Re: [R] Installing package QRMlib
Dear all: well, what Duncan has suggested would work in principle. However, the dependencies of QRMlib as contained in the archive have been deprecated and the package maintainer (cc'ed to this email directly) is pretty close to a re-release of his package on CRAN, whereby primarily the outdated package dependency to fSeries is changed to timeSeries. Hence, before grabbing the deprecated package dependencies on R-Forge and install these, it might be worth waiting for the re-submittance of QRMlib to CRAN, given that it will be made in due course. Scott, do you have any further information whence QRMlib will be made available again on CRAN? Best, Bernhard -Ursprüngliche Nachricht- Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im Auftrag von Duncan Murdoch Gesendet: Montag, 27. Februar 2012 21:16 An: R. Michael Weylandt Cc: r-help@r-project.org; DT54321 Betreff: Re: [R] Installing package QRMlib On 27/02/2012 3:01 PM, R. Michael Weylandt wrote: Do you perhaps need to add install.packages(..., type=src)? Just a (untested) guess... That should be type=source, and that should solve the problem, assuming Deepan has the necessary tools installed. If not, he can get them from CRAN in the bin/windows/Rtools directory. Duncan Murdoch Michael On Mon, Feb 27, 2012 at 12:07 PM, DT54321deepan.tailo...@gmail.com wrote: Hi, I am having real problems downloading the package 'QRMlib'. The tar.gz file is shown here: http://cran.r-project.org/src/contrib/Archive/QRMlib/ I have downloaded this to my local folder and entered the following command: nstall.packages(myLocalFolder/QRMlib_1.4.5.1.tar.gz, repos = NULL) but I am getting the following error message Installing package(s) into 'C:/Program Files/R/R-2.14.1/library' (as 'lib' is unspecified) Warning in install.packages : error 1 in extracting from zip file Warning in install.packages : cannot open compressed file 'QRMlib_1.4.5.1.tar.gz/DESCRIPTION', probable reason 'No such file or directory' Error in install.packages : cannot open the connection What am I doing wrong?? Thanks -- View this message in context: http://r.789695.n4.nabble.com/Installing-package-QRMlib-tp4425269p44 25269.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. * Confidentiality Note: The information contained in this ...{{dropped:10}} __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Principal Components for matrices with NA
On Feb 27, 2012 at 9:30pm Joyous Fisher wrote: Q: is there a way to do princomp or another method where every row has at least one missing column? You have several options. Try function nipals in packages ade4 and plspm. Also look at package pcaMethods (on Bioconductor), where you will find a full range of options for carrying out principal component analysis using matrices with missing values. Regards, Mark. - Mark Difford (Ph.D.) Research Associate Botany Department Nelson Mandela Metropolitan University Port Elizabeth, South Africa -- View this message in context: http://r.789695.n4.nabble.com/Principal-Components-for-matrices-with-NA-tp4425930p4427216.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Conditional density estimation in R and integration
Hey there, I have the following problem: I need to carry out a nonparametric conditional density estimation -- pretty much what can be achieved using the np package. However, I need to be able to use the output of the np package in such a way that i get a continuous function as output, which can be integrated on any given interval between -Inf and Inf. This integration will of have to be across mutliple dimension since I am working with a conditional pdf. Any advise on how to accomplish this? I suspect that I need to use some kind of interpolation package to get from the output of the np package to getting a function that can be integrated, but that is exacly where I am stuck Thanks for your help! -- View this message in context: http://r.789695.n4.nabble.com/Conditional-density-estimation-in-R-and-integration-tp4427356p4427356.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] data analysis
Please assist me with all the tests including risk factor analysis i can use to analyse the enclosed database established from a questionnaire survey to test for the prevalence of tuberculosis in humans . Thank you Nonty -- View this message in context: http://r.789695.n4.nabble.com/data-analysis-tp4427257p4427257.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] macro function
hi, I know how to use the for loop function like: for(i in 1:ncol(mat)){ mat[i]-b[i,2] } but, in this case r1-b[1,1] r2-b[2,1] r3-b[3,1] r4-b[4,1] * * * r3002-b[3002,1] r3003-b[3003,1] - must make vectors how should I make a efficient code for that? Is there anything in R like SAS MACRO function? -- View this message in context: http://r.789695.n4.nabble.com/macro-function-tp4427385p4427385.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error message: object of type 'closure' is not subsettable
Hi All I am trying to use the unlist() in R to a list variable. The following statements are within a function. { denominator - sqrt(s1 / res.em1$n + s2 / res.em2$n) returnValue - l2 / (denominator + 11) attr(returnValue,numerator) - l2 attr(returnValue,denominator) - denominator returnValue } And when I try to unlist the variable returnValue numerators - unlist(testStatistics[numerator,]) denominators - unlist(testStatistics[denominator,]) I get the following error: Error in testStatistics[numerator, ] : object of type 'closure' is not subsettable I read some threads in R help on this error and they had asked to check if we are using the right datatype to the right function. But in my case it is pretty straightforward since I just list it in one function and try to unlist it later. Any suggestions? Thanks for the help :) -- View this message in context: http://r.789695.n4.nabble.com/Error-message-object-of-type-closure-is-not-subsettable-tp3752886p4427399.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data analysis
On Mon, Feb 27, 2012 at 11:04:13PM -0800, nontokozo mhlanga wrote: Please assist me with all the tests including risk factor analysis i can use to analyse the enclosed database established from a questionnaire survey to test for the prevalence of tuberculosis in humans . That's quite a general request. I think you should try to formulate a specific question. Have you read the posting-guide? http://www.R-project.org/posting-guide.html Also, I don't think the list accepts attached files. -- Hans Ekbrand (http://sociologi.cjb.net) h...@sociologi.cjb.net __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] macro function
Hi, There is not anything really like macro functions in R (and honestly, that is probably a good thing). Most the times I have seen people generating thousands of variables (vectors in your case) it is due to a lack of understanding how lists can be utilized to simplify code. If you give us some context on what all these vectors will be used for, we may be able to suggest an easier way to accomplish your goals. If you must create vectors: sapply(1:3003, function(i) { assign(paste(r, i, sep = ''), rnorm(1), envir = .GlobalEnv) }) or with a for loop: for (i in 1:3003) { assign(paste(r, i, sep = ''), rnorm(1), envir = .GlobalEnv) } HTH, Josh On Tue, Feb 28, 2012 at 12:14 AM, mrzung mrzun...@gmail.com wrote: hi, I know how to use the for loop function like: for(i in 1:ncol(mat)){ mat[i]-b[i,2] } but, in this case r1-b[1,1] r2-b[2,1] r3-b[3,1] r4-b[4,1] * * * r3002-b[3002,1] r3003-b[3003,1] - must make vectors how should I make a efficient code for that? Is there anything in R like SAS MACRO function? -- View this message in context: http://r.789695.n4.nabble.com/macro-function-tp4427385p4427385.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error message: object of type 'closure' is not subsettable
Hi Aparna, Can you please post a reproducible example? It is difficult to provide much concrete help without having testStatistics. One thing you might try is looking at: str(testStatistics[numerator,]) is it actually a list? If it is not (most likely given the error) and it is supposed to be, you need to figure out what aspect of the generation of it is going awry. Cheers, Josh On Tue, Feb 28, 2012 at 12:23 AM, Aparna Sampath aparna.sampat...@gmail.com wrote: Hi All I am trying to use the unlist() in R to a list variable. The following statements are within a function. { denominator - sqrt(s1 / res.em1$n + s2 / res.em2$n) returnValue - l2 / (denominator + 11) attr(returnValue,numerator) - l2 attr(returnValue,denominator) - denominator returnValue } And when I try to unlist the variable returnValue numerators - unlist(testStatistics[numerator,]) denominators - unlist(testStatistics[denominator,]) I get the following error: Error in testStatistics[numerator, ] : object of type 'closure' is not subsettable I read some threads in R help on this error and they had asked to check if we are using the right datatype to the right function. But in my case it is pretty straightforward since I just list it in one function and try to unlist it later. Any suggestions? Thanks for the help :) -- View this message in context: http://r.789695.n4.nabble.com/Error-message-object-of-type-closure-is-not-subsettable-tp3752886p4427399.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error message: object of type 'closure' is not subsettable
Hi Aparna, If you have not defined testStatistics in your code, but it is a function, then either you defined it at some earlier point and it was never removed, or some other package you loaded defines it. You could try using your code in a clean R session (make sure that an old workspace is not automatically loaded; one way would be by closing R, deleting the old workspace, and then starting a new instance). Cheers, Josh On Tue, Feb 28, 2012 at 1:37 AM, Aparna Sampath aparna.sampat...@gmail.com wrote: Hi Josh Kindly find below the code as it is executed: test.functional.t - function(res.em1,res.em2,mint,maxt,se.m=0,points=300) { at - seq(mint,maxt,length.out=points) by - at[2] - at[1] mu1 - spline(x=res.em1$tau,y=res.em1$eta,xout=at,method=natural)$y l2 - functional.norm(mu1,mu2,by=by) s1 - mean(sapply(1:res.em1$n, function(i) { v - spline(x=res.em1$tau,y=res.em1$vi[[i]],xout=at,method=natural)$y functional.norm(v,by=by)^2 })) denominator - sqrt(s1 / res.em1$n + s2 / res.em2$n) ### formula for se returnValue - l2 / (denominator + se.m) attr(returnValue,numerator) - l2 attr(returnValue,denominator) - denominator returnValue } moderated.functional.t - function(testStatistics,alpha.step=0.05,quantile.step=0.01) { numerators - unlist(testStatistics[numerator,]) denominators - unlist(testStatistics[denominator,]) } I get my error in the function above moderated.functional.t. testStatistics is shown to be a function(x) when I type it in R console. But there is no function definition for testStatistics in the code. My R understanding is still elementary. Thanks Aparna On Tue, Feb 28, 2012 at 5:25 PM, Joshua Wiley jwiley.ps...@gmail.com wrote: Hi Aparna, Can you please post a reproducible example? It is difficult to provide much concrete help without having testStatistics. One thing you might try is looking at: str(testStatistics[numerator,]) is it actually a list? If it is not (most likely given the error) and it is supposed to be, you need to figure out what aspect of the generation of it is going awry. Cheers, Josh On Tue, Feb 28, 2012 at 12:23 AM, Aparna Sampath aparna.sampat...@gmail.com wrote: Hi All I am trying to use the unlist() in R to a list variable. The following statements are within a function. { denominator - sqrt(s1 / res.em1$n + s2 / res.em2$n) returnValue - l2 / (denominator + 11) attr(returnValue,numerator) - l2 attr(returnValue,denominator) - denominator returnValue } And when I try to unlist the variable returnValue numerators - unlist(testStatistics[numerator,]) denominators - unlist(testStatistics[denominator,]) I get the following error: Error in testStatistics[numerator, ] : object of type 'closure' is not subsettable I read some threads in R help on this error and they had asked to check if we are using the right datatype to the right function. But in my case it is pretty straightforward since I just list it in one function and try to unlist it later. Any suggestions? Thanks for the help :) -- View this message in context: http://r.789695.n4.nabble.com/Error-message-object-of-type-closure-is-not-subsettable-tp3752886p4427399.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/ -- Aparna Sampath Master of Science (Bioinformatics) Nanyang Technological University Mob no : +65 91601854 -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] asking for a script
Hi Dear who may concern, I am sending you this email to ask if I could have someone help me to write this script I am interested in. I am requesting the script for performing variance-stabilizing normalizations on the R language. Could you please help me on that? If I am asking a wrong person, could you please tell me the right email address I should contact? Maybe you shall ask google. I tried variance-stabilizing normalizations and as a second hit I got Variance Stabilization and Normalization using the vsn package in R Is this what you wanted? Regards Petr Appreciated! Thanks! Regards Ruijuan Luo [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Non linear regression with complex equation
Apologies for the phrasing of the question. I've sorted the problem (thanks Bert Gunter) by using the curly brackets {} as below (using a simplified version of my real model). I hope this helps someone else! Jeff --- data Alpha ip X 1 0.7106967 0.3616727 0.006027879 2 2.1678517 5.0615917 0.084359861 3 4.4066250 11.2282945 0.187138242 4 9.8495694 18.0534974 0.300891624 527.7247098 29.2064434 0.486774057 670.6931430 35.3946092 0.589910153 7 133.1240255 46.0347288 0.767245480 8 214.7851844 49.3811149 0.823018582 9 359.5511036 58.5069583 0.975115972 10 748.1840127 57.3744477 0.956240795 11 2129.9844080 60.000 1.0 c-1.83e-9 cFe=c model-nls({Fe1-cFe/(Alpha+1+k*c) + X~Alpha/(Alpha+1+k*c/(1+k*Fe1))},start=list(k=1e10)) summary(model) Formula: X ~ Alpha/(Alpha + 1 + k * c/(1 + k * Fe1)) Parameters: Estimate Std. Error t value Pr(|t|) k 3.491e+10 7.190e+09 4.856 0.000665 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.05589 on 10 degrees of freedom Number of iterations to convergence: 8 Achieved convergence tolerance: 2.393e-06 -- View this message in context: http://r.789695.n4.nabble.com/Non-linear-regression-with-complex-equation-tp4425942p4427617.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Bayesian Hidden Markov Models
Dear James, Basically you just need the values (y) and the positions (in your case it would be the index of the times series). The chromosome argument does not apply to your case so it can be a vector of ones. If the positions are at the same distance between (equally spaced) then the model will be homogeneous. So for example something like this would be enough: library(RJaCGH) y - c(rnorm(100,0,1), rnorm(20, 2, 1), rnorm(50, 0, 1)) Pos - 1:length(y) Chrom - rep(1, length(y)) res - RJaCGH(y=y, Pos=Pos, Chrom=Chrom) summary(res) However, it uses a Reversible Jump algorithm and therefore jumps between models with different hidden states. I would suggest you take a look at the vignette that comes with the package or the paper that is referenced there for specific details of the model it fits. Hope it helps, Oscar On 28/2/12 04:52, monkeylan lanjin...@yahoo.com.cn wrote: Dear Doctor Oscar, Sorry for not noticing that you are the author of the RJaCGH package. But I noticed that hidden Markov model in your package is with non-homogeneous transition probabilities. Here in my work, the HMM is just a first-order homogeneous Markov chain, i.e. the transition matrix is constant. So, Could you please tell me how can I adjust the R functions in your package to implement my analysis? Best Regards, James Allan --- 12年2月27日,周一, Oscar Rueda [via R] ml-node+s789695n4424152...@n4.nabble.com 写道: 发件人: Oscar Rueda [via R] ml-node+s789695n4424152...@n4.nabble.com 主题: Re: Bayesian Hidden Markov Models 收件人: monkeylan lanjin...@yahoo.com.cn 日期: 2012年2月27日,周一,下午6:05 Dear James, Although designed for the analysis of copy number CGH microarrays, RJaCGH uses a Bayesian HMM model. Cheers, Oscar On 27/2/12 08:32, monkeylan [hidden email] wrote: Dear R buddies, Recently, I attempt to model the US/RMB Exchange rate log-return time series with a *Hidden Markov model (first order Markov Chain mixed Normal distributions). * I have applied the RHmm package to accomplish this task, but the results are not so satisfying. So, I would like to try a *Bayesian method *for the parameter estimation of the Hidden Markov model. Could anyone kindly tell me which R package can perform Bayesian estimation of the model? Many thanks for your help and time. Best Regards, James Allan -- View this message in context: http://r.789695.n4.nabble.com/Bayesian-Hidden-Markov-Models-tp4423946p4423946 . html Sent from the R help mailing list archive at Nabble.com. __ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Oscar M. Rueda, PhD. Postdoctoral Research Fellow, Breast Cancer Functional Genomics. Cancer Research UK Cambridge Research Institute. Li Ka Shing Centre, Robinson Way. Cambridge CB2 0RE England NOTICE AND DISCLAIMER This e-mail (including any attachments) is intended for ...{{dropped:16}} __ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. If you reply to this email, your message will be added to the discussion below:http://r.789695.n4.nabble.com/Bayesian-Hidden-Markov-Models-tp4423946p44 24152.html To unsubscribe from Bayesian Hidden Markov Models, click here. NAML -- View this message in context: http://r.789695.n4.nabble.com/Bayesian-Hidden-Markov-Models-tp4423946p4427000. html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] Oscar M. Rueda, PhD. Postdoctoral Research Fellow, Breast Cancer Functional Genomics. Cancer Research UK Cambridge Research Institute. Li Ka Shing Centre, Robinson Way. Cambridge CB2 0RE England NOTICE AND DISCLAIMER This e-mail (including any attachments) is intended for the above-named person(s). If you are not the intended recipient, notify the sender immediately, delete this email from your system and do not disclose or use for any purpose. We may monitor all incoming and outgoing emails in line with current legislation. We have taken steps to ensure that this email and attachments are free from any virus, but it remains your responsibility to ensure that viruses do not adversely affect you. Cancer Research UK Registered in England and Wales Company Registered Number: 4325234. Registered Charity Number: 1089464 and Scotland SC041666 Registered Office Address: Angel Building, 407 St John Street, London EC1V 4AD. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide
Re: [R] aggregating specific parts in zoo index column to perform sliding average
On Mon, Feb 27, 2012 at 10:17 PM, knavero knav...@gmail.com wrote: Here's my code: http://pastebin.com/0yRxEVtm The important parts are uncommented and should be easy to find using the link above. For the following line of code, I plan on looking for a way to offset it up 7 rows so that the 15 minute timestamp would be considered the median of the subset being averaged to find the mean: avgCool = aggregate(intCool, trunc(time(intCool), times(00:15:00)), mean) Currently the issue is that, with the truncate function, it truncates but really rounds down the time series values to the 15 minute time stamp earlier in the series. For example, let's say we have one minute intervals 0:00, 0:01, 0:02,,0:37. It takes 0:00 - 0:14 and replaces that with 0:00. Then it sees 0:15, and changes values from 0:15 - 0:29 to 0:15. In effect, aggregating the values and creating subsets. What I want to do here is change 0:00 - 0:07 to 0:00, change 0:08 - 0:22 to 0:15, and change 0:23 - 0:37 to 0:30 in which 0:15 and 0:30 are the medians of each subset. Anyway, I hope that makes sense. Any ideas on which function will make this an easy job? Much thanks in advance. Add 7.5 minutes before truncating so if x is your times, e.g. x - times(0:15/(24 * 60)), then try this: min15 - times(00:15:00) trunc(x + min15/2, min15) -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] compare two data frames of different dimensions and onlykeep unique rows
TY very much for your setdiffDF(). It does the job perfectly. Arnaud Gaboury A2CT2 Ltd. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Petr Savicky Sent: lundi 27 février 2012 20:41 To: r-help@r-project.org Subject: Re: [R] compare two data frames of different dimensions and onlykeep unique rows On Mon, Feb 27, 2012 at 07:10:57PM +0100, Arnaud Gaboury wrote: No, but I tried your way too. In fact, the only three unique rows are these ones: Product Price Nbr.Lots Cocoa 24405 Cocoa 24501 Cocoa 24406 Here is a dirty working trick I found : df-merge(exportfile,reported,all.y=T) df1-merge(exportfile,reported) dff1-do.call(paste,df) dff-do.call(paste,df) dff1-do.call(paste,df1) df[!dff %in% dff1,] Product Price Nbr.Lots 3 Cocoa 24405 4 Cocoa 24501 My two problems are : I do think it is not so a clean code, then I won't know by advance which of my two df will have the greates dimension (I can add some lines to deal with it, but again, seems very heavy). Hi. Try the following. setdiffDF - function(A, B) { A[!duplicated(rbind(B, A))[nrow(B) + 1:nrow(A)], ] } df1 - setdiffDF(reported, exportfile) df2 - setdiffDF(exportfile, reported) rbind(df1, df2) I obtained Product Price Nbr.Lots 3Cocoa 24405 4Cocoa 24501 31 Cocoa 24406 Is this correct? I see the row Cocoa 2440.006 only in exportfile and not in reported. The trick with paste() is not a bad idea. A variant of it is used also in the base function duplicated.matrix(), since it contains apply(x, MARGIN, function(x) paste(x, collapse = \r)) If speed is critical, then possibly the paste() trick written for the whole columns, for example paste(df[[1]], df[[2]], df[[3]], sep=\r) and then setdiff() can be better. Hope this helps. Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] colour by z value, persp in raster package
Hi all! My question is how to colour pixels by z value in persp plot in raster package. Here is an example: x - seq(-1.95, 1.95, length = 30) y - seq(-1.95, 1.95, length = 35) z - outer(x, y, function(a,b) a*b^2) r1 - raster(nrows=35, ncols=30, xmn=0, xmx=30, ymn = 0, ymx = 35) r1[] - c(z) persp(r1) There already exist some function to produce persp plot for anothe classes, but I have no idea how deal with RasterLayer object. Thanks all! OV [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] kmeans: how to retrieve clusters
Hi, Ok, I understand what you mean. I wanted to get sorted data group by cluster in output ... But I have to do it myself using kres$cluster thanks for your help -- View this message in context: http://r.789695.n4.nabble.com/kmeans-how-to-retrieve-clusters-tp4426427p4427543.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Installing package QRMlib
Thanks for the reply guys. Well, I've tried the following command after installing the package dependancies including timeSeries: install.packages(file_name, type = source, repos = NULL) Ans still no luck...I get the following error message: Installing package(s) into ‘C:/Program Files/R/R-2.14.1/library’ (as ‘lib’ is unspecified) * installing *source* package 'QRMlib' ... ** Creating default NAMESPACE file ** libs ERROR: compilation failed for package 'QRMlib' * removing 'C:/Program Files/R/R-2.14.1/library/QRMlib' * restoring previous 'C:/Program Files/R/R-2.14.1/library/QRMlib' Warning in install.packages : running command 'C:/PROGRA~1/R/R-214~1.1/bin/i386/R CMD INSTALL -l C:/Program Files/R/R-2.14.1/library my_local_folder/QRMlib_1.4.5.1.tar.gz' had status 1 Warning in install.packages : installation of package ‘my_local_folder/QRMlib_1.4.5.1.tar.gz’ had non-zero exit status Any ideas?? -- View this message in context: http://r.789695.n4.nabble.com/Installing-package-QRMlib-tp4425269p4427627.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error message: object of type 'closure' is not subsettable
Hi Josh Kindly find below the code as it is executed: test.functional.t - function(res.em1,res.em2,mint,maxt,se.m=0,points=300) { at - seq(mint,maxt,length.out=points) by - at[2] - at[1] mu1 - spline(x=res.em1$tau,y=res.em1$eta,xout=at,method=natural)$y l2 - functional.norm(mu1,mu2,by=by) s1 - mean(sapply(1:res.em1$n, function(i) { v - spline(x=res.em1$tau,y=res.em1$vi[[i]],xout=at,method=natural)$y functional.norm(v,by=by)^2 })) denominator - sqrt(s1 / res.em1$n + s2 / res.em2$n) ### formula for se returnValue - l2 / (denominator + se.m) attr(returnValue,numerator) - l2 attr(returnValue,denominator) - denominator returnValue } moderated.functional.t - function(testStatistics,alpha.step=0.05,quantile.step=0.01) { numerators - unlist(testStatistics[numerator,]) denominators - unlist(testStatistics[denominator,]) } I get my error in the function above moderated.functional.t. testStatistics is shown to be a function(x) when I type it in R console. But there is no function definition for testStatistics in the code. My R understanding is still elementary. Thanks Aparna On Tue, Feb 28, 2012 at 5:25 PM, Joshua Wiley jwiley.ps...@gmail.comwrote: Hi Aparna, Can you please post a reproducible example? It is difficult to provide much concrete help without having testStatistics. One thing you might try is looking at: str(testStatistics[numerator,]) is it actually a list? If it is not (most likely given the error) and it is supposed to be, you need to figure out what aspect of the generation of it is going awry. Cheers, Josh On Tue, Feb 28, 2012 at 12:23 AM, Aparna Sampath aparna.sampat...@gmail.com wrote: Hi All I am trying to use the unlist() in R to a list variable. The following statements are within a function. { denominator - sqrt(s1 / res.em1$n + s2 / res.em2$n) returnValue - l2 / (denominator + 11) attr(returnValue,numerator) - l2 attr(returnValue,denominator) - denominator returnValue } And when I try to unlist the variable returnValue numerators - unlist(testStatistics[numerator,]) denominators - unlist(testStatistics[denominator,]) I get the following error: Error in testStatistics[numerator, ] : object of type 'closure' is not subsettable I read some threads in R help on this error and they had asked to check if we are using the right datatype to the right function. But in my case it is pretty straightforward since I just list it in one function and try to unlist it later. Any suggestions? Thanks for the help :) -- View this message in context: http://r.789695.n4.nabble.com/Error-message-object-of-type-closure-is-not-subsettable-tp3752886p4427399.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/ -- Aparna Sampath Master of Science (Bioinformatics) Nanyang Technological University Mob no : +65 91601854 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Error in solve.default(res$hessian * n.used) :Lapack routine dgesv: system is exactly singular
Hi there! I´m a noob when it comes to R and I´m using it to run statisc analysis. With the code for ARIMA below I´m getting this error: Error in solve.default(res$hessian * n.used) :Lapack routine dgesv: system is exactly singular The code is: s.ts - ts(x[,7], start = 2004, fre=12) get.best.arima - function (x.ts, maxord=c(1,1,1,1,1,1)) + { + best.aic - 1e8 + n - length(x.ts) + for (p in 0:maxord[1]) for (d in 0:maxord[2]) for (q in 0:maxord[3]) + for (P in 0:maxord[4]) for (D in 0:maxord[5]) for (Q in 0:maxord[6]) + { + fit - arima(x.ts, order=c(p,d,q), + seas = list(order=c(P,D,Q), + frequency(x.ts)), method = CSS) + fit.aic - -2 * fit$loglik + (log(n) + 1) * length(fit$coef) + if (fit.aic best.aic) + { + best.aic - fit.aic + best.fit - fit + best.model - c(p,d,q,P,D,Q) + } + } + list(best.aic,best.fit,best.model) + } best.arima.ss - get.best.arima(log(s.ts),maxord = c(2,2,2,2,2,2)) Error in solve.default(res$hessian * n.used) : Lapack routine dgesv: system is exactly singular s.ts Jan FebMarAprMay Jun Jul Aug 2004 143. 160. 205. 180. 160. 160. 155. 160. 2005 148. 185. 195. 195. 175. 175. 170. 165. 2006 188. 203. 213. 198. 180. 180. 190. 188. 2007 168. 200. 210. 185. 175. 190. 190. 193. 2008 167. 190. 210. 205. 190. 190. 200. 175. Sep Oct Nov Dec 2004 150. 140. 135. 150. 2005 160. 155. 155. 153. 2006 193. 184. 170. 165. 2007 180. 180. 175. 165. 2008 175. 165. 161.6667 161.6667 -- Vinicius Macedo Magalhães (21) 9584-1533 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Interpreting the Results of GLM
Hi, I'm wondering if you can help me, this is a really simple query but I keep getting confused. I have run a GLM to see how boldness varies over time following a particular treatment. The results are as follows... Call: glm(formula = boldtwentyfour ~ treatment + boldcontrol) Deviance Residuals: Min 1Q Median 3Q Max -1.7577 -0.5469 0.0456 0.5515 1.5327 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept)0.8312 0.5444 1.527 0.1363 treatmentPBS 0.1391 0.2842 0.490 0.6277 boldcontrol0.4899 0.2157 2.271 0.0298 * Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for gaussian family taken to be 0.7131691) Null deviance: 27.243 on 35 degrees of freedom Residual deviance: 23.535 on 33 degrees of freedom AIC: 94.862 Number of Fisher Scoring iterations: 2 Basically, where I am having trouble is that I have several GLMs like this and need to display the results in a table and am required to display the intercept, estimates, 95% confidence and the errors. I am confused as to which values are which as for the intercept there are four different values which all seem to relate to everything else I need to report, while everything else would go unreported. From this table of output from R can you help me to identify which value is which? Thanks, Sarah. -- View this message in context: http://r.789695.n4.nabble.com/Interpreting-the-Results-of-GLM-tp4427732p4427732.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to append element at last position in array dynamically
h-array() h [1] NA append(h,9) [1] NA 9 but what it append. h [1] NA -- View this message in context: http://r.789695.n4.nabble.com/how-to-append-element-at-last-position-in-array-dynamically-tp4427893p4427893.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] kmeans: how to retrieve clusters
But it won't be hard at allyou can likely get what you need using the tapply() function (or ave) Michael On Tue, Feb 28, 2012 at 4:33 AM, ikuzar raz...@hotmail.fr wrote: Hi, Ok, I understand what you mean. I wanted to get sorted data group by cluster in output ... But I have to do it myself using kres$cluster thanks for your help -- View this message in context: http://r.789695.n4.nabble.com/kmeans-how-to-retrieve-clusters-tp4426427p4427543.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to get file name/file path
i open a table table-read.table(file.choose(),skip=1) at sometime i forget which file i open,so i want file name/file path -- View this message in context: http://r.789695.n4.nabble.com/how-to-get-file-name-file-path-tp4428126p4428126.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] macro function
Actually, what I really want to do is that, annual productivity data(2011) firts date 2011-01-01 2011-01-02 2011-01-03 2011-01-04 2011-01-05 * * * A2011-02-03 0 0 0 0 0 * * * B2011-01-02 0 10 11 12 13* * * C2011-04-02 00 11 12 13 * * * D2011-02-02 0 0 0 0 0 * * * E2011-11-020 0 0 0 0 * * * * * * I have this annual productivity data(2011) and I need to convert this as D-Day data. That is to say, I need this form. firts date D-DAYD+1 D+2 D+3D+4 * * * A 2011-02-03 0 0 0 0 0 * * * B 2011-01-02 010 11 12 13 * * * C 2011-04-02 00 11 12 13 * * * D 2011-02-02 0 0 0 0 0 * * * E 2011-11-020 0 0 0 0 * * * * * * D-day is the first date and each cells are day-productivity of the day. how can I make a code for that? -- View this message in context: http://r.789695.n4.nabble.com/macro-function-tp4427385p4428131.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to append element at last position in array dynamically
You need to reassign the value of append back to h -- in more technical terms, this is a pass-by-value rather than pass-by-reference behavior: h - append(h, 9) This is not likely to be efficient in production code, however. Michael On Tue, Feb 28, 2012 at 7:15 AM, sagarnikam123 sagarnikam...@gmail.com wrote: h-array() h [1] NA append(h,9) [1] NA 9 but what it append. h [1] NA -- View this message in context: http://r.789695.n4.nabble.com/how-to-append-element-at-last-position-in-array-dynamically-tp4427893p4427893.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] indexing??
Hello All, My algorithm as follows; y - c(1,1,1,0,0,1,0,1,0,0) x - c(1,0,0,1,1,0,0,1,1,0) n - length(x) t - matrix(cbind(y,x), ncol=2) z = x+y for(j in 1:length(x)) { out - vector(list, ) for(i in 1:10) { t.s - t[sample(n,n,replace=T),] y.s - t.s[,1] x.s - t.s[,2] z.s - y.s+x.s out[[i]] - list(ff - (z.s), finding=any (y.s==y[j])) kk - sapply(out, function(x) {x$finding}) ff - out[! kk] } I tried to find the total of the two vectors as statistic by using bootstrap. Finally, I want to get the values which do not contain the y's each elemet. In the algorithm ti is referred to ff. But i get always the same result ; ff list() kk [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE Because, my y vector contains only 2 elements, and probably all of the bootstrap resamples include 1, or all of resamples include 0. So I can not find the true matches. Can anyone help me about how to be? Thanks. -- View this message in context: http://r.789695.n4.nabble.com/indexing-tp4428210p4428210.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to get file name/file path
Use this line instead: table - read.table(fileIchoose - file.choose(), skip = 1) This will have the side effect of creating fileIchoose as a variable containing a file path. Michael On Tue, Feb 28, 2012 at 8:34 AM, sagarnikam123 sagarnikam...@gmail.com wrote: i open a table table-read.table(file.choose(),skip=1) at sometime i forget which file i open,so i want file name/file path -- View this message in context: http://r.789695.n4.nabble.com/how-to-get-file-name-file-path-tp4428126p4428126.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Installing package QRMlib
As stated, you need to install the *deprecated* dependencies of QRMlib as shown in its DESCRIPTION as well as the reverse dependent *deprecated* packages. These can still be fetched from R-Forge (Rmetrics project). The package 'timeSeries' will become a dependency of the to be re-released QRMlib package on CRAN. -Ursprüngliche Nachricht- Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im Auftrag von DT54321 Gesendet: Dienstag, 28. Februar 2012 11:10 An: r-help@r-project.org Betreff: Re: [R] Installing package QRMlib Thanks for the reply guys. Well, I've tried the following command after installing the package dependancies including timeSeries: install.packages(file_name, type = source, repos = NULL) Ans still no luck...I get the following error message: Installing package(s) into ‘C:/Program Files/R/R-2.14.1/library’ (as ‘lib’ is unspecified) * installing *source* package 'QRMlib' ... ** Creating default NAMESPACE file ** libs ERROR: compilation failed for package 'QRMlib' * removing 'C:/Program Files/R/R-2.14.1/library/QRMlib' * restoring previous 'C:/Program Files/R/R-2.14.1/library/QRMlib' Warning in install.packages : running command 'C:/PROGRA~1/R/R-214~1.1/bin/i386/R CMD INSTALL -l C:/Program Files/R/R-2.14.1/library my_local_folder/QRMlib_1.4.5.1.tar.gz' had status 1 Warning in install.packages : installation of package ‘my_local_folder/QRMlib_1.4.5.1.tar.gz’ had non-zero exit status Any ideas?? -- View this message in context: http://r.789695.n4.nabble.com/Installing-package-QRMlib-tp4425269p4427627.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. * Confidentiality Note: The information contained in this message, and any attachments, may contain confidential and/or privileged material. It is intended solely for the person(s) or entity to which it is addressed. Any review, retransmission, dissemination, or taking of any action in reliance upon this information by persons or entities other than the intended recipient(s) is prohibited. If you received this in error, please contact the sender and delete the material from any computer. * __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to append element at last position in array dynamically
'append' returns a value that you have assign back to the object you want, in this case 'h' h - append(h, 9) On Tue, Feb 28, 2012 at 7:15 AM, sagarnikam123 sagarnikam...@gmail.com wrote: h-array() h [1] NA append(h,9) [1] NA 9 but what it append. h [1] NA -- View this message in context: http://r.789695.n4.nabble.com/how-to-append-element-at-last-position-in-array-dynamically-tp4427893p4427893.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error message: object of type 'closure' is not subsettable
Hi Difficult to without knowing what objects you are operating your functions. I get test.functional.t(1:10,1:10,3,4) Error in res.em1$eta : $ operator is invalid for atomic vectors moderated.functional.t(1:10) Error in testStatistics[numerator, ] : incorrect number of dimensions And without suitable objects for those functions to operate it is impossible to come with reasonable suggestion. Anyway there is a function on your system which is called testStatistics and you can not subset functions, hence your error. I did not find this function in R pacages but I can not say that it is really not there. You can get rid of this function by rm(testStatistics) Regards Petr Kindly find below the code as it is executed: test.functional.t - function(res.em1,res.em2,mint,maxt,se.m=0,points=300) { at - seq(mint,maxt,length.out=points) by - at[2] - at[1] mu1 - spline(x=res.em1$tau,y=res.em1$eta,xout=at,method=natural)$y l2 - functional.norm(mu1,mu2,by=by) s1 - mean(sapply(1:res.em1$n, function(i) { v - spline(x=res.em1$tau,y=res.em1$vi[[i]],xout=at,method=natural)$y functional.norm(v,by=by)^2 })) denominator - sqrt(s1 / res.em1$n + s2 / res.em2$n) ### formula for se returnValue - l2 / (denominator + se.m) attr(returnValue,numerator) - l2 attr(returnValue,denominator) - denominator returnValue } moderated.functional.t - function(testStatistics,alpha.step=0.05,quantile.step=0.01) { numerators - unlist(testStatistics[numerator,]) denominators - unlist(testStatistics[denominator,]) } I get my error in the function above moderated.functional.t. testStatistics is shown to be a function(x) when I type it in R console. But there is no function definition for testStatistics in the code. My R understanding is still elementary. Thanks Aparna On Tue, Feb 28, 2012 at 5:25 PM, Joshua Wiley jwiley.ps...@gmail.comwrote: Hi Aparna, Can you please post a reproducible example? It is difficult to provide much concrete help without having testStatistics. One thing you might try is looking at: str(testStatistics[numerator,]) is it actually a list? If it is not (most likely given the error) and it is supposed to be, you need to figure out what aspect of the generation of it is going awry. Cheers, Josh On Tue, Feb 28, 2012 at 12:23 AM, Aparna Sampath aparna.sampat...@gmail.com wrote: Hi All I am trying to use the unlist() in R to a list variable. The following statements are within a function. { denominator - sqrt(s1 / res.em1$n + s2 / res.em2$n) returnValue - l2 / (denominator + 11) attr(returnValue,numerator) - l2 attr(returnValue,denominator) - denominator returnValue } And when I try to unlist the variable returnValue numerators - unlist(testStatistics[numerator,]) denominators - unlist(testStatistics[denominator,]) I get the following error: Error in testStatistics[numerator, ] : object of type 'closure' is not subsettable I read some threads in R help on this error and they had asked to check if we are using the right datatype to the right function. But in my case it is pretty straightforward since I just list it in one function and try to unlist it later. Any suggestions? Thanks for the help :) -- View this message in context: http://r.789695.n4.nabble.com/Error-message-object-of-type-closure-is- not-subsettable-tp3752886p4427399.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/ -- Aparna Sampath Master of Science (Bioinformatics) Nanyang Technological University Mob no : +65 91601854 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] indexing??
Hi My algorithm as follows; y - c(1,1,1,0,0,1,0,1,0,0) x - c(1,0,0,1,1,0,0,1,1,0) n - length(x) t - matrix(cbind(y,x), ncol=2) Do not use t, it is a function for transposing matrix and after you redefine it you can get nasty surprise in future. tt - cbind(y,x) is enough z = x+y for(j in 1:length(x)) { out - vector(list, ) for(i in 1:10) { t.s - t[sample(n,n,replace=T),] t.s - tt[sample(n,n,replace=T),] y.s - t.s[,1] x.s - t.s[,2] z.s - y.s+x.s out[[i]] - list(ff - (z.s), finding=any (y.s==y[j])) Here you compare vector y.s with one element of y as y.s is set of (0,1) values y is either 0 or 1, any tests if there is any match so only in rare case where all values in y.s are 0 and y[something] is 1 you get FALSE kk - sapply(out, function(x) {x$finding}) finding is (almost) always TRUE therefore kk is TRUE ff - out[! kk] } I tried to find the total of the two vectors as statistic by using bootstrap. Finally, I want to get the values which do not contain the y's each elemet. In the algorithm ti is referred to ff. But i get always the same result ; I do not understand your intention so it is difficult to help. What is total of two vectors? sum? What does it mean to get values which do not contain y's each element? Maybe you shall rethink your code and first try to evaluate each line separately to see what it does and if the result is same as you intended. Regards Petr ff list() kk [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE Because, my y vector contains only 2 elements, and probably all of the bootstrap resamples include 1, or all of resamples include 0. So I can not find the true matches. Can anyone help me about how to be? Thanks. -- View this message in context: http://r.789695.n4.nabble.com/indexing- tp4428210p4428210.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] macro function
Hi Your mail is quite messy. what is ***? How your data are structured? str(your.data) Please submit part of your data by dput(your.data) and show what the output shall be. Regards Petr Actually, what I really want to do is that, annual productivity data(2011) firts date 2011-01-01 2011-01-02 2011-01-03 2011-01-04 2011-01-05 * * * A2011-02-03 0 0 0 0 0 * * * B2011-01-02 0 10 11 12 13* * * C2011-04-02 0 0 11 12 13 * * * D2011-02-02 0 0 0 0 0 * * * E2011-11-020 0 0 0 0 * * * * * * I have this annual productivity data(2011) and I need to convert this as D-Day data. That is to say, I need this form. firts date D-DAY D+1 D+2 D+3 D+4 * * * A 2011-02-03 0 0 0 0 0 * * * B 2011-01-02 0 10 11 12 13 * * * C 2011-04-02 0 0 11 12 13 * * * D 2011-02-02 0 0 0 0 0 * * * E 2011-11-020 0 0 0 0 * * * * * * D-day is the first date and each cells are day-productivity of the day. how can I make a code for that? -- View this message in context: http://r.789695.n4.nabble.com/macro- function-tp4427385p4428131.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] indexing??
On Tue, Feb 28, 2012 at 05:59:24AM -0800, helin_susam wrote: Hello All, My algorithm as follows; y - c(1,1,1,0,0,1,0,1,0,0) x - c(1,0,0,1,1,0,0,1,1,0) n - length(x) t - matrix(cbind(y,x), ncol=2) z = x+y for(j in 1:length(x)) { out - vector(list, ) for(i in 1:10) { t.s - t[sample(n,n,replace=T),] y.s - t.s[,1] x.s - t.s[,2] z.s - y.s+x.s out[[i]] - list(ff - (z.s), finding=any (y.s==y[j])) kk - sapply(out, function(x) {x$finding}) ff - out[! kk] } I tried to find the total of the two vectors as statistic by using bootstrap. Finally, I want to get the values which do not contain the y's each elemet. In the algorithm ti is referred to ff. But i get always the same result ; ff list() kk [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE Because, my y vector contains only 2 elements, and probably all of the bootstrap resamples include 1, or all of resamples include 0. So I can not find the true matches. Can anyone help me about how to be? Hi. First of all, there are some unclear points in your code. In particular, i would expect } between the line out[[i]] - list(... and kk - sapply(... Moreover, i do not see, why the loop over j contains the loop over i. I would expect these loops be disjoint, since the loop over i collects all the samples to a list. The following code is a modification, which i suggest as an alternative. y - c(1:5, 1:5) x - c(1,0,0,1,1,0,0,1,1,0) n - length(x) t - matrix(cbind(y,x), ncol=2) z = x+y # generate 10 bootstrap samples and keep z.s, y.s out - vector(list, 10) for(i in 1:10) { t.s - t[sample(n,n,replace=T),] y.s - t.s[,1] x.s - t.s[,2] z.s - y.s+x.s out[[i]] - list(zz = z.s, yy =y.s) } # check, which replications do not contain y[j] in their y.s, # and take the OR of these conditions over j ff - rep(FALSE, times=length(out)) for(j in 1:length(y)) { kk - sapply(out, function(x) {any(x$yy == y[j])}) ff - ff | (! kk) } out[ff] With the original y - c(1,1,1,0,0,1,0,1,0,0), the probability that a bootstrap sample contains only 1's or only 0's is 2 * (1/2)^10, so i replaced the vector y with another, where a missing value is more frequent. I obtained, for example [[1]] [[1]]$zz [1] 2 2 5 2 3 2 3 2 2 6 [[1]]$yy [1] 1 1 5 1 3 2 3 2 1 5 # 4 is missing [[2]] [[2]]$zz [1] 5 5 5 5 3 5 2 5 6 4 [[2]]$yy [1] 4 4 5 4 3 5 2 5 5 3 # 1 is missing [[3]] [[3]]$zz [1] 5 2 5 1 5 1 2 5 5 5 [[3]]$yy [1] 4 2 5 1 5 1 1 4 5 4 # 3 is missing Hope this helps. Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Installing package QRMlib
I think you'll need to install fEcofin from CRAN, and then the following .tar.gz files from the CRAN archives: fUtilities fCalenday fSeries And then try installing the QRMlib .tar.gz. If you're still having problems, try: C:/Program Files/R/R-2.14.0/bin/x64/R CMD check QRMlib_1.4.5.1.tar.gz (or whatever is equivalent for your R installation) from the Windows command prompt . That will output more useful information on errors and warnings. Jeremy Jeremy Hetzel Boston University Pfaff, Bernhard Dr. wrote As stated, you need to install the *deprecated* dependencies of QRMlib as shown in its DESCRIPTION as well as the reverse dependent *deprecated* packages. These can still be fetched from R-Forge (Rmetrics project). The package 'timeSeries' will become a dependency of the to be re-released QRMlib package on CRAN. -Ursprüngliche Nachricht- Von: r-help-bounces@ [mailto:r-help-bounces@] Im Auftrag von DT54321 Gesendet: Dienstag, 28. Februar 2012 11:10 An: r-help@ Betreff: Re: [R] Installing package QRMlib Thanks for the reply guys. Well, I've tried the following command after installing the package dependancies including timeSeries: install.packages(file_name, type = source, repos = NULL) Ans still no luck...I get the following error message: Installing package(s) into ‘C:/Program Files/R/R-2.14.1/library’ (as ‘lib’ is unspecified) * installing *source* package 'QRMlib' ... ** Creating default NAMESPACE file ** libs ERROR: compilation failed for package 'QRMlib' * removing 'C:/Program Files/R/R-2.14.1/library/QRMlib' * restoring previous 'C:/Program Files/R/R-2.14.1/library/QRMlib' Warning in install.packages : running command 'C:/PROGRA~1/R/R-214~1.1/bin/i386/R CMD INSTALL -l C:/Program Files/R/R-2.14.1/library my_local_folder/QRMlib_1.4.5.1.tar.gz' had status 1 Warning in install.packages : installation of package ‘my_local_folder/QRMlib_1.4.5.1.tar.gz’ had non-zero exit status Any ideas?? -- View this message in context: http://r.789695.n4.nabble.com/Installing-package-QRMlib-tp4425269p4427627.html Sent from the R help mailing list archive at Nabble.com. __ R-help@ mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. * Confidentiality Note: The information contained in this message, and any attachments, may contain confidential and/or privileged material. It is intended solely for the person(s) or entity to which it is addressed. Any review, retransmission, dissemination, or taking of any action in reliance upon this information by persons or entities other than the intended recipient(s) is prohibited. If you received this in error, please contact the sender and delete the material from any computer. * __ R-help@ mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Jeremy T. Hetzel Boston University -- View this message in context: http://r.789695.n4.nabble.com/Installing-package-QRMlib-tp4425269p4428517.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] vlookup type function
Hi I''m looking for an Excel Vlookup type function in R. Example: list - c(1,2,3,4,5,6,7) base - c(2.2,3,5.2) What I want is, for each number in base, the highest value in list, which is equal to or less than the number in base So the results would be: base list 2.2 -- 2 3 -- 3 5.2 -- 5 Thanks for your help! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Cluster GUI package worth publishing/enhancing?
For a school course I and a partner developed a GUI in R designed to enable exploration of data via visualization of hierarchical clustering and correlation of cluster partitions with external metadata. The key features were the ability to load in a distance matrix (most GUI-based clustering programs require feature vector input), and the ability to dynamically subset the data via the GUI built using a user-provided meta data file. I didn't think this was sufficient to publish a package, but it did seem like it could make for a good foundation. I so far have not come across other software that provides all of these features. I was hoping to get initial feedback in terms of whether anyone thought this, with or without certain enhancements, might be worthwhile. For a sense of what the existing tool looks like and what it does: http://mason.gmu.edu/~tgillett/R_Cluster_GUI/RClusterGUI.html We intend to extend the tool to enable feature vector input, additional forms of visualization beyond cluster dendrogram, multidimensional scaling of distance matrix input, internal and external cluster statistics (e.g. Davies-Bouldin index, Rand index), random subsampling of data to account for the instability of clusters, and non-hierarchical clustering methods. Obviously a well-organized GUI is important, and we would seek out feedback and perhaps partner developers. I'd appreciate any direct feedback here, or suggestions of more appropriate forums for getting feedback and having a discussion on the topic. Thank you, Todd Gillette __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] vlookup type function
On Tue, Feb 28, 2012 at 09:02:04PM +0530, Priyan Fernando wrote: Hi I''m looking for an Excel Vlookup type function in R. Example: list - c(1,2,3,4,5,6,7) base - c(2.2,3,5.2) What I want is, for each number in base, the highest value in list, which is equal to or less than the number in base So the results would be: base list 2.2 -- 2 3 -- 3 5.2 -- 5 Hi. Try the following. unlist(lapply(base, FUN = function(x) max(list[list = x]))) Hope this helps. Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] vlookup type function
On 28-02-2012, at 16:32, Priyan Fernando wrote: Hi I''m looking for an Excel Vlookup type function in R. Example: list - c(1,2,3,4,5,6,7) base - c(2.2,3,5.2) What I want is, for each number in base, the highest value in list, which is equal to or less than the number in base So the results would be: base list 2.2 -- 2 3-- 3 5.2 -- 5 Don't use list as an object name. It is a standard R function. vlist - c(1,2,3,4,5,6,7) base - c(2.2,3,5.2) findInterval(base, vlist) Berend __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] vlookup type function
findInterval returns the index (into list) of what you want. Use [ to get the numbers at the bottom of the found intervals: list - c(1,2,3,4,5,6,7) base - c(2.2,3,5.2) findInterval(base, list) [1] 2 3 5 findInterval(base+100, list+100) [1] 2 3 5 (list+100)[.Last.value] [1] 102 103 105 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Priyan Fernando Sent: Tuesday, February 28, 2012 7:32 AM To: r-help@r-project.org Subject: [R] vlookup type function Hi I''m looking for an Excel Vlookup type function in R. Example: list - c(1,2,3,4,5,6,7) base - c(2.2,3,5.2) What I want is, for each number in base, the highest value in list, which is equal to or less than the number in base So the results would be: base list 2.2 -- 2 3 -- 3 5.2 -- 5 Thanks for your help! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] vlookup type function
On Tue, Feb 28, 2012 at 09:02:04PM +0530, Priyan Fernando wrote: Hi I''m looking for an Excel Vlookup type function in R. Example: list - c(1,2,3,4,5,6,7) base - c(2.2,3,5.2) What I want is, for each number in base, the highest value in list, which is equal to or less than the number in base So the results would be: base list 2.2 -- 2 3 -- 3 5.2 -- 5 Hi. If base may contain numbers smaller than all numbers in list, the the following modification of the previous suggestion does not generate a warning. list - c(1,2,3,4,5,6,7) base - c(0, 2.2, 3, 5.2, 8) unlist(lapply(base, FUN=function(x) max(list[list = x], -Inf))) [1] -Inf2357 Hope this helps. Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Installing package QRMlib
According to the QRMlib package pdf, http://www.mirrorservice.org/sites/lib.stat.cmu.edu/R/CRAN/doc/packages/QRMlib.pdf the package depennds on methods, fCalendar, fEcofin, mvtnorm, chron,its,Hmisc. I have installed all of these (some were retrieved from archive and some were retrieved from CRAN) and still getting errors. I'm just wondering whether anyone has successfully loaded QRMlib?? -- View this message in context: http://r.789695.n4.nabble.com/Installing-package-QRMlib-tp4425269p4428600.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Volcano Plot
Hi I am using the ggplot2 package for the volcano plot and I am using the following code for the same: g = ggplot(data=data, aes(x=data[11], y=-log10(data[12]), colour=threshold)) + + geom_point(alpha=0.4, size=1.75) + + opts(legend.position = none) + + xlim(c(-10, 10)) + ylim(c(0, 15)) + + xlab(log2 fold change) + ylab(-log10 p-value) data[11] is a column of the fold change values and data[12] contains the P values and I am getting a following error: Error: geom_point requires the following missing aesthetics: x, y What can be done for the same ? And if not what other package may I use for the same in which I don't have to use a lmfit model? Thanks -- View this message in context: http://r.789695.n4.nabble.com/Volcano-Plot-tp4428622p4428622.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Table into a list
Perhaps, the following does it as well. (d - data.frame(x1=letters[2*1:4 - 1], x2=letters[2*1:4])) c(t(d)) Alemtsehai Hello, I am looking for a way to transform an array into a list (or a string). My array has two columns 1 and 2, and I would like to create a list of the values. Let's say I have : x1 x2 1 a b 2 c d 3 e f 4 g h What I would like to obtain is a,b,c,d,e,f,g,h. I tried without success melt and reshape ( reshape(my_array, direction=long, varying=1:2) ) but I cannot get it work. Thanks a lot !!! thomas [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Off Topic: Lowess on Yahoo
Folks: Graphics afficianados might enjoy this link (although conditional probability is also mentioned!): http://news.yahoo.com/blogs/signal/republican-nomination-extends-march-candidates-face-weakening-odds-152347629.html Pay attention to the graph. Never thought I'd see the phrase lowess trend on Yahoo. It looks like the graphs could have actually been drawn in R. Does anyone no whether or not this is so? Cheers, Bert -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Dotplot edition
Hi everyone, Im very new with R and I would like to change the size text (names) in the ylab in a dotplot. I have checked many webpages and R sites, but I have not found any help. This is the same structure of the plot in R graph Gallery: http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=150 thanks in thanks is advance José Mi code is this: ### setup the key k - simpleKey( c( Perdidas , Contestadas ) ) k$points$fill - c(lightblue, lightgreen) k$points$pch - 21 k$points$col - black k$points$cex - 1 cliente100RD- cliente100[order(cliente100$porc_perd),] ### create the plot dotplot( rownames(cliente100RD) ~ perdida + contestada , data = cliente100RD, horiz = T, main=Llamadas Enero 2012: RM,sub=Clientes ROI, par.settings = list( superpose.symbol = list( pch = 21, fill = c( lightblue, lightgreen), cex = 2, col = black ) ) , xlab = n° de llamadas, key = k, panel = function(x, y, ...){ panel.dotplot( x, y, ... ) grid.text( unit( x, native) , unit( y, native) , label = x, gp = gpar( cex = .5 ) ) } ) ### [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Installing package QRMlib
In command prompt, I direct to the following directory: C:\Program Files\R\R-2.14.1\bin Now I enter: R CMD check my_local_folder\QRMlib_1.4.5.1.tar.gz The following appears: * using log directory 'C:/Program Files/R/R-2.14.1/bin/QRMlib.Rcheck' * using R version 2.14.1 (2011-12-22) * using platform: i386-pc-mingw32 (32-bit) * using session charset: ISO8859-1 * checking for file 'QRMlib/DESCRIPTION' ... OK * this is package 'QRMlib' version '1.4.5.1' * checking package dependencies ... OK * checking if this is a source package ... OK * checking if there is a namespace ... NOTE As from R 2.14.0 all packages need a namespace. One will be generated on installation, but it is better to handcraft a NAMESPACE file: R CMD build will produce a suitable starting point. * checking for .dll and .exe files ... OK * checking whether package 'QRMlib' can be installed ... ERROR Installation failed. See 'C:/Program Files/R/R-2.14.1/bin/QRMlib.Rcheck/00install.out' for details. -- View this message in context: http://r.789695.n4.nabble.com/Installing-package-QRMlib-tp4425269p4428988.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] indexing??
Dear Petr Pikal and Petr Savicky thank you for your replies.. If the y vector contains different elements my algorithm gives this result; y - c(1,2,3,4,5,6,7,8,9,10) x - c(1,0,0,1,1,0,0,1,1,0) n - length(x) t - matrix(cbind(y,x), ncol=2) z = x+y for(j in 1:length(x)) { out - vector(list, ) for(i in 1:10) { t.s - t[sample(n,n,replace=T),] y.s - t.s[,1] x.s - t.s[,2] z.s - y.s+x.s out[[i]] - list(ff - (z.s), finding=any (y.s==y[j])) kk - sapply(out, function(x) {x$finding}) ff - out[! kk] } } kk [1] TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE FALSE ff [[1]] [[1]][[1]] [1] 5 7 3 2 2 6 7 2 6 6 [[1]]$finding [1] FALSE [[2]] [[2]][[1]] [1] 7 10 6 2 2 2 6 6 9 3 [[2]]$finding [1] FALSE Here, the two situations are FALSE, that is 5th and 10th bootstrap re-samples do not contain one (or more) element(s) of original vector (y). How can I get the similar result when the y vector includes the only response variable (1 or 0) ? That is y - c(1,1,1,0,0,1,0,1,0,0) Many thanks. -- View this message in context: http://r.789695.n4.nabble.com/indexing-tp4428210p4428746.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Installing package QRMlib
Next, you should check C:/Program Files/R/R-2.14.1/bin/QRMlib.Rcheck/00install.out for details on the error. DT54321 wrote In command prompt, I direct to the following directory: C:\Program Files\R\R-2.14.1\bin Now I enter: R CMD check my_local_folder\QRMlib_1.4.5.1.tar.gz The following appears: * using log directory 'C:/Program Files/R/R-2.14.1/bin/QRMlib.Rcheck' * using R version 2.14.1 (2011-12-22) * using platform: i386-pc-mingw32 (32-bit) * using session charset: ISO8859-1 * checking for file 'QRMlib/DESCRIPTION' ... OK * this is package 'QRMlib' version '1.4.5.1' * checking package dependencies ... OK * checking if this is a source package ... OK * checking if there is a namespace ... NOTE As from R 2.14.0 all packages need a namespace. One will be generated on installation, but it is better to handcraft a NAMESPACE file: R CMD build will produce a suitable starting point. * checking for .dll and .exe files ... OK * checking whether package 'QRMlib' can be installed ... ERROR Installation failed. See 'C:/Program Files/R/R-2.14.1/bin/QRMlib.Rcheck/00install.out' for details. - Jeremy T. Hetzel Boston University -- View this message in context: http://r.789695.n4.nabble.com/Installing-package-QRMlib-tp4425269p4429041.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] macro function
I really appreciate to your attention. I want to change the data form A into B http://r.789695.n4.nabble.com/file/n4428807/1.png http://r.789695.n4.nabble.com/file/n4428807/2.png In fact, there are 1095 column in data A. Specifically, variable publish_day is when some book in the same row is published and D-1, D-2 means the term after the publish day. Of course, the cells in data mean productivity in the day. What can be effecient way to do it? -- View this message in context: http://r.789695.n4.nabble.com/macro-function-tp4427385p4428807.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Installing package QRMlib
I installed fEcofin, fUtilities, fCalenday, and fSeries from the CRAN archives and was able to install and load QRMlib. Which errors does R CMD check QRMlib_1.4.5.1.tar.gz show? DT54321 wrote According to the QRMlib package pdf, http://www.mirrorservice.org/sites/lib.stat.cmu.edu/R/CRAN/doc/packages/QRMlib.pdf the package depennds on methods, fCalendar, fEcofin, mvtnorm, chron,its,Hmisc. I have installed all of these (some were retrieved from archive and some were retrieved from CRAN) and still getting errors. I'm just wondering whether anyone has successfully loaded QRMlib?? - Jeremy T. Hetzel Boston University -- View this message in context: http://r.789695.n4.nabble.com/Installing-package-QRMlib-tp4425269p4428842.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] macro function
Hello, Is it just a columns names issue? Try (DF is your data, 'A' or other data.frame) # colNames - colnames(DF) colNames - c(name, publish day, 2011-01-01, 2011-01-02, 2011-01-03, 2011-01-04, 2011-01-05) x - as.Date(colNames[-(1:2)]) x [1] 2011-01-01 2011-01-02 2011-01-03 2011-01-04 2011-01-05 y - difftime(x[-1], x[1], units=days) y - as.integer(y) y [1] 1 2 3 4 colNames[3] - D-Day colNames[-(1:3)] - paste(D, y, sep=+) colNames [1] namepublish day D-Day D+1 D+2 [6] D+3 D+4 # colnames(DF) - colNames Hope this helps, Rui Barradas -- View this message in context: http://r.789695.n4.nabble.com/macro-function-tp4427385p4429055.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] indexing??
On Tue, Feb 28, 2012 at 08:50:45AM -0800, helin_susam wrote: Dear Petr Pikal and Petr Savicky thank you for your replies.. If the y vector contains different elements my algorithm gives this result; y - c(1,2,3,4,5,6,7,8,9,10) x - c(1,0,0,1,1,0,0,1,1,0) n - length(x) t - matrix(cbind(y,x), ncol=2) z = x+y for(j in 1:length(x)) { out - vector(list, ) for(i in 1:10) { t.s - t[sample(n,n,replace=T),] y.s - t.s[,1] x.s - t.s[,2] z.s - y.s+x.s out[[i]] - list(ff - (z.s), finding=any (y.s==y[j])) kk - sapply(out, function(x) {x$finding}) ff - out[! kk] } } Hi. It is hard to debug a code, which we do not understand. Both me and Petr Pikal expressed objections against your code. It would help us to reply your question, if you take our objections and suggestions into account or explain, what we do not understand well. Can you comment on the suggestions from the previous emails? I would like to add one more. Why do you use ff - (z.s) inside out[[i]] - list(ff - (z.s), finding=any (y.s==y[j])) ? This expression includes the value into the list, but not under the name ff and rewrites the global variable ff instead. If you want to include (z.s) as a component named as ff, then use list(ff = (z.s), finding... Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Dotplot edition
There seems to be a clash of character sets so I can't see just what you're trying to do, but I think you'll find out how to adjust the text size with a scales list. Check out how it's used in the help for xyplot. HTH On Tue, 28-Feb-2012 at 06:05PM +, Jose Bustos Melo wrote: | Hi everyone, | Im very new with R and I would like to change the size text (names) in the ylab in a dotplot. I have checked many webpages and R sites,? but I have not found any help. | This is the same structure of the plot in R graph Gallery: http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=150 | | thanks in thanks is advance | Jos? | | | | Mi code is this: | | ### setup the key | k - simpleKey( c( Perdidas ,? Contestadas ) ) | k$points$fill - c(lightblue, lightgreen) | k$points$pch - 21 | k$points$col - black | k$points$cex - 1 | | cliente100RD- cliente100[order(cliente100$porc_perd),] | | ### create the plot | dotplot( rownames(cliente100RD) ~ perdida + contestada , data = cliente100RD, horiz = T, | ?main=Llamadas Enero 2012: RM,sub=Clientes ROI, | ??? par.settings = list( | ??? ??? superpose.symbol = list( | ??? ??? ??? pch = 21, | ??? ??? ??? fill = c( lightblue, lightgreen), | ??? ??? ??? cex = 2, | ??? ??? ??? col = black? | ??? ??? ) | ??? ?) , xlab = n? de llamadas, key = k, | ??? ?panel = function(x, y, ...){ | ??? ?? panel.dotplot( x, y, ... ) | ??? ?? grid.text( | ??? ?? ??? ??? unit( x, native) , unit( y, native) , | ??? ??? ??? label = x, gp = gpar( cex = .5 ) ) | ??? ?} ) | | ### | [[alternative HTML version deleted]] | | __ | R-help@r-project.org mailing list | https://stat.ethz.ch/mailman/listinfo/r-help | PLEASE do read the posting guide http://www.R-project.org/posting-guide.html | and provide commented, minimal, self-contained, reproducible code. -- ~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~. ___Patrick Connolly {~._.~} Great minds discuss ideas _( Y )_ Average minds discuss events (:_~*~_:) Small minds discuss people (_)-(_) . Eleanor Roosevelt ~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] update.formula has 512 char buffer?
Hello, I am trying to paste together a formula to use in the mob function of party. This means the formula will be of the form y ~ x1+ ...+xM | z1+..zN. I am doing some preliminary fits of y ~ x1+ ...+xM, then want to add the conditional part of the equation using update(). Here's the test code: var1 - 1:78 x1 - paste(x, var1, sep=) f1 - paste(f, var1[1:10], sep=) # use first 77 variables fmla - as.formula( paste(y ~ , paste(x1[1:77], collapse= + , sep=), sep=)) fmla2 - update(fmla, paste(. ~ . | , paste(f1, collapse= + ), sep=)) # CHANGE x to all 78 variables fmla - as.formula( paste(y ~ , paste(x1, collapse= + , sep=), sep=)) fmla2 - update(fmla, paste(. ~ . | , paste(f1, collapse= + ), sep=)) I have run this in Windows and Linux (64 bit) and both fail when using all 78 terms (and anything more than 78 terms). The error message contains Error in parse(text = x) : :1:514: unexpected ')'. Changing the length of the names of the x variables will break the update() with fewer variables, but always with an error referring to just more than 512 characters. There is nothing special about 77 or 78 variables here; I want to do this with hundreds of variables. Is there a workaround to this? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] indexing??
On Tue, Feb 28, 2012 at 08:50:45AM -0800, helin_susam wrote: Dear Petr Pikal and Petr Savicky thank you for your replies.. If the y vector contains different elements my algorithm gives this result; y - c(1,2,3,4,5,6,7,8,9,10) x - c(1,0,0,1,1,0,0,1,1,0) n - length(x) t - matrix(cbind(y,x), ncol=2) z = x+y for(j in 1:length(x)) { out - vector(list, ) for(i in 1:10) { t.s - t[sample(n,n,replace=T),] y.s - t.s[,1] x.s - t.s[,2] z.s - y.s+x.s out[[i]] - list(ff - (z.s), finding=any (y.s==y[j])) kk - sapply(out, function(x) {x$finding}) ff - out[! kk] } } kk [1] TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE FALSE ff [[1]] [[1]][[1]] [1] 5 7 3 2 2 6 7 2 6 6 [[1]]$finding [1] FALSE [[2]] [[2]][[1]] [1] 7 10 6 2 2 2 6 6 9 3 [[2]]$finding [1] FALSE Here, the two situations are FALSE, that is 5th and 10th bootstrap re-samples do not contain one (or more) element(s) of original vector (y). Hi. Your code generates a new list out for each j. This means that you generate a list out, test the presence of y[1] in its components, then delete out, replace it by a new list and test the presence of y[2] in this new list, then out is deleted and replaced by another out, etc. This is probably not, what you want. Is this correct? Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data analysis
On Feb 28, 2012, at 4:16 AM, Hans Ekbrand wrote: On Mon, Feb 27, 2012 at 11:04:13PM -0800, nontokozo mhlanga wrote: Please assist me with all the tests including risk factor analysis i can use to analyse the enclosed database established from a questionnaire survey to test for the prevalence of tuberculosis in humans . That's quite a general request. I think you should try to formulate a specific question. Have you read the posting-guide? http://www.R-project.org/posting-guide.html Yes. Also, I don't think the list accepts attached files. That last statement is incorrect, but there are specific requirements. Generally mail clients will send .txt, .png, .pdf. and .ps files with the proper mime type so that the mail server will accept. -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Error in read.table(file = file, header = header, sep = sep, quote = quote, : more columns than column names
Hey, I just googled my error and many things came up. I followed the leads and read the ?read.delim page; I tried changing header = TRUE, and row.names = TRUE-- but I've still been having trouble fixing it, so I would greatly appreciate any help you can provide. Here is my code: rm(list=ls()) source(../../functions.R) uncurated - read.csv(../uncurated/GSE3141_full_pdata.csv, as.is =TRUE,row.names=1) celfile.dir - ../../../DATA/GSE3141/RAW ##initial creation of curated dataframe curated - initialCuratedDF(rownames(uncurated), template.filename=template.csv) The error occurs when I run this line: curated - initialCuratedDF(rownames(uncurated), template.filename=template.csv) Error in read.table(file = file, header = header, sep = sep, quote = quote, : more columns than column name I would greatly appreciate any help. Thanks! Ben [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] update.formula has 512 char buffer?
Your code seems to be screwed up. There are no (small) size limitations on formulas (afaik). The following worked fine for me: x - paste(x,1:100,sep=,collapse = + ) part1 - paste(y,x,sep= ~ ) part2 - paste(f,1:50, sep=,collapse = + ) fml - formula(paste(part1,part2,sep= | )) Cheers, Bert On Tue, Feb 28, 2012 at 11:11 AM, Chris Hane christopher.a.h...@gmail.com wrote: Hello, I am trying to paste together a formula to use in the mob function of party. This means the formula will be of the form y ~ x1+ ...+xM | z1+..zN. I am doing some preliminary fits of y ~ x1+ ...+xM, then want to add the conditional part of the equation using update(). Here's the test code: var1 - 1:78 x1 - paste(x, var1, sep=) f1 - paste(f, var1[1:10], sep=) # use first 77 variables fmla - as.formula( paste(y ~ , paste(x1[1:77], collapse= + , sep=), sep=)) fmla2 - update(fmla, paste(. ~ . | , paste(f1, collapse= + ), sep=)) # CHANGE x to all 78 variables fmla - as.formula( paste(y ~ , paste(x1, collapse= + , sep=), sep=)) fmla2 - update(fmla, paste(. ~ . | , paste(f1, collapse= + ), sep=)) I have run this in Windows and Linux (64 bit) and both fail when using all 78 terms (and anything more than 78 terms). The error message contains Error in parse(text = x) : :1:514: unexpected ')'. Changing the length of the names of the x variables will break the update() with fewer variables, but always with an error referring to just more than 512 characters. There is nothing special about 77 or 78 variables here; I want to do this with hundreds of variables. Is there a workaround to this? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] colour by z value, persp in raster package
On Feb 28, 2012, at 4:34 AM, Omphalodes Verna wrote: Hi all! My question is how to colour pixels by z value in persp plot in raster package. Here is an example: x - seq(-1.95, 1.95, length = 30) y - seq(-1.95, 1.95, length = 35) z - outer(x, y, function(a,b) a*b^2) r1 - raster(nrows=35, ncols=30, xmn=0, xmx=30, ymn = 0, ymx = 35) r1[] - c(z) persp(r1) There already exist some function to produce persp plot for anothe classes, but I have no idea how deal with RasterLayer object. Is that really an example? After going to the trouble of correcting the error in this code from failing to load the raster package, I now see that there is no connection between the values of x, y, or x with the raster-call. r1[] - as.matrix(z) Error in .local(x, values) : cannot use a matrix with these dimensions Error in .local(x, i, j, ..., value) : cannot replace values on this raster (it is too large persp(r1) Error: hasValues(x) is not TRUE # Post some useful code. -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] memory management
My basic worry is that the GC does not work properly, i.e., the unreachable data is never collected. * Bert Gunter thagre.ore...@trar.pbz [2012-02-27 14:35:14 -0800]: This appears to be the sort of query that (with apologies to other R gurus) only Brian Ripley or Luke Tierney could figure out. R generally passes by value into function calls (but not *always*), so often multiple copies of objects are made during the course of calls. I would speculate that this is what might be going on below -- maybe even that's what you meant. Just a guess on my part, of course, so treat accordingly. -- Bert On Mon, Feb 27, 2012 at 1:03 PM, Sam Steingold s...@gnu.org wrote: It appears that the intermediate data in functions is never GCed even after the return from the function call. R's RSS is 4 Gb (after a gc()) and sum(unlist(lapply(lapply(ls(),get),object.size))) [1] 1009496520 (less than 1 GB) how do I figure out where the 3GB of uncollected garbage is hiding? -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://www.childpsy.net/ http://jihadwatch.org http://memri.org http://palestinefacts.org http://truepeace.org http://iris.org.il I may be getting older, but I refuse to grow up! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] memory management
On Tue, Feb 28, 2012 at 11:57 AM, Sam Steingold s...@gnu.org wrote: My basic worry is that the GC does not work properly, i.e., the unreachable data is never collected. Highly unlikely. Such basic inner R code has been well tested over 20 years. I believe that you merely don't understand the inner guts of what R is doing here, which is the essence of my response. (Clearly, I make no claim that I do either). I suggest you move on. -- Bert * Bert Gunter thagre.ore...@trar.pbz [2012-02-27 14:35:14 -0800]: This appears to be the sort of query that (with apologies to other R gurus) only Brian Ripley or Luke Tierney could figure out. R generally passes by value into function calls (but not *always*), so often multiple copies of objects are made during the course of calls. I would speculate that this is what might be going on below -- maybe even that's what you meant. Just a guess on my part, of course, so treat accordingly. -- Bert On Mon, Feb 27, 2012 at 1:03 PM, Sam Steingold s...@gnu.org wrote: It appears that the intermediate data in functions is never GCed even after the return from the function call. R's RSS is 4 Gb (after a gc()) and sum(unlist(lapply(lapply(ls(),get),object.size))) [1] 1009496520 (less than 1 GB) how do I figure out where the 3GB of uncollected garbage is hiding? -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://www.childpsy.net/ http://jihadwatch.org http://memri.org http://palestinefacts.org http://truepeace.org http://iris.org.il I may be getting older, but I refuse to grow up! -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] macro function
On Feb 28, 2012, at 12:10 PM, mrzung wrote: I really appreciate to your attention. I want to change the data form A into B http://r.789695.n4.nabble.com/file/n4428807/1.png http://r.789695.n4.nabble.com/file/n4428807/2.png Please use text copied versions of dput(a) and dput(b) rather than posting bitmapped images of your screen. In fact, there are 1095 column in data A. Specifically, variable publish_day is when some book in the same row is published and D-1, D-2 means the term after the publish day. Of course, the cells in data mean productivity in the day. What can be effecient way to do it? Specify what you want tot do in English. Asking how to create 1000 separate variable is almost certainly to be the wrong way to go. If you wanted a list with names like that we can probably help. -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] memory management
Look into environments that may be stored with your data. object.size(obj) does not report on the size of the environment(s) associated with obj. E.g., f - function(n) { +d - data.frame(y=rnorm(n), x1=rnorm(n), x2=rnorm(n)) +terms(data=d, y~.) + } z - f(1e6) object.size(z) 1760 bytes eapply(environment(z), object.size) $d 24000520 bytes $n 32 bytes That happens because formula objects (like function objects) contain a reference to the environment in which they were created and that environmentwill not be destroyed until the last reference to it is gone. You might be able write code using, e.g., the codetools package to walk through your objects looking for all distinct environments that they reference (directly and indirectly, via ancestors of environments directly referenced). Then you can add up the sizes of things in those environments. Another possible reason for your problem is that by using ls() instead of ls(all=TRUE) you are not looking at datasets whose names start with a dot. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Sam Steingold Sent: Tuesday, February 28, 2012 11:58 AM To: r-help@r-project.org; Bert Gunter Subject: Re: [R] memory management My basic worry is that the GC does not work properly, i.e., the unreachable data is never collected. * Bert Gunter thagre.ore...@trar.pbz [2012-02-27 14:35:14 -0800]: This appears to be the sort of query that (with apologies to other R gurus) only Brian Ripley or Luke Tierney could figure out. R generally passes by value into function calls (but not *always*), so often multiple copies of objects are made during the course of calls. I would speculate that this is what might be going on below -- maybe even that's what you meant. Just a guess on my part, of course, so treat accordingly. -- Bert On Mon, Feb 27, 2012 at 1:03 PM, Sam Steingold s...@gnu.org wrote: It appears that the intermediate data in functions is never GCed even after the return from the function call. R's RSS is 4 Gb (after a gc()) and sum(unlist(lapply(lapply(ls(),get),object.size))) [1] 1009496520 (less than 1 GB) how do I figure out where the 3GB of uncollected garbage is hiding? -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://www.childpsy.net/ http://jihadwatch.org http://memri.org http://palestinefacts.org http://truepeace.org http://iris.org.il I may be getting older, but I refuse to grow up! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] update.formula has 512 char buffer?
On 28/02/2012 2:46 PM, Bert Gunter wrote: Your code seems to be screwed up. There are no (small) size limitations on formulas (afaik). The following worked fine for me: x- paste(x,1:100,sep=,collapse = + ) part1- paste(y,x,sep= ~ ) part2- paste(f,1:50, sep=,collapse = + ) fml- formula(paste(part1,part2,sep= | )) I think the problem is in R, not in what Chris did. It seems to be in update.formula, or terms.formula, or some related function. I'll try to track it down. Duncan Cheers, Bert On Tue, Feb 28, 2012 at 11:11 AM, Chris Hane christopher.a.h...@gmail.com wrote: Hello, I am trying to paste together a formula to use in the mob function of party. This means the formula will be of the form y ~ x1+ ...+xM | z1+..zN. I am doing some preliminary fits of y ~ x1+ ...+xM, then want to add the conditional part of the equation using update(). Here's the test code: var1- 1:78 x1- paste(x, var1, sep=) f1- paste(f, var1[1:10], sep=) # use first 77 variables fmla- as.formula( paste(y ~ , paste(x1[1:77], collapse= + , sep=), sep=)) fmla2- update(fmla, paste(. ~ . | , paste(f1, collapse= + ), sep=)) # CHANGE x to all 78 variables fmla- as.formula( paste(y ~ , paste(x1, collapse= + , sep=), sep=)) fmla2- update(fmla, paste(. ~ . | , paste(f1, collapse= + ), sep=)) I have run this in Windows and Linux (64 bit) and both fail when using all 78 terms (and anything more than 78 terms). The error message contains Error in parse(text = x) : :1:514: unexpected ')'. Changing the length of the names of the x variables will break the update() with fewer variables, but always with an error referring to just more than 512 characters. There is nothing special about 77 or 78 variables here; I want to do this with hundreds of variables. Is there a workaround to this? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Cleaning up messy Excel data
Unfortunately, some data I need to work with was delivered in a rather messy Excel file. I want to import into R and clean up some things so that I can do my analysis. Pulling in a CSV from Excel is the easy part. My current challenge is dealing with some text mixed in the values. i.e. 118 5.7 2.0 3.7 Since this column in Excel has a 2.0 value, then R reads the column as a factor with levels. Ideally, I want to convert it a normal vector of scalars and code code the 2.0 as 0. Can anyone suggest an easy way to do this? Thanks! -- Noah Silverman UCLA Department of Statistics 8117 Math Sciences Building Los Angeles, CA 90095 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cleaning up messy Excel data
First of all when reading in the CSV file, use 'as.is = TRUE' to prevent the changing to factors. Now that things are character in that column, you can use some pattern expressions (gsub, regex, ...) to search for and change your data. E.g., sub(.*, 0, yourCol) should do it for you. On Tue, Feb 28, 2012 at 4:27 PM, Noah Silverman noahsilver...@ucla.edu wrote: Unfortunately, some data I need to work with was delivered in a rather messy Excel file. I want to import into R and clean up some things so that I can do my analysis. Pulling in a CSV from Excel is the easy part. My current challenge is dealing with some text mixed in the values. i.e. 118 5.7 2.0 3.7 Since this column in Excel has a 2.0 value, then R reads the column as a factor with levels. Ideally, I want to convert it a normal vector of scalars and code code the 2.0 as 0. Can anyone suggest an easy way to do this? Thanks! -- Noah Silverman UCLA Department of Statistics 8117 Math Sciences Building Los Angeles, CA 90095 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cleaning up messy Excel data
-Original Message- From: Noah Silverman Sent: Tuesday, February 28, 2012 3:27 PM To: r-help Subject: [R] Cleaning up messy Excel data Unfortunately, some data I need to work with was delivered in a rather messy Excel file. I want to import into R and clean up some things so that I can do my analysis. Pulling in a CSV from Excel is the easy part. My current challenge is dealing with some text mixed in the values. i.e. 118 5.7 2.0 3.7 Since this column in Excel has a 2.0 value, then R reads the column as a factor with levels. Ideally, I want to convert it a normal vector of scalars and code code the 2.0 as 0. Can anyone suggest an easy way to do this? -- ?as.character will show you how to change the factor column into a character column. Then, you can replace text using any of a number of procedures. see for example ?gsub finally, you can use as.numeric if you want numbers. Coding is best done in the context of factors, so you might want to consider where replacing 2 with NA is more appropriate than replacing with 0. In this end, the choice might be context sensitive. Rob -- Robert W. Baer, Ph.D. Professor of Physiology Kirksville College of Osteopathic Medicine A. T. Still University of Health Sciences 800 W. Jefferson St. Kirksville, MO 63501 660-626-2322 FAX 660-626-2965 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cleaning up messy Excel data
That's exactly what I need. Thank You!! -- Noah Silverman UCLA Department of Statistics 8117 Math Sciences Building Los Angeles, CA 90095 On Feb 28, 2012, at 1:42 PM, jim holtman wrote: First of all when reading in the CSV file, use 'as.is = TRUE' to prevent the changing to factors. Now that things are character in that column, you can use some pattern expressions (gsub, regex, ...) to search for and change your data. E.g., sub(.*, 0, yourCol) should do it for you. On Tue, Feb 28, 2012 at 4:27 PM, Noah Silverman noahsilver...@ucla.edu wrote: Unfortunately, some data I need to work with was delivered in a rather messy Excel file. I want to import into R and clean up some things so that I can do my analysis. Pulling in a CSV from Excel is the easy part. My current challenge is dealing with some text mixed in the values. i.e. 118 5.7 2.0 3.7 Since this column in Excel has a 2.0 value, then R reads the column as a factor with levels. Ideally, I want to convert it a normal vector of scalars and code code the 2.0 as 0. Can anyone suggest an easy way to do this? Thanks! -- Noah Silverman UCLA Department of Statistics 8117 Math Sciences Building Los Angeles, CA 90095 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] memory management
* William Dunlap jqha...@gvopb.pbz [2012-02-28 20:19:06 +]: Look into environments that may be stored with your data. thanks, but I see nothing like that: for (n in ls(all.names = TRUE)) { o - get(n) print(object.size(o), units=Kb) e - environment(o) if (!identical(e,NULL) !identical(e,.GlobalEnv)) { print(e) print(eapply(e,object.size)) } } 25.8 Kb 0.5 Kb 49.1 Kb 0.1 Kb 30.8 Kb 13.6 Kb 17.4 Kb 59.4 Kb 52.2 Kb 0.1 Kb 3.9 Kb 49.1 Kb 21.2 Kb 0.1 Kb 0.1 Kb 51 Kb 13.2 Kb 53.5 Kb 18.1 Kb 64.3 Kb 25.8 Kb 33.5 Kb 0.1 Kb 0.1 Kb 8 Kb 10 Kb 15.7 Kb 15.6 Kb 9.9 Kb 401672.7 Kb 19.1 Kb 76 Kb 12 Kb 32.4 Kb 156.3 Kb 13.1 Kb 20.5 Kb 21.8 Kb 10.8 Kb sum(unlist(lapply(lapply(ls(all.names = TRUE),get),object.size))) [1] 412351928 i.e., total of data is about 400MB. why does the process take in access of 1GB? top: 1235m 1.1g 4452 S0 14.6 7:12.27 R -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://www.childpsy.net/ http://pmw.org.il http://camera.org http://dhimmi.com http://palestinefacts.org http://ffii.org Fighting for peace is like screwing for virginity. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] memory management
You need to walk through the objects, checking for environments on each component or attribute of an object. You also have to look at the parent.env of each environment found. E.g., f - function(n) { + d - data.frame(y = rnorm(n), x = rnorm(n)) + lm(y ~ poly(x, 4), data=d) + } z - f(1e5) environment(z) NULL object.size(z) 21610708 bytes sapply(z, object.size) coefficients residuals effects 384 4400104 1200336 rank fitted.valuesassign 32 440010456 qr df.residual xlevels 760123232 104 call terms model 508 2804 4004276 environment(z$terms) environment: 0x0abb86e4 eapply(environment(z$terms), object.size) $d 1600448 bytes $n 32 bytes Coding this is tedious; the codetools package may make it easier. Summing the sizes may well give an overestimate of the memory actually used, since several objects may share the same memory. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: Sam Steingold [mailto:sam.steing...@gmail.com] On Behalf Of Sam Steingold Sent: Tuesday, February 28, 2012 2:56 PM To: r-help@r-project.org; William Dunlap Subject: Re: memory management * William Dunlap jqha...@gvopb.pbz [2012-02-28 20:19:06 +]: Look into environments that may be stored with your data. thanks, but I see nothing like that: for (n in ls(all.names = TRUE)) { o - get(n) print(object.size(o), units=Kb) e - environment(o) if (!identical(e,NULL) !identical(e,.GlobalEnv)) { print(e) print(eapply(e,object.size)) } } 25.8 Kb 0.5 Kb 49.1 Kb 0.1 Kb 30.8 Kb 13.6 Kb 17.4 Kb 59.4 Kb 52.2 Kb 0.1 Kb 3.9 Kb 49.1 Kb 21.2 Kb 0.1 Kb 0.1 Kb 51 Kb 13.2 Kb 53.5 Kb 18.1 Kb 64.3 Kb 25.8 Kb 33.5 Kb 0.1 Kb 0.1 Kb 8 Kb 10 Kb 15.7 Kb 15.6 Kb 9.9 Kb 401672.7 Kb 19.1 Kb 76 Kb 12 Kb 32.4 Kb 156.3 Kb 13.1 Kb 20.5 Kb 21.8 Kb 10.8 Kb sum(unlist(lapply(lapply(ls(all.names = TRUE),get),object.size))) [1] 412351928 i.e., total of data is about 400MB. why does the process take in access of 1GB? top: 1235m 1.1g 4452 S0 14.6 7:12.27 R -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://www.childpsy.net/ http://pmw.org.il http://camera.org http://dhimmi.com http://palestinefacts.org http://ffii.org Fighting for peace is like screwing for virginity. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cleaning up messy Excel data
Just replace that value with zero. If you provide some reproducible code I could probably give you a solution. ?dput good luck, Stephen On 02/28/2012 03:27 PM, Noah Silverman wrote: Unfortunately, some data I need to work with was delivered in a rather messy Excel file. I want to import into R and clean up some things so that I can do my analysis. Pulling in a CSV from Excel is the easy part. My current challenge is dealing with some text mixed in the values. i.e. 118 5.72.0 3.7 Since this column in Excel has a 2.0 value, then R reads the column as a factor with levels. Ideally, I want to convert it a normal vector of scalars and code code the 2.0 as 0. Can anyone suggest an easy way to do this? Thanks! -- Noah Silverman UCLA Department of Statistics 8117 Math Sciences Building Los Angeles, CA 90095 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick ** Auburn University Biological Sciences 331 Funchess Hall Auburn, Alabama 36849 ** sas0...@auburn.edu http://www.auburn.edu/~sas0025 ** Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] group calculations with other columns for the ride
Hello, I can get the median for each factor, but I'd like another column to go with each factor. The nm column is a long name for the lvls column. So unique work except for the order can get messed up. Example: x = data.frame(val=1:10,lvls=c('cat2',rep(cat1,4),rep(cat2,4),'cat1'),nm=c('longname2',rep(longname1,4),rep(longname2,4),'longname1')) x val lvlsnm 11 cat2 longname2 22 cat1 longname1 33 cat1 longname1 44 cat1 longname1 55 cat1 longname1 66 cat2 longname2 77 cat2 longname2 88 cat2 longname2 99 cat2 longname2 10 10 cat1 longname1 unique doesn't work in data.frame: mdn = do.call(rbind,lapply(split(x[,1], x[,2]), median)) data.frame(mdn,ln=as.character(unique(x[,3]))) mdnln cat1 4 longname2 cat2 7 longname1 I want: mdnln cat1 4 longname1 cat2 7 longname2 Thank you very much! PS - looking for simple'ish solutions. I know I can do it with loops and merges, but is there an option I am not using here? Ben [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] group calculations with other columns for the ride
aggregate(val~lvls+nm,data=x,FUN='median') On Tue, Feb 28, 2012 at 4:43 PM, Ben quant ccqu...@gmail.com wrote: Hello, I can get the median for each factor, but I'd like another column to go with each factor. The nm column is a long name for the lvls column. So unique work except for the order can get messed up. Example: x = data.frame(val=1:10,lvls=c('cat2',rep(cat1,4),rep(cat2,4),'cat1'),nm=c('longname2',rep(longname1,4),rep(longname2,4),'longname1')) x val lvls nm 1 1 cat2 longname2 2 2 cat1 longname1 3 3 cat1 longname1 4 4 cat1 longname1 5 5 cat1 longname1 6 6 cat2 longname2 7 7 cat2 longname2 8 8 cat2 longname2 9 9 cat2 longname2 10 10 cat1 longname1 unique doesn't work in data.frame: mdn = do.call(rbind,lapply(split(x[,1], x[,2]), median)) data.frame(mdn,ln=as.character(unique(x[,3]))) mdn ln cat1 4 longname2 cat2 7 longname1 I want: mdn ln cat1 4 longname1 cat2 7 longname2 Thank you very much! PS - looking for simple'ish solutions. I know I can do it with loops and merges, but is there an option I am not using here? Ben [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] group calculations with other columns for the ride
Excellent! I wonder why I haven't seen aggregate before. Thanks! ben On Tue, Feb 28, 2012 at 4:51 PM, ilai ke...@math.montana.edu wrote: aggregate(val~lvls+nm,data=x,FUN='median') On Tue, Feb 28, 2012 at 4:43 PM, Ben quant ccqu...@gmail.com wrote: Hello, I can get the median for each factor, but I'd like another column to go with each factor. The nm column is a long name for the lvls column. So unique work except for the order can get messed up. Example: x = data.frame(val=1:10,lvls=c('cat2',rep(cat1,4),rep(cat2,4),'cat1'),nm=c('longname2',rep(longname1,4),rep(longname2,4),'longname1')) x val lvlsnm 11 cat2 longname2 22 cat1 longname1 33 cat1 longname1 44 cat1 longname1 55 cat1 longname1 66 cat2 longname2 77 cat2 longname2 88 cat2 longname2 99 cat2 longname2 10 10 cat1 longname1 unique doesn't work in data.frame: mdn = do.call(rbind,lapply(split(x[,1], x[,2]), median)) data.frame(mdn,ln=as.character(unique(x[,3]))) mdnln cat1 4 longname2 cat2 7 longname1 I want: mdnln cat1 4 longname1 cat2 7 longname2 Thank you very much! PS - looking for simple'ish solutions. I know I can do it with loops and merges, but is there an option I am not using here? Ben [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error in read.table(file = file, header = header, sep = sep, quote = quote, : more columns than column names
What is initialCuratedDF? I'm not seeing it in any standard location so there's no way to debug it... Also see inline. Michael On Tue, Feb 28, 2012 at 2:41 PM, Ben Ganzfried ben.ganzfr...@gmail.com wrote: Hey, I just googled my error and many things came up. I followed the leads and read the ?read.delim page; I tried changing header = TRUE, and row.names = TRUE-- but I've still been having trouble fixing it, so I would greatly appreciate any help you can provide. Here is my code: PLEASE don't ever do this -- it's potentially very destructive to anyone who tries to copy your script to help you. David also mentioned this when you posted the same code ~ 6 months ago. ## rm(list=ls()) ## ALWAYS WRAP IN COMMENTS IF YOU ARE GOING TO INSIST ON INCLUDING IT source(../../functions.R) uncurated - read.csv(../uncurated/GSE3141_full_pdata.csv, as.is =TRUE,row.names=1) celfile.dir - ../../../DATA/GSE3141/RAW ##initial creation of curated dataframe curated - initialCuratedDF(rownames(uncurated), template.filename=template.csv) The error occurs when I run this line: curated - initialCuratedDF(rownames(uncurated), template.filename=template.csv) Error in read.table(file = file, header = header, sep = sep, quote = quote, : more columns than column name I would greatly appreciate any help. Thanks! Ben [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] aggregating specific parts in zoo index column to perform sliding average
Thanks Gabor. Answering all my noob questions haha. Interesting strategy to adjust the input raw data as opposed to Excel where one offsets the output data instead. I'll remember that. -- View this message in context: http://r.789695.n4.nabble.com/aggregating-specific-parts-in-zoo-index-column-to-perform-sliding-average-tp4426798p4429257.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Error: could not find function hier.part
Error: could not find function hier.part things I have tried: 1. reinstall R (lastest version, on windows XP) 2. install package gtools 3. include: library(gtools) require(gtools) 4. how I call this function: hier.part(c$Y, xdata, fam = gaussian, gof = Rsqu) 5.when I try to check what's in the package gtools, I get (hier.part is not included): gtools:: TAB gtools::addLast gtools::ask gtools::assert gtools::binsearch gtools::capture gtools::checkRVersion gtools::combinationsgtools::ddirichlet gtools::defmacro gtools::evengtools::foldchange gtools::foldchange2logratio gtools::inv.logit gtools::invalid gtools::keywords gtools::logit gtools::logratio2foldchange gtools::mixedorder gtools::mixedsort gtools::odd gtools::permutations gtools::permute gtools::quantcutgtools::rdirichlet gtools::running gtools::scat gtools::setTCPNoDelay gtools::smartbind gtools::sprint gtools::strmacro I am new to R, and I googled everything about the error message. I don't really understand what is requires the gtools package in the gregmisc bundle. tried to install packagegregmisc, did not help. Could someone please help? Thank you very much, -- View this message in context: http://r.789695.n4.nabble.com/Error-could-not-find-function-hier-part-tp4429644p4429644.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Installing package QRMlib
Yes, I did. But my Windows did not know what program to open it with... -- View this message in context: http://r.789695.n4.nabble.com/Installing-package-QRMlib-tp4425269p4429275.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Installing the package Rcplex
Hello, I have posted this question earlier and got a lot of helps. Appreciate all those helps. After intensive learning and researching, this is to summurize what I have now. Target: Install the package Rcplex OS: windows 7 64bit The Story: Installed the Rtool, and setted the path to all relevant directories. Downloaded the tar.gz file to directory of c:/temp and rename it to be Rcplex.tar.gz and then used tar command to unzip it under the same directory. Also installed the software of cplex from IBM with license under the directory of c:/IBM, and edited the file of makevars.win in the R package folder as the following PKG_CPPFLAGS=-I C:/IBM/ILOG/CPLEX_Studio_Academic124/cplex/include PKG_LIBS=-L C:/IBM/ILOG/CPLEX_Studio_Academic124/cplex/lib/x64_windows_vs2010/stat_mda -l cplex124 -lm Then I go to directory of {installation path for R}\bin\x64 under cmd and typed R CMD INSTALL -l {the path to my personal library} c:/temp/Rcplex I got the error message with a bunch of lines similar to Rcplex_QCP.o:Rcplex_QCP.c:.text+0x217 undefined reference to '_imp_CPXcreateprob@12' Rcplex_QCP.o:Rcplex_QCP.c:.text+0x217 undefined reference to '_imp_CPXcopylp@60' . . Some suspects: Not sure the following information would be relevant Originally, I used the command of R CMD INSTALL c:/temp/Rcplex directly and was told I have no permission to install to the main library. So I used the personal library which was automatically created while I installed other packages. I googled about this error message of undefined reference to and some people said The problem may occur when the library (a file with the .h extension) in which the function is defined is missing or mislocated. I checked the include folder for software cplex and it has files cplex.h and ilocplex.h which I thought to be needed. Any ideas? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] indexing??
Dear Petr Savicky, Actually, this is based on jackknife after bootstrap algorithm. In summary, I have a data set, and I want to compute some values by using this algorithm. Firstly, using bootstrap, I create some bootstrap re-samples. This step O.K. Then, for each data point within these re-samples, I want to get a subset which do not contain that data point ( this point would be any point of the original data set), in general, if B is the number of bootstrap-resamples, there are B/e resamples obtained for each data point. And finally, I want to calculate some values for each of this re samples. Explanation of my algorithm; #My data set: (x and y) y - c(1,2,3,4,5,6,7,8,9,10) x - c(1,0,0,1,1,0,0,1,1,0) n - length(x) t - matrix(cbind(y,x), ncol=2) z = x+y for(j in 1:length(x)) { out - vector(list, ) for(i in 1:10) { t.s - t[sample(n,n,replace=T),] # Here is the bootstrap step y.s - t.s[,1] x.s - t.s[,2] z.s - y.s+x.s nn - sum (z.s) # For example, I want to calculate this value out[[i]] - list(ff - (nn), finding=any (y.s==y[j])) # I get the mentioned subset in here kk - sapply(out, function(x) {x$finding}) ff - out[! kk] } } I obtained the following results of an experiment; kk [1] FALSE TRUE TRUE FALSE TRUE TRUE FALSE TRUE TRUE TRUE ff [[1]] [[1]][[1]] [1] 47 [[1]]$finding [1] FALSE [[2]] [[2]][[1]] [1] 46 [[2]]$finding [1] FALSE [[3]] [[3]][[1]] [1] 52 [[3]]$finding [1] FALSE It is easy to do when y contains different elements. out[[i]] - list(ff - (nn), finding=any (y.s==y[j])) But, when y contains the same element, doing this process can be confusing confusing.. Because, (y - c(1,1,1,0,0,1,0,1,0,0)) for y[j] when j= 1 there are some other 1 in the y. Is there something special about the y to an j ? Thanks -- View this message in context: http://r.789695.n4.nabble.com/indexing-tp4428210p4429280.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] igraph find the path with the smallest weights
Hi all, I am trying to analyze some graphs in R. Given a weighted directed network with 7 nodes. I want to find a path from node 1 to node 7, so the sum of all the edges on the path is the smallest comparing to all the other paths. I am wondering if I can do this in igraph. After I identify the path, can I colour code the path? Thank you in advance. Wendy -- View this message in context: http://r.789695.n4.nabble.com/igraph-find-the-path-with-the-smallest-weights-tp4429343p4429343.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Quantile scores as dependent variables.. an R and general method question
I have a dataset that does not include native scores, but only serial quantile rankings for a set of units. Clearly these observations are dependent (in that you can't alter one observation without also altering others). Are there methods for dealing with quantile dependent variables. My atempt to find such methods has not bee successful. Any leads to theory, texts or R code would be most appeciated. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ts.plot and x axes customization
Dear List, I would be pleased if someone can help me with the following issue: I'm about to plot two time series in one plot via ts.plot which looks like: ts.plot(series1, series2, main=main, xlab=xlab, ylab=ylab, col=c(green, red, blue), lwd=2) The problem is, that R automatically sets the x axes labels in 5-year-intervalls. Every 5 years there's one tick and one label with the respective year. Now I would like to customize the axes in a way that there is a label every year and a tick every quarter. In the previous code I've determined the time series with series1 = ts(x, start=c(2000,1), frequency=4) series2 = ts(y, start=c(2000,1), frequency=4) What I've tried before is deleting the X axes via gpars=list(xaxt=n) in the ts.plot-code. But after that I was not aible to add the customized axes via axis()... In advance, thank you very much for your help and hints! Regards, Jochem Ihr WEB.DE Postfach immer dabei: die kostenlose WEB.DE Mail App für iPhone und Android. [1]https://produkte.web.de/freemail_mobile_startseite/ References 1. https://produkte.web.de/freemail_mobile_startseite/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Database
Hi all, I´m new using Access. I see that many things that you can do on Access you can do on CRAN R but not on contrary. My question is: Is there any manual with examples comparing how to do data base analysis on access and making the same on CRAN R? Imagine I want to compare two columns Name of two different data bases. I want to see if there are identical names on both files. It is better to use Access? Or it is better to use cran r (importing data and work on CRAN R)? This is only an example. I know CRAN R is more specialized on statistics and data analysis but I ´m trying not to learn Access and SQL so on. I cannot explain better I hope you comprehed me. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error: could not find function hier.part
It would seem that hier.part is not in the gtools packagedo you have some reference that suggests otherwise? I don't know the function, but I might suggest you check the hier.part package (also on CRAN). Michael On Tue, Feb 28, 2012 at 4:27 PM, haiyan tucsonaug...@gmail.com wrote: Error: could not find function hier.part things I have tried: 1. reinstall R (lastest version, on windows XP) 2. install package gtools 3. include: library(gtools) require(gtools) 4. how I call this function: hier.part(c$Y, xdata, fam = gaussian, gof = Rsqu) 5.when I try to check what's in the package gtools, I get (hier.part is not included): gtools:: TAB gtools::addLast gtools::ask gtools::assert gtools::binsearch gtools::capture gtools::checkRVersion gtools::combinations gtools::ddirichlet gtools::defmacro gtools::even gtools::foldchange gtools::foldchange2logratio gtools::inv.logit gtools::invalid gtools::keywords gtools::logit gtools::logratio2foldchange gtools::mixedorder gtools::mixedsort gtools::odd gtools::permutations gtools::permute gtools::quantcut gtools::rdirichlet gtools::running gtools::scat gtools::setTCPNoDelay gtools::smartbind gtools::sprint gtools::strmacro I am new to R, and I googled everything about the error message. I don't really understand what is requires the gtools package in the gregmisc bundle. tried to install packagegregmisc, did not help. Could someone please help? Thank you very much, -- View this message in context: http://r.789695.n4.nabble.com/Error-could-not-find-function-hier-part-tp4429644p4429644.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Installing package QRMlib
Please quote context. You can open it with any plain text editor. (i.e., notepad) Michael On Tue, Feb 28, 2012 at 2:42 PM, DT54321 deepan.tailo...@gmail.com wrote: Yes, I did. But my Windows did not know what program to open it with... -- View this message in context: http://r.789695.n4.nabble.com/Installing-package-QRMlib-tp4425269p4429275.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Siegel-Tukey test for equal variability (code)
The previously posted code contains bugs. The code below should work: x: a vector of data y: Group indicator (if id.col=TRUE); data of the second group (if id.col=FALSE). If y is the group indicator it MUST take 0 or 1 to indicate the groups, and x must contain the data for both groups. id.col: If TRUE (default), then x is the data column and y is the ID column, indicating the groups. If FALSE, x and y are both data columns. id.col must be FALSE only if both data columns are of the same length. adjust.median: Should between-group differences in medians be leveled before performing the test? In certain cases, the Siegel-Tukey test is susceptible to median differences and may indicate significant differences in variability that, in reality, stem from differences in medians. rnd: Should the data be rounded and, if so, to which decimal? The default (-1) uses the data as is. Otherwise, rnd must be a non-negative integer. Typically, this option is not needed. However, occasionally, differences in the precision with which certain functions return values cause the merging of two data frames to fail within the siegel.tukey function. Only then rounding is necessary. This operation should not be performed if it affects the ranks of observations. … arguments passed on to the Wilcoxon test. See ?wilcox.test Value: Among other output, the function returns the data, the Siegel-Tukey ranks, the associated Wilcoxon’s W and the p-value for a Wilcoxon test on tie-adjusted Siegel-Tukey ranks (i.e., it performs and returns a Siegel-Tukey test). If significant, the group with the smaller rank sum has greater variability. References: Sidney Siegel and John Wilder Tukey (1960) “A nonparametric sum of ranks procedure for relative spread in unpaired samples.” Journal of the American Statistical Association. See also, David J. Sheskin (2004) ”Handbook of parametric and nonparametric statistical procedures.” 3rd edition. Chapman and Hall/CRC. Boca Raton, FL. Notes: The Siegel-Tukey test has relatively low power and may, under certain conditions, indicate significance due to differences in medians rather than differences in variabilities (consider using the argument adjust.median). Output (in this order) 1. Group medians (after median adjustment if specified) 2. Wilcoxon-test for between-group differences in medians (after the median adjustment if specified) 3. Data, group membership, and the Siegel-Tukey ranks 4. Mean Siegel-Tukey rank by group (smaller values indicate greater variability) 5. Siegel-Tukey test (Wilcoxon test on tie-adjusted Siegel-Tukey ranks) siegel.tukey=function(x,y,id.col=TRUE,adjust.median=F,rnd=-1,alternative=two.sided,mu=0,paired=FALSE,exact=FALSE,correct=TRUE,conf.int=FALSE,conf.level=0.95){ if(id.col==FALSE){ data=data.frame(c(x,y),rep(c(0,1),c(length(x),length(y } else { data=data.frame(x,y) } names(data)=c(x,y) data=data[order(data$x),] if(rnd-1){data$x=round(data$x,rnd)} if(adjust.median==T){ cat(\n,Adjusting medians...,\n,sep=) data$x[data$y==0]=data$x[data$y==0]-(median(data$x[data$y==0])) data$x[data$y==1]=data$x[data$y==1]-(median(data$x[data$y==1])) } cat(\n,Median of group 1 = ,median(data$x[data$y==0]),\n,sep=) cat(Median of group 2 = ,median(data$x[data$y==1]),\n,\n,sep=) cat(Testing median differences...,\n) print(wilcox.test(data$x[data$y==0],data$x[data$y==1])) cat(Performing Siegel-Tukey rank transformation...,\n,\n) sort.x-sort(data$x) sort.id-data$y[order(data$x)] data.matrix-data.frame(sort.x,sort.id) base1-c(1,4) iterator1-matrix(seq(from=1,to=length(x),by=4))-1 rank1-apply(iterator1,1,function(x) x+base) iterator2-matrix(seq(from=2,to=length(x),by=4)) base2-c(0,1) rank2-apply(iterator2,1,function(x) x+base2) #print(rank1) #print(rank2) if(length(rank1)==length(rank2)){ rank-c(rank1[1:floor(length(x)/2)],rev(rank2[1:ceiling(length(x)/2)])) } else{ rank-c(rank1[1:ceiling(length(x)/2)],rev(rank2[1:floor(length(x)/2)])) } unique.ranks-tapply(rank,sort.x,mean) unique.x-as.numeric(as.character(names(unique.ranks))) rank.matrix-data.frame(unique.x,unique.ranks) ST.matrix-merge(data.matrix,rank.matrix,by.x=sort.x,by.y=unique.x) print(ST.matrix) cat(\n,Performing Siegel-Tukey test...,\n,sep=) ranks0-ST.matrix$unique.ranks[ST.matrix$sort.id==0] ranks1-ST.matrix$unique.ranks[ST.matrix$sort.id==1] cat(\n,Mean rank of group 0: ,mean(ranks0),\n,sep=) cat(Mean rank of group 1: ,mean(ranks1),\n,sep=) print(wilcox.test(ranks0,ranks1,alternative=alternative,mu=mu,paired=paired,exact=exact,correct=correct,conf.int=conf.int,conf.level=conf.level)) } Examples: x - c(33, 62, 84, 85, 88, 93, 97, 4, 16, 48, 51, 66, 98) id - c(0,0,0,0,0,0,0,1,1,1,1,1,1) siegel.tukey(x,id,adjust.median=F,exact=T) x-c(0,0,1,4,4,5,5,6,6,9,10,10) id-c(0,0,0,1,1,1,1,1,1,0,0,0) siegel.tukey(x,id) x - c(85,106,96, 105, 104, 108, 86) id-c(0,0,1,1,1,1,1) siegel.tukey(x,id)
Re: [R] Error message: object of type 'closure' is not subsettable
Hi Josh, Petr I checked that testStatistics is a function but it is not defined in the code anywhere else. But if I am able to remove it and give the input by my own to the function which calls testStatistics, it might work. The hurdle here for me is that the input is a set* of values and each value has 2 attributes. Each variable is generated per iteration and has the numerator, denominator as its attributes and a value. But when I read it in a loop, i face two problems: 1. only the last i value is stored and the rest are NULL. 2. the one stored does not contain the attribute values. Each value looks like this: t [1] 3.897434 attr(,numerator) [1] 0.0002134457 attr(,denominator) [1] 5.47657e-05 Eg: for (i in 1:5) { t- test.functional.t(res.em1,res.em2,mint,maxt,se.m=0,points=300) ### function calling } t [[1]] NULL [[2]] NULL [[3]] NULL [[4]] NULL [1] 3.897434 Any suggestions. Thanks for all the help. :) Regards Aparna On Tue, Feb 28, 2012 at 11:01 PM, Petr PIKAL petr.pi...@precheza.cz wrote: Hi Difficult to without knowing what objects you are operating your functions. I get test.functional.t(1:10,1:10,3,4) Error in res.em1$eta : $ operator is invalid for atomic vectors moderated.functional.t(1:10) Error in testStatistics[numerator, ] : incorrect number of dimensions And without suitable objects for those functions to operate it is impossible to come with reasonable suggestion. Anyway there is a function on your system which is called testStatistics and you can not subset functions, hence your error. I did not find this function in R pacages but I can not say that it is really not there. You can get rid of this function by rm(testStatistics) Regards Petr Kindly find below the code as it is executed: test.functional.t - function(res.em1,res.em2,mint,maxt,se.m=0,points=300) { at - seq(mint,maxt,length.out=points) by - at[2] - at[1] mu1 - spline(x=res.em1$tau,y=res.em1$eta,xout=at,method=natural)$y l2 - functional.norm(mu1,mu2,by=by) s1 - mean(sapply(1:res.em1$n, function(i) { v - spline(x=res.em1$tau,y=res.em1$vi[[i]],xout=at,method=natural)$y functional.norm(v,by=by)^2 })) denominator - sqrt(s1 / res.em1$n + s2 / res.em2$n) ### formula for se returnValue - l2 / (denominator + se.m) attr(returnValue,numerator) - l2 attr(returnValue,denominator) - denominator returnValue } moderated.functional.t - function(testStatistics,alpha.step=0.05,quantile.step=0.01) { numerators - unlist(testStatistics[numerator,]) denominators - unlist(testStatistics[denominator,]) } I get my error in the function above moderated.functional.t. testStatistics is shown to be a function(x) when I type it in R console. But there is no function definition for testStatistics in the code. My R understanding is still elementary. Thanks Aparna On Tue, Feb 28, 2012 at 5:25 PM, Joshua Wiley jwiley.ps...@gmail.comwrote: Hi Aparna, Can you please post a reproducible example? It is difficult to provide much concrete help without having testStatistics. One thing you might try is looking at: str(testStatistics[numerator,]) is it actually a list? If it is not (most likely given the error) and it is supposed to be, you need to figure out what aspect of the generation of it is going awry. Cheers, Josh On Tue, Feb 28, 2012 at 12:23 AM, Aparna Sampath aparna.sampat...@gmail.com wrote: Hi All I am trying to use the unlist() in R to a list variable. The following statements are within a function. { denominator - sqrt(s1 / res.em1$n + s2 / res.em2$n) returnValue - l2 / (denominator + 11) attr(returnValue,numerator) - l2 attr(returnValue,denominator) - denominator returnValue } And when I try to unlist the variable returnValue numerators - unlist(testStatistics[numerator,]) denominators - unlist(testStatistics[denominator,]) I get the following error: Error in testStatistics[numerator, ] : object of type 'closure' is not subsettable I read some threads in R help on this error and they had asked to check if we are using the right datatype to the right function. But in my case it is pretty straightforward since I just list it in one function and try to unlist it later. Any suggestions? Thanks for the help :) -- View this message in context: http://r.789695.n4.nabble.com/Error-message-object-of-type-closure-is- not-subsettable-tp3752886p4427399.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal,
[R] How to replace the values in a column
Dear All, I've been searching relevant topics about replacing values, none seemed to be applicable to me... I have a file with many many varieties, and want to replace some of them into different names. I tried various of ways, still don't know how to do that most efficiently.. Here is part of the example data: GenRep A_1 1 A_1 2 A_2 1 A_2 2 B_1 1 B_1 2 B_3 1 B_3 2 OP1_1 1 OP1_1 2 OP1_5 1 OP1_5 2 For example, I want to replace A_1, B_3, OP1_1 into different name Wynda So that the expected file should become: Gen Rep Wynda 1 Wynda2 A_2 1 A_2 2 B_1 1 B_1 2 Wynda1 Wynda2 Wynda1 Wynda2 OP1_5 1 OP1_5 2 I have created a link file, which contains two rows, translating which Gen correlating to which Name. Not sure if this file helps or not, example as below: Column1(Gen)Column2(Name) A_1 Wynda A_2 A_2 B_1 B_1 B_3 Wynda OP1_1 Wynda OP1_5OP1_5 Though I can replace one by one in excel, since there are too many files and too many reps, it'll be very time-consuming also easy to make mistakes. Please give me any guidance or help in terms of finish this with R. Thanks so much ! Sincerely Hannah -- View this message in context: http://r.789695.n4.nabble.com/How-to-replace-the-values-in-a-column-tp4430448p4430448.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Using a FOR LOOP to name objects
Hello, I am trying to use a for loop to name objects in each iteraction. As in the following example (which doesn't work quite well) my_list-c(A,B,C,D,E,F) for(i in c(1:length(my_list))){ url- http://finance.yahoo.com; doc = htmlTreeParse(url, useInternalNodes = T) tab_nodes = xpathApply(doc, //table[@cellpadding = '3']) *my_list[i]*=lapply(tab_nodes, readHTMLTable) #problem is in this line names(*my_list[i]*)=c(Ins,outs) } The problem is that in iteraction #1, I need the info to be stored at an object called A; At iteraction #2 at object called B... and so on Any idea/help? thank you in advance! -- View this message in context: http://r.789695.n4.nabble.com/Using-a-FOR-LOOP-to-name-objects-tp4430454p4430454.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Bayesian Hidden Markov Models
Dear Oscar,  I am extremely grateful to your help and detailed explanation of the use of RJaCGH package. But, when runing the sample codes you listed, another issue I am a little confused is as following: After runing summary(res), I have got the estimation of the random matrix Beta: Parameters of the transition functions:       Normal Gain Normal 0.000 4.258 Gain   2.001 0.000  But, the transition probabilty matrix Q based on the aboving Beta is more concerned in my modeling. Here, I am not sure how can I get the matrix Q. I did try the Q.NH functions.However, Shoud I set the distance parameter x be 1 or 0? I am not sure.   If 1( according to my own understanding), the following result seems not reseanable.  tran-matrix(c(0,2.001,4.528,0),2,2) Q.NH(beta=tran, x=1)     [,1] [,2] [1,] 0.5 0.5 [2,] 0.5 0.5  Many thanks for your further help and time.  James Allan --- 12å¹´2æ28æ¥ï¼å¨äº, Oscar Rueda [via R] ml-node+s789695n4427760...@n4.nabble.com åéï¼ å件人: Oscar Rueda [via R] ml-node+s789695n4427760...@n4.nabble.com 主é¢: Re: Bayesian Hidden Markov Models æ¶ä»¶äºº: monkeylan lanjin...@yahoo.com.cn æ¥æ: 2012å¹´2æ28æ¥,å¨äº,ä¸å7:02 Dear James, Basically you just need the values (y) and the positions (in your case it would be the index of the times series). The chromosome argument does not apply to your case so it can be a vector of ones. If the positions are at the same distance between (equally spaced) then the model will be homogeneous. So for example something like this would be enough: library(RJaCGH) y - c(rnorm(100,0,1), rnorm(20, 2, 1), rnorm(50, 0, 1)) Pos - 1:length(y) Chrom - rep(1, length(y)) res - RJaCGH(y=y, Pos=Pos, Chrom=Chrom) summary(res) However, it uses a Reversible Jump algorithm and therefore jumps between models with different hidden states. I would suggest you take a look at the vignette that comes with the package or the paper that is referenced there for specific details of the model it fits. Hope it helps, Oscar  On 28/2/12 04:52, monkeylan [hidden email] wrote: Dear Doctor Oscar,  Sorry for not noticing that you are the author of the RJaCGH package. But I noticed that hidden Markov model in your package is with non-homogeneous transition probabilities. Here in my work, the HMM is just a first-order homogeneous Markov chain, i.e. the  transition  matrix is constant.  So, Could you please tell me how can I adjust the R functions in your package to implement my analysis?  Best Regards,  James Allan --- 12å¹´2æ27æ¥ï¼å¨ä¸, Oscar Rueda [via R] [hidden email] åéï¼ å件人: Oscar Rueda [via R] [hidden email] 主é¢: Re: Bayesian Hidden Markov Models æ¶ä»¶äºº: monkeylan [hidden email] æ¥æ: 2012å¹´2æ27æ¥,å¨ä¸,ä¸å6:05 Dear James, Although designed for the analysis of copy number CGH microarrays, RJaCGH uses a Bayesian HMM model. Cheers, Oscar On 27/2/12 08:32, monkeylan [hidden email] wrote: Dear R buddies, Recently, I attempt to model the US/RMB Exchange rate log-return time series with a *Hidden Markov model (first order Markov Chain mixed Normal distributions). * I have applied the RHmm package to accomplish this task, but the results are not so satisfying. So, I would like to try a *Bayesian method *for the parameter estimation of the Hidden Markov model. Could anyone kindly tell me which R package can perform Bayesian estimation of the model? Many thanks for your help and time. Best Regards, James Allan -- View this message in context: http://r.789695.n4.nabble.com/Bayesian-Hidden-Markov-Models-tp4423946p4423946 . html Sent from the R help mailing list archive at Nabble.com. __ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Oscar M. Rueda, PhD. Postdoctoral Research Fellow, Breast Cancer Functional Genomics. Cancer Research UK Cambridge Research Institute. Li Ka Shing Centre, Robinson Way. Cambridge CB2 0RE England NOTICE AND DISCLAIMER This e-mail (including any attachments) is intended for ...{{dropped:16}} __ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. If you reply to this email, your message will be added to the discussion below:http://r.789695.n4.nabble.com/Bayesian-Hidden-Markov-Models-tp4423946p44 24152.html To unsubscribe from Bayesian Hidden Markov Models, click here. NAML -- View this message in
Re: [R] colour by z value, persp in raster package
David thanks! Maybe is better ''orginal'' example from raster package: r - raster(system.file(external/test.grd, package=raster)) persp(r) class(r) It is not a problem to colour plot by z value in perspective plot (persp) in package graphic - there is an example in help files: par(bg = white) x - seq(-1.95, 1.95, length = 30) y - seq(-1.95, 1.95, length = 35) z - outer(x, y, function(a,b) a*b^2) nrz - nrow(z) ncz - ncol(z) # Create a function interpolating colors in the range of specified colors jet.colors - colorRampPalette( c(blue, green) ) # Generate the desired number of colors from this palette nbcol - 100 color - jet.colors(nbcol) # Compute the z-value at the facet centres zfacet - z[-1, -1] + z[-1, -ncz] + z[-nrz, -1] + z[-nrz, -ncz] # Recode facet z-values into color indices facetcol - cut(zfacet, nbcol) persp(x, y, z, col=color[facetcol], phi=30, theta=-30) par(op) But question is how specify right sequences of colours (e.g. terrain.colors) for RasterLayer object. The idea of code is persp(r, col = terrain.color(n)), where r is class raster. Thanks all! OV From: David Winsemius dwinsem...@comcast.net Cc: r-help@r-project.org r-help@r-project.org Sent: Tuesday, February 28, 2012 8:47 PM Subject: Re: [R] colour by z value, persp in raster package On Feb 28, 2012, at 4:34 AM, Omphalodes Verna wrote: Hi all! My question is how to colour pixels by z value in persp plot in raster package. Here is an example: x - seq(-1.95, 1.95, length = 30) y - seq(-1.95, 1.95, length = 35) z - outer(x, y, function(a,b) a*b^2) r1 - raster(nrows=35, ncols=30, xmn=0, xmx=30, ymn = 0, ymx = 35) r1[] - c(z) persp(r1) There already exist some function to produce persp plot for anothe classes, but I have no idea how deal with RasterLayer object. Is that really an example? After going to the trouble of correcting the error in this code from failing to load the raster package, I now see that there is no connection between the values of x, y, or x with the raster-call. r1[] - as.matrix(z) Error in .local(x, values) : cannot use a matrix with these dimensions Error in .local(x, i, j, ..., value) : cannot replace values on this raster (it is too large persp(r1) Error: hasValues(x) is not TRUE # Post some useful code. --David Winsemius, MD Heritage Laboratories West Hartford, CT [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to replace the values in a column
Hannah, If Gen is a factor you can simply build the new factor on top of it: dataframe$Gen- factor( c('Wynda' , 'A_2' , 'B_1' , 'Wynda' , 'Wynda' , 'OP1_5')[Gen] ) Just make sure the replacement labels are in the same order as levels(Gen). Cheers On Tue, Feb 28, 2012 at 8:39 PM, hannahmaohuang hannahmaohu...@gmail.com wrote: Dear All, I've been searching relevant topics about replacing values, none seemed to be applicable to me... I have a file with many many varieties, and want to replace some of them into different names. I tried various of ways, still don't know how to do that most efficiently.. Here is part of the example data: Gen Rep A_1 1 A_1 2 A_2 1 A_2 2 B_1 1 B_1 2 B_3 1 B_3 2 OP1_1 1 OP1_1 2 OP1_5 1 OP1_5 2 For example, I want to replace A_1, B_3, OP1_1 into different name Wynda So that the expected file should become: Gen Rep Wynda 1 Wynda 2 A_2 1 A_2 2 B_1 1 B_1 2 Wynda 1 Wynda 2 Wynda 1 Wynda 2 OP1_5 1 OP1_5 2 I have created a link file, which contains two rows, translating which Gen correlating to which Name. Not sure if this file helps or not, example as below: Column1(Gen) Column2(Name) A_1 Wynda A_2 A_2 B_1 B_1 B_3 Wynda OP1_1 Wynda OP1_5 OP1_5 Though I can replace one by one in excel, since there are too many files and too many reps, it'll be very time-consuming also easy to make mistakes. Please give me any guidance or help in terms of finish this with R. Thanks so much ! Sincerely Hannah -- View this message in context: http://r.789695.n4.nabble.com/How-to-replace-the-values-in-a-column-tp4430448p4430448.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ts.plot and x axes customization
On Tue, Feb 28, 2012 at 12:57 PM, Jochem Schuster jochem.schus...@web.de wrote: ts.plot(series1, series2, main=main, xlab=xlab, ylab=ylab, col=c(green, red, blue), lwd=2) What I've tried before is deleting the X axes via gpars=list(xaxt=n) in the ts.plot-code. But after that I was not aible to add the customized axes via axis()... So what was the problem? A reproducible example with your attempt at annotating the ?axis would help. In advance, thank you very much for your help and hints! Regards, Jochem Ihr WEB.DE Postfach immer dabei: die kostenlose WEB.DE Mail App für iPhone und Android. [1]https://produkte.web.de/freemail_mobile_startseite/ References 1. https://produkte.web.de/freemail_mobile_startseite/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to replace the values in a column
Hi. You can do something like this: df - read.table(textConnection( GenRep A_1 1 A_1 2 A_2 1 A_2 2 B_1 1 B_1 2 B_3 1 B_3 2 OP1_1 1 OP1_1 2 OP1_5 1 OP1_5 2),header=TRUE,stringsAsFactors=FALSE) str(df) #adding a new column cbind(df,ncol = ifelse(df$Gen==A_1,Wynda,ifelse(df$Gen==B_3,Wynda,df$Gen))) #replacing exisiting values df$Gen[df$Gen==A_1 | df$Gen==B_3] - Wynda df Andrija On Wed, Feb 29, 2012 at 7:06 AM, ilai ke...@math.montana.edu wrote: Hannah, If Gen is a factor you can simply build the new factor on top of it: dataframe$Gen- factor( c('Wynda' , 'A_2' , 'B_1' , 'Wynda' , 'Wynda' , 'OP1_5')[Gen] ) Just make sure the replacement labels are in the same order as levels(Gen). Cheers On Tue, Feb 28, 2012 at 8:39 PM, hannahmaohuang hannahmaohu...@gmail.com wrote: Dear All, I've been searching relevant topics about replacing values, none seemed to be applicable to me... I have a file with many many varieties, and want to replace some of them into different names. I tried various of ways, still don't know how to do that most efficiently.. Here is part of the example data: Gen Rep A_1 1 A_1 2 A_2 1 A_2 2 B_1 1 B_1 2 B_3 1 B_3 2 OP1_1 1 OP1_1 2 OP1_5 1 OP1_5 2 For example, I want to replace A_1, B_3, OP1_1 into different name Wynda So that the expected file should become: Gen Rep Wynda 1 Wynda 2 A_2 1 A_2 2 B_1 1 B_1 2 Wynda 1 Wynda 2 Wynda 1 Wynda 2 OP1_5 1 OP1_5 2 I have created a link file, which contains two rows, translating which Gen correlating to which Name. Not sure if this file helps or not, example as below: Column1(Gen) Column2(Name) A_1 Wynda A_2 A_2 B_1 B_1 B_3 Wynda OP1_1 Wynda OP1_5 OP1_5 Though I can replace one by one in excel, since there are too many files and too many reps, it'll be very time-consuming also easy to make mistakes. Please give me any guidance or help in terms of finish this with R. Thanks so much ! Sincerely Hannah -- View this message in context: http://r.789695.n4.nabble.com/How-to-replace-the-values-in-a-column-tp4430448p4430448.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] SOUTH ACTRESS PHOTOS VIDEOS
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[R] Cannot use negative argument in function
Hi, today i wrote a function in R of the type: index.refraction - function(Temp,Press, RH, CO2) When i try to plug a negative number in Temp, i got this type of error: n - index.refraction(Temp= -40,100,80,CO2) Messages d'avis : 1: In Ops.ordered(left, right) : '-' is not meaningful for ordered factors 2: In Ops.factor(left, right) : - not meaningful for factors 3: In Ops.factor(left, right) : - not meaningful for factors 4: In Ops.ordered(left, right) : '-' is not meaningful for ordered factors 5: In Ops.factor(left, right) : - not meaningful for factors 6: In Ops.factor(left, right) : - not meaningful for factors Do you have any idea what can be the reason? Thanks Simon -- View this message in context: http://r.789695.n4.nabble.com/Cannot-use-negative-argument-in-function-tp4430667p4430667.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Coverage probability for the normal distribution in plot.spec.coherency function.
Hello to the members of the list, I am using the 'spectrum' function from 'stats' package, to calculate the squared coherency between two time series. The function 'plot.spec.coherency' provides information for the coverage probability for the normal distribution. It seems that the calculation is based on Enochson and Goodman, 1965, Gaussian approximations to the distribution of sample coherence. AFFDL-TR-65-57, Whright-Patterson Air Force base. Could someone tell me if this reference was really used or another one? In advance, thank you. Pascal __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.