody. I've found in a previous post too: merge(merge(j1,j2),j3) BR Antonio // Prof Brian Ripley escribió: merge() merges *two* data frames, as is very clearly stated on its help page. In your first version, j3 is matching by.x (and the second is wrong, as j does not exist and names() applies to an object). You can do j4 - merge(j1, j2, by=Var1, all=TRUE) j5 - merge(j4, j3, all=TRUE) j5 Var1 Freq.x Freq.y Freq 1 1988-01-13 1 11 2 1988-01-16 1 11 3 1988-01-20 3 33 4 1988-01-25 2 22 5 1988-01-30 1 11 6 1988-02-01 5 44 7 1988-02-08 4 44 8 1988-02-14 1 NA NA 9 1988-02-16 1 11 10 1988-02-18 4 44 11 1988-02-24 NA 2 NA 12 1988-03-04 NA 1 NA 13 1988-03-07 NA 1 NA (or omit all=TRUE if you only want columns which are in all 3 data frames). On Wed, 22 Nov 2006, antonio rodriguez wrote: Hi, Having 3 dataframes with different row numbers, but equal column names (see below) I want to merge them by Var1 so I've tried: merge(j1,j2,j3,by=Var1) merge(j,j1,j2,by=names(Var1)) But always got the same message: Erro en fix.by(by.x, x) : 'by' must specify column(s) as numbers, names or logical What I'm doing wrong? Thanks, Antonio j1 Var1 Freq 1 1988-01-131 2 1988-01-161 3 1988-01-203 4 1988-01-252 5 1988-01-301 6 1988-02-015 7 1988-02-084 8 1988-02-141 9 1988-02-161 10 1988-02-184 j2 Var1 Freq 1 1988-01-131 2 1988-01-161 3 1988-01-203 4 1988-01-252 5 1988-01-301 6 1988-02-014 7 1988-02-084 8 1988-02-161 9 1988-02-184 10 1988-02-242 11 1988-03-041 12 1988-03-071 j3 Var1 Freq 1 1988-01-131 2 1988-01-161 3 1988-01-203 4 1988-01-252 5 1988-01-301 6 1988-02-014 7 1988-02-084 8 1988-02-161 9 1988-02-184 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] advanced plotting
downunder wrote: Hi all. I need some help. I have to plot so many observation in a coordinate system that you can't see really much. Is there any possiblilty in R to reduce the size of a plotted point? In the plot command i could find a solution. plot(,type = p ,..) thanks in advance lars Hi Lars, You might be interested in the count.overplot function in the plotrix package that displays a single count for each cluster of points. You can adjust the size of the cluster using the tol argument. Jim __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] differences between aov and lme
Hi, we have a split-plot experiment in which we measured the yield of crop fields. The factors we studied were: B : 3 blocks I : 2 main plots for presence of Irrigation V : 2 plots for Varieties N : 3 levels of Nitrogen Each block contains two plots (irrigated or not) . Each plot is divided into two secondary parcels for the two varieties. Each of these parcels is divided into three subplots corresponding to three ordered levels of nitrogen. We found in Venables Ripley (Modern Applied Statistics with S-plus, 3rd edition) the multistratum model for the same type of dataset but for three levels (without the Irrigation partition): aov(Y~N*V+Error(B/V), qr=T) which we adapted to our model: aov(Y~N*V*I+Error(B/V/I)) In Pinheiro Bates (Mixed-effect models in S and S-plus) and as we saw in the message Re: lme and lmer syntax from Ronaldo Reis-Jr, Wed 26 Oct 2005, we fitted also the mixed model : lme(Y~N*V*I, random~1|B/V/I) On a random simulated response Y, we didn't obtain similar results - only one factor with the same F-value - from the aov function and the lme one (oppositely to the example used by Venables Ripley and Pinheiro Bates). Is there a mistake in one of our two models or an explanation of this difference? Thanks a lot in advance. Caroline Domerg and Frederic Chiroleu UMR 53 PVBMT (Peuplements Vegetaux et Bio-agresseurs en Milieu Tropical) CIRAD Pôle de Protection des Plantes (3P) - Saint-Pierre Ile de la Réunion __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] differences between aov and lme
The way you describe this, it would appear to be B/I/V, not B/V/I. From the footer: PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Had you given us access to the data, it would have been much easier to answer the question: if you need further help please do so. On Wed, 22 Nov 2006, domerg wrote: Hi, we have a split-plot experiment in which we measured the yield of crop fields. The factors we studied were: B : 3 blocks I : 2 main plots for presence of Irrigation V : 2 plots for Varieties N : 3 levels of Nitrogen Each block contains two plots (irrigated or not) . Each plot is divided into two secondary parcels for the two varieties. Each of these parcels is divided into three subplots corresponding to three ordered levels of nitrogen. We found in Venables Ripley (Modern Applied Statistics with S-plus, 3rd edition) the multistratum model for the same type of dataset but for three levels (without the Irrigation partition): aov(Y~N*V+Error(B/V), qr=T) which we adapted to our model: aov(Y~N*V*I+Error(B/V/I)) In Pinheiro Bates (Mixed-effect models in S and S-plus) and as we saw in the message Re: lme and lmer syntax from Ronaldo Reis-Jr, Wed 26 Oct 2005, we fitted also the mixed model : lme(Y~N*V*I, random~1|B/V/I) On a random simulated response Y, we didn't obtain similar results - only one factor with the same F-value - from the aov function and the lme one (oppositely to the example used by Venables Ripley and Pinheiro Bates). Is there a mistake in one of our two models or an explanation of this difference? Thanks a lot in advance. Caroline Domerg and Frederic Chiroleu UMR 53 PVBMT (Peuplements Vegetaux et Bio-agresseurs en Milieu Tropical) CIRAD Pôle de Protection des Plantes (3P) - Saint-Pierre Ile de la Réunion __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595__ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problem with the plm package
Hi all, I have a problem in installing and using the plm package using R 2.2.0 on windows xp. I installed it from a .zip file downloaded from the CRAN. Apparently everything is ok: utils:::menuInstallLocal() package 'plm' successfully unpacked and MD5 sums checked updating HTML package descriptions However, when I try to load it: library(plm) Errore in lazyLoadDBfetch(key, datafile, compressed, envhook) : ReadItem: tipo 241 sconosciuto Inoltre: Warning message: il pacchetto 'plm' è stato creato con R versione 2.4.0 Errore: caricamento pacchetto/namespace fallito per 'plm' Sorry I installed R in Italian, but that should be easy to understand. Errore means, of course, error and sconosciuto means unknown. Basically the package cannot be loaded because there is an unknowwn item (at least, that's what I understand). Does anybody have an idea of the problem? Thank you, Giangiacomo __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] odd behaviour of %%?
Dear R Helpers, I am trying to extract the modulus from divisions by a sequence of fractions. I noticed that %% seems to behave inconsistently (to my untutored eye), thus: 0.1%%0.1 [1] 0 0.2%%0.1 [1] 0 0.3%%0.1 [1] 0.1 0.4%%0.1 [1] 0 0.5%%0.1 [1] 0.1 0.6%%0.1 [1] 0.1 0.7%%0.1 [1] 0.1 0.8%%0.1 [1] 0 0.9%%0.1 The modulus for 0.1, 0.2, 0.4 and 0.8 is zero, as I'd expect. But, the modulus for 0.3, 0.6, 0.7 and 0.9 is 0.1 - which I did not expect. I can see no obvious rule that predicts whether x%%0.1 will give an answer of 0 or 0.1. I could find no explanation of the way that %% works in the R manuals. So, I have 3 questions:- 1) Why is the modulus of 0.3%%0.1 (and 0.5%%0.1 and 0.6%%0.1...) not zero? 2) Are there any algorithms in R that use the %% operator with fractional divisors in this way, and do they know about its apparently inconsistent behaviour? 3) If %% is not intended for use with fractional divisors, then would it be a good idea to trap attempts to use them? Thanks, in advance, for your help Jonathan Williams __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with the plm package
You downloaded a Windows binary package for R 2.4.0 and tried to load it into R 2.2.0. That does not work, as you found out. You need to install from the source package, or (better) update your version of R (as the posting guide asked you to do before posting). On Wed, 22 Nov 2006, Giangiacomo Bravo wrote: Hi all, I have a problem in installing and using the plm package using R 2.2.0 on windows xp. I installed it from a .zip file downloaded from the CRAN. Apparently everything is ok: utils:::menuInstallLocal() package 'plm' successfully unpacked and MD5 sums checked updating HTML package descriptions However, when I try to load it: library(plm) Errore in lazyLoadDBfetch(key, datafile, compressed, envhook) : ReadItem: tipo 241 sconosciuto Inoltre: Warning message: il pacchetto 'plm' è stato creato con R versione 2.4.0 Errore: caricamento pacchetto/namespace fallito per 'plm' Sorry I installed R in Italian, but that should be easy to understand. Errore means, of course, error and sconosciuto means unknown. Basically the package cannot be loaded because there is an unknowwn item (at least, that's what I understand). Does anybody have an idea of the problem? Thank you, Giangiacomo __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595__ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Mango Solutions Announces R Public Training Course in Paris
Mango Solutions are pleased to announce the above course in Paris as part of our schedule for Q1 2007. --- Introduction to R and R Programming - 12th February 2008-14th February --- (Please find a french version of this announcement from http://www.mango-solutions.com/services/rtraining/r_paris_training_07_french.html) * Who Should Attend ? This is a course suitable for beginners and improvers in the R language and is ideal for people wanting an all round introduction to R * Course Goals - To allow attendees to understand the technology behind the R package - Improve attendees programming style and confidence - To enable users to access a wide range of available functionality - To enable attendees to program in R within their own environment - To understand how to embed R routines within other applications * Course Outline 1. Introduction to the R language and the R community 2. The R Environment 3. R data objects 4. Using R functions 5. The apply family of functions 6. Writing R functions 7. Standard Graphics 8. Advanced Graphics 9. R Statistics 10. R Applications The cost of these courses is €1800 for commercial attendees and €850 for academic attendees. A €100 discount will be offered to members of the R Foundation (http://www.r-project.org/foundation/main.html). Should your organization have more than 3 possible attendees why not talk to us about hosting a customized and focused course delivered at your premises? Details of further courses in alternative locations are available at http://www.mango-solutions.com/services/training.html More details about this course : - in french : http://www.mango-solutions.com/services/rtraining/r_paris_training_07_french.html - in english : http://www.mango-solutions.com/services/rtraining/r_paris_training_07.html Should you want to book a place on this course or have any questions please contact [EMAIL PROTECTED] Cordialement, Romain Francois -- Mango Solutions Tel +44 (0)1249 467 467 Mob +44 (0)7813 526 123 Fax +44 (0)1249 467 468 data analysis that delivers __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] odd behaviour of %%?
Jonathan Williams [EMAIL PROTECTED] writes: Dear R Helpers, I am trying to extract the modulus from divisions by a sequence of fractions. I noticed that %% seems to behave inconsistently (to my untutored eye), thus: 0.1%%0.1 [1] 0 0.2%%0.1 [1] 0 0.3%%0.1 [1] 0.1 0.4%%0.1 [1] 0 0.5%%0.1 [1] 0.1 0.6%%0.1 [1] 0.1 0.7%%0.1 [1] 0.1 0.8%%0.1 [1] 0 0.9%%0.1 The modulus for 0.1, 0.2, 0.4 and 0.8 is zero, as I'd expect. But, the modulus for 0.3, 0.6, 0.7 and 0.9 is 0.1 - which I did not expect. I can see no obvious rule that predicts whether x%%0.1 will give an answer of 0 or 0.1. I could find no explanation of the way that %% works in the R manuals. So, I have 3 questions:- 1) Why is the modulus of 0.3%%0.1 (and 0.5%%0.1 and 0.6%%0.1...) not zero? 2) Are there any algorithms in R that use the %% operator with fractional divisors in this way, and do they know about its apparently inconsistent behaviour? 3) If %% is not intended for use with fractional divisors, then would it be a good idea to trap attempts to use them? These are not fractions but floating point numbers. See FAQ 7.31 http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f and the references therein for reasons why 0.3 is not an integer multiple of 0.1 in binary. etc. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] odd behaviour of %%?
To answer part 1) of your question, see point 7.31 in the R FAQ (http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these- numbers-are-equal_003f). 0.3 - 3*0.1 [1] -5.551115e-17 It always amazes me in how many different guises the problem of floating point representation crops up. HTH -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Jonathan Williams Sent: 22 November 2006 12:15 PM To: Ethz. Ch Subject: [R] odd behaviour of %%? Dear R Helpers, I am trying to extract the modulus from divisions by a sequence of fractions. I noticed that %% seems to behave inconsistently (to my untutored eye), thus: 0.1%%0.1 [1] 0 0.2%%0.1 [1] 0 0.3%%0.1 [1] 0.1 0.4%%0.1 [1] 0 0.5%%0.1 [1] 0.1 0.6%%0.1 [1] 0.1 0.7%%0.1 [1] 0.1 0.8%%0.1 [1] 0 0.9%%0.1 The modulus for 0.1, 0.2, 0.4 and 0.8 is zero, as I'd expect. But, the modulus for 0.3, 0.6, 0.7 and 0.9 is 0.1 - which I did not expect. I can see no obvious rule that predicts whether x%%0.1 will give an answer of 0 or 0.1. I could find no explanation of the way that %% works in the R manuals. So, I have 3 questions:- 1) Why is the modulus of 0.3%%0.1 (and 0.5%%0.1 and 0.6%%0.1...) not zero? 2) Are there any algorithms in R that use the %% operator with fractional divisors in this way, and do they know about its apparently inconsistent behaviour? 3) If %% is not intended for use with fractional divisors, then would it be a good idea to trap attempts to use them? Thanks, in advance, for your help Jonathan Williams __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Nedbank Limited Reg No 1951/09/06. The following link displays the names of the Nedbank Board of Directors and Company Secretary. [ http://www.nedbank.co.za/terms/DirectorsNedbank.htm ] This email is confidential and is intended for the addressee only. The following link will take you to Nedbank's legal notice. [ http://www.nedbank.co.za/terms/EmailDisclaimer.htm ] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Dealing with NAs
Dear R Users,
Once the function svm ( ) in library{e1071} like any function in R is run
with some variables including missing values ( NA ), it is needed to include
na.pass=TRUE arguement in function in order to deal with NAs. The arguement
na.omit is not desired, since it redueces the length of variables.
A problem when running the function svm including na.pass, is that the
result of fitting consists a part of seemingly rigth fiitted values from the
first of the length of fitted values to somewhere in a series and another part
of fitted values which are NAs comes to the end of length of fitted values in
a series.
But we expect fitted values without any NA.
How it can be corrected ?
Thank you so much for help.
Amir
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Statistical Software Comparison
Some comments that I hope may be of use - Yes the STATA manuals are very good. At $495 per set they should be. Buying individual volumes is not an option as there are many cross-references between different volumes and the price of the set is discounted relative to buying individual volumes. The online manuals are just a basic summary of the syntax of the commands and are terrible. Other commercial packages that I know (e.g. RATS, EVIEWS, MATLAB, GAUSS, ) have much better help files and distribute their manuals in pdf format along with the product. I have switched to R for most of my econometrics partly for reason of $$ but also for the simple reason that I can get my analysis done in R. For simple econometrics I do return to simpler products or a product that has a specific purpose (eg GRETL for basic econometrics , RATS for time series analysis etc.). R and STATA are both superb products. Both are so large that most people will be familiar with only a small part of each. Both are programmable. Modern or experimental methods can be implemented in both. Which programming language you prefer is probably a matter of taste. If you can find a person who is working in your field and is familiar with one of the languages and is willing to help you then my advice would be to use his expertise regardless of the package that he is using. One area where R is much better than STATA would be in finance where R has the advantage of the superb package Rmetrics. Any person choosing between R and STATA should look for similar packages in their expertise. The Task Views on CRAN are very good summaries or R facilities in the areas of econometrics and finance. I have added empirical content to a statistics course for mathematics masters students using R. I teach empirical econometrics using STATA and MATLAB (for institutional reasons) but would prefer yo use R. On 22/11/06, Robert Duval [EMAIL PROTECTED] wrote: I'll answer the ones I know of: 4. Manuals (theory included in the manuals). Stata manuals are superb. The online help manuals are really minimal, but the complete set of manuals for sale is really good. Not only they discuss the Stata implementetation, but they give a concise theoretical discussion of what the statistical methods are actually doing. While they don't get to talk much about the inner workings of Stata, (as some of the R manuals do) I like them much better to the R ones. Many of the statistical procedures are illustrated with examples using the datasets included with the software 5. Support (in this aspect there is no comparison with R, the R list is the best known support). R list has a better support than Statalist, but still Statalist is quite active and helpful. Plus they are more polite... no RTFM or stuff like that. If you own a Stata license, you can get direct support from somebody at StataCorp (in addition to Statalist). This is specially relevant if you have questions on how Stata is estimating something, bugs, etc. 6. Numerical stability. Quite stable. The only glitch I've observed is that after new releases their routines are not very reliable... meaning they sometimes change the way something is being computed and might they mess up something that previously was running fine. Right now Stata 9 is pretty stable, but if Stata 10 would come up in the market now, I would probably wait for a couple of months and make sure everything is well tested. One last thing, while I abandoned the Stata world to move to R (due to $$), I have to say that the only thing I really miss about it is its ability to handle large datasets. Stata comes with great Data management routines, and it can hold large amount of data in its memory. Here R is light years behind. This is particularly relevant if you have to clean-up large datasets before you actually start doing statistics. hope this helps robert On 11/21/06, Kenneth Cabrera [EMAIL PROTECTED] wrote: Hi R users: I want to know if any of you had used Stata or Statgraphics. What are the advantages and disadvantages with respect to R on the following aspects? 1. Statistical functions or options for advanced experimental design (fractional, mixed models, greco-latin squares, split-plot, etc). 2. Bayesian approach to experimental design. 3. Experimental design planing options. 4. Manuals (theory included in the manuals). 5. Support (in this aspect there is no comparison with R, the R list is the best known support). 6. Numerical stability. 7. Implementation of modern statistical approaches. Thank you for your help. Kenneth -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problem with RWeka-rJava packages
Hello: I´m trying to execute Apriori(file.arff) command of RWeka package. I´m working with: Operating System: Windows XP home R-2.4.0 RWeka_0.2-11 rJava_0.4-11 classpath= .;C:\Archivos de programa\Java\jdk1.5.0\lib;C:\Archivos de programa\R\R- 2.4.0\library\RWeka\jar An error occurs when .jnew command is executed, on class weka/core/Instances : .jnew(weka/core/Instances) Failed to create object of class ´weka/core/Instances´ and a warning message: createObject.GetMethodID(weka/core/Instances,()V) failed What can I do? Thanks everybody. Alejandra Malberti [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] odd behaviour of %%?
On 22-Nov-06 Jonathan Williams wrote: Dear R Helpers, I am trying to extract the modulus from divisions by a sequence of fractions. I noticed that %% seems to behave inconsistently (to my untutored eye), thus: 0.1%%0.1 [1] 0 0.2%%0.1 [1] 0 0.3%%0.1 [1] 0.1 0.4%%0.1 [1] 0 0.5%%0.1 [1] 0.1 0.6%%0.1 [1] 0.1 0.7%%0.1 [1] 0.1 0.8%%0.1 [1] 0 0.9%%0.1 This is yet another manifestation of the fact that R (like most computer languages) stores numbers as binary representations of fixed length. This is OK for inetegers up to a certain maximum value, but relatively few factions (those which are multiples of powers of 1/2) will be stored exactly. In particular, 0.1 is not stored exactly. When you do 0.2 %% 0.1, you will be getting the remainder of twice the binary representation of 0.1, modulo 0.1, which will be zero; and similarly for 0.4 and 0.8, since these are going up by powers of 2 and the relationships between their binary representations will match what you would expect. However, the binary representation of 0.3 does not correspond exactly to 3 times the binary representation of 0.1: 0.3 - 3*0.1 [1] -5.551115e-17 so 3*0.1 is represented by a binary fraction slightly greater than the binary representation of 0.3, so 0.1 only goes into 0.3 twice, with a remainder slightly less than 0.1 which rounds to 0.1 when displayed as a result: 0.3 %% 0.1 [1] 0.1 0.1 - (0.3 %% 0.1) [1] 2.775558e-17 Similarly for 0.5, 0.6, 0.7 and 0.9. The modulus for 0.1, 0.2, 0.4 and 0.8 is zero, as I'd expect. But, the modulus for 0.3, 0.6, 0.7 and 0.9 is 0.1 - which I did not expect. I can see no obvious rule that predicts whether x%%0.1 will give an answer of 0 or 0.1. I could find no explanation of the way that %% works in the R manuals. So, I have 3 questions:- 1) Why is the modulus of 0.3%%0.1 (and 0.5%%0.1 and 0.6%%0.1...) not zero? See above. 2) Are there any algorithms in R that use the %% operator with fractional divisors in this way, I don't know -- though others will! and do they know about its apparently inconsisten behaviour? People who write algorithms should be aware of such effects of binary representation, so a well-written algorithm will take account of this danger. 3) If %% is not intended for use with fractional divisors, then would it be a good idea to trap attempts to use them? The best protection against this kind of thing is circumspection on the part of the programmer -- who, therefore should do his own trapping. An alternative aproach to the calculations you made could be 0.3 - 0.1*(0.3/0.1) [1] 0 which is algebraically equivalent, and gives the right answer: identical(0.3 - 0.1*(0.3/0.1),0) [1] TRUE However, this is not always going to work either: 101.1 - (101.1/0.1)*0.1 [1] -1.164153e-10 so it's prudent to wrap it in a suitable round(): round(101.1 - (101.1/0.1)*0.1, digits=9) [1] 0 but you still have to be aware of what value to set the number of digits to round to: round(101.1 - (101.1/0.1)*0.1, digits=10) [1] -1e-10 Hoping this helps, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 22-Nov-06 Time: 11:33:05 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] XEmacs: substitute \n by real linefeeds
Hi, This may be a useful tip for R users on emacs: Problem: substitute \n by a real linefeed, so that aa \n bb \n ccc becomes aa bb ccc ? Solution: M-% RT \n RT C-q 12 RT It works, because ASCII 12 = RT Enjoy! Christian -- Dr. Christian W. Hoffmann, Swiss Federal Research Institute WSL Zuercherstrasse 111, CH-8903 Birmensdorf, Switzerland Tel +41-44-7392-277 (office), -111(exchange), -215 (fax) [EMAIL PROTECTED], www.wsl.ch/staff/christian.hoffmann __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lme - plot - labels
Hello there, I am using the 'nlme' package to analyse among group and in between group variances. I do struggle a little using the respective plot-functions. I created a grouped object: group.lme - groupedData(obsday ~ oro | id, data=read.table(data-lme.txt, header=T), labels=list(x = Day of Year, y = ID number)) When I plot, however plot(group.lme) the y-labels appear on the x-axis and the x-labels do not are not drawn at all. Any advice? I am using R 2.2.1 on Windows XP Thanks, Daniel __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Latent models
Hai, Can anyone help me with some literature (R related) about latent models? Thanx, Wilfred __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] RBloomberg Multi-ticker problem
Bloomberg recommends getting only one ticker at a time for intraday
data.
David L. Reiner
Rho Trading Securities, LLC
Chicago IL 60605
312-362-4963
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Shubha Vishwanath
Karanth
Sent: Wednesday, November 22, 2006 1:09 AM
To: r-help@stat.math.ethz.ch; [EMAIL PROTECTED]
Subject: [R] RBloomberg Multi-ticker problem
Hi,
I am trying to download data from Bloomberg through R. If I try to
download intraday data for multiple tickers and only one field, I get
the error, written below in red. How do I get rid of this error?
dat-blpGetData(conn, c(NOK1V FH Equity,AUA AV Equity),
LAST_PRICE,
start=as.chron(as.Date(2006-9-13,%Y-%m-%d)),barfields=VOLUME,
barsize=10,retval=data.frame)
Error in if (typ[n] == character) { : missing value where TRUE/FALSE
needed
* blpDisconnect(conn)
Thank you,
Shubha.
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Re: [R] Fitting mixed-effects models with lme with fixed error term variances
On 11/21/06, Gregor Gorjanc [EMAIL PROTECTED] wrote:
Douglas Bates bates at stat.wisc.edu writes:
...
Can you be more specific about which parameters you want to fix and
which you want to vary in the optimization?
It would be nice to have the ability to fix all variances i.e. variances of
random effects.
That gets tricky in terms of the parameterization of the
variance-covariance matrices for vector-valued random effects. These
matrices are not expressed in the conventional parameterization of
variances and covariances or even variances and correlation because
the conditions for the resulting matrix to be positive definite are
not simple bounds or easily expressed transformations then the matrix
is larger than 2 by 2. I suppose what I could do is to allow these
matrices to be specified in the parameterization that is used in the
optimization and provide a utility function to map from the
conventional parameters to these. That would mean that you couldn't
fix ,say, the variance of the intercept term for vector-valued random
effects but allow the variance of a slope for the same grouping factor
to be estimated. Well, you could but only in the fortune(Yoda)
sense.
By the way, if you fix all the variances then what are you optimizing
over? The fixed effects? In that case the solution can be calculated
explicitly for a linear mixed model. The conditional estimates of the
fixed effects given the variance components are the solution to a
penalized linear least squares problem. (Yes, the solution can also
be expressed as a generalized linear least squares problem but there
are advantages to using the penalized least squares representation.
See
@Article{bates04:_linear,
author = {Douglas M. Bates and Saikat DebRoy},
title ={Linear mixed models and penalized least squares},
journal = {Journal of Multivariate Analysis},
year = 2004,
volume = 91,
number = 1,
pages ={1--17}
}
)
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Re: [R] Latent models
Is ltm what you want? packageDescription(ltm) Package: ltm Title: Latent Trait Models under IRT Version: 0.6-1 Date: 2006-10-02 Author: Dimitris Rizopoulos [EMAIL PROTECTED] Maintainer: Dimitris Rizopoulos [EMAIL PROTECTED] Description: Analysis of multivariate dichotomous and polytomous data using latent trait models under the Item Response Theory approach. It includes the Rasch, the Two-Parameter Logistic, the Birnbaum's Three-Parameter, and the Graded Response Models. Depends: R(= 2.3.0), MASS, gtools, msm LazyLoad: yes LazyData: yes License: GPL version 2 or later URL: http://wiki.r-project.org/rwiki/doku.php?id=packages:cran:ltm Packaged: Tue Oct 3 09:07:38 2006; dimitris Built: R 2.4.0; ; 2006-10-03 17:16:04; windows If you are looking function for Latent Class Analysis (LCA), then lca in package e1071 is what you want. On 11/22/06, Bi-Info (http://members.home.nl/bi-info) [EMAIL PROTECTED] wrote: Hai, Can anyone help me with some literature (R related) about latent models? Thanx, Wilfred __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ronggui Huang Department of Sociology Fudan University, Shanghai, China 黄荣贵 复旦大学社会学系 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] What training algorithm does nnet package use?
Greetings list, I've just swapped from the neural package to the nnet package and I've noticed that the training is orders of magnitude faster, and the results are way more accurate. This leads me to wonder, what training algorithm is nnet using? Is it a modification on the standard backpropagation? Or a completely different algorithm? I'm trying to account for the speed differences between neural and nnet, and the documentation on the nnet package is rather sparse on what training algorithm is used (either that, or I'm getting blind and missed it totally). Any help would be much appreciated. Regards, Wee-Jin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problem with the plm package (2)
Thanks a lot for the quick answerd I received and that helped me to solve my first (simple) problem with the plm package. Unfortunatley, here's another one: I do not know why, but I'm unable to estimate a simple model in the form y ~ 1 + x When I try zz - plm(y ~ 1 + x , data=mydata) I obtain Errore in X.m[, coef.within, drop = F] : numero di dimensioni errato which means that the number of dimensions is wrong in X.m[, coef.within, drop = F]. However, I have no problem to estimate a more complex model, e.g. zz - plm(y ~ 1 + x + I(x^2), data=mydata) in this case everything is ok. What's wrong? Thank you, Giangiacomo __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Latent models
Those are only a few of the possible latent models. Factor analysis and SEM are also latent variable models, as indeed are the mixed models of nlme, lme4, mmlcr, polr models Then there are the graphical models of (at least) packages ggm and latentnet, polytomous LDA in package polLCA help.search(latent, agrep=FALSE) on a comprehensive installation helped me find some of these. On Wed, 22 Nov 2006, ronggui wrote: Is ltm what you want? packageDescription(ltm) Package: ltm Title: Latent Trait Models under IRT Version: 0.6-1 Date: 2006-10-02 Author: Dimitris Rizopoulos [EMAIL PROTECTED] Maintainer: Dimitris Rizopoulos [EMAIL PROTECTED] Description: Analysis of multivariate dichotomous and polytomous data using latent trait models under the Item Response Theory approach. It includes the Rasch, the Two-Parameter Logistic, the Birnbaum's Three-Parameter, and the Graded Response Models. Depends: R(= 2.3.0), MASS, gtools, msm LazyLoad: yes LazyData: yes License: GPL version 2 or later URL: http://wiki.r-project.org/rwiki/doku.php?id=packages:cran:ltm Packaged: Tue Oct 3 09:07:38 2006; dimitris Built: R 2.4.0; ; 2006-10-03 17:16:04; windows If you are looking function for Latent Class Analysis (LCA), then lca in package e1071 is what you want. On 11/22/06, Bi-Info (http://members.home.nl/bi-info) [EMAIL PROTECTED] wrote: Hai, Can anyone help me with some literature (R related) about latent models? Thanx, Wilfred __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problems with garchFit
Hi all,
I post it on both r-help and r-finance since I don't know where is most
appropriate for this topic. Sorry if it bothers you.
I did garch fitting on SP500 monthly returns with garchFit from fSeries. I
got same coefficients from all cond.dist except normal. I thought that is
probabaly usual for the data. But when I play with it, I got another
question.
I plot skew normal with skew = 1 and a standard normal, they overlap
eachother, so I think they are the same. Skew = 1 means no skewness (I can
not find the paper defining the distribution).
library(fSeries)
curve(dsnorm(x, 0, 1, 1), -2, 2, add = F, col = 'red') #skew normal with
skew 1
curve(dnorm(x, 0, 1), -2, 2, add = T, col = 'blue') #normal
Then I try them as innovations,
#normal innovation
garch_norm - garchFit(series = logr, include.mean = F)
#skew normal innovation
#this line do not include skew, so it got same result as normal
garch_snorm - garchFit(series = logr, cond.dist = 'dsnorm', include.mean =
F, include.skew = F)
#this line includes skew, but use default skew = 1, and it got results
different from normal, which I don't understand
garch_snorm - garchFit(series = logr, cond.dist = 'dsnorm', include.mean =
F, include.skew = T)
Have I done something wrong? I am attaching the code, thank you.
Tian
#GARCH analysis of monthly return
rm(list=ls(all=TRUE))
sp500 - read.csv('sp_m90.csv', header=TRUE)
sp500 - sp500[,2] #only adjusted close
n - length(sp500)
logr - log(sp500[1:n-1] / sp500[2:n])
acf(logr)
ar5 - arima(logr, order = c(5, 0, 0), include.mean = T)
logr- ar5$res
acf(logr)
#fit GARCH distribution
hist(logr, freq = F, ylim = c(0, 12), breaks = 'FD')
norm_fit - normFit(logr)
curve(dnorm(x, norm_fit$est[1], norm_fit$est[2]), -.15, .15, add = TRUE,
col=2)
t_fit - stdFit(logr)
curve(dstd(x, t_fit$est[1], t_fit$est[2], t_fit$est[3]), -.15, .15, add =
TRUE, col=6)
snorm_fit - snormFit(logr)
curve(dsnorm(x, snorm_fit$est[1], snorm_fit$est[2], snorm_fit$est[3]), -.25,
.15, add = TRUE, col=4)
st_fit - sstdFit(logr)
curve(dsstd(x, st_fit$est[1], st_fit$est[2], st_fit$est[3], st_fit$est[4]),
-.25, .15, add = TRUE, col=3)
library(fSeries)
#normal innovation
garch_norm - garchFit(series = logr, include.mean = F)
#t inovation
garch_t - garchFit(series = logr, cond.dist = 'dstd', include.mean = F,
include.shape = T)
garch_t1 - garchFit(series = logr, cond.dist = 'dstd', include.mean = F,
shape = t_fit$est[3], include.shape = T)
#skew normal innovation
garch_snorm - garchFit(series = logr, cond.dist = 'dsnorm', include.mean =
F, include.skew = T)
garch_snorm1 - garchFit(series = logr, cond.dist = 'dsnorm', include.mean =
F, skew = snorm_fit$est[3], include.skew = T)
#skew t innovation
garch_st - garchFit(series = logr, cond.dist = 'dsstd', include.mean = F,
include.skew = T, include.shape = T)
garch_st1 - garchFit(series = logr, cond.dist = 'dsstd', include.mean = F,
skew = st_fit$est[4], shape = st_fit$est[3], include.skew = T, include.shape= T)
vix - read.csv('D:/Documents and Settings/Mu Tian/Desktop/8780/8780
project/vix_m.csv', header=TRUE)
vix - (vix[,2]/100) / (12^.5)
plot_sd - function(x, ylim = null, col = null, ...) {
xcsd = [EMAIL PROTECTED]
plot(xcsd, type = l, col = col, ylab = x,
main = Conditional SD, ylim = ylim)
abline(h = 0, col = grey, lty = 3)
grid()
}
plot_sd(garch_norm, ylim = c(0.02, 0.13), col = 2)
xcsd = [EMAIL PROTECTED]
lines(xcsd, col = 3)
lines(1:n, vix)
#predict
predict(garch_norm)
predict(garch_t)
#demonstration of skew distributions
#skew normal
curve(dsnorm(x, 0, 1, .1), -2, 2, add = F, col = 'green')
curve(dsnorm(x, 0, 1, snorm_fit$est[3]), type = 'l', col = 'blue', add = T)
curve(dsnorm(x, 0, 1, 1), -2, 2, add = T, col = 'red') #normal
#skew t
curve(dsstd(x, 0, 1, 4, 1), -2, 2, add = F, col = 'red')
curve(dsstd(x, 0, 1, st_fit$est[3], st_fit$est[4]), type = 'l', col =
'blue', add = T)
curve(dsstd(x, 0, 1, 100, .5), -2, 2, add = T, col = 'green')
#t
curve(dstd(x, 0, 1, 4), -2, 2, add = T, col = 'red')
curve(dstd(x, 0, 1, t_fit$est[3]), type = 'l', col = 'blue', add = T)
curve(dstd(x, 0, 1, 100), -2, 2, add = T, col = 'green')
curve(dsnorm(x, 0, 1, 1), -2, 2, add = F, col = 'red') #normal
curve(dnorm(x, 0, 1), -2, 2, add = T, col = 'blue') #normal
curve(dsnorm(x, 0, 1, .1), -2, 2, add = T, col = 'green') #normal
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fitting mixed-effects models with lme with fixed error term variances
Douglas Bates wrote: On 11/21/06, Gregor Gorjanc [EMAIL PROTECTED] wrote: Douglas Bates bates at stat.wisc.edu writes: ... Can you be more specific about which parameters you want to fix and which you want to vary in the optimization? It would be nice to have the ability to fix all variances i.e. variances of random effects. That gets tricky in terms of the parameterization of the variance-covariance matrices for vector-valued random effects. These matrices are not expressed in the conventional parameterization of variances and covariances or even variances and correlation because the conditions for the resulting matrix to be positive definite are not simple bounds or easily expressed transformations then the matrix is larger than 2 by 2. I suppose what I could do is to allow these matrices to be specified in the parameterization that is used in the optimization and provide a utility function to map from the conventional parameters to these. That would mean that you couldn't fix ,say, the variance of the intercept term for vector-valued random effects but allow the variance of a slope for the same grouping factor to be estimated. Well, you could but only in the fortune(Yoda) sense. Yes, I agree here. Thank you for the detailed answer. By the way, if you fix all the variances then what are you optimizing over? The fixed effects? In that case the solution can be calculated explicitly for a linear mixed model. The conditional estimates of the fixed effects given the variance components are the solution to a penalized linear least squares problem. (Yes, the solution can also be expressed as a generalized linear least squares problem but there are advantages to using the penalized least squares representation. Yup. It would really be great to be able to do that nicely in R, say use lmer() once and since this might take some time use estimates of variance components next time to get fixed and random effects. Is this possible with lmer or any related function - not in fortune(Yoda) sense ;) -- Lep pozdrav / With regards, Gregor Gorjanc -- University of Ljubljana PhD student Biotechnical Faculty Zootechnical Department URI: http://www.bfro.uni-lj.si/MR/ggorjan Groblje 3 mail: gregor.gorjanc at bfro.uni-lj.si SI-1230 Domzale tel: +386 (0)1 72 17 861 Slovenia, Europefax: +386 (0)1 72 17 888 -- One must learn by doing the thing; for though you think you know it, you have no certainty until you try. Sophocles ~ 450 B.C. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Probit analysis
Respected Sir/Madam, I have a question regarding calculation of LD50 (Lethal Dose) and IC50 (50% inhibitory concentration) of an antimicrobial experiment. I have used a compound isolated from a plant and observed its effect on the fungus *Fusarium oxysporum* by the food poisoning method. Solutions of the compound at concentrations of 0, 50, 100, 150, 200 and 250µg/ ml were added to sterilized medium in Petri dish (9cm dia.) and after transferring the mycelia of fungal strains, the dishes were incubated in the dark. When the mycelium of fungi reached the edges of the control (0µg/ ml) dishes, the antifungal indices were calculated. Each test was repeated three times and the average was calculated. The formula to calculate antifungal index (AI) was AI % = (1- Da / Db) * 100 Where Da = diameter of growth zone in experiment (cm); Db = diameter of growth zone in control (cm). The results are as follows Concentration (µg/ ml) 0 50 100 150 200 250 AI% (average of 3 replicates) 0 10 21.59 33.89 47.96 59.93 I have the Version 2.3.1 (2006-06-01) of R loaded in my computer and want to calculate LD50 and if possible IC50 values with the data. I have checked the Help menu---Manuals in PDF---An introduction to R. There is an example of calculating LD50 in Chapter 11, page no. 63. But I cannot fit my data in the modelsome error messages are coming. Please suggest some functions on R or the method by which I can do the calculations. I am not at all used to programming language so a detailed solution would be very much helpful. Thanking you in anticipation. Sincerely yours. -- Dipjyoti Chakraborty C/o Dr Adinpunya Mitra Natural Products Biotechnology Lab Agriculture Food Technology Depart. IIT-KGP Kharagpur-721 302-India [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Summary, was Re: Confidence interval for relative risk
At 14:43 10/11/2006, Michael Dewey wrote: After considerable help from list members and some digging of my own I have prepared a summary of the findings which I have posted (see link below). Broadly there were four suggestions 1 - Wald-type intervals, 2 - transforming the odds ratio confidence interval, 3 - method based on score test, 4 - method based on likelihood. Method 3 was communicated to me off-list === I haven't followed all of the thread either but has someone already pointed out to you confidence intervals that use the score method? For example Agresti (Categorical Data Analysis 2nd edition, p. 77-78) note that 'although computationally more complex, these methods often perform better'. However, they also note that 'currently they are not available in standard software'. But then again, R is not standard software: the code (riskscoreci) can be found from here: http://www.stat.ufl.edu/~aa/cda/R/two_sample/R2/index.html best regards, Jukka Jokinen and so I reproduce it here. Almost a candidate for the fortunes package there. The other three can be found from the archive under the same subject although not all in the same thread. Methods 3 and 4 seem to have more going for them as far as I can judge. Thanks to David Duffy, Spencer Graves, Jukka Jokinen, Terry Therneau, and Wolfgan Viechtbauer. The views and calculations presented here and in the summary are my own and any errors are my responsibility not theirs. The summary document is available from here http://www.zen103156.zen.co.uk/rr.pdf Original post follows. The concrete problem is that I am refereeing a paper where a confidence interval is presented for the risk ratio and I do not find it credible. I show below my attempts to do this in R. The example is slightly changed from the authors'. I can obtain a confidence interval for the odds ratio from fisher.test of course === fisher.test example === outcome - matrix(c(500, 0, 500, 8), ncol = 2, byrow = TRUE) fisher.test(outcome) Fisher's Exact Test for Count Data data: outcome p-value = 0.00761 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 1.694792 Inf sample estimates: odds ratio Inf === end example === but in epidemiology authors often prefer to present risk ratios. Using the facility on CRAN to search the site I find packages epitools and Epi which both offer confidence intervals for the risk ratio === Epi example === library(Epi) twoby2(outcome[c(2,1),c(2,1)]) 2 by 2 table analysis: -- Outcome : Col 1 Comparing : Row 1 vs. Row 2 Col 1 Col 2P(Col 1) 95% conf. interval Row 1 8 500 0.01570.0079 0.0312 Row 2 0 500 0.0. NaN 95% conf. interval Relative Risk:Inf NaN Inf Sample Odds Ratio:Inf NaN Inf Conditional MLE Odds Ratio:Inf1.6948 Inf Probability difference: 0.01570.0027 0.0337 Exact P-value: 0.0076 Asymptotic P-value: NaN -- === end example === So Epi gives me a lower limit of NaN but the same confidence interval and p-value as fisher.test === epitools example === library(epitools) riskratio(outcome) $data Outcome Predictor Disease1 Disease2 Total Exposed1 5000 500 Exposed2 5008 508 Total10008 1008 $measure risk ratio with 95% C.I. Predictor estimate lower upper Exposed11NANA Exposed2 Inf NaN Inf $p.value two-sided Predictor midp.exact fisher.exact chi.square Exposed1 NA NA NA Exposed2 0.00404821 0.007610478 0.004843385 $correction [1] FALSE attr(,method) [1] Unconditional MLE normal approximation (Wald) CI Warning message: Chi-squared approximation may be incorrect in: chisq.test(xx, correct = correction) === end example === And epitools also gives a lower limit of NaN. === end all examples === I would prefer not to have to tell the authors of the paper I am refereeing that I think they are wrong unless I can help them with what they should have done. Is there another package I should have tried? Is there some other way of doing this? Am I doing something fundamentally wrong-headed? Michael Dewey http://www.aghmed.fsnet.co.uk __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Questions regarding integrate function
A follow-up. Thanks to Matthias, the codes worked very well.
Le
On 11/18/06, Matthias Kohl [EMAIL PROTECTED] wrote:
Hello,
your integrand needs to be a function which accepts a numeric vector as
first argument and returns a vector of the same length (see ?integrate).
Your function does not fulfill this requirement. Hence, you have to
rewrite your function or use sapply, apply or friends; something like
newintegrand - function(x) sapply(x, integrand)
By the way you use a very old R version and I would recommend to update
to R 2.4.0.
hth
Matthias
Le Wang schrieb:
Hi there. Thanks for your time in advance.
I am using R 2.2.0 and OS: Windows XP.
My final goal is to calculate 1/2*integral of
(f1(x)^1/2-f2(x)^(1/2))^2dx (Latex codes:
$\frac{1}{2}\int^{{\infty}}_{\infty}
(\sqrt{f_1(x)}-\sqrt{f_2(x)})^2dx $.) where f1(x) and f2(x) are two
marginal densities.
My problem:
I have the following R codes using adapt package. Although adapt
function is mainly designed for more than 2 dimensions, the manual
says it will also call up integrate if the number of dimension
equals one. I feed in the data x1 and x2 and bandwidths h1 and h2.
These codes worked well when my final goal was to take double
integrals.
integrand - function(x) {
# x input is evaluation point for x1 and x2, a 2x1 vector
x1.eval - x[1]
x2.eval - x[2]
# n is the number of observations
n - length(x1)
# x1 and x2 are the vectors read from data.dat
# Compute the marginal densities
f.x1 - sum(dnorm((x1.eval-x1)/h1))/(n*h1)
f.x2 - sum(dnorm((x2.eval-x2)/h2))/(n*h2)
# Return the integrand #
return((sqrt(f.x1)-sqrt(f.x2))**2)
}
estimate-0.5*adapt(1, lo=lo.default, up=up.default,
minpts=minpts.default, maxpts=maxpts.default,
functn=integrand, eps=eps.default, x1, x2,h1,h2)$value
But when I used it for one-dimension, it failed. Some of my
colleagues suggested getting rid of x2.eval in the integrand
because it is only one integral. But after I changed it, it still
didn't work. R gave the error msg: evaluation of function gave a
result of wrong length
I am not a frequent R user..although I looked up the mailing list
for a while and there were few postings asking similar questions, I
can't still figure out why my codes won't work. Any help will be
appreciated.
Le
-
~~
Le Wang, Ph.D
Population Center
University of Minnesota
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--
Dr. rer. nat. Matthias Kohl
E-Mail: [EMAIL PROTECTED]
Home: www.stamats.de
--
~~
Le Wang, Ph.D
Population Center
University of Minnesota
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[R] Sued Tuerkei.
Sehr geehrte Damen und Herren, Als SI GROUP Für Herbst und Winter haben wir Ihnen ein preislich unschlagbares Angebot an die türkische Riviera-BELEK Halb Pansion 5 Sterne Hotels anzubieten. Ab Zuerich und Basel. INCLUSIV FLUG Hin und Zurück und 8-tägige Ferien an der Türkischen Riviera-Antalya-5 Sterne Halb Pansion(Frühstück und Abendessen) TERMINE UND PREISE- 06.01. bis 13.01.2007 Euro 295 13.01. bis 20.01.2007 Euro 295 20.01. bis 27.01.2007 Euro 295 27.01. bis 03.02.2007 Euro 295 03.02. bis 10.02.2007 Euro 295 10.02. bis 17.02.2007 Euro 295 17.02. bis 24.02.2007 Euro 295 24.02. bis 03.03.2007 Euro 295 03.03. bis 10.03.2007 Euro 295 10.03. bis 17.03.2007 Euro 295 17.03. bis 24.03.2007 Euro 295 24.03. bis 31.03.2007 Euro 295 31.03. bis 07.04.2007 Euro 295 07.04. bis 14.04.2007 Euro 295 14.04. bis 21.04.2007 Euro 295 21.04. bis 28.04.2007 Euro 395 28.04. bis 05.05.2007 Euro 390 Inklusivleistungen: 6 Übernachtungen im 5 - Sterne - Hotel im Doppelzimmer. 1 Übernachtung in einem Thermalhotel in Pamukkale. 7 x Halbpension. Begrüßungscocktail. 1 eintägiger Ausflug nach Antalya mit Bootsfahrt. Alle Eintritte während der Reise. Klimatisierte Busse für 2 Ausflüge. Reiseleitung. Mit Herzlichen Grüssen. SI GROUP www.si-groupreisen.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Statistical Software Comparison
On Tue, 21 Nov 2006, Kenneth Cabrera wrote: Hi R users: I want to know if any of you had used Stata or Statgraphics. We use Stata for teaching courses aimed at graduate students in other departments, and also (as a consequence) on a lot of medical/public health research projects. It is easier to learn than R, and has good support for all the methods we teach in the service courses [unlike, eg, SPSS or Minitab]. Part of the reason it is easier to learn is that there is a very regular syntax. [There is also a GUI, now, but it isn't a very good one and we were using Stata for teaching before it had a GUI]. What are the advantages and disadvantages with respect to R on the following aspects? 1. Statistical functions or options for advanced experimental design (fractional, mixed models, greco-latin squares, split-plot, etc). Stata is not very good at this sort of thing. Neither is R, yet, since lme() is really for longitudinal data and lmer() is still developing. 2. Bayesian approach to experimental design. Not much here, either, in Stata 3. Experimental design planing options. Or here. 4. Manuals (theory included in the manuals). Stata is excellent. They usually give formulas as well as references (and sometimes algorithms and computational notes that are not in the references). The only problem is they keep growing and dividing, so the cost of a complete set goes up quite rapidly with each release (and the volume that you want is always on the other side of the room or lent out to someone). The online help is also good. It suffers relative to R from the examples not necessarily being directly executable. 5. Support (in this aspect there is no comparison with R, the R list is the best known support). The Stata list is pretty good, too. You can see it at http://www.hsph.harvard.edu/statalist/ 6. Numerical stability. For most purposes this is not really an issue and I haven't pushed Stata to the edge. I haven't seen any problems. Stata does have a smaller range of built-in optimizers, and they seem to have stopped at the Marquadt algorithm. This has only once been a problem for me (in fitting log-binomial generalized linear models), but could be a problem in implementing new methods. 7. Implementation of modern statistical approaches. It depends on the area. It's not bad at all in biostatistics and in some areas of econometrics. As with R there is also a lot of user-written code, some of it of excellent quality. The Stata language is better than it looks, but some things can be easily programmed in it and some can't. The last two versions of Stata have introduced language changes in order to be able to implement better graphics and linear mixed models, and you can also now call C code from Stata, so things are improving. Algorithms that are suited to a `one rectangular dataset' view of the world are often very fast in Stata, but the penalty for not vectorizing is even stiffer than in R. -thomas Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problems with garchFit
Hi all,
Thank you for responses. If any one need the data, I can email it to you. I
don't think I can attach it to R-help. It is only SP 500 monthly returns I
downloaded from Yahoo finance, with only date and adj. close kept.
Thank you,
Tian
On 11/22/06, T Mu [EMAIL PROTECTED] wrote:
Hi all,
I post it on both r-help and r-finance since I don't know where is most
appropriate for this topic. Sorry if it bothers you.
I did garch fitting on SP500 monthly returns with garchFit from fSeries.
I got same coefficients from all cond.dist except normal. I thought that
is probabaly usual for the data. But when I play with it, I got another
question.
I plot skew normal with skew = 1 and a standard normal, they overlap
eachother, so I think they are the same. Skew = 1 means no skewness (I can
not find the paper defining the distribution).
library(fSeries)
curve(dsnorm(x, 0, 1, 1), -2, 2, add = F, col = 'red') #skew normal with
skew 1
curve(dnorm(x, 0, 1), -2, 2, add = T, col = 'blue') #normal
Then I try them as innovations,
#normal innovation
garch_norm - garchFit(series = logr, include.mean = F)
#skew normal innovation
#this line do not include skew, so it got same result as normal
garch_snorm - garchFit(series = logr, cond.dist = 'dsnorm', include.mean= F,
include.skew = F)
#this line includes skew, but use default skew = 1, and it got results
different from normal, which I don't understand
garch_snorm - garchFit(series = logr, cond.dist = 'dsnorm', include.mean= F,
include.skew = T)
Have I done something wrong? I am attaching the code, thank you.
Tian
#GARCH analysis of monthly return
rm(list=ls(all=TRUE))
sp500 - read.csv('sp_m90.csv', header=TRUE)
sp500 - sp500[,2] #only adjusted close
n - length(sp500)
logr - log(sp500[1:n-1] / sp500[2:n])
acf(logr)
ar5 - arima(logr, order = c(5, 0, 0), include.mean = T)
logr- ar5$res
acf(logr)
#fit GARCH distribution
hist(logr, freq = F, ylim = c(0, 12), breaks = 'FD')
norm_fit - normFit(logr)
curve(dnorm(x, norm_fit$est[1], norm_fit$est[2]), -.15, .15, add = TRUE,
col=2)
t_fit - stdFit(logr)
curve(dstd(x, t_fit$est[1], t_fit$est[2], t_fit$est[3]), -.15, .15, add =
TRUE, col=6)
snorm_fit - snormFit(logr)
curve(dsnorm(x, snorm_fit$est[1], snorm_fit$est[2], snorm_fit$est[3]),
-.25, .15, add = TRUE, col=4)
st_fit - sstdFit(logr)
curve(dsstd(x, st_fit$est[1], st_fit$est[2], st_fit$est[3],
st_fit$est[4]), -.25, .15, add = TRUE, col=3)
library(fSeries)
#normal innovation
garch_norm - garchFit(series = logr, include.mean = F)
#t inovation
garch_t - garchFit(series = logr, cond.dist = 'dstd', include.mean = F,
include.shape = T)
garch_t1 - garchFit(series = logr, cond.dist = 'dstd', include.mean = F,
shape = t_fit$est[3], include.shape = T)
#skew normal innovation
garch_snorm - garchFit(series = logr, cond.dist = 'dsnorm', include.mean= F,
include.skew = T)
garch_snorm1 - garchFit(series = logr, cond.dist = 'dsnorm', include.mean=
F, skew = snorm_fit$est[3],
include.skew = T)
#skew t innovation
garch_st - garchFit(series = logr, cond.dist = 'dsstd', include.mean = F,
include.skew = T, include.shape = T)
garch_st1 - garchFit(series = logr, cond.dist = 'dsstd', include.mean =
F, skew = st_fit$est[4], shape = st_fit$est[3], include.skew = T,
include.shape = T)
vix - read.csv('D:/Documents and Settings/Mu Tian/Desktop/8780/8780
project/vix_m.csv', header=TRUE)
vix - (vix[,2]/100) / (12^.5)
plot_sd - function(x, ylim = null, col = null, ...) {
xcsd = [EMAIL PROTECTED]
plot(xcsd, type = l, col = col, ylab = x,
main = Conditional SD, ylim = ylim)
abline(h = 0, col = grey, lty = 3)
grid()
}
plot_sd(garch_norm, ylim = c(0.02, 0.13), col = 2)
xcsd = [EMAIL PROTECTED]
lines(xcsd, col = 3)
lines(1:n, vix)
#predict
predict(garch_norm)
predict(garch_t)
#demonstration of skew distributions
#skew normal
curve(dsnorm(x, 0, 1, .1), -2, 2, add = F, col = 'green')
curve(dsnorm(x, 0, 1, snorm_fit$est[3]), type = 'l', col = 'blue', add =
T)
curve(dsnorm(x, 0, 1, 1), -2, 2, add = T, col = 'red') #normal
#skew t
curve(dsstd(x, 0, 1, 4, 1), -2, 2, add = F, col = 'red')
curve(dsstd(x, 0, 1, st_fit$est[3], st_fit$est[4]), type = 'l', col =
'blue', add = T)
curve(dsstd(x, 0, 1, 100, .5), -2, 2, add = T, col = 'green')
#t
curve(dstd(x, 0, 1, 4), -2, 2, add = T, col = 'red')
curve(dstd(x, 0, 1, t_fit$est[3]), type = 'l', col = 'blue', add = T)
curve(dstd(x, 0, 1, 100), -2, 2, add = T, col = 'green')
curve(dsnorm(x, 0, 1, 1), -2, 2, add = F, col = 'red') #normal
curve(dnorm(x, 0, 1), -2, 2, add = T, col = 'blue') #normal
curve(dsnorm(x, 0, 1, .1), -2, 2, add = T, col = 'green') #normal
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Re: [R] Latent models
Hai, Wrong button. The email went out too quick... This might be the one, Brian. The problem was, and is, that the data is composed of two interrelated sets of variables, from two completely different ways of measurement, which I have to connect through some latent structure. Most variables are dichotomous / slightly scaled (1,2,3) and only one is numerically scaled. It's not reasonable to assume a linear relation, because I looked at the Rank linearity condition and it is not satisfied, probably none of the relations is linear. The statistics book advised me to look for latent models. I'll take a look at these packages. Additional suggestions are welcome. Thanks, Wilfred Prof Brian Ripley schreef: Those are only a few of the possible latent models. Factor analysis and SEM are also latent variable models, as indeed are the mixed models of nlme, lme4, mmlcr, polr models Then there are the graphical models of (at least) packages ggm and latentnet, polytomous LDA in package polLCA help.search(latent, agrep=FALSE) on a comprehensive installation helped me find some of these. On Wed, 22 Nov 2006, ronggui wrote: Is ltm what you want? packageDescription(ltm) Package: ltm Title: Latent Trait Models under IRT Version: 0.6-1 Date: 2006-10-02 Author: Dimitris Rizopoulos [EMAIL PROTECTED] Maintainer: Dimitris Rizopoulos [EMAIL PROTECTED] Description: Analysis of multivariate dichotomous and polytomous data using latent trait models under the Item Response Theory approach. It includes the Rasch, the Two-Parameter Logistic, the Birnbaum's Three-Parameter, and the Graded Response Models. Depends: R(= 2.3.0), MASS, gtools, msm LazyLoad: yes LazyData: yes License: GPL version 2 or later URL: http://wiki.r-project.org/rwiki/doku.php?id=packages:cran:ltm Packaged: Tue Oct 3 09:07:38 2006; dimitris Built: R 2.4.0; ; 2006-10-03 17:16:04; windows If you are looking function for Latent Class Analysis (LCA), then lca in package e1071 is what you want. On 11/22/06, Bi-Info (http://members.home.nl/bi-info) [EMAIL PROTECTED] wrote: Hai, Can anyone help me with some literature (R related) about latent models? Thanx, Wilfred __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] NEWBIE: Help explaining use of lm()?
James W. MacDonald wrote: Hi Kevin, The hint here would be the fact that you are coding your 'Groups' variable as a factor, which implies (at least in this situation) that the Groups are unordered levels. Your understanding of a linear model fit is correct when the independent varible is continuous (like the amount of hormone administered). However, in this case, with unordered factor levels, a linear model is the same as fitting an analysis of variance (ANOVA) model. Note that the statistic for an ANOVA (the F-statistic) is a generalization of the t-statistic to more than two groups. In essence the question you are asking with the lm() fit is 'Is the mean age at walking different from a baseline mean for any of these groups?' as compared to the t-test which asks 'Is the mean age at walking different between Group x and Group y?'. The way you have set up the analysis uses the 'active' group as a baseline and compares all others to that. In the output you supply, the intercept is the mean age that the active group started walking, and it appears that there is moderate evidence that the 'ctr.8w' group started walking later (the estimate being a little over 2 months later). If you were to do individual t-tests making the same comparisons you would end up with the same conclusion, but slightly different p-values, etc, because the degrees of freedom would be different. One concept that this explanation seems to skip lightly over (and that took me awhile to absorb) is what it means to substitute a factor in place of a continuous independent variable in lm(). I don't claim to be an expert in the field, though, so comments on my elaboration are welcome. The use of continuous independent variables in lm() leads to a linear combination of these variables (m_1*x_1 + m_2*x_2 + ... + m_n*x_n) added to an intercept (b). If you replace one continuous variable with a factor, (x_n - f_1) you divide the independent variable space ([x_1,x_2,...,x_n-1]) into separate compartments each with their own intercept. That intercept is represented as a combination of an intercept specific to the applicable value of the factor on top of an overall intercept: m_1*x_1+m_2*x_2+...+m_n-1*x_n-1 + b_f_1 + b so that the term that used to represent the influence of a continuous variable (m_n*x_n) is replaced with one of a list of constants (b_a, b_b, b_c if factor f_1 has three levels a, b, and c). In Kevin's example, no continuous variables are included at all, so the solution coefficients simply consist of a list of intercepts, of which the first is present by default because the model doesnt specify a -1 term, and the remaining ones are the list of intercepts b_f_1. It is important to understand clearly that a single scaling factor m_n applied to a continuous variable has been replaced by a specific value defined for each level of the factor. The loss of continuity means that the influence of the variable no longer shows up in a simple scaling factor (a single number) and must be called out individually. The compartmentalization suggests that separate lm() fits could be applied for each level of the factor, combining the two intercept terms for each fit but the use of a single fit permits the uncertainty in separate components of the intercept to be identified with the individual factor levels as well as overall, which (I believe) explains Jim's assertion that your example amounts to an ANOVA. I, too, HTH :) HTH, Jim Zembower, Kevin wrote: I'm attempting the heruclean task of teaching myself Introductory Statistics and R at the same time. I'm working through Peter Dalgaard's Introductory Statistics with R, but don't understand why the answer to one of the exercises works. I'm hoping someone will have the patience to explain the answer to me, both in the statistics and R areas. Exercise 6.1 says: The zelazo data are in the form of a list of vectors, one for each of the four groups. Convert the data to a form suitable for the use of lm, and calculate the relevant tests. ... This stumped me right from the beginning. I thought I understood that linear models tried to correlate an independent variable (such as the amount of a hormone administered) to a dependent variable (such as the height of a cornstalk). Its output was a model that could state, for every 10% increase in the hormone, the height increased by X%. The zelazo data are the ages at walking (in months) of four groups of infants, two controls and two experimentals subjected to different exercise regimens. I don't understand why lm() can be used at all in this circumstance. My initial attempt was to use t.test(), which the answer key does also. I would have never thought to use lm() except for the requirement in the problem. I've pasted in the output of the exercise below, for those without the dataset. Would someone explain why lm() is appropriate to use in this situation, and what the results mean 'in
Re: [R] Fitting mixed-effects models with lme with fixed error term variances
On 11/22/06, Gregor Gorjanc [EMAIL PROTECTED] wrote: Douglas Bates wrote: On 11/21/06, Gregor Gorjanc [EMAIL PROTECTED] wrote: Douglas Bates bates at stat.wisc.edu writes: ... Can you be more specific about which parameters you want to fix and which you want to vary in the optimization? It would be nice to have the ability to fix all variances i.e. variances of random effects. That gets tricky in terms of the parameterization of the variance-covariance matrices for vector-valued random effects. These matrices are not expressed in the conventional parameterization of variances and covariances or even variances and correlation because the conditions for the resulting matrix to be positive definite are not simple bounds or easily expressed transformations then the matrix is larger than 2 by 2. I suppose what I could do is to allow these matrices to be specified in the parameterization that is used in the optimization and provide a utility function to map from the conventional parameters to these. That would mean that you couldn't fix ,say, the variance of the intercept term for vector-valued random effects but allow the variance of a slope for the same grouping factor to be estimated. Well, you could but only in the fortune(Yoda) sense. Yes, I agree here. Thank you for the detailed answer. By the way, if you fix all the variances then what are you optimizing over? The fixed effects? In that case the solution can be calculated explicitly for a linear mixed model. The conditional estimates of the fixed effects given the variance components are the solution to a penalized linear least squares problem. (Yes, the solution can also be expressed as a generalized linear least squares problem but there are advantages to using the penalized least squares representation. Yup. It would really be great to be able to do that nicely in R, say use lmer() once and since this might take some time use estimates of variance components next time to get fixed and random effects. Is this possible with lmer or any related function - not in fortune(Yoda) sense ;) Not quite. There is now a capability in lmer to specify starting estimates for the relative precision matrices which means that you can use the estimates from one fit as starting estimates for another fit. It looks like fm1 - lmer(...) fm2 - lmer(y ~ x + (...), start = [EMAIL PROTECTED]) I should say that in my experience this has not been as effective as I had hoped it would be. What I see in the optimization is that the first few iterations reduce the deviance quite quickly but the majority of the time is spent refining the estimates near the optimum. So, for example, if it took 40 iterations to converge from the rather crude starting estimates calculated within lmer it might instead take 35 iterations if you give it good starting estimates. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Probit Analysis
Respected Sir/Madam, I have a question regarding calculation of LD50 (Lethal Dose) and IC50 (50% inhibitory concentration) of an antimicrobial experiment. I have used a compound isolated from a plant and observed its effect on the fungus *Fusarium oxysporum* by the food poisoning method. Solutions of the compound at concentrations of 0, 50, 100, 150, 200 and 250µg/ ml were added to sterilized medium in Petri dish (9cm dia.) and after transferring the mycelia of fungal strains, the dishes were incubated in the dark. When the mycelium of fungi reached the edges of the control (0µg/ ml) dishes, the antifungal indices were calculated. Each test was repeated three times and the average was calculated. The formula to calculate antifungal index (AI) was AI % = (1- Da / Db) * 100 Where Da = diameter of growth zone in experiment (cm); Db = diameter of growth zone in control (cm). The results are as follows Concentration (µg/ ml) 0 50 100 150 200 250 AI% (average of 3 replicates) 0 10 21.59 33.89 47.96 59.93 I have the Version 2.3.1 (2006-06-01) of R loaded in my computer and want to calculate LD50 and if possible IC50 values with the data. I have checked the Help menu---Manuals in PDF---An introduction to R. There is an example of calculating LD 50 in Chapter 11, page no. 63. Please suggest some functions on R or the method by which I can do the calculations. I am not at all used to programming language so a detailed solution would be very much helpful. I am sending the full analysis that I am doing just below this letter. I am also sending it as attachment. There is a warning message that i have marked in red--if it is not in color in your mail---then just see the message after the comand fmp - glm(Ymat~x,family = binomial(link=probit), data = Fusarium. It says non integer counts in a binomial glm! in: eval(expr, envir, enclos) Actually I am not quite sure I am doing the correct thing for probit analysis. You can see that I do get a value of ld50 at the last step, but I do not know whether my methodology is correct. Fusarium - data.frame(x = c(0,50,100,150,200,250), n = rep(90,6), + y = c(90,81,75.4,58.5,51.3,65.1)) Fusarium$Ymat - cbind(Fusarium$y, Fusarium$n - Fusarium$y) fmp - glm(Ymat ~ x, family = binomial(link=probit), data = Fusarium) Warning message: non-integer counts in a binomial glm! in: eval(expr, envir, enclos) summary(fmp) Call: glm(formula = Ymat ~ x, family = binomial(link = probit), data = Fusarium) Deviance Residuals: 1 2 3 4 5 6 3.26268 -0.06454 -0.15297 -2.48265 -1.99479 3.13791 Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) 1.5767662 0.1345939 11.715 2e-16 *** x -0.0056714 0.0007908 -7.172 7.41e-13 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 84.795 on 5 degrees of freedom Residual deviance: 30.662 on 4 degrees of freedom AIC: 58.283 Number of Fisher Scoring iterations: 5 summary(fml) Call: glm(formula = Ymat ~ x, family = binomial, data = kalythos) Deviance Residuals: 1 2 3 4 5 1.386 0.838 -2.021 -2.151 2.398 Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) 2.069867 0.284782 7.268 3.64e-13 *** x -0.006615 0.001591 -4.157 3.22e-05 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 35.262 on 4 degrees of freedom Residual deviance: 17.083 on 3 degrees of freedom AIC: 44.426 Number of Fisher Scoring iterations: 4 ld50 - function(b) -b[1]/b[2] ldp - ld50(coef(fmp)); ldl - ld50(coef(fml)); c(ldp, ldl) (Intercept) (Intercept) 278.0220312.8854 Thanking you in anticipation. Sincerely yours. -- Dipjyoti Chakraborty C/o Dr Adinpunya Mitra Natural Products Biotechnology Lab Agriculture Food Technology Depart. IIT-KGP Kharagpur-721 302-India __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie problem ... Forest plot
At 15:16 22/11/2006, you wrote: When I use studlab parameter I don`t have (don't know how to put) names inside graph, X and Y axis, and I have Black/White forest plot, and I need colored. 1 - please cc to r-help so that others can learn from your experience 2 - when you use the studlab parameter what happens, and how does that differ from what you wanted to happen? Michael Dewey http://www.aghmed.fsnet.co.uk __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] plotting a groupedData object
Hello all, I am plotting a groupedData object and the key at the top of the graph runs off the page. I don't want to set key=F because it is useful (or would be if I could see it). Is it possible to explicitly cause the key to wrap? I have used this function before with no trouble but now I have just 5 groups with rather long descriptions (which I can't meaningfully shorten). Thank you for your help, Lisa Avery [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problems with garchFit
Hi all, Got a few inquiries for data, thank you. I attach code and data below. Copy it to your R script and it should be runnable. Now the Snorm with skew = 1 does have same result with Norm (for this set of data). But all other innovations still got same coefficients. Thank you, Tian #GARCH analysis of monthly return rm(list=ls(all=TRUE)) logr - c(0.0107098575567282, 0.0238439568947517, 0.0170914961728296, 0.0138732445655177, -0.00210495583051609, -0.00709127573708259, -0.0385827936083661, 0.00490449365266998, 0.00385685394698455, -0.00672488595832236, 0.0179700812080439, -0.00813073003671577, 0.0274033576545142, -0.0250778743362483, -0.000252972595615477, -0.0184633479948336, 0.0281585718422547, -0.00732056749937242, 0.0223343433626314, -0.0274913993701448, -0.0264806322770031, 0.0115490548518633, -0.0327936279909903, 0.0247646111707177, 0.0306908994386586, 0.00673907585573934, 0.00214245677973209, -0.00489315945076054, -0.0420701182378048, 0.0106513092268381, 0.00483314418308997, -0.0241112735170185, -0.0236721006924634, 0.00495722080665149, 0.00995100224049281, 0.0423410192839968, -7.4654666429242e-05, 0.0463263884424721, -0.0191941125904601, 0.0105374637681618, 0.00891562645629928, 0.00408074598837779, 0.0424677854668133, 0.0707494700070023, 0.00114499423035526, -0.0243277242813142, -0.0349753736938974, -0.0694071571523498, 0.0483227046536378, 0.0757365410062828, -0.123739048470341, - 0.00230833563215873, -0.0894777518603734, -0.0823922232983805, -0.0163008223196004, -0.0705625628068879, 0.028902196230284, -0.0281627666976421, -0.0228742536894077, 0.000367412011421635, 0.0653064704108988, 0.0107593083066410, -0.0924344952694643, -0.0734334859099414, -0.0179761012283746, -0.0325320322718984, -0.00210059232386598, 0.066829130533506, -0.0735364239526057, -0.104008970439016, 0.0268723664276674, -0.00313268679975119, -0.0906340137330617, -0.0121396649961373, -0.0621441723072229, 0.0517502775657374, -0.0236541332868366, 0.0164737511342563, -0.029336578252438, -0.0384578572805523, 0.0851459320997492, -0.0274909426329228, -0.0594227029752583, 0.0490549196316311, 0.0117045927385518, 0.0534838867245415, - 0.036145147104809, -0.0134516584002713, -0.0397486953757553, 0.0458303597691911, -0.0324653458761952, 0.0300639359263621, 0.0308827205446822, -0.0399929706878522, 0.0330129512566845, 0.0476656477094465, 0.0502661919846773, 0.0700555274728492, 0.0533484191777131, - 0.164763950096769, -0.0188612611651085, 0.0315025148659792, -0.026183523443084, 0.00185764564623504, 0.0415605496104671, 0.0609005489249022, 0.00292109288076270, 0.00843129126020518, 0.0364435424654694, -0.0422639215772581, 0.0446111390279706, -0.0663607453035695, 0.0680648616436016, 0.0353571745939384, 0.0497475635284525, 0.0495856850474945, -0.0507265004938753, -0.00126783225527763, 0.052332767500398, -0.0289178666588803, 0.0636310849271901, 0.0185883019590537, 0.0456074218283171, 0.0114612960614454, -0.0540054008094362, -0.00492372717398046, 0.0154182797009216, 0.0061641662077013, 0.000707504517465771, -0.000268063662287813, 0.0249188086513136, 0.0101155993444056, 0.0330509893647089, -0.0121696993613180, 0.0321366073123792, -0.00749818182068220, 0.0241037519137133, 0.013877482054341, 0.0284900964619952, 0.0203986640653641, 0.0197845872016367, 0.0282608312416226, 0.0168097602528238, 0.00504624695886614, -0.047483970806859, 0.0134418482374845, -0.0344232233373743, 0.0297312626164135, 0.0238263428912005, -0.0343340943399759, 0.00514305696494622, 0.00428675908876259, -0.0540037546954394, -0.0376835393117851, 0.0248059444691968, 0.00286271109440267, -0.0201726197962942, 0.0120294090576544, -0.0172160135048806, 0.0266745775509609, -0.0125191786357240, -0.00642292839233706, 0.0152853839659078, -0.0329232528272489, 0.0113467545039784, 0.00325116163241795, -0.000156615943755373, 0.00287917888713352, 0.0226350405134303, -0.00507382415825686, 0.00188653413193280, -0.0314680606930909, 0.0314404828152606, -0.0246891655066154, -0.00621437278185084, 0.0203328943393480, -0.0292515600983759, 0.00236594272243002, -0.0273027918584042, 0.0986116252116223, -0.0520744999798322, 0.00458681078225353, -0.0265072106468637, 0.0122803714060437, 0.036704412456513, -0.0562553901943863, 0.0307005847712896, -0.00685811877481276, 0.0147820750313374, 0.057936537214, 0.0335011692685766, 0.0173466769183716, 0.0510289606341445, - 0.0138986659133316, -0.0597185581683447, -0.106240705003888, -0.0124147403601571, -0.0161039038326663, 0.0808230314002266, -0.0344330483221435, 0.0167876627118019, 0.00132482639172523) hist(logr, freq = F, ylim = c(0, 13), breaks = 'FD') norm_fit - normFit(logr) curve(dnorm(x, norm_fit$est[1], norm_fit$est[2]), -.15, .15, add = TRUE, col=2) t_fit - stdFit(logr) curve(dstd(x, t_fit$est[1], t_fit$est[2], t_fit$est[3]), -.15, .15, add = TRUE, col=6) snorm_fit - snormFit(logr) curve(dsnorm(x, snorm_fit$est[1], snorm_fit$est[2], snorm_fit$est[3]), -.25, .15, add = TRUE, col=4) st_fit - sstdFit(logr) curve(dsstd(x, st_fit$est[1], st_fit$est[2],
Re: [R] lme - plot - labels
On 11/22/06, Doktor, Daniel [EMAIL PROTECTED] wrote: Hello there, I am using the 'nlme' package to analyse among group and in between group variances. I do struggle a little using the respective plot-functions. I created a grouped object: group.lme - groupedData(obsday ~ oro | id, data=read.table(data-lme.txt, header=T), labels=list(x = Day of Year, y = ID number)) When I plot, however plot(group.lme) the y-labels appear on the x-axis and the x-labels do not are not drawn at all. I can't reproduce the problem. Try library(nlme) data(sleepstudy, package = lme4) sleep - groupedData(Reaction ~ Days|Subject, sleepstudy, +labels = list(x = Days of sleep deprivation, y = Average reaction time (ms))) plot(sleep) Under R-2.4.0 that produces a plot with correctly labeled axes. Any advice? Well there is always the old read the manual approach. The formula for a groupedData object is of the form response ~ covariate | groups I am using R 2.2.1 on Windows XP __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plotting a groupedData object
On 11/22/06, Lisa Avery [EMAIL PROTECTED] wrote: Hello all, I am plotting a groupedData object and the key at the top of the graph runs off the page. I don't want to set key=F because it is useful (or would be if I could see it). Is it possible to explicitly cause the key to wrap? I have used this function before with no trouble but now I have just 5 groups with rather long descriptions (which I can't meaningfully shorten). Plotting a groupedData object is just a convenient way of accessing the 'xyplot' function from the lattice package. If you need finer control over the plot than is available through groupedData then I suggest that you call xyplot directly. Wading through the descriptions of all the options in xyplot is a formidable task but the results are worth it. In particular it took me a long time to work out how to use the auto.key argument to get exactly what I wanted but now I can generally get what I want on the first or second try. The nlme package was initially written for S and the graphics used the trellis formulation. Later it was ported to R and lattice graphics. With Deepayan's continuing development of the lattice package it is now far beyond the capabilities of trellis graphics so I would recommend using lattice directly. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] dataframe manipulation
Hi, Having a dataframe 'l1' (dput output is below): dim(l1) 1274 2 l1[1:12,] Var1 Freq 1 1988-01-131 2 1988-01-161 3 1988-01-203 4 1988-01-252 5 1988-01-301 6 1988-02-015 7 1988-02-084 8 1988-02-141 9 1988-02-161 10 1988-02-184 11 1988-02-242 12 1988-03-041 I want to extract the times 1988-01, 1988-02,...,2005-12 are present, in order to know, for example, that 1988-01 = 5 days 1988-02 = 6 days and so on. The Freq column doesn't matter in this case, just the Var1 column I've been trying with: regexpr(2005-03-,l1$Var1) But the result is not satisfactory since I need to find and count where these values are. Thanks Antonio dput(l1,control=all) structure(list(Var1 = structure(as.integer(c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244, 245, 246, 247, 248, 249, 250, 251, 252, 253, 254, 255, 256, 257, 258, 259, 260, 261, 262, 263, 264, 265, 266, 267, 268, 269, 270, 271, 272, 273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287, 288, 289, 290, 291, 292, 293, 294, 295, 296, 297, 298, 299, 300, 301, 302, 303, 304, 305, 306, 307, 308, 309, 310, 311, 312, 313, 314, 315, 316, 317, 318, 319, 320, 321, 322, 323, 324, 325, 326, 327, 328, 329, 330, 331, 332, 333, 334, 335, 336, 337, 338, 339, 340, 341, 342, 343, 344, 345, 346, 347, 348, 349, 350, 351, 352, 353, 354, 355, 356, 357, 358, 359, 360, 361, 362, 363, 364, 365, 366, 367, 368, 369, 370, 371, 372, 373, 374, 375, 376, 377, 378, 379, 380, 381, 382, 383, 384, 385, 386, 387, 388, 389, 390, 391, 392, 393, 394, 395, 396, 397, 398, 399, 400, 401, 402, 403, 404, 405, 406, 407, 408, 409, 410, 411, 412, 413, 414, 415, 416, 417, 418, 419, 420, 421, 422, 423, 424, 425, 426, 427, 428, 429, 430, 431, 432, 433, 434, 435, 436, 437, 438, 439, 440, 441, 442, 443, 444, 445, 446, 447, 448, 449, 450, 451, 452, 453, 454, 455, 456, 457, 458, 459, 460, 461, 462, 463, 464, 465, 466, 467, 468, 469, 470, 471, 472, 473, 474, 475, 476, 477, 478, 479, 480, 481, 482, 483, 484, 485, 486, 487, 488, 489, 490, 491, 492, 493, 494, 495, 496, 497, 498, 499, 500, 501, 502, 503, 504, 505, 506, 507, 508, 509, 510, 511, 512, 513, 514, 515, 516, 517, 518, 519, 520, 521, 522, 523, 524, 525, 526, 527, 528, 529, 530, 531, 532, 533, 534, 535, 536, 537, 538, 539, 540, 541, 542, 543, 544, 545, 546, 547, 548, 549, 550, 551, 552, 553, 554, 555, 556, 557, 558, 559, 560, 561, 562, 563, 564, 565, 566, 567, 568, 569, 570, 571, 572, 573, 574, 575, 576, 577, 578, 579, 580, 581, 582, 583, 584, 585, 586, 587, 588, 589, 590, 591, 592, 593, 594, 595, 596, 597, 598, 599, 600, 601, 602, 603, 604, 605, 606, 607, 608, 609, 610, 611, 612, 613, 614, 615, 616, 617, 618, 619, 620, 621, 622, 623, 624, 625, 626, 627, 628, 629, 630, 631, 632, 633, 634, 635, 636, 637, 638, 639, 640, 641, 642, 643, 644, 645, 646, 647, 648, 649, 650, 651, 652, 653, 654, 655, 656, 657, 658, 659, 660, 661, 662, 663, 664, 665, 666, 667, 668, 669, 670, 671, 672, 673, 674, 675, 676, 677, 678, 679, 680, 681, 682, 683, 684, 685, 686, 687, 688, 689, 690, 691, 692, 693, 694, 695, 696, 697, 698, 699, 700, 701, 702, 703, 704, 705, 706, 707, 708, 709, 710, 711, 712, 713, 714, 715, 716, 717, 718, 719, 720, 721, 722, 723, 724, 725, 726, 727, 728, 729, 730, 731, 732, 733, 734, 735, 736, 737, 738, 739, 740, 741, 742, 743, 744, 745, 746, 747, 748, 749, 750, 751, 752, 753, 754, 755, 756, 757, 758, 759, 760, 761, 762, 763, 764, 765, 766, 767, 768, 769, 770, 771, 772, 773, 774, 775, 776, 777, 778, 779, 780, 781, 782, 783, 784, 785, 786, 787, 788, 789, 790, 791, 792, 793, 794, 795, 796, 797, 798, 799, 800, 801, 802, 803, 804, 805, 806, 807, 808, 809, 810, 811, 812, 813, 814, 815, 816, 817, 818, 819, 820, 821, 822, 823, 824, 825, 826, 827, 828, 829, 830, 831, 832, 833, 834, 835, 836, 837, 838, 839, 840, 841, 842, 843, 844, 845, 846, 847, 848, 849, 850, 851, 852, 853, 854, 855, 856, 857, 858, 859, 860, 861, 862, 863,
Re: [R] dataframe manipulation
On Wed, 2006-11-22 at 20:27 +0100, antonio rodriguez wrote: Hi, Having a dataframe 'l1' (dput output is below): dim(l1) 1274 2 l1[1:12,] Var1 Freq 1 1988-01-131 2 1988-01-161 3 1988-01-203 4 1988-01-252 5 1988-01-301 6 1988-02-015 7 1988-02-084 8 1988-02-141 9 1988-02-161 10 1988-02-184 11 1988-02-242 12 1988-03-041 I want to extract the times 1988-01, 1988-02,...,2005-12 are present, in order to know, for example, that 1988-01 = 5 days 1988-02 = 6 days and so on. The Freq column doesn't matter in this case, just the Var1 column I've been trying with: regexpr(2005-03-,l1$Var1) But the result is not satisfactory since I need to find and count where these values are. Thanks Antonio Using just the example data: table(substr(DF$Var1, 1, 7)) 1988-01 1988-02 1988-03 5 6 1 See ?substr for more information. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] dataframe manipulation
If x is the object at the end of your post: library(zoo) z - zoo(x$Freq, as.Date(x$Var1)) aggregate(z, as.yearmon, length) If out contains the result of aggregate then coredata(out) are the counts and time(out) are the corresponding year-months. See ?yearmon On 11/22/06, antonio rodriguez [EMAIL PROTECTED] wrote: Hi, Having a dataframe 'l1' (dput output is below): dim(l1) 1274 2 l1[1:12,] Var1 Freq 1 1988-01-131 2 1988-01-161 3 1988-01-203 4 1988-01-252 5 1988-01-301 6 1988-02-015 7 1988-02-084 8 1988-02-141 9 1988-02-161 10 1988-02-184 11 1988-02-242 12 1988-03-041 I want to extract the times 1988-01, 1988-02,...,2005-12 are present, in order to know, for example, that 1988-01 = 5 days 1988-02 = 6 days and so on. The Freq column doesn't matter in this case, just the Var1 column I've been trying with: regexpr(2005-03-,l1$Var1) But the result is not satisfactory since I need to find and count where these values are. Thanks Antonio dput(l1,control=all) structure(list(Var1 = structure(as.integer(c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244, 245, 246, 247, 248, 249, 250, 251, 252, 253, 254, 255, 256, 257, 258, 259, 260, 261, 262, 263, 264, 265, 266, 267, 268, 269, 270, 271, 272, 273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287, 288, 289, 290, 291, 292, 293, 294, 295, 296, 297, 298, 299, 300, 301, 302, 303, 304, 305, 306, 307, 308, 309, 310, 311, 312, 313, 314, 315, 316, 317, 318, 319, 320, 321, 322, 323, 324, 325, 326, 327, 328, 329, 330, 331, 332, 333, 334, 335, 336, 337, 338, 339, 340, 341, 342, 343, 344, 345, 346, 347, 348, 349, 350, 351, 352, 353, 354, 355, 356, 357, 358, 359, 360, 361, 362, 363, 364, 365, 366, 367, 368, 369, 370, 371, 372, 373, 374, 375, 376, 377, 378, 379, 380, 381, 382, 383, 384, 385, 386, 387, 388, 389, 390, 391, 392, 393, 394, 395, 396, 397, 398, 399, 400, 401, 402, 403, 404, 405, 406, 407, 408, 409, 410, 411, 412, 413, 414, 415, 416, 417, 418, 419, 420, 421, 422, 423, 424, 425, 426, 427, 428, 429, 430, 431, 432, 433, 434, 435, 436, 437, 438, 439, 440, 441, 442, 443, 444, 445, 446, 447, 448, 449, 450, 451, 452, 453, 454, 455, 456, 457, 458, 459, 460, 461, 462, 463, 464, 465, 466, 467, 468, 469, 470, 471, 472, 473, 474, 475, 476, 477, 478, 479, 480, 481, 482, 483, 484, 485, 486, 487, 488, 489, 490, 491, 492, 493, 494, 495, 496, 497, 498, 499, 500, 501, 502, 503, 504, 505, 506, 507, 508, 509, 510, 511, 512, 513, 514, 515, 516, 517, 518, 519, 520, 521, 522, 523, 524, 525, 526, 527, 528, 529, 530, 531, 532, 533, 534, 535, 536, 537, 538, 539, 540, 541, 542, 543, 544, 545, 546, 547, 548, 549, 550, 551, 552, 553, 554, 555, 556, 557, 558, 559, 560, 561, 562, 563, 564, 565, 566, 567, 568, 569, 570, 571, 572, 573, 574, 575, 576, 577, 578, 579, 580, 581, 582, 583, 584, 585, 586, 587, 588, 589, 590, 591, 592, 593, 594, 595, 596, 597, 598, 599, 600, 601, 602, 603, 604, 605, 606, 607, 608, 609, 610, 611, 612, 613, 614, 615, 616, 617, 618, 619, 620, 621, 622, 623, 624, 625, 626, 627, 628, 629, 630, 631, 632, 633, 634, 635, 636, 637, 638, 639, 640, 641, 642, 643, 644, 645, 646, 647, 648, 649, 650, 651, 652, 653, 654, 655, 656, 657, 658, 659, 660, 661, 662, 663, 664, 665, 666, 667, 668, 669, 670, 671, 672, 673, 674, 675, 676, 677, 678, 679, 680, 681, 682, 683, 684, 685, 686, 687, 688, 689, 690, 691, 692, 693, 694, 695, 696, 697, 698, 699, 700, 701, 702, 703, 704, 705, 706, 707, 708, 709, 710, 711, 712, 713, 714, 715, 716, 717, 718, 719, 720, 721, 722, 723, 724, 725, 726, 727, 728, 729, 730, 731, 732, 733, 734, 735, 736, 737, 738, 739, 740, 741, 742, 743, 744, 745, 746, 747, 748, 749, 750, 751, 752, 753, 754, 755, 756, 757, 758, 759, 760, 761, 762, 763, 764, 765, 766, 767, 768, 769, 770, 771, 772, 773, 774, 775, 776, 777, 778, 779, 780, 781, 782, 783,
Re: [R] dataframe manipulation
Marc Schwartz escribió: Using just the example data: table(substr(DF$Var1, 1, 7)) 1988-01 1988-02 1988-03 5 6 1 This is just what I wanted! Many thanks Marc! BR Antonio See ?substr for more information. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] dataframe manipulation[Solved]
Gabor Grothendieck escribió: If x is the object at the end of your post: library(zoo) z - zoo(x$Freq, as.Date(x$Var1)) aggregate(z, as.yearmon, length) Gabor, Very nice output of this option. I get: ene 1988 feb 1988 mar 1988 abr 1988 may 1988 jun 1988 jul 1988 ago 1988 56565374 sep 1988 oct 1988 nov 1988 dic 1988 ene 1989 feb 1989 mar 1989 abr 1989 78666346 Thanks! Antonio If out contains the result of aggregate then coredata(out) are the counts and time(out) are the corresponding year-months. See ?yearmon On 11/22/06, antonio rodriguez [EMAIL PROTECTED] wrote: Hi, Having a dataframe 'l1' (dput output is below): dim(l1) 1274 2 l1[1:12,] Var1 Freq 1 1988-01-131 2 1988-01-161 3 1988-01-203 4 1988-01-252 5 1988-01-301 6 1988-02-015 7 1988-02-084 8 1988-02-141 9 1988-02-161 10 1988-02-184 11 1988-02-242 12 1988-03-041 I want to extract the times 1988-01, 1988-02,...,2005-12 are present, in order to know, for example, that 1988-01 = 5 days 1988-02 = 6 days and so on. The Freq column doesn't matter in this case, just the Var1 column I've been trying with: regexpr(2005-03-,l1$Var1) But the result is not satisfactory since I need to find and count where these values are. Thanks Antonio dput(l1,control=all) structure(list(Var1 = structure(as.integer(c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244, 245, 246, 247, 248, 249, 250, 251, 252, 253, 254, 255, 256, 257, 258, 259, 260, 261, 262, 263, 264, 265, 266, 267, 268, 269, 270, 271, 272, 273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287, 288, 289, 290, 291, 292, 293, 294, 295, 296, 297, 298, 299, 300, 301, 302, 303, 304, 305, 306, 307, 308, 309, 310, 311, 312, 313, 314, 315, 316, 317, 318, 319, 320, 321, 322, 323, 324, 325, 326, 327, 328, 329, 330, 331, 332, 333, 334, 335, 336, 337, 338, 339, 340, 341, 342, 343, 344, 345, 346, 347, 348, 349, 350, 351, 352, 353, 354, 355, 356, 357, 358, 359, 360, 361, 362, 363, 364, 365, 366, 367, 368, 369, 370, 371, 372, 373, 374, 375, 376, 377, 378, 379, 380, 381, 382, 383, 384, 385, 386, 387, 388, 389, 390, 391, 392, 393, 394, 395, 396, 397, 398, 399, 400, 401, 402, 403, 404, 405, 406, 407, 408, 409, 410, 411, 412, 413, 414, 415, 416, 417, 418, 419, 420, 421, 422, 423, 424, 425, 426, 427, 428, 429, 430, 431, 432, 433, 434, 435, 436, 437, 438, 439, 440, 441, 442, 443, 444, 445, 446, 447, 448, 449, 450, 451, 452, 453, 454, 455, 456, 457, 458, 459, 460, 461, 462, 463, 464, 465, 466, 467, 468, 469, 470, 471, 472, 473, 474, 475, 476, 477, 478, 479, 480, 481, 482, 483, 484, 485, 486, 487, 488, 489, 490, 491, 492, 493, 494, 495, 496, 497, 498, 499, 500, 501, 502, 503, 504, 505, 506, 507, 508, 509, 510, 511, 512, 513, 514, 515, 516, 517, 518, 519, 520, 521, 522, 523, 524, 525, 526, 527, 528, 529, 530, 531, 532, 533, 534, 535, 536, 537, 538, 539, 540, 541, 542, 543, 544, 545, 546, 547, 548, 549, 550, 551, 552, 553, 554, 555, 556, 557, 558, 559, 560, 561, 562, 563, 564, 565, 566, 567, 568, 569, 570, 571, 572, 573, 574, 575, 576, 577, 578, 579, 580, 581, 582, 583, 584, 585, 586, 587, 588, 589, 590, 591, 592, 593, 594, 595, 596, 597, 598, 599, 600, 601, 602, 603, 604, 605, 606, 607, 608, 609, 610, 611, 612, 613, 614, 615, 616, 617, 618, 619, 620, 621, 622, 623, 624, 625, 626, 627, 628, 629, 630, 631, 632, 633, 634, 635, 636, 637, 638, 639, 640, 641, 642, 643, 644, 645, 646, 647, 648, 649, 650, 651, 652, 653, 654, 655, 656, 657, 658, 659, 660, 661, 662, 663, 664, 665, 666, 667, 668, 669, 670, 671, 672, 673, 674, 675, 676, 677, 678, 679, 680, 681, 682, 683, 684, 685, 686, 687, 688, 689, 690, 691, 692, 693, 694, 695, 696, 697, 698, 699, 700, 701, 702, 703, 704, 705,
[R] data in form of a date
Dear all, I often use dates and times in analyses. I just can't figure out how to format my date or time column in R. So, apparently R sees the date as something other than date (character). Let's say I am opening a CSV file, one of the columns of which is a date or time. How do I specify that when opening the file? Thanks for the help, Jim -- - James J. Roper, Ph.D. Universidade Federal do Paraná Depto. de Zoologia Caixa Postal 19020 81531-990 Curitiba, Paraná, Brasil = E-mail: [EMAIL PROTECTED] Phone/Fone/Teléfono: 55 41 33611764 celular: 55 41 99870543 = Zoologia na UFPR http://zoo.bio.ufpr.br/zoologia/ Ecologia e Conservação na UFPR http://www.bio.ufpr.br/ecologia/ - http://jjroper.sites.uol.com.br Currículo Lattes http://lattes.cnpq.br/2553295738925812 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data in form of a date
there are probably many other ways but check out read.zoo. I cans end you of
read.zoo also if you like.
Let me know.
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of James J. Roper
Sent: Wednesday, November 22, 2006 4:18 PM
To: r-help@stat.math.ethz.ch
Subject: [R] data in form of a date
Dear all,
I often use dates and times in analyses. I just can't figure out how to format
my date or time column in R. So, apparently R sees the date as something other
than date (character). Let's say I am opening a CSV file, one of the columns
of which is a date or time. How do I specify that when opening the file?
Thanks for the help,
Jim
--
-
James J. Roper, Ph.D.
Universidade Federal do Paraná
Depto. de Zoologia
Caixa Postal 19020
81531-990 Curitiba, Paraná, Brasil
=
E-mail: [EMAIL PROTECTED]
Phone/Fone/Teléfono: 55 41 33611764
celular: 55 41 99870543
=
Zoologia na UFPR
http://zoo.bio.ufpr.br/zoologia/
Ecologia e Conservação na UFPR
http://www.bio.ufpr.br/ecologia/
-
http://jjroper.sites.uol.com.br
Currículo Lattes
http://lattes.cnpq.br/2553295738925812
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
This is not an offer (or solicitation of an offer) to buy/se...{{dropped}}
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] What training algorithm does nnet package use?
Just to add to this, I also need to know what language is the nnet package written in? Is it in pure R or is it a wrapper for a C library. It really is performing very quickly, going through 200 epochs in seconds when it took neural minutes, and neural is written in R. It all may sound like a trivial question, but it is important as I need to know all this for the analysis I'm doing for a paper. Cheers, Wee-Jin On 22 Nov 2006, at 15:41, Wee-Jin Goh wrote: Greetings list, I've just swapped from the neural package to the nnet package and I've noticed that the training is orders of magnitude faster, and the results are way more accurate. This leads me to wonder, what training algorithm is nnet using? Is it a modification on the standard backpropagation? Or a completely different algorithm? I'm trying to account for the speed differences between neural and nnet, and the documentation on the nnet package is rather sparse on what training algorithm is used (either that, or I'm getting blind and missed it totally). Any help would be much appreciated. Regards, Wee-Jin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] installation step for RSperl
Hi, I try to use R within perl. however, I have a bit difficulty to install RSperl. I followd steps from http://www.omegahat.org/RSPerl/. but still can' t make it work. could someone list a fe w clean steps that I can follow to install it? cheers -- Yuandan Zhang å¼ å æ¦ . [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] question about the solve function in library Matrix
Hi:
I have some problems when I use the function solve function in a loop. In
the following code, I have a diagonal martix ttt whose elements change in
every iteration in a loop. I defined a dpoMatrixclass before the loop so I
do not need to define this class every time in the loop. The reason is to save
some computing time. The code is below. The inverse of the matrix ttt
should change according to the change of ttt in the loop. However, the
values in sigma.dpo.solve, which is the inverse of ttt does not change.
Can anybody figure out what is wrong with my code? Thanks a lot in advance!
liu
library(Matrix)
ttt-diag(1,2)
sigma.dpo-as(ttt, dpoMatrix)
for(i in 1:5)
{ ttt-diag(i,2)
[EMAIL PROTECTED]-as.vector(ttt)
sigma.dpo.solve-solve(sigma.dpo)
print(-)
print(sigma.dpo)
print(sigma.dpo.solve)
}
##
-
[[alternative HTML version deleted]]
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Re: [R] data in form of a date
Read the help desk article in R News 4/1. On 11/22/06, James J. Roper [EMAIL PROTECTED] wrote: Dear all, I often use dates and times in analyses. I just can't figure out how to format my date or time column in R. So, apparently R sees the date as something other than date (character). Let's say I am opening a CSV file, one of the columns of which is a date or time. How do I specify that when opening the file? Thanks for the help, Jim -- - James J. Roper, Ph.D. Universidade Federal do Paraná Depto. de Zoologia Caixa Postal 19020 81531-990 Curitiba, Paraná, Brasil = E-mail: [EMAIL PROTECTED] Phone/Fone/Teléfono: 55 41 33611764 celular: 55 41 99870543 = Zoologia na UFPR http://zoo.bio.ufpr.br/zoologia/ Ecologia e Conservação na UFPR http://www.bio.ufpr.br/ecologia/ - http://jjroper.sites.uol.com.br Currículo Lattes http://lattes.cnpq.br/2553295738925812 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] DSC 2007 registration and scientific program
Hi DSC 2007, a conference on systems and environments for statistical computing, will take place in Auckland, New Zealand on February 15 16, 2007. This is a reminder that the deadline for early bird DSC registration is December 1, 2006. Also, the accepted abstracts for DSC are now available on the web site http://www.stat.auckland.ac.nz/dsc-2007/abstracts/abstracts.html Finally, a reminder that there is a small amount of financial assistance being offered to help people attend DSC. For details, please see http://www.stat.auckland.ac.nz/dsc-2007/#funding Please visit the conference web page at http://www.stat.auckland.ac.nz/dsc-2007/ Paul (on behalf of the Organising Committee) -- 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-announce __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fitting mixed-effects models with lme with fixed error term variances
Hello! Douglas Bates wrote: On 11/22/06, Gregor Gorjanc [EMAIL PROTECTED] wrote: Douglas Bates wrote: On 11/21/06, Gregor Gorjanc [EMAIL PROTECTED] wrote: Douglas Bates bates at stat.wisc.edu writes: ... Can you be more specific about which parameters you want to fix and which you want to vary in the optimization? It would be nice to have the ability to fix all variances i.e. variances of random effects. ... effects but allow the variance of a slope for the same grouping factor to be estimated. Well, you could but only in the fortune(Yoda) sense. Yes, I agree here. Thank you for the detailed answer. By the way, if you fix all the variances then what are you optimizing over? The fixed effects? In that case the solution can be calculated explicitly for a linear mixed model. The conditional estimates of the fixed effects given the variance components are the solution to a penalized linear least squares problem. (Yes, the solution can also be expressed as a generalized linear least squares problem but there are advantages to using the penalized least squares representation. Yup. It would really be great to be able to do that nicely in R, say use lmer() once and since this might take some time use estimates of variance components next time to get fixed and random effects. Is this possible with lmer or any related function - not in fortune(Yoda) sense ;) Not quite. There is now a capability in lmer to specify starting estimates for the relative precision matrices which means that you can use the estimates from one fit as starting estimates for another fit. It looks like fm1 - lmer(...) fm2 - lmer(y ~ x + (...), start = [EMAIL PROTECTED]) I should say that in my experience this has not been as effective as I had hoped it would be. What I see in the optimization is that the first few iterations reduce the deviance quite quickly but the majority of the time is spent refining the estimates near the optimum. So, for example, if it took 40 iterations to converge from the rather crude starting estimates calculated within lmer it might instead take 35 iterations if you give it good starting estimates. Yes, I also have the same experience with other programs, say VCE[1]. What I was hopping for was just solutions from linear mixed model i.e. either via GLS or PLS representations and no optimization if values for variance components are passed as input - there should be a way to distinguish that from just passing starting values.. [1]http://vce.tzv.fal.de/index.pl Gregor __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] integrate R code to WinEdt file
I want to integrate R code to my WinEdt file. Can someone tell me how to do this? I copy R code to my WinEdt file, but it does't work. Aimin Yan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Sweave question
I try Sweave and get Sweave-test-1.tex but hot to run LaTeX on 'Sweave-test-1.tex'? I am using WinEdt. thanks, Aimin Sweave(testfile) Writing to file Sweave-test-1.tex Processing code chunks ... 1 : print term verbatim 2 : term hide 3 : echo print term verbatim 4 : term verbatim 5 : echo term verbatim 6 : echo term verbatim eps pdf 7 : echo term verbatim eps pdf You can now run LaTeX on 'Sweave-test-1.tex' LaTeX() Error: could not find function LaTeX __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sweave question
Hello, R actually creates a latexfile. I am actually using Miktex and the editor Texniccenter so I open the *.tex file there and then I can do whatever I want. Usually I use latex=PDF to create pdf files. So I think you have to handle your .tex file with another application. (How do you usually create something out of latexfiles? hope that helped.. Markus Schweitzer http://www.pokertips.tk -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Aimin Yan Sent: Donnerstag, 23. November 2006 07:16 To: r-help@stat.math.ethz.ch Subject: [R] Sweave question I try Sweave and get Sweave-test-1.tex but hot to run LaTeX on 'Sweave-test-1.tex'? I am using WinEdt. thanks, Aimin Sweave(testfile) Writing to file Sweave-test-1.tex Processing code chunks ... 1 : print term verbatim 2 : term hide 3 : echo print term verbatim 4 : term verbatim 5 : echo term verbatim 6 : echo term verbatim eps pdf 7 : echo term verbatim eps pdf You can now run LaTeX on 'Sweave-test-1.tex' LaTeX() Error: could not find function LaTeX __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data in form of a date
James J. Roper escribió: Dear all, I often use dates and times in analyses. I just can't figure out how to format my date or time column in R. So, apparently R sees the date as something other than date (character). Let's say I am opening a CSV file, one of the columns of which is a date or time. How do I specify that when opening the file? Thanks for the help, Jim Jim, Suppose a matrix F (6575,189) where rows are days and columns some grid points: First we set the starting date of the series (6575 is the total number of days): library(zoo) x.Date-as.Date(1987-12-31)+ c(1:6575) F.zoo - zoo(F, x.Date) #then take a look F.zoo[,1] HTH Antonio __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] installation step for RSperl
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 You have to give us some details about what went wrong and details about the platform/operating system you are using. D. Yuandan Zhang wrote: Hi, I try to use R within perl. however, I have a bit difficulty to install RSperl. I followd steps from http://www.omegahat.org/RSPerl/. but still can' t make it work. could someone list a fe w clean steps that I can follow to install it? cheers __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - -- Duncan Temple Lang[EMAIL PROTECTED] Department of Statistics work: (530) 752-4782 4210 Mathematical Sciences Building fax: (530) 752-7099 One Shields Ave. University of California at Davis Davis, CA 95616, USA -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.3 (Darwin) iD8DBQFFZUvo9p/Jzwa2QP4RAqYtAJ98rZE2w2RSIIdNShh1xf1OX2Z4jgCfSryK nrrDdgBNhMNH9Gi/iUTalWM= =euUt -END PGP SIGNATURE- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] What training algorithm does nnet package use?
Wee-Jin Goh wjgoh at brookes.ac.uk writes: Just to add to this, I also need to know what language is the nnet package written in? Is it in pure R or is it a wrapper for a C library. As usual, you can download the full source to find out what you want, but it's a bit hidden. Simply said, nnet (R+C) is part of package MASS is part of bundle VR, and can be downloaded as a tar.gz from http://cran.at.r-project.org/src/contrib/Descriptions/VR.html (No private flames, please, in case I should have mixed up package and bundle). /* nnet/nnet.c by W. N. Venables and B. D. Ripley Copyright (C) 1992-2002 * * weights are stored in order of their destination unit. * the array Conn gives the source unit for the weight (0 = bias unit) * the array Nconn gives the number of first weight connecting to each unit, * so the weights connecting to unit i are Nconn[i] ... Nconn[i+1] - 1. * */ #include R.h #include R_ext/Applic.h __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
