[R] fitting an interaction term
Hello! IŽm fitting a model with glm(family binomial). The best model counts 9 Variables and includes an interaction term that was generated by the product of to continuous variables (a*b). All variables are correlated under a value of 0.7 (Spearman rank order) While the estimates of both main effects are negativ, the resulting interaction term is positiv. This change of sign makes it difficult to interpret the model and above all, is this perhaps due to a bad variable choice ? Thanks a lot for helping Christian __ 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] Substring and strsplit
you can also use substring(), e.g., substring(x3, 1:nchar(x3), 1:nchar(x3)) Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: Erin Hodgess [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Wednesday, August 30, 2006 12:25 AM Subject: [R] Substring and strsplit Dear R People: I am trying to split a character vector into a set of individual letters: Ideal: x3 - c(dog) d o g I tried the following: strsplit(x3) Error in strsplit(x3) : argument split is missing, with no default strsplit(x3,1) [[1]] [1] dog I know that this is incredibly simple, but what am I doing wrong? Either Windows or Linux 2.3.1 Thanks in advance! Sincerely, Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help on apply() function
Respected Sir/Madam, I have a problem with apply function. I have to two matrices of dimension of one column but n rows. I have to check whether one matrix is greater than other by going thru each row (ie) using if condition to check one matrix with another matrix. I like to use apply() function to this approach. That is apply function between two matrices. I searched for examples online but I couldn't find any. I don't know how to loop thru the matrices. Please help with this problem. Any help is appreciated. My mailid is [EMAIL PROTECTED] Thanks, Anusha. __ 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 of SrcStatConnectorSrv on Windows
I am trying to install SrcStatConnectorSrv(2) and rcom from local zip files. I have successfully downloaded both files (from CRAN for rcom CRAN Other for SrcStatConnectorSrv) and installed rcom. However, I get the following error message when I try to install SrcStatConnectorSrv (the version dated 21-Aug-2006: utils:::menuInstallLocal() Error in gzfile(file, r) : unable to open connection In addition: Warning message: cannot open compressed file 'SrcStatConnectorSrv(2)/DESCRIPTION' I have also run RSiteSearch(SrcStatConnectorSrv) but this failed to yield any results. I am running Windows XP Professional and version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 3.1 year 2006 month 06 day01 svn rev38247 language R version.string Version 2.3.1 (2006-06-01) Could anyone shed any light on what I am doing wrong? Many thanks in anticipation Gordon HSBC Bank plc may be solicited in the course of its placement efforts for a new issue, by investment clients of the firm for whom the Bank as a firm already provides other services. It may equally decide to allocate to its own proprietary book or with an associate of HSBC Group. This represents a potential conflict of interest. HSBC Bank plc has internal arrangements designed to ensure that the firm would give unbiased and full advice to the corporate finance client about the valuation and pricing of the offering as well as internal systems, controls and procedures to identify and manage conflicts of interest. HSBC Bank plc Registered Office: 8 Canada Square, London E14 5HQ, United Kingdom Registered in England - Number 14259 Authorised and regulated by the Financial Services Authority. HSBC Bank plc may be solicited in the course of its placement efforts for a new issue, by investment clients of the firm for whom the Bank as a firm already provides other services. It may equally decide to allocate to its own proprietary book or with an associate of HSBC Group. This represents a potential conflict of interest. HSBC Bank plc has internal arrangements designed to ensure that the firm would give unbiased and full advice to the corporate finance client about the valuation and pricing of the offering as well as internal systems, controls and procedures to identify and manage conflicts of interest. HSBC Bank plc Registered Office: 8 Canada Square, London E14 5HQ, United Kingdom Registered in England - Number 14259 Authorised and regulated by the Financial Services Authority. - SAVE PAPER - THINK BEFORE YOU PRINT! This transmission has been issued by a member of the HSBC Group HSBC for the information of the addressee only and should not be reproduced and/or distributed to any other person. Each page attached hereto must be read in conjunction with any disclaimer which forms part of it. Unless otherwise stated, this transmission is neither an offer nor the solicitation of an offer to sell or purchase any investment. Its contents are based on information obtained from sources believed to be reliable but HSBC makes no representation and accepts no responsibility or liability as to its completeness or accuracy. __ 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] Substring and strsplit
If you are using 'only' English then str - dog strsplit(str,NULL)[[1]] works perfectly and it is fast. But if you also dealing with Unicode character have a look at http://wiki.r-project.org/rwiki/doku.php?id=tips:data- strings:decomposestring Cheers, Hans you can also use substring(), e.g., substring(x3, 1:nchar(x3), 1:nchar(x3)) Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: Erin Hodgess [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Wednesday, August 30, 2006 12:25 AM Subject: [R] Substring and strsplit Dear R People: I am trying to split a character vector into a set of individual letters: Ideal: x3 - c(dog) d o g I tried the following: strsplit(x3) Error in strsplit(x3) : argument split is missing, with no default strsplit(x3,1) [[1]] [1] dog I know that this is incredibly simple, but what am I doing wrong? Either Windows or Linux 2.3.1 Thanks in advance! Sincerely, Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] MCMClogit
Hi, I am using MCMCpack and the MCMClogit function to create logistic regression models in a medical (adverse event) study. My question is, is there a way where I can directly create the estimated probabilities of the adverse outcome, moreover the confidence interval for the estimated probabilities? Thank you for your help! Andras __ 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 put title Vertically
Dear all R users, Suppose, Dear all R users, Suppose, pauto.cor = pacf(lh, plot=F) max.lag = max(pauto.cor$lag) min.lag = min(pauto.cor$lag) centre = (max.lag - min.lag)/2 pauto.cor = pauto.cor$acf pauto.cor = pauto.cor[-1] par(mar=c(3,0,1,1)) barplot(pauto.cor, axes=F,xlim=c(max(pauto.cor), min(pauto.cor)), space=0, col=green4,border=green,horiz=T) #This plots PACF vertically Now I want to put a title of above plot but NOT horizontally rather Vertically. Can anyone please tell me how to do that? Thanks and regards, stat thanks in advance __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Barplot
Dear all, I have a dataset. I want to make barplot from this data. Zero1 - V1 V2 V3 V4 V5 V6 V7 V8 V9 1 1 0 0 0 1 0 0 0 Positive 2 0 0 1 0 1 0 1 1 Negative 3 0 0 1 0 0 0 1 1 Positive 4 0 1 0 1 1 1 0 1 Negative 5 0 0 1 0 1 1 0 0 Positive 6 0 1 0 0 1 1 1 1 Negative 7 1 0 1 1 1 1 1 1 Negative 8 0 0 0 0 1 0 0 1 Negative 9 0 1 1 1 1 0 0 1 Negative 10 0 0 0 1 1 0 1 0 Positive 11 0 0 0 0 1 0 0 1 Negative 12 0 0 1 1 1 1 1 0 Positive 13 0 1 1 0 1 1 1 1 Negative z1 - read.table(textConnection(Zero1), header=TRUE) z1 str(z1) A simple way I can use mosaic plot mosaicplot(table(z1)) library(vcd) mosaic(table(z1)) I have tried to learn ?xtabs ?table and ?ftable but I can't figure out. I need a barplot for all variables and the result maybe like | | | | | | | | | || | | |pos|neg| |pos|neg||pos|neg| | | | | | || | | - -- v1v2v3 v7 v8 Thanks you for any helps. Regards, Muhammad Subianto __ 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 of SrcStatConnectorSrv on Windows
That is not an R package. It looks like the sources of the server described on http://cran.r-project.org/contrib/extra/dcom/RSrv200.html which tells you about the approriate mailing list to ask about it. You probably want to use RSrv200.exe to install the server (and other tools and examples). On Wed, 30 Aug 2006, [EMAIL PROTECTED] wrote: I am trying to install SrcStatConnectorSrv(2) and rcom from local zip files. I have successfully downloaded both files (from CRAN for rcom CRAN Other for SrcStatConnectorSrv) and installed rcom. However, I get the following error message when I try to install SrcStatConnectorSrv (the version dated 21-Aug-2006: utils:::menuInstallLocal() Error in gzfile(file, r) : unable to open connection In addition: Warning message: cannot open compressed file 'SrcStatConnectorSrv(2)/DESCRIPTION' I have also run RSiteSearch(SrcStatConnectorSrv) but this failed to yield any results. I am running Windows XP Professional and version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 3.1 year 2006 month 06 day01 svn rev38247 language R version.string Version 2.3.1 (2006-06-01) Could anyone shed any light on what I am doing wrong? Many thanks in anticipation Gordon HSBC Bank plc may be solicited in the course of its placement efforts for a new issue, by investment clients of the firm for whom the Bank as a firm already provides other services. It may equally decide to allocate to its own proprietary book or with an associate of HSBC Group. This represents a potential conflict of interest. HSBC Bank plc has internal arrangements designed to ensure that the firm would give unbiased and full advice to the corporate finance client about the valuation and pricing of the offering as well as internal systems, controls and procedures to identify and manage conflicts of interest. HSBC Bank plc Registered Office: 8 Canada Square, London E14 5HQ, United Kingdom Registered in England - Number 14259 Authorised and regulated by the Financial Services Authority. HSBC Bank plc may be solicited in the course of its placement efforts for a new issue, by investment clients of the firm for whom the Bank as a firm already provides other services. It may equally decide to allocate to its own proprietary book or with an associate of HSBC Group. This represents a potential conflict of interest. HSBC Bank plc has internal arrangements designed to ensure that the firm would give unbiased and full advice to the corporate finance client about the valuation and pricing of the offering as well as internal systems, controls and procedures to identify and manage conflicts of interest. HSBC Bank plc Registered Office: 8 Canada Square, London E14 5HQ, United Kingdom Registered in England - Number 14259 Authorised and regulated by the Financial Services Authority. - SAVE PAPER - THINK BEFORE YOU PRINT! This transmission has been issued by a member of the HSBC Group HSBC for the information of the addressee only and should not be reproduced and/or distributed to any other person. Each page attached hereto must be read in conjunction with any disclaimer which forms part of it. Unless otherwise stated, this transmission is neither an offer nor the solicitation of an offer to sell or purchase any investment. Its contents are based on information obtained from sources believed to be reliable but HSBC makes no representation and accepts no responsibility or liability as to its completeness or accuracy. __ 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] Barplot
barplot(t(sapply(split(z1[,1:8], z1$V9),colSums)), beside=T) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Muhammad Subianto a écrit : Dear all, I have a dataset. I want to make barplot from this data. Zero1 - V1 V2 V3 V4 V5 V6 V7 V8 V9 1 1 0 0 0 1 0 0 0 Positive 2 0 0 1 0 1 0 1 1 Negative 3 0 0 1 0 0 0 1 1 Positive 4 0 1 0 1 1 1 0 1 Negative 5 0 0 1 0 1 1 0 0 Positive 6 0 1 0 0 1 1 1 1 Negative 7 1 0 1 1 1 1 1 1 Negative 8 0 0 0 0 1 0 0 1 Negative 9 0 1 1 1 1 0 0 1 Negative 10 0 0 0 1 1 0 1 0 Positive 11 0 0 0 0 1 0 0 1 Negative 12 0 0 1 1 1 1 1 0 Positive 13 0 1 1 0 1 1 1 1 Negative z1 - read.table(textConnection(Zero1), header=TRUE) z1 str(z1) A simple way I can use mosaic plot mosaicplot(table(z1)) library(vcd) mosaic(table(z1)) I have tried to learn ?xtabs ?table and ?ftable but I can't figure out. I need a barplot for all variables and the result maybe like | | | | | | | | | || | | |pos|neg| |pos|neg||pos|neg| | | | | | || | | - -- v1v2v3 v7 v8 Thanks you for any helps. Regards, Muhammad Subianto __ 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] MCMClogit
Hi, I am using MCMCpack and the MCMClogit function to create logistic regression models in a medical (adverse event) study. My question is, is there a way where I can directly create the estimated probabilities of the adverse outcome, and the confidence interval for the estimated probabilities? Or is there another package I should use instead? Thank you for your help! Andras __ 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] density() with from, to or cut and comparrison of density()
Hi the function density() does normally integrate to one - I've checked it and it works and I also read the previous threads. But I realised that it does not integrate to one if I use from, to or cut. My scenario: I simulated densities of a plants originating from an sseed source at distance zero. Therefore the density of the plants will be highest close to zero. Is there anything I can do to have this pattern? If I use 'from' or 'cut', the resulting densities do not integrate to one which I need as I want to compare different density curves. Ny second question is concerning the bandwidth. An I correct in saying that if I want to compare different density estimates that the bandwidth should be the same for all of them? Thanks in advance for your help, Rainer -- Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT) Department of Conservation Ecology and Entomology University of Stellenbosch Matieland 7602 South Africa Tel:+27 - (0)72 808 2975 (w) Fax:+27 - (0)21 808 3304 Cell: +27 - (0)83 9479 042 email: [EMAIL PROTECTED] [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help on apply() function
On 8/30/06, Mark Lyman [EMAIL PROTECTED] wrote: I have a problem with apply function. I have to two matrices of dimension of one column but n rows. I have to check whether one matrix is greater than other by going thru each row (ie) using if condition to check one matrix with another matrix. I like to use apply() function to this approach. That is apply function between two matrices. I searched for examples online but I couldn't find any. I don't know how to loop thru the matrices. You can use the functions all.equal with the function isTRUE (see ?all.equal) to check if two objects are nearly equal (within a certain tolerance). Or you can use identical (see ?identical) to check if they are exactly the same. See the examples in the help for identical. A possible solution could be: x - 1:5 x [1] 1 2 3 4 5 y - 2:6 y [1] 2 3 4 5 6 all((y-x)0) [1] TRUE The result TRUE means that each element of y is greater than the homologous element in x. Paul __ 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] Barplot
Muhammad Subianto wrote: ... I have tried to learn ?xtabs ?table and ?ftable but I can't figure out. I need a barplot for all variables and the result maybe like | | | | | | | | | || | | |pos|neg| |pos|neg||pos|neg| | | | | | || | | - -- v1v2v3 v7 v8 barplot(sapply(z1[1:8],by,z1[9],sum),beside=TRUE) 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.
Re: [R] Firefox extension fo R Site Search
May be it's not a bug, but I tried to search for the package rpanel and I was not find. At the r-project's site that package is available. How to explain it? Rob __ 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] Firefox extension fo R Site Search
Robert Mcfadden a écrit : May be it's not a bug, but I tried to search for the package rpanel and I was not find. At the r-project's site that package is available. How to explain it? Rob Hi Rob, If it's not there : http://finzi.psych.upenn.edu/R/library/ it's not on the extension Cheers, Romain __ 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 put title Vertically
stat stat wrote: Dear all R users, Suppose, Dear all R users, Suppose, pauto.cor = pacf(lh, plot=F) max.lag = max(pauto.cor$lag) min.lag = min(pauto.cor$lag) centre = (max.lag - min.lag)/2 pauto.cor = pauto.cor$acf pauto.cor = pauto.cor[-1] par(mar=c(3,0,1,1)) barplot(pauto.cor, axes=F,xlim=c(max(pauto.cor), min(pauto.cor)), space=0, col=green4,border=green,horiz=T) #This plots PACF vertically Now I want to put a title of above plot but NOT horizontally rather Vertically. Can anyone please tell me how to do that? Use mtext() Uwe Ligges Thanks and regards, stat thanks in advance __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ 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] density() with from, to or cut and comparrison of density()
Rainer M Krug wrote: Hi the function density() does normally integrate to one - I've checked it and it works and I also read the previous threads. But I realised that it does not integrate to one if I use from, to or cut. My scenario: I simulated densities of a plants originating from an sseed source at distance zero. Therefore the density of the plants will be highest close to zero. Is there anything I can do to have this pattern? If I use 'from' or 'cut', the resulting densities do not integrate to one which I need as I want to compare different density curves. The kernel chosen might be not the ideal one for such a restriction. If the density outside the cut range is extremely small, you might want to do a dirty transformation so that the values sum up to 1 again. Ny second question is concerning the bandwidth. An I correct in saying that if I want to compare different density estimates that the bandwidth should be the same for all of them? Yes. Uwe Ligges Thanks in advance for your help, Rainer __ 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] Firefox extension fo R Site Search
Romain Francois wrote: Robert Mcfadden a écrit : May be it's not a bug, but I tried to search for the package rpanel and I was not find. At the r-project's site that package is available. How to explain it? Rob Hi Rob, If it's not there : http://finzi.psych.upenn.edu/R/library/ it's not on the extension ... because that side is update once a month, hence simply wait for a couple of days until it will show up. Uwe Ligges Cheers, Romain __ 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] Firefox extension fo R Site Search
On 08/30/06 11:55, Romain Francois wrote: Robert Mcfadden a ?crit : May be it's not a bug, but I tried to search for the package rpanel and I was not find. At the r-project's site that package is available. How to explain it? Rob Hi Rob, If it's not there : http://finzi.psych.upenn.edu/R/library/ it's not on the extension The list is updated monthly, and rpanel is new. Jon -- Jonathan Baron, Professor of Psychology, University of Pennsylvania Home page: http://www.sas.upenn.edu/~baron __ 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] Barplot - thanks
Dear all, Many Thanks to Jacques VESLOT and Jim Lemon for their helps. Best, Muhammad Subianto #Jacques VESLOT barplot(t(sapply(split(z1[,1:8], z1$V9),colSums)), beside=T) #Jim Lemon barplot(sapply(z1[1:8],by,z1[9],sum),beside=TRUE) On this day 30/08/2006 11:43, Muhammad Subianto wrote: Dear all, I have a dataset. I want to make barplot from this data. Zero1 - V1 V2 V3 V4 V5 V6 V7 V8 V9 1 1 0 0 0 1 0 0 0 Positive 2 0 0 1 0 1 0 1 1 Negative 3 0 0 1 0 0 0 1 1 Positive 4 0 1 0 1 1 1 0 1 Negative 5 0 0 1 0 1 1 0 0 Positive 6 0 1 0 0 1 1 1 1 Negative 7 1 0 1 1 1 1 1 1 Negative 8 0 0 0 0 1 0 0 1 Negative 9 0 1 1 1 1 0 0 1 Negative 10 0 0 0 1 1 0 1 0 Positive 11 0 0 0 0 1 0 0 1 Negative 12 0 0 1 1 1 1 1 0 Positive 13 0 1 1 0 1 1 1 1 Negative z1 - read.table(textConnection(Zero1), header=TRUE) z1 str(z1) A simple way I can use mosaic plot mosaicplot(table(z1)) library(vcd) mosaic(table(z1)) I have tried to learn ?xtabs ?table and ?ftable but I can't figure out. I need a barplot for all variables and the result maybe like | | | | | | | | | || | | |pos|neg| |pos|neg||pos|neg| | | | | | || | | - -- v1v2v3 v7 v8 Thanks you for any helps. Regards, Muhammad Subianto __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] Version 1.2-0 of the Rcmdr package
I've submitted a new, and substantially enhanced, version (1.2-0) of the Rcmdr package to CRAN. Some highlights (from the CHANGES) file: o Added ability to import from Excel, Access or dBase files (contributed by Samir Messad, Renaud Lancelot and Matthieu Lesnoff). o Added ability to read data from the clipboard (suggested by Graham Smith). o Added Data - Active data set - Stack variables in active data set (suggested by Richard Heiberger). o Added many probability distributions, courtesy of Miroslav Ristic and G. Jay Kerns, Andy Chang, and Theophilius Boye. o Added dialogs to sample from probability distributions; divided distributions menus into Continuous and Discrete. o Added totPercents function. Added total percentages and components of chi-square to two-way tables (suggested by Richard Heiberger). o Added Statistics - Summaries - Count missing observations. o Added Statistics - Summaries - Correlation test, contributed by Stefano Calza. o The numerical summaries, table of statistics, frequency distributions, numeric to factor, and recode dialogs will now process multiple variables in parallel (suggested by Bo Sommerlund). o Added function numSummaries for neater output from Statistics - Summaries - Numerical summaries (motivated by comments from Bo Sommerlund). o XY conditioning plots dialog added, courtesy of Richard Heiberger. o Enhancements to 3D scatterplots. o Small interface improvements (e.g., pressing letter key in a list box moves to next entry starting with that letter). o New Rcmdr options (etc and etcMenus) determine where the Commander will look for its configuration and menu files (motivated by a suggestion by Richard Heiberger). Rcmdr version 1.2-0 includes revised Catalan and Japanese translations (courtesy of Manel Salamero and Takaharu Araki); as other revised translations become available, I'll submit updated versions of the package to CRAN. Version 1.2-1 will also have a revised introductory manual. As usual, bug reports, comments, and suggestions are welcome. John John Fox Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox ___ R-packages mailing list R-packages@stat.math.ethz.ch https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Datetime
Hi all, I'm getting confused by date handling. I wish to read a date from a file wich is a number of seconds since 1970 (POSIXct). Then i wish to convert this date to a human readable form (POSIXlt) By example : ctDate-1132963200 #Wed Aug 30 14:24:37 2006 is(ctDate) [1] numeric vector ctDate isn't a numeric vector, it is a POSIXct... none of the clues inside the docs has given me the answer, i can't cast it, i can't converti t to POSIXlt because it is wrong class (as.POSIX** don't want to work...) What is the trick ?? Any ideas would be great thks COMTE Guillaume [[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] Handling realisations in geoRglm
Chris Barker cmbarker at ucdavis.edu writes: However, after reading the packages' documentation and searching the mailing lists and other sources, it seems that the functions in geoRglm do not acknowledge the existence of multiple realisations per site. I see that the likfit() function in geoR has an argument for realisations, but I cannot find anything similar in geoRglm. Is it possible to model data with repeated realisations from the same sites using geoRglm? If so, how can this be done and is there a way to model dependence among the realisations over time? I don't know; I would be interested in the conclusions you reach. I, too, found the support for multiple realisations to be relatively skimpy in geoR (only likfit, and not any of the variogram-based methods, support realisations); the code for likfit.glsm in the geoRglm packages also reveals no hint of realisation code. Even so, what there is in geoR says quite strongly that realisations are expected to be independent. My guess is that for complex space/time dependence, and wanting an answer to a practical problem, you would be better off with a GEE approach ... Writing the package maintainer directly would probably be more useful. Ben Bolker __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] MCMClogit
Andras, At this point you need to write your own function to take the posterior density sample (stored in a coda mcmc object) and covariates of interest to get a Monte Carlo estimate of these probabilities. Best, ADM On Aug 30, 2006, at 5:10 AM, Andras Treszl wrote: Hi, I am using MCMCpack and the MCMClogit function to create logistic regression models in a medical (adverse event) study. My question is, is there a way where I can directly create the estimated probabilities of the adverse outcome, and the confidence interval for the estimated probabilities? Or is there another package I should use instead? Thank you for your help! Andras __ 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. -- Andrew D. Martin, Ph.D. Professor and CERL Director, School of Law Professor, Department of Political Science Washington University in St. Louis (314) 935-5863 (Office) (314) 935-5150 (Fax) Office: Anheuser-Busch 470 Email: [EMAIL PROTECTED] WWW: http://adm.wustl.edu __ 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] Create a vector from another vector
Dear list Suppose I have the following vector: x - c(3,4,2,5,6) Obviously, this sums to 20. Now, I want to have a second vector, call it x2, that sums to x where 5 = x = 20, but there are constraints. 1) The new vector must be same length as x 2) No element of the new vector can be 0 3) Element x2[i] of the new vector cannot be larger than element x[i] of the original vector 4) Ordering is not important The following would be an example of what I would want if the user wanted the vector x2 to sum to 19 x2 - c(2,4,2,5,6) Or, because ordering is not important, this is acceptable x2 - c(3,3,2,5,6) Whereas this would not be appropriate x3 - c(4, 2,2,5,6) Because element x3[1] is larger than x[1] even though it sums to 19. Ideally, the function would take as input the original vector, x, and the number that the new vector would sum to. In this example, the vector could sum to any number 5 through 20. For example, myFun - function(x, sumto) ... details ... Is there a preexisiting function that would already do this? I have spent too much (unsuccessful) time trying to write a function of my own but can't seem to get this to work properly. Any hints would be greatly appreciated. Harold [[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] Create a vector from another vector
Dear Harold package partitions does almost this: library(partitions) x - 1+restrictedparts(15,5) x[,1:10] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [1,] 16 15 14 13 12 11 10 [2,]1234567 [3,]1111111 [4,]1111111 [5,]1111111 [,8] [,9] [,10] [1,]9 1413 [2,]82 3 [3,]12 2 [4,]11 1 [5,]11 1 HTH Robin On 30 Aug 2006, at 14:49, Doran, Harold wrote: Dear list Suppose I have the following vector: x - c(3,4,2,5,6) Obviously, this sums to 20. Now, I want to have a second vector, call it x2, that sums to x where 5 = x = 20, but there are constraints. 1) The new vector must be same length as x 2) No element of the new vector can be 0 3) Element x2[i] of the new vector cannot be larger than element x [i] of the original vector 4) Ordering is not important The following would be an example of what I would want if the user wanted the vector x2 to sum to 19 x2 - c(2,4,2,5,6) Or, because ordering is not important, this is acceptable x2 - c(3,3,2,5,6) Whereas this would not be appropriate x3 - c(4, 2,2,5,6) Because element x3[1] is larger than x[1] even though it sums to 19. Ideally, the function would take as input the original vector, x, and the number that the new vector would sum to. In this example, the vector could sum to any number 5 through 20. For example, myFun - function(x, sumto) ... details ... Is there a preexisiting function that would already do this? I have spent too much (unsuccessful) time trying to write a function of my own but can't seem to get this to work properly. Any hints would be greatly appreciated. Harold [[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. -- Robin Hankin Uncertainty Analyst National Oceanography Centre, Southampton European Way, Southampton SO14 3ZH, UK tel 023-8059-7743 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Optimization
Dear R-list, I'm trying to estimate the relative importance of 6 environmental variables in determining clam yield. To estimate clam yield a previous work used the function Yield = (A^a*B^b*C^c...)^1/(a+b+c+...) where A,B,C... are the values of the environmental variables and the weights a,b,c... have not been calibrated on data but taken from literature. Now I'd like to estimate the weights a,b,c... by using a dataset with 110 observations of yield and values of the environmental variables. I'm wondering if it is feasible or if the number of observation is too low, if some data transformation is needed and which R function is the most appropriate to try to estimate the weights. Any help would be greatly appreciated. Simone Vincenzi _ Simone Vincenzi, PhD Student Department of Environmental Sciences University of Parma Parco Area delle Scienze, 33/A, 43100 Parma, Italy Phone: +39 0521 905696 Fax: +39 0521 906611 e.mail: [EMAIL PROTECTED] -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] converting decimal - hexadecimal
Hi, do you know, a method to convert an decimal value (integer) to the corresponding hexadecimal value ? thinks for help. Romain -- Lorrillière Romain UMR 8079 Laboratoire Ecologie, Systématique et Evolution Bât. 362 Université Paris-Sud 91405 Orsay cedex France tel : 01 69 15 56 85 fax : 01 69 15 56 96 mobile : 06 81 70 90 70 email : [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Romain Lorrillière UMR 8079 Laboratoire Ecologie, Systématique et Evolution Bât. 362 Université Paris-Sud 91405 Orsay cedex France tel : 01 69 15 56 85 fax : 01 69 15 56 96 mobile : 06 81 70 90 70 email : [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Create a vector from another vector
On 8/30/2006 9:49 AM, Doran, Harold wrote: Dear list Suppose I have the following vector: x - c(3,4,2,5,6) Obviously, this sums to 20. Now, I want to have a second vector, call it x2, that sums to x where 5 = x = 20, but there are constraints. 1) The new vector must be same length as x 2) No element of the new vector can be 0 3) Element x2[i] of the new vector cannot be larger than element x[i] of the original vector 4) Ordering is not important The following would be an example of what I would want if the user wanted the vector x2 to sum to 19 x2 - c(2,4,2,5,6) Or, because ordering is not important, this is acceptable x2 - c(3,3,2,5,6) Whereas this would not be appropriate x3 - c(4, 2,2,5,6) Because element x3[1] is larger than x[1] even though it sums to 19. I don't think it's really clear what you mean by ordering is not important. Would x2 - c(6,5,2,4,2) be acceptable (a re-ordering of your first two examples), even though x2[1] x1[1]? It's also not clear what result you want in the usual case where there are multiple possible answers. Do you want a randomly chosen one (from what distribution?), or would any value satisfying the constraints be okay? (The latter would be easiest: just start with x1, and decrement entries until the desired sum is achieved. Whether a random value is easy or not really depends on the desired distribution.) Do entries need to be integer valued? Ideally, the function would take as input the original vector, x, and the number that the new vector would sum to. In this example, the vector could sum to any number 5 through 20. For example, myFun - function(x, sumto) ... details ... Is there a preexisiting function that would already do this? I have spent too much (unsuccessful) time trying to write a function of my own but can't seem to get this to work properly. I doubt it, because this doesn't look like a standard problem. Duncan Murdoch Any hints would be greatly appreciated. Harold [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Create a vector from another vector
I think I've got it now.
If I understand your question, try:
x - do.call(cbind,lapply(5:7,function(i){restrictedparts(i,
5,include.zero=FALSE)}))
acceptable - function(x){all(x=c(3,4,5,2,6))}
x[,apply(x,2,acceptable)]
[,1] [,2] [,3] [,4]
[1,]1232
[2,]1112
[3,]1111
[4,]1111
[5,]1111
rksh
On 30 Aug 2006, at 14:49, Doran, Harold wrote:
Dear list
Suppose I have the following vector:
x - c(3,4,2,5,6)
Obviously, this sums to 20. Now, I want to have a second vector,
call it
x2, that sums to x where 5 = x = 20, but there are constraints.
1) The new vector must be same length as x
2) No element of the new vector can be 0
3) Element x2[i] of the new vector cannot be larger than element x
[i] of
the original vector
4) Ordering is not important
The following would be an example of what I would want if the user
wanted the vector x2 to sum to 19
x2 - c(2,4,2,5,6)
Or, because ordering is not important, this is acceptable
x2 - c(3,3,2,5,6)
Whereas this would not be appropriate
x3 - c(4, 2,2,5,6)
Because element x3[1] is larger than x[1] even though it sums to 19.
Ideally, the function would take as input the original vector, x, and
the number that the new vector would sum to. In this example, the
vector
could sum to any number 5 through 20.
For example,
myFun - function(x, sumto) ... details ...
Is there a preexisiting function that would already do this? I have
spent too much (unsuccessful) time trying to write a function of my
own
but can't seem to get this to work properly.
Any hints would be greatly appreciated.
Harold
[[alternative HTML version deleted]]
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--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
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Re: [R] Barplot
Try this. First we reduce the data to a frequency matrix and then plot it using classic and then lattice graphics: zm - as.matrix(rowsum(z1[-9], z1[,9])) barplot(zm, beside = TRUE, col = grey.colors(2)) legend(topleft, legend = levels(z1[,9]), fill = grey.colors(2)) library(lattice) barchart(Freq ~ Var2, as.data.frame.table(zm), groups = Var1, origin = 0, auto.key = TRUE) On 8/30/06, Muhammad Subianto [EMAIL PROTECTED] wrote: Dear all, I have a dataset. I want to make barplot from this data. Zero1 - V1 V2 V3 V4 V5 V6 V7 V8 V9 1 1 0 0 0 1 0 0 0 Positive 2 0 0 1 0 1 0 1 1 Negative 3 0 0 1 0 0 0 1 1 Positive 4 0 1 0 1 1 1 0 1 Negative 5 0 0 1 0 1 1 0 0 Positive 6 0 1 0 0 1 1 1 1 Negative 7 1 0 1 1 1 1 1 1 Negative 8 0 0 0 0 1 0 0 1 Negative 9 0 1 1 1 1 0 0 1 Negative 10 0 0 0 1 1 0 1 0 Positive 11 0 0 0 0 1 0 0 1 Negative 12 0 0 1 1 1 1 1 0 Positive 13 0 1 1 0 1 1 1 1 Negative z1 - read.table(textConnection(Zero1), header=TRUE) z1 str(z1) A simple way I can use mosaic plot mosaicplot(table(z1)) library(vcd) mosaic(table(z1)) I have tried to learn ?xtabs ?table and ?ftable but I can't figure out. I need a barplot for all variables and the result maybe like | | | | | | | | | || | | |pos|neg| |pos|neg||pos|neg| | | | | | || | | - -- v1v2v3 v7 v8 Thanks you for any helps. Regards, Muhammad Subianto __ 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] converting decimal - hexadecimal
Try: sprintf(%x, 109) On 9/30/06, Romain Lorrilliere [EMAIL PROTECTED] wrote: Hi, do you know, a method to convert an decimal value (integer) to the corresponding hexadecimal value ? thinks for help. Romain -- Lorrillière Romain UMR 8079 Laboratoire Ecologie, Systématique et Evolution Bât. 362 Université Paris-Sud 91405 Orsay cedex France tel : 01 69 15 56 85 fax : 01 69 15 56 96 mobile : 06 81 70 90 70 email : [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Romain Lorrillière UMR 8079 Laboratoire Ecologie, Systématique et Evolution Bât. 362 Université Paris-Sud 91405 Orsay cedex France tel : 01 69 15 56 85 fax : 01 69 15 56 96 mobile : 06 81 70 90 70 email : [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] converting decimal - hexadecimal
use sprintf sprintf(%x,123) [1] 7b On 9/30/06, Romain Lorrilliere [EMAIL PROTECTED] wrote: Hi, do you know, a method to convert an decimal value (integer) to the corresponding hexadecimal value ? thinks for help. Romain -- Lorrillière Romain UMR 8079 Laboratoire Ecologie, Systématique et Evolution Bât. 362 Université Paris-Sud 91405 Orsay cedex France tel : 01 69 15 56 85 fax : 01 69 15 56 96 mobile : 06 81 70 90 70 email : [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Romain Lorrillière UMR 8079 Laboratoire Ecologie, Systématique et Evolution Bât. 362 Université Paris-Sud 91405 Orsay cedex France tel : 01 69 15 56 85 fax : 01 69 15 56 96 mobile : 06 81 70 90 70 email : [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ 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] converting decimal - hexadecimal
?sprintf
ex :
sprintf('%X',10)
[1] A
hih
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[R] function which gives the hessian matrix of the log-likelihood of a nonlinear mixed model?
Hi, Can anyone tell me which function in R gives the hessian matrix of the log-likelihood of a nonlinear mixed model? fdHess is for scarlar function only. Thanks in advance! Hongmei [[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] converting decimal - hexadecimal
how about learning to use help.search? (In may you already asked a similiar question...) help.search(hexadecimal) which would lead you to format.hexmode ?format.hexmode Romain Lorrilliere schrieb: Hi, do you know, a method to convert an decimal value (integer) to the corresponding hexadecimal value ? thinks for help. Romain __ 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] working with summarized data
The data sets I am working with all have a weight variable--e.g., each row doesn't mean 1 observation. With that in mind, nearly all of the graphs and summary statistics are incorrect for my data, because they don't take into account the weight. For example median is incorrect, as the quantiles aren't calculated with weights: sum( weights[X median(X)] ) / sum(weights) This should be 0.5... of course it's not. Unfortunately, it seems that most(all?) of R's graphics and summary statistic functions don't take a weight or frequency argument. (Fortunately the models do...) Am I completely missing how to do this? One way would be to replicate each row proportional to the weight (e.g. if the weight was 4, we would 3 additional copies) but this will get prohibitive pretty quickly as the dataset grows. Thanks in advance! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Create a vector from another vector
Hi Duncan
Here is a bit more detail, this is a bit tough to explain, sorry for not
being clear. Ordering is not important because the vector I am creating
is used as a sufficient statistic in an optimization routine to get some
MLEs. So, any combination of the vector that sums to X is OK. But, the
condition that x2[i] = x[i] must be maintained. So, the example below
would not work because x2[1] x[1] as you note below.
I don't think it's really clear what you mean by ordering is
not important. Would
x2 - c(6,5,2,4,2)
be acceptable (a re-ordering of your first two examples),
even though x2[1] x1[1]?
To be concrete, the following is the optimization function. This is a
psychometric problem where the goal is to get the MLE for a test taker
conditional on their response pattern (i.e., number of points on the
test) and the item parameters.
pcm.max3 - function(score, d){
pcm - function(theta, d, score)
exp(sum(theta-d[1:score]))/sum(exp(cumsum(theta-d)))
opt - function(theta) -sum(log(mapply(pcm, d, theta = theta, score=
score )))
start_val - log(sum(score-1)/(length(score-1)/sum(score-1)))
out - optim(start_val, opt, method = BFGS, hessian = TRUE)
cat('theta is about', round(out$par, 2), ', se',
1/sqrt(out$hes),'\n')
}
Suppose we have a three item test. I store the item parameters in a list
as
items - list(c(0,.5,1), c(0,1), c(0, -1, .5, 1))
We can get the total possible number correct as
(x - sapply(items, length))
[1] 3 2 4
But, you cannot actually get the MLE for this because the likelihood is
unbounded in this case.
So, let's say the student scored in the following categories for each
item:
x2 - c(3,1,4)
By x2[i] = x[i], I mean that there are 3 possible categories for item 1
above. So, a student can only score in categories 1,2 or 3. He cannot
score in category 4. This is why the condition that x2[i] = x[i] is
critical.
But, because total score is a sufficient statistic, (i.e., ordering is
not important) we could either vector in the function pcm.
x3 - c(3,2,3)
Using the function
pcm.max3(x2, items)
pcm.max3(x3, items)
Gives the same MLE.
But, the vector
X_bad - c(4,1,3)
Would not work. You can see that the elements of this vector actually
serve as indices denoting which category a test taker scored in for each
item in the list items
I hope this is helpful and appreciate your time.
Harold
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Re: [R] converting decimal - hexadecimal
An other way would be:
a - 123
class(a) - hexmode
a
[1] 7b
On 30 Aug 2006, at 16:26, mel wrote:
?sprintf
ex :
sprintf('%X',10)
[1] A
hih
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**
Hans-Joerg Bibiko
Max Planck Institute for Evolutionary Anthropology
Department of Linguistics
Deutscher Platz 6 phone: +49 (0) 341 3550 341
D-04103 Leipzig fax: +49 (0) 341 3550 333
Germany e-mail: [EMAIL PROTECTED]
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Re: [R] converting decimal - hexadecimal
Romain Lorrilliere [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] Hi, do you know, a method to convert an decimal value (integer) to the corresponding hexadecimal value ? Starting in R 2.1.0, sprintf can be used: x - c(0, 65535, 65536, 305419896, 2^31-1) y - sprintf(0x%X, x) y [1] 0x00x 0x10x12345678 0x7FFF as.numeric(y) [1] 0 65535 65536 305419896 2147483647 efg Earl F. Glynn Scientific Programmer Stowers Institute for Medical Research __ 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] working with summarized data
In each case, look around (help.search, RSiteSearch) to see if you can find a function that handles weights. For the case you mention, medians, it can be done via quantile regression: x - w - 1:5 library(quantreg) coef(rq(x ~ 1, weight = w)) On 8/30/06, Rick Bischoff [EMAIL PROTECTED] wrote: The data sets I am working with all have a weight variable--e.g., each row doesn't mean 1 observation. With that in mind, nearly all of the graphs and summary statistics are incorrect for my data, because they don't take into account the weight. For example median is incorrect, as the quantiles aren't calculated with weights: sum( weights[X median(X)] ) / sum(weights) This should be 0.5... of course it's not. Unfortunately, it seems that most(all?) of R's graphics and summary statistic functions don't take a weight or frequency argument. (Fortunately the models do...) Am I completely missing how to do this? One way would be to replicate each row proportional to the weight (e.g. if the weight was 4, we would 3 additional copies) but this will get prohibitive pretty quickly as the dataset grows. Thanks in advance! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ 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] Create a vector from another vector
OK,
With this extra detail, a
third solution follows, which may be closer in spirit
to your application.
It may or may not be faster than the other two,
depending on the exact parameters used:
library(partitions)
1+blockparts(n=15,y=c(3,4,2,5,6)-1,include.fewer=T)
(720 distinct solutions)
rksh
On 30 Aug 2006, at 15:49, Doran, Harold wrote:
Hi Duncan
Here is a bit more detail, this is a bit tough to explain, sorry
for not
being clear. Ordering is not important because the vector I am
creating
is used as a sufficient statistic in an optimization routine to get
some
MLEs. So, any combination of the vector that sums to X is OK. But, the
condition that x2[i] = x[i] must be maintained. So, the example below
would not work because x2[1] x[1] as you note below.
I don't think it's really clear what you mean by ordering is
not important. Would
x2 - c(6,5,2,4,2)
be acceptable (a re-ordering of your first two examples),
even though x2[1] x1[1]?
To be concrete, the following is the optimization function. This is a
psychometric problem where the goal is to get the MLE for a test taker
conditional on their response pattern (i.e., number of points on the
test) and the item parameters.
pcm.max3 - function(score, d){
pcm - function(theta, d, score)
exp(sum(theta-d[1:score]))/sum(exp(cumsum(theta-d)))
opt - function(theta) -sum(log(mapply(pcm, d, theta = theta,
score=
score )))
start_val - log(sum(score-1)/(length(score-1)/sum(score-1)))
out - optim(start_val, opt, method = BFGS, hessian = TRUE)
cat('theta is about', round(out$par, 2), ', se',
1/sqrt(out$hes),'\n')
}
Suppose we have a three item test. I store the item parameters in a
list
as
items - list(c(0,.5,1), c(0,1), c(0, -1, .5, 1))
We can get the total possible number correct as
(x - sapply(items, length))
[1] 3 2 4
But, you cannot actually get the MLE for this because the
likelihood is
unbounded in this case.
So, let's say the student scored in the following categories for each
item:
x2 - c(3,1,4)
By x2[i] = x[i], I mean that there are 3 possible categories for
item 1
above. So, a student can only score in categories 1,2 or 3. He cannot
score in category 4. This is why the condition that x2[i] = x[i] is
critical.
But, because total score is a sufficient statistic, (i.e.,
ordering is
not important) we could either vector in the function pcm.
x3 - c(3,2,3)
Using the function
pcm.max3(x2, items)
pcm.max3(x3, items)
Gives the same MLE.
But, the vector
X_bad - c(4,1,3)
Would not work. You can see that the elements of this vector actually
serve as indices denoting which category a test taker scored in for
each
item in the list items
I hope this is helpful and appreciate your time.
Harold
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--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
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Re: [R] Create a vector from another vector
maybe something like this could be of help:
max.score - c(3,4,3) # max score for each item
all.pats - as.matrix(expand.grid(lapply(max.score, :, 1)))
all.pats[rowSums(all.pats) == 5, ]
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: Doran, Harold [EMAIL PROTECTED]
To: Duncan Murdoch [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Wednesday, August 30, 2006 4:49 PM
Subject: Re: [R] Create a vector from another vector
Hi Duncan
Here is a bit more detail, this is a bit tough to explain, sorry for
not
being clear. Ordering is not important because the vector I am
creating
is used as a sufficient statistic in an optimization routine to get
some
MLEs. So, any combination of the vector that sums to X is OK. But,
the
condition that x2[i] = x[i] must be maintained. So, the example
below
would not work because x2[1] x[1] as you note below.
I don't think it's really clear what you mean by ordering is
not important. Would
x2 - c(6,5,2,4,2)
be acceptable (a re-ordering of your first two examples),
even though x2[1] x1[1]?
To be concrete, the following is the optimization function. This is
a
psychometric problem where the goal is to get the MLE for a test
taker
conditional on their response pattern (i.e., number of points on the
test) and the item parameters.
pcm.max3 - function(score, d){
pcm - function(theta, d, score)
exp(sum(theta-d[1:score]))/sum(exp(cumsum(theta-d)))
opt - function(theta) -sum(log(mapply(pcm, d, theta = theta,
score=
score )))
start_val - log(sum(score-1)/(length(score-1)/sum(score-1)))
out - optim(start_val, opt, method = BFGS, hessian = TRUE)
cat('theta is about', round(out$par, 2), ', se',
1/sqrt(out$hes),'\n')
}
Suppose we have a three item test. I store the item parameters in a
list
as
items - list(c(0,.5,1), c(0,1), c(0, -1, .5, 1))
We can get the total possible number correct as
(x - sapply(items, length))
[1] 3 2 4
But, you cannot actually get the MLE for this because the likelihood
is
unbounded in this case.
So, let's say the student scored in the following categories for
each
item:
x2 - c(3,1,4)
By x2[i] = x[i], I mean that there are 3 possible categories for
item 1
above. So, a student can only score in categories 1,2 or 3. He
cannot
score in category 4. This is why the condition that x2[i] = x[i] is
critical.
But, because total score is a sufficient statistic, (i.e., ordering
is
not important) we could either vector in the function pcm.
x3 - c(3,2,3)
Using the function
pcm.max3(x2, items)
pcm.max3(x3, items)
Gives the same MLE.
But, the vector
X_bad - c(4,1,3)
Would not work. You can see that the elements of this vector
actually
serve as indices denoting which category a test taker scored in for
each
item in the list items
I hope this is helpful and appreciate your time.
Harold
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code.
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Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
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Re: [R] working with summarized data
Unfortunately, it seems that most(all?) of R's graphics and summary statistic functions don't take a weight or frequency argument. (Fortunately the models do...) I have been been meaning to add this functionality to my graphics package ggplot (http://had.co.nz/ggplot), but unfortunately haven't had time yet. I'm guessing you want something like: * scatterplot: scale size of point according to weight (can do) * bar chart: bars should have height proportional to weight (can do) * histogram: area proportion to weighting variable (have some half finished code to do) * smoothers: should automatically use weights * boxplot: use weighted quantiles/letter statistics (is there a function for that?) What else is there? densityplot is the only other one I can think of at the moment... With the rest of those, I could certainly live without it though! Thanks! __ 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] working with summarized data
On Wed, 30 Aug 2006, Rick Bischoff wrote: Unfortunately, it seems that most(all?) of R's graphics and summary statistic functions don't take a weight or frequency argument. (Fortunately the models do...) I have been been meaning to add this functionality to my graphics package ggplot (http://had.co.nz/ggplot), but unfortunately haven't had time yet. I'm guessing you want something like: * scatterplot: scale size of point according to weight (can do) * bar chart: bars should have height proportional to weight (can do) * histogram: area proportion to weighting variable (have some half finished code to do) * smoothers: should automatically use weights * boxplot: use weighted quantiles/letter statistics (is there a function for that?) What else is there? densityplot is the only other one I can think of at the moment... With the rest of those, I could certainly live without it though! Density plots, scatterplot smoothers, hexbin plots, bubble plots, histograms, and boxplots are available in the survey package. These are probability-weighted rather than frequency-weighted but it doesn't matter for graphics. You could use them as is (which requires setting up a survey design object) or rip the internals out of them. -thomas __ 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] lmer applied to a wellknown (?) example
Dear all, During my pre-R era I tried (yes, tried) to understand mixed models by working through the 'rat example' in Sokal and Rohlfs Biometry (2000) 3ed p 288-292. The same example was later used by Crawley (2002) in his Statistical Computing p 363-373 and I have seen the same data being used elsewhere in the litterature. Because this example is so thoroughly described, I thought it would be very interesting to analyse it also using lmer and to see how the different approaches and outputs differs - from the more or less manual old-school (?) approach in Sokal, aov in Crawley, and to mixed models by lmer. In the example, three treatments (Treatment) with two rats (Rat) each (i.e six unique rats in total). Three liver preparations (Liver) are taken from each rat (i.e 18 unique liver preparations), and two glycogen readings (Glycogen) are taken from each liver preparation (36 readings). We want to test if treatments has affected the glycogen levels. The readings are nested in preparation and the preparations nested in rats. The data can be found here (or on p. 289 in Sokal): http://www.bio.ic.ac.uk/research/mjcraw/statcomp/data/rats.txt // I was hoping to use the rat example as some kind of reference on my way to understand mixed models and using lmer. However, first I wish someone could check my suggested models! My suggestions: attach(rats) rats$Glycogen - as.numeric(Glycogen) rats$Treatment - as.factor(Treatment) rats$Rat - as.factor(Rat) rats$Liver - as.factor(Liver) str(rats) model1 - lmer(Glycogen ~ Treatment + (1|Liver) + (1|Rat), data=rats) summary(model1) Was that it? I also tried to make the 'liver-in-rat' nesting explicit (as suggested in 'Examples from...') model2 - lmer(Glycogen ~ Treatment + (1|Rat:Liver) + (1|Rat), data=rats) summary(model2) but then the random effects differs from model1. Does the non-unique coding of rats and preparations in the original data set mess things up? Do I need to recode the ids to unique levels... rats$rat2 - as.factor(rep(1:6, each=6)) rats$liver2 - as.factor(rep(1:18, each=2)) str(rats) ...and then: model3 - lmer(Glycogen ~ Treatment + (1|liver2) + (1|rat2), data=rats) # or maybe model3 - lmer(Glycogen ~ Treatment + (1|rat2:liver2) + (1|rat2), data=rats) Can anyone help me to get this right! Thanks in advance! P.S. Thanks to all contributors to lme/lmer topic on the list (yes, I have searched for 'lmer nested'...) and also the examples provided by John Fox' 'Linear mixed models, Appendix to An R and S-PLUS companion...' and Douglas Bates' 'Examples from Multilevel Software...' and R-news 5/1. Very helpful, but as usually I bet I missed something...Sorry. Regards, Henrik -- Henrik Pärn Department of Biology NTNU 7491 Trondheim Norway +47 735 96282 (office) +47 909 89 255 (mobile) +47 735 96100 (fax) __ 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] bootstrap for group and subgroup
Dear R-friends, I have a table data structured by group, subgroups, records and attributes. For each group and subgroup I have differente number of records (more than 200). I need bootstrap 100 records for each group/subgroup combinations and repeat it a big number of times. Could someone of you help me on this hard (almost for me) task. Kind regards, miltinho - Música para ver e ouvir: You're Beautiful, do James Blunt [[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] Bootstraping for groups and subgroups and joing with other table
I have a table with following collumns: State, SamplePlot, Species and
BodySize. I sampled bird species at
34 SamplePlots and 5 States (regions) monthly during two years. On each bird
record I measured bodysize
and identified the species. So I have many records of each species (about 150
species) at each SamplePlot
and each Region (State).
Now I would like bootstrap these data, selecting 50 records for each
State/SamplePlot combinations and
count how many species (richness) were sampled at bootstrap. I need to do this
1.000 times.
After that and need join the number of species [obtained at each bootstrap
and for each State/SamplePlot
combination] with a dataframe that have other attributes for SamplePlot (like
Area, Perimeter etc).
I asked a similar question earlier... IF you have data frame birds and
bird.plots, maybe something like this.
#initialize empty variable
boot-NULL
for(i in 1:100)
{
## split on state and site and create list a
a -split(birds, paste(birds$State,birds$SampleSite), drop=T)
# sample 50 rows each OR by number of observations (better?)
# b-lapply( a, function(x) x[sample(nrow(x), 50, replace=T),])
b-lapply( a, function(x) x[sample(nrow(x), replace=T),])
## count number of unique species or other statistic?
### and add row to boot matrix
boot-rbind(boot, unlist( lapply(b, function (x) length(unique(x$Species)) )
))
}
## mean
y-apply(boot, 2, mean)
## convert to data frame for merge
y-data.frame(y)
names(y)-boot.count
## add row names to bird.plots for easy join
rownames(bird.plots)-paste(bird.plots$State,bird.plots$SampleSite)
merge(bird.plots,y, by=0)
Row.names State SampleSite Area boot.count
1 Bahia Site1 Bahia Site1 10 1.00
2 Bahia Site2 Bahia Site2 25 1.96
3 Bahia Site3 Bahia Site3 70 1.72
4 Bahia Site4 Bahia Site4 15 1.73
5 Bahia Site5 Bahia Site55 1.42
6 RioJaneiro Site1 RioJaneiro Site1 32 2.49
7 RioJaneiro Site2 RioJaneiro Site2 45 1.63
8 RioJaneiro Site3 RioJaneiro Site3 10 2.37
9SaoPaulo Site1 SaoPaulo Site1 23 2.41
10 SaoPaulo Site2 SaoPaulo Site2 45 2.57
Chris Stubben
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Re: [R] Barplot
Dear all, To Gabor Grothendieck, (again) thanks you very much for your help. Now, I can play around with lattice package. Best, Muhammad Subianto #Gabor #reduce the data to a frequency matrix and #then plot it using classic and then lattice graphics: zm - as.matrix(rowsum(z1[-9], z1[,9])) barplot(zm, beside = TRUE, col = grey.colors(2)) legend(topleft, legend = levels(z1[,9]), fill = grey.colors(2)) library(lattice) barchart(Freq ~ Var2, as.data.frame.table(zm), groups = Var1, origin = 0, auto.key = TRUE) On this day 30/08/2006 16:18, Gabor Grothendieck wrote: Try this. First we reduce the data to a frequency matrix and then plot it using classic and then lattice graphics: zm - as.matrix(rowsum(z1[-9], z1[,9])) barplot(zm, beside = TRUE, col = grey.colors(2)) legend(topleft, legend = levels(z1[,9]), fill = grey.colors(2)) library(lattice) barchart(Freq ~ Var2, as.data.frame.table(zm), groups = Var1, origin = 0, auto.key = TRUE) On 8/30/06, Muhammad Subianto [EMAIL PROTECTED] wrote: Dear all, I have a dataset. I want to make barplot from this data. Zero1 - V1 V2 V3 V4 V5 V6 V7 V8 V9 1 1 0 0 0 1 0 0 0 Positive 2 0 0 1 0 1 0 1 1 Negative 3 0 0 1 0 0 0 1 1 Positive 4 0 1 0 1 1 1 0 1 Negative 5 0 0 1 0 1 1 0 0 Positive 6 0 1 0 0 1 1 1 1 Negative 7 1 0 1 1 1 1 1 1 Negative 8 0 0 0 0 1 0 0 1 Negative 9 0 1 1 1 1 0 0 1 Negative 10 0 0 0 1 1 0 1 0 Positive 11 0 0 0 0 1 0 0 1 Negative 12 0 0 1 1 1 1 1 0 Positive 13 0 1 1 0 1 1 1 1 Negative z1 - read.table(textConnection(Zero1), header=TRUE) z1 str(z1) A simple way I can use mosaic plot mosaicplot(table(z1)) library(vcd) mosaic(table(z1)) I have tried to learn ?xtabs ?table and ?ftable but I can't figure out. I need a barplot for all variables and the result maybe like | | | | | | | | | || | | |pos|neg| |pos|neg||pos|neg| | | | | | || | | - -- v1v2v3 v7 v8 Thanks you for any helps. Regards, Muhammad Subianto __ 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] Ranking and selection statistical procedure
Dear R helpers I would like to know if the Ranking and Selection statistical procedure has been implemented in R. I made a quick search in the R packages list but I could not find it. Thanks in advance Prasanna __ 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] Producing R demos
One option is to use VNC along with vncrec to do the recording (see the website: http://www.sodan.org/~penny/vncrec/). I think there are some other recorders also available for vnc, so you might try a google search. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Gregor Gorjanc Sent: Tuesday, August 29, 2006 3:03 PM To: r-help@stat.math.ethz.ch Subject: [R] Producing R demos Hello! I have found terrific demo or R package functionality at http://had.co.nz/reshape/french-fries-demo.html Author has told me off-list that he is using SnapzProX (on mac), and ghostwriter (http://had.co.nz/ghostwriter/) to automate the typing. Unfortunatelly, I do not have mac ;) Can anyone on the list suggest, how such a demo (video + automated typing of a script) could be produced under either Windows or Linux OS? I would like to create such a demo for presentation as I will not be able to install R on the machine, but I could play a movie. Thanks! -- 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-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] density() with from, to or cut and comparrison of density()
You may want to look at the logspline package, it uses a different technique than density does, but it estimates densities and allows you to tell the routine that there is a minimum value and that the density does not extend beyond there. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Rainer M Krug Sent: Wednesday, August 30, 2006 4:27 AM To: R help list Subject: [R] density() with from, to or cut and comparrison of density() Hi the function density() does normally integrate to one - I've checked it and it works and I also read the previous threads. But I realised that it does not integrate to one if I use from, to or cut. My scenario: I simulated densities of a plants originating from an sseed source at distance zero. Therefore the density of the plants will be highest close to zero. Is there anything I can do to have this pattern? If I use 'from' or 'cut', the resulting densities do not integrate to one which I need as I want to compare different density curves. Ny second question is concerning the bandwidth. An I correct in saying that if I want to compare different density estimates that the bandwidth should be the same for all of them? Thanks in advance for your help, Rainer -- Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT) Department of Conservation Ecology and Entomology University of Stellenbosch Matieland 7602 South Africa Tel:+27 - (0)72 808 2975 (w) Fax:+27 - (0)21 808 3304 Cell: +27 - (0)83 9479 042 email: [EMAIL PROTECTED] [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list 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] working with summarized data
There are functions to do weighted summary statistics in the Hmisc package (wtd.quantile, ...). For more complicated analyses (but not plots yet) the biglm package has a bigglm function that expects the data in chunks, you could write a function that expand parts of the dataset at a time. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Rick Bischoff Sent: Wednesday, August 30, 2006 8:28 AM To: r-help@stat.math.ethz.ch Subject: [R] working with summarized data The data sets I am working with all have a weight variable--e.g., each row doesn't mean 1 observation. With that in mind, nearly all of the graphs and summary statistics are incorrect for my data, because they don't take into account the weight. For example median is incorrect, as the quantiles aren't calculated with weights: sum( weights[X median(X)] ) / sum(weights) This should be 0.5... of course it's not. Unfortunately, it seems that most(all?) of R's graphics and summary statistic functions don't take a weight or frequency argument. (Fortunately the models do...) Am I completely missing how to do this? One way would be to replicate each row proportional to the weight (e.g. if the weight was 4, we would 3 additional copies) but this will get prohibitive pretty quickly as the dataset grows. Thanks in advance! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ 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] Debugging with gdb
I tried to run gdb in linux with emacs But could not even run a simple example in the writing extensions tutorial. The execution history is as follows. Gdb worked fine for other debugging such as C++ codes. Thanks Han I started R at echo of emacs by typing (also tried other methods mentioned in the tutorial both in emacs and xterm.) M-x gdb== R -d gdb (gdb) run Starting program: /home/gcmio/local.20060808/lib/R/bin/exec/R -cd /home/a409791/R/R-Test/ -fullname [Thread debugging using libthread_db enabled] [New Thread -1218514176 (LWP 11086)] WARNING: unknown option '-cd' ARGUMENT '/home/a409791/R/R-Test/' __ignored__ WARNING: unknown option '-fullname' R : Copyright 2006, The R Foundation for Statistical Computing Version 2.3.1 (2006-06-01) ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. [Previously saved workspace restored] Program received signal SIGINT, Interrupt. [Switching to Thread -1218514176 (LWP 11086)] 0x00366c58 in ___newselect_nocancel () from /lib/tls/libc.so.6 (gdb) b do_get Breakpoint 1 at 0x80ca4f5: file envir.c, line 1615. (gdb) signal 0 Continuing with no signal. x - 1 get(x) Breakpoint 1, do_get (call=0x9443878, op=0x934bd54, args=0x9416408, rho=0x9417a54) at envir.c:1615 1615checkArity(op, args); (gdb) p $1 History has not yet reached $1. (gdb) p R_PV(x) No symbol x in current context. (gdb) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice and several groups
hadley wickham a écrit : I would like to use the lattice library to show several groups on the same graph. Here's my example : ## the data f1 - factor(c(mod1,mod2,mod3),levels=c(mod1,mod2,mod3)) f1 - rep(f1,3) f2 - factor(rep(c(g1,g2,g3),each=3),levels=c(g1,g2,g3)) df - data.frame(val=c(4,3,2,5,4,3,6,5,4), x=rep(c(1,2,3),3),f1=f1,f2=f2) It's pretty easy to do this with ggplot: install.packages(ggplot, repos=http://ggobi.org/r/;) library(ggplot) qplot(x, val, data=df, shape=f2, colour=f1) Hadley Great, Hadley, but the code was not exactly the good one. The code below works fine for me : install.packages(ggplot, repos=http://ggobi.org/r/;) library(ggplot) qplot(x, val, data=df, glyph=f1, col=f2) In fact, my problem is to fit the data for every level of the f2 factor, showing the levels of the f1 factor and that for several surveys . Here's an example closer to my actual data : ## the data n - 18 x1 - seq(1,n) val1 - -2*x1+50 val2 - (-2*(x1-8)^2)+100 val3 - (-2*(x1-8)^2)+50 y - c(val1,val2,val3) x - rep(x1,3) f1 - rep(c(mod1,mod2,mod3),each=n/3) f1 - rep(f1,3) f2 - rep(c(g1,g2,g3),each=n) df - data.frame(x=x,y=y,f1=f1,f2=f2) surveys - factor(c(rep(survey1,n*3),rep(survey2,n*3),rep(survey3,n*3))) df - rbind(df,df,df) df - data.frame(df,surveys=surveys) ### library(lattice) para.liste - trellis.par.get() superpose.symbol - para.liste$superpose.symbol superpose.symbol$pch - c(1,2,3) trellis.par.set(superpose.symbol,superpose.symbol) xyplot( y~x | surveys, data=df, group=f1, auto.key=list(space=right) ) xyplot( y~x | surveys , data=df, type=l, group=f2, auto.key=list(space=right,points=FALSE,lines=TRUE) ) Can I use the ggplot library ? __ 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] .Rprofile under Windoze.
I am (for my sins) having to do some work using R under Windoze. I wanted to set up a .Rprofile to control my set-up. The docs on .Rprofile say that it can/should be placed in ``the user's home directory''. ``An Introduction to R'' observes lucidly that this concept needs to be clarified under Windoze. Following the suggestions in An Introduction to R, I tried putting a .Rprofile in C:\Documents and Settings\rolf\My Documents When that didn't work, I tried putting it in the starting directory (and confirmed that I'd got that right by checking with getwd() and list.files(all.files=TRUE) ). The last invocation indicated that the name of the file was *really* ``.Rprofile.txt'' --- although I'd tried to save it as (simply) ``.Rprofile''. Is that the problem? If so, how can I persuade Windoze NOT to stick that damned .txt tag on the end? (Gawd, but I ***hate*** Windoze!!!) If that's not the problem, can you suggest what *is* the problem? All that .Rprofile(.txt) has in it at the moment is options(prompt=Wheee! ) so that I can easily tell whether it's working. If I execute source(.Rprofile.txt) the prompt does indeed get changed to ``Wheee! '' as it should. I would appreciate enlightenment. Ta. cheers, Rolf Turner [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice and several groups
Gabor Grothendieck a écrit : Try this: xyplot(val ~ x, data = df, type = p, col = as.numeric(df$f1), pch = as.numeric(df$f2)) key1 - list(border = TRUE, colums = 2, text = list(levels(df$f1)), points = list(pch = 1:nlevels(df$f1)) ) key2 - list(border = TRUE, colums = 2, text = list(levels(df$f2)), points = list(pch = 20, col = 1:nlevels(df$f2)) ) trellis.focus(panel, 1, 1) draw.key(key1, draw = TRUE, vp = viewport(.9, .9)) draw.key(key2, draw = TRUE, vp = viewport(.75, .9)) trellis.unfocus() On 8/29/06, Laurent Rhelp [EMAIL PROTECTED] wrote: Dear R-list, I would like to use the lattice library to show several groups on the same graph. Here's my example : ## the data f1 - factor(c(mod1,mod2,mod3),levels=c(mod1,mod2,mod3)) f1 - rep(f1,3) f2 - factor(rep(c(g1,g2,g3),each=3),levels=c(g1,g2,g3)) df - data.frame(val=c(4,3,2,5,4,3,6,5,4), x=rep(c(1,2,3),3),f1=f1,f2=f2) # library(lattice) para.liste - trellis.par.get() superpose.symbol - para.liste$superpose.symbol superpose.symbol$pch - c(1,2,3) trellis.par.set(superpose.symbol,superpose.symbol) # Now I can see the group according to the f1 factor (with a different symbol for every modality) xyplot( val~x, data=df, group=f1, auto.key=list(space=right) ) # or I can see the group according to the f2 factor xyplot( val~x, data=df, type=l, group=f2, auto.key=list(space=right,points=FALSE,lines=TRUE) ) How can I do to highlight both the f1 and f2 factors on one panel with the legends, using the lattice function ? Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Thank you, Gabor. The way to put the two legends is very interesting. For the graphs, in fact, my problem is to fit the data for every level of the f2 factor, showing the levels of the f1 factor in each panel and that for several surveys . Here's an example closer to my actual data : ## the data n - 18 x1 - seq(1,n) val1 - -2*x1+50 val2 - (-2*(x1-8)^2)+100 val3 - (-2*(x1-8)^2)+50 y - c(val1,val2,val3) x - rep(x1,3) f1 - rep(c(mod1,mod2,mod3),each=n/3) f1 - rep(f1,3) f2 - rep(c(g1,g2,g3),each=n) df - data.frame(x=x,y=y,f1=f1,f2=f2) surveys - factor(c(rep(survey1,n*3),rep(survey2,n*3),rep(survey3,n*3))) df - rbind(df,df,df) df - data.frame(df,surveys=surveys) ### library(lattice) para.liste - trellis.par.get() superpose.symbol - para.liste$superpose.symbol superpose.symbol$pch - c(1,2,3) trellis.par.set(superpose.symbol,superpose.symbol) xyplot( y~x | surveys, data=df, group=f1, auto.key=list(space=right) ) xyplot( y~x | surveys , data=df, type=l, group=f2, auto.key=list(space=right,points=FALSE,lines=TRUE) ) Certainly, I have to use the panel function but I don't know how to mark the f1 factor in each panel ! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice and several groups
Gabor Grothendieck a écrit : Note that before entering this you need: library(lattice) library(grid) # to access the viewport function On 8/29/06, Gabor Grothendieck [EMAIL PROTECTED] wrote: Try this: xyplot(val ~ x, data = df, type = p, col = as.numeric(df$f1), pch = as.numeric(df$f2)) key1 - list(border = TRUE, colums = 2, text = list(levels(df$f1)), points = list(pch = 1:nlevels(df$f1)) ) key2 - list(border = TRUE, colums = 2, text = list(levels(df$f2)), points = list(pch = 20, col = 1:nlevels(df$f2)) ) trellis.focus(panel, 1, 1) draw.key(key1, draw = TRUE, vp = viewport(.9, .9)) draw.key(key2, draw = TRUE, vp = viewport(.75, .9)) trellis.unfocus() On 8/29/06, Laurent Rhelp [EMAIL PROTECTED] wrote: Dear R-list, I would like to use the lattice library to show several groups on the same graph. Here's my example : ## the data f1 - factor(c(mod1,mod2,mod3),levels=c(mod1,mod2,mod3)) f1 - rep(f1,3) f2 - factor(rep(c(g1,g2,g3),each=3),levels=c(g1,g2,g3)) df - data.frame(val=c(4,3,2,5,4,3,6,5,4), x=rep(c(1,2,3),3),f1=f1,f2=f2) # library(lattice) para.liste - trellis.par.get() superpose.symbol - para.liste$superpose.symbol superpose.symbol$pch - c(1,2,3) trellis.par.set(superpose.symbol,superpose.symbol) # Now I can see the group according to the f1 factor (with a different symbol for every modality) xyplot( val~x, data=df, group=f1, auto.key=list(space=right) ) # or I can see the group according to the f2 factor xyplot( val~x, data=df, type=l, group=f2, auto.key=list(space=right,points=FALSE,lines=TRUE) ) How can I do to highlight both the f1 and f2 factors on one panel with the legends, using the lattice function ? Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Thank you, Gabor. The way to put the two legends is very interesting. For the graphs, in fact, my problem is to fit the data for every level of the f2 factor, showing the levels of the f1 factor in each panel and that for several surveys . Here's an example closer to my actual data : ## the data n - 18 x1 - seq(1,n) val1 - -2*x1+50 val2 - (-2*(x1-8)2)+100 val3 - (-2*(x1-8)2)+50 y - c(val1,val2,val3) x - rep(x1,3) f1 - rep(c(mod1,mod2,mod3),each=n/3) f1 - rep(f1,3) f2 - rep(c(g1,g2,g3),each=n) df - data.frame(x=x,y=y,f1=f1,f2=f2) surveys - factor(c(rep(survey1,n*3),rep(survey2,n*3),rep(survey3,n*3))) df - rbind(df,df,df) df - data.frame(df,surveys=surveys) ### library(lattice) para.liste - trellis.par.get() superpose.symbol - para.liste$superpose.symbol superpose.symbol$pch - c(1,2,3) trellis.par.set(superpose.symbol,superpose.symbol) xyplot( y~x | surveys, data=df, group=f1, auto.key=list(space=right) ) xyplot( y~x | surveys , data=df, type=l, group=f2, auto.key=list(space=right,points=FALSE,lines=TRUE) ) Certainly, I have to use the panel function but I don't know how to mark the f1 factor in each panel (I want to fit the values according to the f2 factor) ! __ 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] .Rprofile under Windoze.
Under Windows mine is located here C:\Program Files\R\R-2.3.1\library\base\R The file name, however is not .Rprofile, but rather Rprofile Charles Annis, P.E. [EMAIL PROTECTED] phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Rolf Turner Sent: Wednesday, August 30, 2006 5:50 PM To: r-help@stat.math.ethz.ch Subject: [R] .Rprofile under Windoze. I am (for my sins) having to do some work using R under Windoze. I wanted to set up a .Rprofile to control my set-up. The docs on .Rprofile say that it can/should be placed in ``the user's home directory''. ``An Introduction to R'' observes lucidly that this concept needs to be clarified under Windoze. Following the suggestions in An Introduction to R, I tried putting a .Rprofile in C:\Documents and Settings\rolf\My Documents When that didn't work, I tried putting it in the starting directory (and confirmed that I'd got that right by checking with getwd() and list.files(all.files=TRUE) ). The last invocation indicated that the name of the file was *really* ``.Rprofile.txt'' --- although I'd tried to save it as (simply) ``.Rprofile''. Is that the problem? If so, how can I persuade Windoze NOT to stick that damned .txt tag on the end? (Gawd, but I ***hate*** Windoze!!!) If that's not the problem, can you suggest what *is* the problem? All that .Rprofile(.txt) has in it at the moment is options(prompt=Wheee! ) so that I can easily tell whether it's working. If I execute source(.Rprofile.txt) the prompt does indeed get changed to ``Wheee! '' as it should. I would appreciate enlightenment. Ta. cheers, Rolf Turner [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] .Rprofile under Windoze.
On 8/30/06, Rolf Turner [EMAIL PROTECTED] wrote: I am (for my sins) having to do some work using R under Windoze. I wanted to set up a .Rprofile to control my set-up. The docs on .Rprofile say that it can/should be placed in ``the user's home directory''. ``An Introduction to R'' observes lucidly that this concept needs to be clarified under Windoze. Saving .Rprofile to the R root directory works for me. On my system that is C:\Program Files\R\R-2.3.1\ jab -- John Bollinger, CFA, CMT www.BollingerBands.com If you advance far enough, you arrive at the beginning. __ 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] lattice and several groups
In fact, my problem is to fit the data for every level of the f2 factor, showing the levels of the f1 factor and that for several surveys . Here's an example closer to my actual data : Then maybe you want: qplot(x, y, . ~ surveys, data=df, type=line, colour=f1, id=f2, size=f2) (which doesn't produce a very nice legend) Or to build up piece by piece p - ggplot(df, . ~ surveys, aes=list(x=x, y=y, colour=f1, id=f2)) p - ggline(p) ggpoint(p, aes=list(shape=f2)) (you might want to flip shape and colour around to get what you want) Hadley __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] .Rprofile under Windoze.
And... If you have/use shortcuts to R, you may also save an .Rprofile to whatever directory you name in the Start in: field of the shortcut. This allows one to have many profiles. jab -- John Bollinger, CFA, CMT www.BollingerBands.com If you advance far enough, you arrive at the beginning. __ 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 read just a column
Hi, how can I read, using for example read.table() or scan(), just one column from a text file that has more columns without any header? Thanks, bye. __ 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] .Rprofile under Windoze.
-Original Message- From: [EMAIL PROTECTED] [mailto:r-help- [EMAIL PROTECTED] On Behalf Of Rolf Turner Sent: Wednesday, August 30, 2006 2:50 PM To: r-help@stat.math.ethz.ch Subject: [R] .Rprofile under Windoze. I am (for my sins) having to do some work using R under Windoze. I wanted to set up a .Rprofile to control my set-up. The docs on .Rprofile say that it can/should be placed in ``the user's home directory''. ``An Introduction to R'' observes lucidly that this concept needs to be clarified under Windoze. Following the suggestions in An Introduction to R, I tried putting a .Rprofile in C:\Documents and Settings\rolf\My Documents When that didn't work, I tried putting it in the starting directory (and confirmed that I'd got that right by checking with getwd() and list.files(all.files=TRUE) ). The last invocation indicated that the name of the file was *really* ``.Rprofile.txt'' --- although I'd tried to save it as (simply) ``.Rprofile''. Is that the problem? If so, how can I persuade Windoze NOT to stick that damned .txt tag on the end? (Gawd, but I ***hate*** Windoze!!!) If that's not the problem, can you suggest what *is* the problem? snip Rolf, Whether an extension is automagically added (and if so what) is usually a function of the program writing the file out. In MS Windows programs, there is usually an option in the Save/SaveAs menu called something like Save As Type. To save without an extension you want to make sure that the value is 'All Files (*.*)', otherwise the program will usually tag on a default extension. Hope this is helpful, Dan Daniel J. Nordlund Research and Data Analysis Washington State Department of Social and Health Services Olympia, WA 98504-5204 __ 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] .Rprofile under Windoze.
On 8/30/2006 5:49 PM, Rolf Turner wrote: I am (for my sins) having to do some work using R under Windoze. I wanted to set up a .Rprofile to control my set-up. The docs on .Rprofile say that it can/should be placed in ``the user's home directory''. ``An Introduction to R'' observes lucidly that this concept needs to be clarified under Windoze. Following the suggestions in An Introduction to R, I tried putting a .Rprofile in C:\Documents and Settings\rolf\My Documents That's probably the right place. You can confirm in the R Edit|GUI preferences dialog: it defaults to trying to save or load from the same place it would look for .Rprofile. When that didn't work, I tried putting it in the starting directory (and confirmed that I'd got that right by checking with getwd() and list.files(all.files=TRUE) ). The last invocation indicated that the name of the file was *really* ``.Rprofile.txt'' --- although I'd tried to save it as (simply) ``.Rprofile''. Is that the problem? If so, how can I persuade Windoze NOT to stick that damned .txt tag on the end? You can put it in double quotes. It's really your editor doing this; most reasonable editors (but not the ones that come with Windows) are configurable to never add the extension. Another change to Windows defaults that's essential to maintain sanity is to tell it to display full filenames, not to hide the .txt in the first place. You do this in an Explorer window by clicking on Tools|Folder options...|View|Hide extensions for known file types (where each of the 4 components in that path is a different kind of interface element. I love the rich Windows user interface!) Duncan Murdoch (Gawd, but I ***hate*** Windoze!!!) If that's not the problem, can you suggest what *is* the problem? All that .Rprofile(.txt) has in it at the moment is options(prompt=Wheee! ) so that I can easily tell whether it's working. If I execute source(.Rprofile.txt) the prompt does indeed get changed to ``Wheee! '' as it should. I would appreciate enlightenment. Ta. cheers, Rolf Turner [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to read just a column
you can read them all in and delete the ones you don't want. Or check out 'what' on 'scan' or colClasses on 'read.table' On 8/30/06, Carlo Trimarchi [EMAIL PROTECTED] wrote: Hi, how can I read, using for example read.table() or scan(), just one column from a text file that has more columns without any header? Thanks, bye. __ 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ 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] lattice and several groups
To handle conditioning on survey we provide a panel function that subsets col and pch: # define test data - df # note that your val2 and val3 lines had a syntax # so we have commented them out and # replaced them as shown. n - 18 x1 - seq(1,n) val1 - -2*x1+50 # val2 - (-2*(x1-8)2)+100 val2 - (-2*(x1-8))+100 # val3 - (-2*(x1-8)2)+50 val3 - (-2*(x1-8))+50 y - c(val1,val2,val3) x - rep(x1,3) f1 - rep(c(mod1,mod2,mod3),each=n/3) f1 - rep(f1,3) f2 - rep(c(g1,g2,g3),each=n) df - data.frame(x=x,y=y,f1=f1,f2=f2) surveys - factor(c(rep(survey1,n*3),rep(survey2,n*3),rep(survey3,n*3))) df - rbind(df,df,df) df - data.frame(df,surveys=surveys) # create xyplot library(lattice) library(grid) pnl - function(x, y, groups, subscripts, col, pch, ...) panel.xyplot(x, y, col = col[subscripts], pch = pch[subscripts], ...) xyplot(y ~ x | surveys, data = df, col = as.numeric(df$f1), pch = as.numeric(df$f2), panel = pnl) key1 - list(border = TRUE, colums = 2, text = list(levels(df$f1)), points = list(pch = 1:nlevels(df$f1)) ) key2 - list(border = TRUE, colums = 2, text = list(levels(df$f2)), points = list(pch = 20, col = 1:nlevels(df$f2)) ) # add legend draw.key(key1, draw = TRUE, vp = viewport(.9, .9)) draw.key(key2, draw = TRUE, vp = viewport(.75, .9)) On 8/30/06, Laurent Rhelp [EMAIL PROTECTED] wrote: Gabor Grothendieck a écrit : Note that before entering this you need: library(lattice) library(grid) # to access the viewport function On 8/29/06, Gabor Grothendieck [EMAIL PROTECTED] wrote: Try this: xyplot(val ~ x, data = df, type = p, col = as.numeric(df$f1), pch = as.numeric(df$f2)) key1 - list(border = TRUE, colums = 2, text = list(levels(df$f1)), points = list(pch = 1:nlevels(df$f1)) ) key2 - list(border = TRUE, colums = 2, text = list(levels(df$f2)), points = list(pch = 20, col = 1:nlevels(df$f2)) ) trellis.focus(panel, 1, 1) draw.key(key1, draw = TRUE, vp = viewport(.9, .9)) draw.key(key2, draw = TRUE, vp = viewport(.75, .9)) trellis.unfocus() On 8/29/06, Laurent Rhelp [EMAIL PROTECTED] wrote: Dear R-list, I would like to use the lattice library to show several groups on the same graph. Here's my example : ## the data f1 - factor(c(mod1,mod2,mod3),levels=c(mod1,mod2,mod3)) f1 - rep(f1,3) f2 - factor(rep(c(g1,g2,g3),each=3),levels=c(g1,g2,g3)) df - data.frame(val=c(4,3,2,5,4,3,6,5,4), x=rep(c(1,2,3),3),f1=f1,f2=f2) # library(lattice) para.liste - trellis.par.get() superpose.symbol - para.liste$superpose.symbol superpose.symbol$pch - c(1,2,3) trellis.par.set(superpose.symbol,superpose.symbol) # Now I can see the group according to the f1 factor (with a different symbol for every modality) xyplot( val~x, data=df, group=f1, auto.key=list(space=right) ) # or I can see the group according to the f2 factor xyplot( val~x, data=df, type=l, group=f2, auto.key=list(space=right,points=FALSE,lines=TRUE) ) How can I do to highlight both the f1 and f2 factors on one panel with the legends, using the lattice function ? Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Thank you, Gabor. The way to put the two legends is very interesting. For the graphs, in fact, my problem is to fit the data for every level of the f2 factor, showing the levels of the f1 factor in each panel and that for several surveys . Here's an example closer to my actual data : ## the data n - 18 x1 - seq(1,n) val1 - -2*x1+50 val2 - (-2*(x1-8)2)+100 val3 - (-2*(x1-8)2)+50 y - c(val1,val2,val3) x - rep(x1,3) f1 - rep(c(mod1,mod2,mod3),each=n/3) f1 - rep(f1,3) f2 - rep(c(g1,g2,g3),each=n) df - data.frame(x=x,y=y,f1=f1,f2=f2) surveys - factor(c(rep(survey1,n*3),rep(survey2,n*3),rep(survey3,n*3))) df - rbind(df,df,df) df - data.frame(df,surveys=surveys) ### library(lattice) para.liste - trellis.par.get() superpose.symbol - para.liste$superpose.symbol superpose.symbol$pch - c(1,2,3) trellis.par.set(superpose.symbol,superpose.symbol) xyplot( y~x | surveys, data=df, group=f1, auto.key=list(space=right) ) xyplot( y~x | surveys , data=df, type=l, group=f2, auto.key=list(space=right,points=FALSE,lines=TRUE) )
Re: [R] .Rprofile under Windoze.
Rolf The last invocation indicated that the name of the file was *really* ``.Rprofile.txt'' --- although I'd tried to save it as (simply) ``.Rprofile''. Is that the problem? If so, how can I persuade Windoze NOT to stick that damned .txt tag on the end? The easiest way is to use a smarter editor. Emacs is perfectly happy to name a file .Rprofile and wouldn't dream of appending an extension to the filename that you specify. Rich __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] NaN when using dffits, stemming from lm.influence call
Hi all I'm getting a NaN returned on using dffits, as explained below. To me, there seems no obvious (or non-obvious reason for that matter) reason why a NaN appears. Before I start digging further, can anyone see why dffits might be failing? Is there a problem with the data? Consider: # Load data dep - read.table(http://www.sci.usq.edu.au/staff/dunn/Datasets/Books/Hand/Hand-R/factor1-R.dat;, header=TRUE) attach(dep) dep # Fit Poisson glm dep.glm2 - glm( Counts ~ factor(Depression) + factor(SLE) + factor(Children) + factor(Depression):factor(SLE), family=poisson( link=log) ) # Compute dffits dffits( dep.glm2 ) This produces the output: 1 2 3 4 5 6 1.4207746 -0.1513808NaN 0.9079142 -0.1032664 -1.0860289 7 8 0.4853797 3.8560863 NaN exists for Observation 3. I cannot understand why: there's nothing grossly unusual or bad about it. I look a bit closer, and it falls over in lm.influence when computing the deletion statistic sigma: lm.influence(dep.glm2)$sigma 12345678 0.914829 2.134279 NaN 2.186707 2.224885 1.934539 2.225115 1.957111 The rest of the results from lm.influence are OK; for example: lm.influence(dep.glm2)$wt.res 1 2 3 4 5 6 2.62840627 -0.88476903 -1.09492912 0.20247856 -0.23114458 -0.95123387 7 8 0.07521515 0.30208051 Use of debug( lm.influence ) indicates the NaN appears in this line of lm.influence: res - .Fortran(lminfl, model$qr$qr, n, n, k, as.integer(do.coef), model$qr$qraux, wt.res = e, hat = double(n), coefficients = if (do.coef) matrix(0, n, k) else double(0), sigma = double(n), tol = 10 * .Machine$double.eps, DUP = FALSE, PACKAGE = base)[c(hat, coefficients, sigma, wt.res)] I don't particularly wish to dig around in the Fortran if someone else can look at it and see my problem easily. But if I must... The appearance of the NaN seems odd, since (as I understand it) lm.influence(dep.glm2)$sigma computes sigma when each observation is removed in turn. So if I remove Observation 3 and try fitting the model, there are no problems or complaints: dep.glm3 - glm( Counts ~ factor(Depression) + factor(SLE) + factor(Children) + factor(Depression):factor(SLE), family=poisson( link=log), subset=(-3) ) This produces: dep.glm3 Call: glm(formula = Counts ~ factor(Depression) + factor(SLE) + factor(Children) + factor(Depression):factor(SLE), family = poisson(link = log), subset = (-3)) Coefficients: (Intercept) factor(Depression)1 5.4389 -4.1392 factor(SLE)1 factor(Children)1 -0.6503 -2.4036 factor(Depression)1:factor(SLE)1 3.9513 Degrees of Freedom: 6 Total (i.e. Null); 2 Residual Null Deviance: 695.9 Residual Deviance: 0.8535 AIC: 41.25 No problems, errors, or any signs of potential problems. In the changes to R 2.3.0 (in the NEWS file, eg http://mirror.aarnet.edu.au/pub/CRAN/src/base/NEWS) I find this: oInfluence measures such as rstandard() and cooks.distance() could return infinite values rather than NaN for a case which was fitted exactly. Similarly, plot.lm() could fail on such examples. plot.lm(which = 5) had to be modified to only plot cases with hat 1. (PR#8367) lm.influence() was incorrectly reporting 'coefficients' and 'sigma' as NaN for cases with hat = 1, and on some platforms not detecting hat = 1 correctly. The last sentence identifies NaN being reported for sigma, as I find with my data. But my data do not have hat = 1, but the hat diagonals are large. The troublesome Observation 3 does not have the largest hat value in the data though: hatvalues(dep.glm2) 1 2 3 4 5 6 7 8 0.1689061 0.1064651 0.9098542 0.9030814 0.3799079 0.6382790 0.9327408 0.9607654 And besides, I am using the most recent version of R (see below). BTW, the NaNs appear in the previous version of R also. So I'm flummoxed. As always, help appreciated. P. version _ platform i386-pc-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major 2 minor 3.1 year 2006 month 06 day01 svn rev38247 language R version.string Version 2.3.1 (2006-06-01) -- Dr Peter Dunn | Email: dunn at usq.edu.au Faculty of Sciences, University of Southern Queensland and the Australian Centre for Sustainable Catchments CRICOS: QLD 00244B | NSW 02225M | VIC 02387D | WA 02521C __
Re: [R] lattice and several groups
Or maybe this is what you are looking for where pnl below was
created by modifying source to the panel.plot.default in the zoo
package (there might be a simpler way):
pnl - function (x, y, subscripts, groups, col, pch, type, ...) {
for (g in levels(groups)) {
idx - g == groups[subscripts]
if (any(idx))
panel.xyplot(x[idx], y[idx], ..., col = col[subscripts][idx],
pch = pch[subscripts][idx], type = type)
}
}
xyplot(y ~ x | surveys, data = df, groups = df$f2, type = b,
col = as.numeric(df$f2), pch = as.numeric(df$f1), panel = pnl)
key1 - list(border = TRUE, colums = 2, text = list(levels(df$f1)),
points = list(pch = 1:nlevels(df$f1))
)
key2 - list(border = TRUE, colums = 2, text = list(levels(df$f2)),
points = list(pch = 20, col = 1:nlevels(df$f2))
)
draw.key(key1, draw = TRUE, vp = viewport(.9, .9))
draw.key(key2, draw = TRUE, vp = viewport(.75, .9))
On 8/30/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
To handle conditioning on survey we provide a panel function
that subsets col and pch:
# define test data - df
# note that your val2 and val3 lines had a syntax
# so we have commented them out and
# replaced them as shown.
n - 18
x1 - seq(1,n)
val1 - -2*x1+50
# val2 - (-2*(x1-8)2)+100
val2 - (-2*(x1-8))+100
# val3 - (-2*(x1-8)2)+50
val3 - (-2*(x1-8))+50
y - c(val1,val2,val3)
x - rep(x1,3)
f1 - rep(c(mod1,mod2,mod3),each=n/3)
f1 - rep(f1,3)
f2 - rep(c(g1,g2,g3),each=n)
df - data.frame(x=x,y=y,f1=f1,f2=f2)
surveys -
factor(c(rep(survey1,n*3),rep(survey2,n*3),rep(survey3,n*3)))
df - rbind(df,df,df)
df - data.frame(df,surveys=surveys)
# create xyplot
library(lattice)
library(grid)
pnl - function(x, y, groups, subscripts, col, pch, ...)
panel.xyplot(x, y, col = col[subscripts], pch = pch[subscripts], ...)
xyplot(y ~ x | surveys, data = df,
col = as.numeric(df$f1), pch = as.numeric(df$f2), panel = pnl)
key1 - list(border = TRUE, colums = 2, text = list(levels(df$f1)),
points = list(pch = 1:nlevels(df$f1))
)
key2 - list(border = TRUE, colums = 2, text = list(levels(df$f2)),
points = list(pch = 20, col = 1:nlevels(df$f2))
)
# add legend
draw.key(key1, draw = TRUE, vp = viewport(.9, .9))
draw.key(key2, draw = TRUE, vp = viewport(.75, .9))
On 8/30/06, Laurent Rhelp [EMAIL PROTECTED] wrote:
Gabor Grothendieck a écrit :
Note that before entering this you need:
library(lattice)
library(grid) # to access the viewport function
On 8/29/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try this:
xyplot(val ~ x, data = df, type = p,
col = as.numeric(df$f1), pch = as.numeric(df$f2))
key1 - list(border = TRUE, colums = 2, text = list(levels(df$f1)),
points = list(pch = 1:nlevels(df$f1))
)
key2 - list(border = TRUE, colums = 2, text = list(levels(df$f2)),
points = list(pch = 20, col = 1:nlevels(df$f2))
)
trellis.focus(panel, 1, 1)
draw.key(key1, draw = TRUE, vp = viewport(.9, .9))
draw.key(key2, draw = TRUE, vp = viewport(.75, .9))
trellis.unfocus()
On 8/29/06, Laurent Rhelp [EMAIL PROTECTED] wrote:
Dear R-list,
I would like to use the lattice library to show several groups on
the same graph. Here's my example :
## the data
f1 - factor(c(mod1,mod2,mod3),levels=c(mod1,mod2,mod3))
f1 - rep(f1,3)
f2 - factor(rep(c(g1,g2,g3),each=3),levels=c(g1,g2,g3))
df - data.frame(val=c(4,3,2,5,4,3,6,5,4), x=rep(c(1,2,3),3),f1=f1,f2=f2)
#
library(lattice)
para.liste - trellis.par.get()
superpose.symbol - para.liste$superpose.symbol
superpose.symbol$pch - c(1,2,3)
trellis.par.set(superpose.symbol,superpose.symbol)
# Now I can see the group according to the f1 factor (with a different
symbol for every modality)
xyplot( val~x,
data=df,
group=f1,
auto.key=list(space=right)
)
# or I can see the group according to the f2 factor
xyplot( val~x,
data=df,
type=l,
group=f2,
auto.key=list(space=right,points=FALSE,lines=TRUE)
)
How can I do to highlight both the f1 and f2 factors on one panel with
the legends, using the lattice function ?
Thanks
__
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.
Thank you, Gabor. The way to put the two legends is very
Re: [R] lattice and several groups
In thinking about this a bit more we can use
panel.superpose/panel.groups to shorten it:
# define data -- df
# note that your val2 and val3 lines had a syntax
# so we have commented them out and
# replaced them as shown.
n - 18
x1 - seq(1,n)
val1 - -2*x1+50
# val2 - (-2*(x1-8)2)+100
val2 - (-2*(x1-8))+100
# val3 - (-2*(x1-8)2)+50
val3 - (-2*(x1-8))+50
y - c(val1,val2,val3)
x - rep(x1,3)
f1 - rep(c(mod1,mod2,mod3),each=n/3)
f1 - rep(f1,3)
f2 - rep(c(g1,g2,g3),each=n)
df - data.frame(x=x,y=y,f1=f1,f2=f2)
surveys -
factor(c(rep(survey1,n*3),rep(survey2,n*3),rep(survey3,n*3)))
df - rbind(df,df,df)
df - data.frame(df,surveys=surveys)
# create xyplot
library(lattice)
library(grid)
# set custom col and pch here
my.col - 1:nlevels(df$f2)
my.pch - 1:nlevels(df$f1)
pnl - function(x, y, subscripts, pch, type, ...)
panel.xyplot(x, y, type = type, pch = my.pch[df[subscripts, f1]], ...)
xyplot(y ~ x | surveys, data = df, groups = df$f2, type = b,
panel = panel.superpose,
panel.groups = pnl,
par.settings = list(superpose.line = list(col = my.col),
superpose.symbol = list(col = my.col))
)
key1 - list(border = TRUE, colums = 2, text = list(levels(df$f1)),
points = list(pch = my.pch)
)
key2 - list(border = TRUE, colums = 2, text = list(levels(df$f2)),
lines = list(col = my.col)
)
draw.key(key1, draw = TRUE, vp = viewport(.9, .9))
draw.key(key2, draw = TRUE, vp = viewport(.75, .9))
On 8/30/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Or maybe this is what you are looking for where pnl below was
created by modifying source to the panel.plot.default in the zoo
package (there might be a simpler way):
pnl - function (x, y, subscripts, groups, col, pch, type, ...) {
for (g in levels(groups)) {
idx - g == groups[subscripts]
if (any(idx))
panel.xyplot(x[idx], y[idx], ..., col = col[subscripts][idx],
pch = pch[subscripts][idx], type = type)
}
}
xyplot(y ~ x | surveys, data = df, groups = df$f2, type = b,
col = as.numeric(df$f2), pch = as.numeric(df$f1), panel = pnl)
key1 - list(border = TRUE, colums = 2, text = list(levels(df$f1)),
points = list(pch = 1:nlevels(df$f1))
)
key2 - list(border = TRUE, colums = 2, text = list(levels(df$f2)),
points = list(pch = 20, col = 1:nlevels(df$f2))
)
draw.key(key1, draw = TRUE, vp = viewport(.9, .9))
draw.key(key2, draw = TRUE, vp = viewport(.75, .9))
On 8/30/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
To handle conditioning on survey we provide a panel function
that subsets col and pch:
# define test data - df
# note that your val2 and val3 lines had a syntax
# so we have commented them out and
# replaced them as shown.
n - 18
x1 - seq(1,n)
val1 - -2*x1+50
# val2 - (-2*(x1-8)2)+100
val2 - (-2*(x1-8))+100
# val3 - (-2*(x1-8)2)+50
val3 - (-2*(x1-8))+50
y - c(val1,val2,val3)
x - rep(x1,3)
f1 - rep(c(mod1,mod2,mod3),each=n/3)
f1 - rep(f1,3)
f2 - rep(c(g1,g2,g3),each=n)
df - data.frame(x=x,y=y,f1=f1,f2=f2)
surveys -
factor(c(rep(survey1,n*3),rep(survey2,n*3),rep(survey3,n*3)))
df - rbind(df,df,df)
df - data.frame(df,surveys=surveys)
# create xyplot
library(lattice)
library(grid)
pnl - function(x, y, groups, subscripts, col, pch, ...)
panel.xyplot(x, y, col = col[subscripts], pch = pch[subscripts], ...)
xyplot(y ~ x | surveys, data = df,
col = as.numeric(df$f1), pch = as.numeric(df$f2), panel = pnl)
key1 - list(border = TRUE, colums = 2, text = list(levels(df$f1)),
points = list(pch = 1:nlevels(df$f1))
)
key2 - list(border = TRUE, colums = 2, text = list(levels(df$f2)),
points = list(pch = 20, col = 1:nlevels(df$f2))
)
# add legend
draw.key(key1, draw = TRUE, vp = viewport(.9, .9))
draw.key(key2, draw = TRUE, vp = viewport(.75, .9))
On 8/30/06, Laurent Rhelp [EMAIL PROTECTED] wrote:
Gabor Grothendieck a écrit :
Note that before entering this you need:
library(lattice)
library(grid) # to access the viewport function
On 8/29/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try this:
xyplot(val ~ x, data = df, type = p,
col = as.numeric(df$f1), pch = as.numeric(df$f2))
key1 - list(border = TRUE, colums = 2, text = list(levels(df$f1)),
points = list(pch = 1:nlevels(df$f1))
)
key2 - list(border = TRUE, colums = 2, text = list(levels(df$f2)),
points = list(pch = 20, col = 1:nlevels(df$f2))
)
trellis.focus(panel, 1, 1)
draw.key(key1, draw = TRUE, vp = viewport(.9, .9))
draw.key(key2, draw = TRUE, vp = viewport(.75, .9))
trellis.unfocus()
On 8/29/06, Laurent Rhelp [EMAIL PROTECTED] wrote:
Dear R-list,
I would like to use the lattice library to show several groups on
the same graph. Here's my example :
## the data
f1 -
[R] [R-pkgs] New package 'random' for non-deterministic random number
Dear useRs, A few days ago, the initial version 0.1.0 of a new package 'random' was uploaded to CRAN. The random packages provides convenient access to the non-deterministic random numbers provided by the random.org site created by Mads Haahr (http://www.random.org). While certain hardware and software solutions that provide access to non-deterministic random-numbers exist, few if any are portable across all the hardware platforms R supports. Retrieving non-deterministic random numbers may be beneficial to seed parallel simulations with independent draws, to obtain portable initializations for other RNGs, to validate simulation with non-deterministic RNGs, or simply for fun and experimentations. The package contains five simple functions randomNumber (random integeres between min, max w/ duplicates) randomSequence (random sequences between min, max w/o duplicates) randomBytes(in hex, dec, oct or bin) randomBufferStatus (to query the server status) sufficientBits (boolean test of randomBufferStats vs rec'ed value) Also included are two vignettes that can be accessed via vignette(random-intro, package = random) vignette(random-essay, package = random) 'random-intro' provides some background on the package as well as initial test results using the dieharder suite by Robert G. Brown. I hope to expand on these tests in the near future. 'random-essay' is a transcript of the web essay at http://www.random.org/essay.html and provides some background on the random.org service. Some information is / will be at http://dirk.eddelbuettel.com/code/random.html will Comments or suggestions are more than welcome! Regards, Dirk -- Hell, there are no rules here - we're trying to accomplish something. -- Thomas A. Edison ___ R-packages mailing list R-packages@stat.math.ethz.ch https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Combine 'overlapping' dataframes, respecting row names
Hi, I've examined the archives and found quite a few questions on concatenating dataframes, but none that really addressed my issue, I'm afraid. I've also examined the cbind and rbind documentation but nonetheless here I am writing to r-help ;) This is what I have (the row names are dates used for conversion to an irregular time series with the its package): cvsFrame cvsactions 2002-11-15 4 2002-12-15 9 2003-01-15 5 2003-02-15 5 downloadsFrame downloads 2002-09-15 1 2002-10-15 2 2002-11-15 12 2002-12-15 8 (notice how the dates are overlapping?) The output I'd like is: cvsaction downloads 2002-09-15 NA 1 2002-10-15 NA 2 2002-11-15 4 12 2002-12-15 9 8 2003-01-15 5 NA 2003-02-15 5 NA ie. merge the data.frames, respecting the row.names and inserting NAs where a frame didn't contain info for a row in the final frame. This is the closest I gotten (I'm sure cbind is doing what it's meant to do but it's obviously not what I need) cbind(downloadsFrame,cvsFrame) downloads cvsactions 2002-09-15 1 4 2002-10-15 2 9 2002-11-15 12 5 2002-12-15 8 5 It takes the row.names from the first frame given and then just adds the data in rows 1 through 4, regardless of their row.name. And it doesn't work at all if the column lengths are different. (Yes, it would be nice if the 'its' class had a way to merge 'its' objects, but the question seemed general enough to ask on list.) Thanks, James __ 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] Combine 'overlapping' dataframes, respecting row names
If you are converting them to 'its' anyways then after the conversion to 'its' use the 'its' union command. On 8/31/06, James Howison [EMAIL PROTECTED] wrote: Hi, I've examined the archives and found quite a few questions on concatenating dataframes, but none that really addressed my issue, I'm afraid. I've also examined the cbind and rbind documentation but nonetheless here I am writing to r-help ;) This is what I have (the row names are dates used for conversion to an irregular time series with the its package): cvsFrame cvsactions 2002-11-15 4 2002-12-15 9 2003-01-15 5 2003-02-15 5 downloadsFrame downloads 2002-09-15 1 2002-10-15 2 2002-11-15 12 2002-12-15 8 (notice how the dates are overlapping?) The output I'd like is: cvsaction downloads 2002-09-15 NA 1 2002-10-15 NA 2 2002-11-15 4 12 2002-12-15 9 8 2003-01-15 5 NA 2003-02-15 5 NA ie. merge the data.frames, respecting the row.names and inserting NAs where a frame didn't contain info for a row in the final frame. This is the closest I gotten (I'm sure cbind is doing what it's meant to do but it's obviously not what I need) cbind(downloadsFrame,cvsFrame) downloads cvsactions 2002-09-15 1 4 2002-10-15 2 9 2002-11-15 12 5 2002-12-15 8 5 It takes the row.names from the first frame given and then just adds the data in rows 1 through 4, regardless of their row.name. And it doesn't work at all if the column lengths are different. (Yes, it would be nice if the 'its' class had a way to merge 'its' objects, but the question seemed general enough to ask on list.) Thanks, James __ 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] Combine 'overlapping' dataframes, respecting row names
'merge' is the key here. You say you want to merge, but it seems did not try merge() (res - merge(cvsFrame, downloadsFrame, by=row.names, all=TRUE)) Row.names cvsactions downloads 1 2002-11-15 412 2 2002-12-15 9 8 3 2003-01-15 5NA 4 2003-02-15 5NA 5 2002-09-15 NA 1 6 2002-10-15 NA 2 You can sort on Row.names later: say res[order(as.character(res$Row.names)), ] On Thu, 31 Aug 2006, James Howison wrote: Hi, I've examined the archives and found quite a few questions on concatenating dataframes, but none that really addressed my issue, I'm afraid. I've also examined the cbind and rbind documentation but nonetheless here I am writing to r-help ;) This is what I have (the row names are dates used for conversion to an irregular time series with the its package): cvsFrame cvsactions 2002-11-15 4 2002-12-15 9 2003-01-15 5 2003-02-15 5 downloadsFrame downloads 2002-09-15 1 2002-10-15 2 2002-11-15 12 2002-12-15 8 (notice how the dates are overlapping?) The output I'd like is: cvsaction downloads 2002-09-15 NA 1 2002-10-15 NA 2 2002-11-15 4 12 2002-12-15 9 8 2003-01-15 5 NA 2003-02-15 5 NA ie. merge the data.frames, respecting the row.names and inserting NAs where a frame didn't contain info for a row in the final frame. This is the closest I gotten (I'm sure cbind is doing what it's meant to do but it's obviously not what I need) cbind(downloadsFrame,cvsFrame) downloads cvsactions 2002-09-15 1 4 2002-10-15 2 9 2002-11-15 12 5 2002-12-15 8 5 It takes the row.names from the first frame given and then just adds the data in rows 1 through 4, regardless of their row.name. And it doesn't work at all if the column lengths are different. (Yes, it would be nice if the 'its' class had a way to merge 'its' objects, but the question seemed general enough to ask on list.) Thanks, James __ 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 read just a column
Carlo Trimarchi a écrit : Hi, how can I read, using for example read.table() or scan(), just one column from a text file that has more columns without any header? Thanks, bye. afaik, you have to read all the table and then you select the column you want. eg read.table(blabla)[3] to get the 3rd column. You can read partially rows (see nrows) but not columns. Please somebody correct me if I am wrong. (Of course a trick could be to transpose your table before writing it, etc) hih __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
