Re: [R] How to install older version of R?
Dear Jia-Chiun Pan, First of all, why install an older version of R? In my experience the backward compatibility of R is quite good. So using R 2.10.0 should be ok. But if you really want this you can build R from source. The source can be downloaded from CRAN: http://cran.r-project.org/src/base/R-2/R-2.7.2.tar.gz Extract, ./configure, make, make install and you have R 2.7.2 installed on your system. cheers, Paul Pan, Jia-chiun/潘家群 wrote: Dear list This is much like a linux problem, but I can't find any reference for it. My OS is ubuntu 9.04 and a version of 2.9.2 of R has been already installed in. Now, I need to install the version of 2.7.1. I google a lot of websites and it seems like without a painless way provided me to do it. If any one offers me some suggestions/reference, I will appreciate. Jia-Chiun Pan __ 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. -- Drs. Paul Hiemstra Department of Physical Geography Faculty of Geosciences University of Utrecht Heidelberglaan 2 P.O. Box 80.115 3508 TC Utrecht Phone: +3130 274 3113 Mon-Tue Phone: +3130 253 5773 Wed-Fri http://intamap.geo.uu.nl/~paul __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] name of a name
I have quite a complicated problem that's hard to describe. Suppose I have a dataframe d. I want to access the vector d$var, where var is one of the variables in d. Just take for granted that there's a good reason I want to do this as follows. var-DeOxyA xx-paste(d$,var, sep=) mean(xx) [1] NA Warning message: In mean.default(xx) : argument is not numeric or logical: returning NA mean(d$DeOxyA) [1] 21.98904 How can I convert xx so I can do mean(xx), for example? Currently xx is a character, not a variable. as.numeric doesn't do it. I want R to see mean(DeOxyA), not mean(DeOxyA) I'm stumped. Thanks very much for any help. Bill __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Overlaying multiple ROC curves using ROCR
Hello, I was following an example on The ROCR Package pdf, learning to overlay ROC curves on the same plot using the add = TRUE statement. I used this one: data(ROCR.hiv) attach(ROCR.hiv) pred.svm - prediction(hiv.svm$predictions, hiv.svm$labels) perf.svm - performance(pred.svm, 'tpr', 'fpr') pred.nn - prediction(hiv.nn$predictions, hiv.svm$labels) perf.nn - performance(pred.nn, 'tpr', 'fpr') plot(perf.svm, lty=3, col=red,main=SVMs and NNs for prediction of HIV-1 coreceptor usage) plot(perf.nn, lty=3, col=blue,add=TRUE) plot(perf.svm, avg=vertical, lwd=3, col=red, spread.estimate=stderror,plotCI.lwd=2,add=TRUE) plot(perf.nn, avg=vertical, lwd=3, col=blue, spread.estimate=stderror,plotCI.lwd=2,add=TRUE) legend(0.6,0.6,c('SVM','NN'),col=c('red','blue'),lwd=3) The problem that I face is that the ROC plots get replaced even with the add = TRUE statement. I've done this example several times, but the result is the same; the ROC curves are not overlaid on the same plot. Is there any reason why these ROC curves are not overlaying on the same plot. Is there a work around for this, or a setting that I'm unaware of? Thank you for any assistance. -Matt [[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] Unnecesary code?
On Thu, 19 Nov 2009 00:13:27 +0100 Duncan Murdoch murd...@stats.uwo.ca wrote: hunsynte...@hush.com wrote: Dear R-ers, While browsing the R sources, I found the following piece of code in src\main\memory.c: static void reset_pp_stack(void *data) { R_size_t *poldpps = data; R_PPStackSize = *poldpps; } To me, it looks like the poldpps pointer is a nuissance; can't you just cast the data pointer and derefer it at once? Say, static void reset_pp_stack(void *data) { R_PPStackSize = * (R_size_t *) data; } What would you gain by this change? Duncan Murdoch Seriously? What would you gain by rejecting the change? I think the gain is obvious, even if not essential: the code is cleaner. If there is a choice between two different pieces of code that have the same effect, choosing the simpler makes it easier to maintain the code, and easier for a casual user to understand what's going on. Anyone looking at the original code for the first time will have to realise that poldpps is a nuissance variable with no practical importance and no gain whatsoever, the change cuts this need. There is also a negligible loss in performance when the inessential stack variable is allocated. -- Hun __ 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] 3D plot, rotatable and with adjustable symbols
Hi all, I've tried to make a 3D plot, but have run into some problems. I'd like to have a plot that I can rotate interactively using the mouse, which is possible using the plots3d {R.basic}. However, I would like to change the symbols used as the points, but there's no pch in plot3d(). If I use the Scatterplot3d package, I'm able to change this, but not able to rotate the plot interactively. Does anyone know a solution to this? Maybe another package is better? Best regards, Joel _ Nya Windows 7 gör allt lite enklare. Hitta en dator som passar dig! http://windows.microsoft.com/shop [[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] name of a name
Hi, Try this, d - data.frame(a=1:4, b=3:6) var - a mean(d[var]) ## or, if you are not aware of ## fortune(parse) xx - paste(d$,var, sep=) mean(eval(parse(text=xx))) HTH, baptiste 2009/11/19 William Simpson william.a.simp...@gmail.com: I have quite a complicated problem that's hard to describe. Suppose I have a dataframe d. I want to access the vector d$var, where var is one of the variables in d. Just take for granted that there's a good reason I want to do this as follows. var-DeOxyA xx-paste(d$,var, sep=) mean(xx) [1] NA Warning message: In mean.default(xx) : argument is not numeric or logical: returning NA mean(d$DeOxyA) [1] 21.98904 How can I convert xx so I can do mean(xx), for example? Currently xx is a character, not a variable. as.numeric doesn't do it. I want R to see mean(DeOxyA), not mean(DeOxyA) I'm stumped. Thanks very much for any help. Bill __ 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] Presentation of data in Graphical format
On 11/19/2009 03:13 AM, Sunita Patil wrote: Hello Sir I have got 150 observations, got 10 posts/ 6 departments/ tasks vary from 5 to 10, A few of the variables are crossed specially in case of Office boy, where the tasks are like open the door, put on the lights, Yes time variable I have used Chron package, so that it works well My aim for this study is to check the amount of time and its variability for groups of tasks Its my project work so need to work this out myself if it doesnt work then I will have to consult a statistician Thanks for guiding me to put up the question in more clearer way, I will sure take care next time Hi Sunita, You seem to have two aims, one to display the tasks, and the other to summarize the times. I have been looking at the plot.dendrite function and it might perform the first task with a bit of rewriting (which it needs anyway). The second task might be handled by the hierobarp function. I'll try to work out whether these will do the job in the next day or two. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] name of a name
Thanks very much Phil and Baptiste! d[[var]] is exactly what I wanted. Sorry for being so inarticulate. I still couldn't describe my problem if I wanted to! Anyway, now I have the solution. Cheers Bill On Thu, Nov 19, 2009 at 9:39 AM, baptiste auguie baptiste.aug...@googlemail.com wrote: Hi, Try this, d - data.frame(a=1:4, b=3:6) var - a mean(d[var]) ## or, if you are not aware of ## fortune(parse) xx - paste(d$,var, sep=) mean(eval(parse(text=xx))) HTH, baptiste 2009/11/19 William Simpson william.a.simp...@gmail.com: I have quite a complicated problem that's hard to describe. Suppose I have a dataframe d. I want to access the vector d$var, where var is one of the variables in d. Just take for granted that there's a good reason I want to do this as follows. var-DeOxyA xx-paste(d$,var, sep=) mean(xx) [1] NA Warning message: In mean.default(xx) : argument is not numeric or logical: returning NA mean(d$DeOxyA) [1] 21.98904 How can I convert xx so I can do mean(xx), for example? Currently xx is a character, not a variable. as.numeric doesn't do it. I want R to see mean(DeOxyA), not mean(DeOxyA) I'm stumped. Thanks very much for any help. Bill __ 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] PLoS, Arial, R linux
Greetings All. According to the PLoS (Public Library of Science) Guidelines for Figure preparation, if a figure is submitted as EPS rather than TIFF then Figure text must be in Arial font -- see: [1] http://www.plosntds.org/static/figureGuidelines.action#text and also other sections in that web-page [2] http://www.plosntds.org/static/figureGuidelines.action Now, Arial font is not (as a rule) available in Linux, and EPS diagrams prepared using R graphics will call for Helvetica (by default). Of course the Helvetica family is very similar to Arial, but the in-file font references will call for Helvetica so a production system which is expecting calls for Arial may be thrown off the rails by an EPS file which calls for Helvetica. While there is a section (Enable the use of Arial in R) in the Guidelines (URL [2]), the instructions assume the presence of Arial .ttf files, not usually the case with Linux. The PLoS Guidelines state: Figure text that requires a font family other than Arial (math symbols, etc.) must have the font information embedded in the figure file, or be converted to outlines. (same URL as [1] above). Of course, the Helvetica fonts are amongst the Standard Adobe set, and are assumed to be available on any PostScript-capable rendering device/system (either in the official Adobe font-definition form, or as a simulacrum which can be evoked by the same name), so as a rule the issue of embedding fiont definitions does not arise. Therefore I am wondering whether [A] An EPS which simply uses Helvetica will be accepted by PLoS (the substitution of Arial being automatic, as an alias); or [B] One has to take special measures when preparing a diagram for PLoS using R in Linux when Arial is not available; or [C] One should proceed in quite a different way! Note: I am not keen on the PLoS preferred alternative of submitting a TIFF file, since this gives a bit-mapped result whicvh could render poorly. On the other hand, the Guidelines state (with resepct to LaTeX files): PLoS does not accept vector EPS figures generated using LaTeX. We only accept LaTeX generated figures in TIFF format. which suggests that vector-graphics formats (however generated) may not be acceptable anyway! I would very much welcome any guidance on these questions, especially from people with experience of publishing with PLoS who use R with Linux. With thanks, Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 19-Nov-09 Time: 11:00:15 -- XFMail -- __ 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] 3D plot, rotatable and with adjustable symbols
Try the rgl package. r - Remko Duursma Post-Doctoral Fellow Centre for Plants and the Environment University of Western Sydney Hawkesbury Campus Richmond NSW 2753 Dept of Biological Science Macquarie University North Ryde NSW 2109 Australia Mobile: +61 (0)422 096908 www.remkoduursma.com 2009/11/19 Joel Fürstenberg-Hägg joel_furstenberg_h...@hotmail.com: Hi all, I've tried to make a 3D plot, but have run into some problems. I'd like to have a plot that I can rotate interactively using the mouse, which is possible using the plots3d {R.basic}. However, I would like to change the symbols used as the points, but there's no pch in plot3d(). If I use the Scatterplot3d package, I'm able to change this, but not able to rotate the plot interactively. Does anyone know a solution to this? Maybe another package is better? Best regards, Joel _ Nya Windows 7 gör allt lite enklare. Hitta en dator som passar dig! http://windows.microsoft.com/shop [[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] randomForest: impact of bug fixed in version 4.5-13
The NEWS of the randomForest R library mention that version 4.5-13 fixed a bug in predict.randomForest() when newdata is a matrix with no rownames. I think it corresponds to the difference in files predict.randomForest.R which is the new line if (is.null(rn)) rn - keep As I've been using version 4.5-12 to make predictions for newdata with no row labels, could anyone tell me how this bug affected the predictions? Thanks, jaslqa [[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] How to analyse/visualize a city budget ?
Hello all, I happen to get a (legitimate) hold of a city budget for the (4) years: 2006,2007,2008,2009 The budget holds over 12,000 rows of budget sections with numbers being Zero's positive and negatives. I would like to find something interesting in this dataset. I don't have a clear definition of what this interesting might be, nor how to find it. But my aim is to find where the city council did something fishy (again, no clear definition). My hope is to try and use the time element to catch something on the variables. My initial idea was to try to use each section 4 (time) data points, and maybe check 1) correlations and clusters within the section. to find suspicious similar sections. 2) Also, I was hoping to make a small model for each section, and see if it had major 1 outlier relative to the other 3 data points it has. (I feel that is serious stretching of the data though...) I would love for any interesting ideas (analysis or visualization vise). Best, Tal -- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | * www.r-statistics.com*/ (English) [[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] Unnecesary code?
On 19/11/2009 4:23 AM, Hun S. Tesatte wrote: On Thu, 19 Nov 2009 00:13:27 +0100 Duncan Murdoch murd...@stats.uwo.ca wrote: hunsynte...@hush.com wrote: Dear R-ers, While browsing the R sources, I found the following piece of code in src\main\memory.c: static void reset_pp_stack(void *data) { R_size_t *poldpps = data; R_PPStackSize = *poldpps; } To me, it looks like the poldpps pointer is a nuissance; can't you just cast the data pointer and derefer it at once? Say, static void reset_pp_stack(void *data) { R_PPStackSize = * (R_size_t *) data; } What would you gain by this change? Duncan Murdoch Seriously? What would you gain by rejecting the change? I would save about an hour spent making the change, testing and committing it. I think the gain is obvious, even if not essential: the code is cleaner. If there is a choice between two different pieces of code that have the same effect, choosing the simpler makes it easier to maintain the code, and easier for a casual user to understand what's going on. Anyone looking at the original code for the first time will have to realise that poldpps is a nuissance variable with no practical importance and no gain whatsoever, the change cuts this need. But it makes the expression more complex, and doesn't give a hint about what's going on. The name poldpps adds a bit of explanation of what the assumption is about what's being passed in data. There is also a negligible loss in performance when the inessential stack variable is allocated. There is likely no variable allocated. Compilers are reasonably smart these days. Duncan Murdoch -- Hun __ 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] PLoS, Arial, R linux
Hello On Thu, Nov 19, 2009 at 11:00 AM, Ted Harding ted.hard...@manchester.ac.uk wrote: While there is a section (Enable the use of Arial in R) in the Guidelines (URL [2]), the instructions assume the presence of Arial .ttf files, not usually the case with Linux. If you have a Windows installation, then you have the appropriate license to use Arial, on Windows or on Linux. From what I remember, in this case it could be relatively easy to copy the font files on the Linux partition and install them. With some distributions it is possible to install MS fonts from a third-party source (on Debian ttf-mscorefonts-installer, on Gentoo some package also called corefonts). Regards Liviu __ 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] PLoS, Arial, R linux
On Thu, Nov 19, 2009 at 4:30 PM, Ted Harding ted.hard...@manchester.ac.uk wrote: Greetings All. According to the PLoS (Public Library of Science) Guidelines for Figure preparation, if a figure is submitted as EPS rather than TIFF then Figure text must be in Arial font -- see: [1] http://www.plosntds.org/static/figureGuidelines.action#text and also other sections in that web-page [2] http://www.plosntds.org/static/figureGuidelines.action Now, Arial font is not (as a rule) available in Linux, and EPS diagrams prepared using R graphics will call for Helvetica (by default). Of course the Helvetica family is very similar to Arial, but the in-file font references will call for Helvetica so a production system which is expecting calls for Arial may be thrown off the rails by an EPS file which calls for Helvetica. While there is a section (Enable the use of Arial in R) in the The fact that there is such a section suggests that they accept the resulting files, doesn't it? Guidelines (URL [2]), the instructions assume the presence of Arial .ttf files, not usually the case with Linux. But easy enough to fix (completely legally, thanks to a licensing oversight by Microsoft): http://corefonts.sourceforge.net/ In Debian (and I assume Ubuntu): $ sudo apt-get install msttcorefonts -Deepayan [...] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problem with sqlSave
Hello, the sqlSave function is used in order to write a dataframe on excel. This function creates worksheets using the attribute tablename and writes the data.frame in it. What I want to do is to create this data.frame but being able in case this worksheet already exists to delete the former datas and write the new ones in it. I used the safer and append attributes. When you set safer to false, from what I understood, it should delete the former datas. Then what I did is setting safer as false and append as true so it can append to the empty dataframe the new dataframe but it didn´t work. It appends in the existing worksheets but doesn´ t delete the existing datas. Any idea? - Anna Lippel new in R so be careful I should be asking a lt of questions!:teeth: -- View this message in context: http://old.nabble.com/Problem-with-sqlSave-tp26421303p26421303.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] PLoS, Arial, R linux
On 19-Nov-09 11:52:57, Deepayan Sarkar wrote: On Thu, Nov 19, 2009 at 4:30 PM, Ted Harding ted.hard...@manchester.ac.uk wrote: Greetings All. According to the PLoS (Public Library of Science) Guidelines for Figure preparation, if a figure is submitted as EPS rather than TIFF then Figure text must be in Arial font -- see: [1] http://www.plosntds.org/static/figureGuidelines.action#text and also other sections in that web-page [2] http://www.plosntds.org/static/figureGuidelines.action Now, Arial font is not (as a rule) available in Linux, and EPS diagrams prepared using R graphics will call for Helvetica (by default). Of course the Helvetica family is very similar to Arial, but the in-file font references will call for Helvetica so a production system which is expecting calls for Arial may be thrown off the rails by an EPS file which calls for Helvetica. While there is a section (Enable the use of Arial in R) in the The fact that there is such a section suggests that they accept the resulting files, doesn't it? Guidelines (URL [2]), the instructions assume the presence of Arial .ttf files, not usually the case with Linux. But easy enough to fix (completely legally, thanks to a licensing oversight by Microsoft): http://corefonts.sourceforge.net/ In Debian (and I assume Ubuntu): $ sudo apt-get install msttcorefonts -Deepayan Many thanks, Deepayan! I wasn't aware of that route. Using synaptic on Debian (Etch), with Search: msttcorefonts the font files were installed quite painlessly. I now have: /usr/share/fonts/truetype/msttcorefonts/arialbd.ttf /usr/share/fonts/truetype/msttcorefonts/arialbi.ttf /usr/share/fonts/truetype/msttcorefonts/ariali.ttf /usr/share/fonts/truetype/msttcorefonts/arial.ttf so should be able to make progress as described in the Guidelines. Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 19-Nov-09 Time: 12:17:51 -- XFMail -- __ 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] [Follow-up] Re: PLoS, Arial, R linux
See Addendum at end. On 19-Nov-09 12:17:54, Ted Harding wrote: On 19-Nov-09 11:52:57, Deepayan Sarkar wrote: On Thu, Nov 19, 2009 at 4:30 PM, Ted Harding ted.hard...@manchester.ac.uk wrote: Greetings All. According to the PLoS (Public Library of Science) Guidelines for Figure preparation, if a figure is submitted as EPS rather than TIFF then Figure text must be in Arial font -- see: [1] http://www.plosntds.org/static/figureGuidelines.action#text and also other sections in that web-page [2] http://www.plosntds.org/static/figureGuidelines.action Now, Arial font is not (as a rule) available in Linux, and EPS diagrams prepared using R graphics will call for Helvetica (by default). Of course the Helvetica family is very similar to Arial, but the in-file font references will call for Helvetica so a production system which is expecting calls for Arial may be thrown off the rails by an EPS file which calls for Helvetica. While there is a section (Enable the use of Arial in R) in the The fact that there is such a section suggests that they accept the resulting files, doesn't it? Guidelines (URL [2]), the instructions assume the presence of Arial .ttf files, not usually the case with Linux. But easy enough to fix (completely legally, thanks to a licensing oversight by Microsoft): http://corefonts.sourceforge.net/ In Debian (and I assume Ubuntu): $ sudo apt-get install msttcorefonts -Deepayan Many thanks, Deepayan! I wasn't aware of that route. Using synaptic on Debian (Etch), with Search: msttcorefonts the font files were installed quite painlessly. I now have: /usr/share/fonts/truetype/msttcorefonts/arialbd.ttf /usr/share/fonts/truetype/msttcorefonts/arialbi.ttf /usr/share/fonts/truetype/msttcorefonts/ariali.ttf /usr/share/fonts/truetype/msttcorefonts/arial.ttf so should be able to make progress as described in the Guidelines. Ted. Addendum: Well, the fonts were installed as above, indeed. However, according to the PLoS Guidelines one must generate .afm files before using them in R, along the lines of: First, convert the Arial .ttf files to afm: ttf2afm /usr/share/fonts/msttcorefonts/arial.ttf ~/arial.afm ttf2afm /usr/share/fonts/msttcorefonts/ariali.ttf ~/ariali.afm ttf2afm /usr/share/fonts/msttcorefonts/arialbd.ttf ~/arialbd.afm ttf2afm /usr/share/fonts/msttcorefonts/arialbi.ttf ~/arialbi.afm and then one can use them in R's postscript() with the family argument on the lines of postscript(file=try.ps, horizontal=F, onefile=F, width=4, height=4, family=c(/home/stephen/arial.afm, /home/stephen/arialbd.afm, /home/stephen/ariali.afm, /home/stephen/arialbi.afm), pointsize=12) However, (a) I do not have the 'ttf2afm' program; (b) Synaptic cannot find it in the repositories. So, I'm off on another hunt! Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 19-Nov-09 Time: 12:58:32 -- XFMail -- __ 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] Accessing list names in lapply
Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? A trivial example: df1 - split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num [1:10] -1.14 -1.773 2.512 -2.072 -0.904 ... # $ d: num [1:10] 2.109 1.243 0.627 -2.343 -6.071 ... #... par(mfcol=c(5,2)) lapply(df1, plot) This plots each element of the list, but the label on the vertical axis is X[[0L]] (as expected from the documentation in ?lapply). I'd like the heading for each plot to be the name of that item in the list. This can be achieved by using a for-loop: for (i in names(df1)) plot(df1[[i]], ylab=i) but can it somehow be achieved bu using lapply? I would be hoping for something like lapply(df1, function(x) plot(x, ylab=parent.index())) or some way to parse the index number out of the call, using match.call() or something like that. Thanks in advance for any comments, Bjarke Christensen __ 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] Accessing list names in lapply
lapply over the list names rather than the list itself: junk - lapply(names(df1), function(nm) plot(df1[[nm]], ylab = nm)) On Thu, Nov 19, 2009 at 7:27 AM, Bjarke Christensen bjarke.christen...@sydbank.dk wrote: Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? A trivial example: df1 - split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num [1:10] -1.14 -1.773 2.512 -2.072 -0.904 ... # $ d: num [1:10] 2.109 1.243 0.627 -2.343 -6.071 ... #... par(mfcol=c(5,2)) lapply(df1, plot) This plots each element of the list, but the label on the vertical axis is X[[0L]] (as expected from the documentation in ?lapply). I'd like the heading for each plot to be the name of that item in the list. This can be achieved by using a for-loop: for (i in names(df1)) plot(df1[[i]], ylab=i) but can it somehow be achieved bu using lapply? I would be hoping for something like lapply(df1, function(x) plot(x, ylab=parent.index())) or some way to parse the index number out of the call, using match.call() or something like that. Thanks in advance for any comments, Bjarke Christensen __ 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] Accessing list names in lapply
On 19/11/2009 7:27 AM, Bjarke Christensen wrote: Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? No, but you can loop over the indices in lapply, not just in a for loop. For example, lapply(names(df1), function(x) plot(df1[[x]], ylab=x)) Duncan Murdoch A trivial example: df1 - split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num [1:10] -1.14 -1.773 2.512 -2.072 -0.904 ... # $ d: num [1:10] 2.109 1.243 0.627 -2.343 -6.071 ... #... par(mfcol=c(5,2)) lapply(df1, plot) This plots each element of the list, but the label on the vertical axis is X[[0L]] (as expected from the documentation in ?lapply). I'd like the heading for each plot to be the name of that item in the list. This can be achieved by using a for-loop: for (i in names(df1)) plot(df1[[i]], ylab=i) but can it somehow be achieved bu using lapply? I would be hoping for something like lapply(df1, function(x) plot(x, ylab=parent.index())) or some way to parse the index number out of the call, using match.call() or something like that. Thanks in advance for any comments, Bjarke Christensen __ 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] [Follow-up] Re: PLoS, Arial, R linux
On 19 November 2009 at 13:00, (Ted Harding) wrote: | However, | (a) I do not have the 'ttf2afm' program; | (b) Synaptic cannot find it in the repositories. | | So, I'm off on another hunt! i) You should try to get over the 'Synaptic is the only interface to package' syndrome. It limits your ability to use your system to the fullest. ii) E.g. consider http://packages.debian.org/ which has the handy shortcut http://packages.debian.org/file:path for the search for paths ending in the keyword. iii) This was another question for r-sig-debian, not r-help iv) You want the texlive-font-utils package as per ii) and http://packages.debian.org/search?searchon=contentskeywords=ttf2afmmode=filenamesuite=unstablearch=any Dirk -- Three out of two people have difficulties with fractions. __ 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] [Follow-up] Re: PLoS, Arial, R linux
On 11/19/09, Ted Harding ted.hard...@manchester.ac.uk wrote: However, (a) I do not have the 'ttf2afm' program; (b) Synaptic cannot find it in the repositories. http://packages.debian.org/search?searchon=contentskeywords=ttf2afmmode=exactfilenamesuite=testingarch=any Liviu __ 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] t-criterion calculation using means and SE
Dear all, is there any functions which allow to calculate Student t-criterion using means, their SE and sample size? I've seek for, but bulit-in t-criterion works only with initial sample... Best regards, A.Morkovin __ 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] Accessing list names in lapply
Maybe this : http://tolstoy.newcastle.edu.au/R/e4/help/08/04/8720.html Romain On 11/19/2009 01:27 PM, Bjarke Christensen wrote: Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? A trivial example: df1- split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num [1:10] -1.14 -1.773 2.512 -2.072 -0.904 ... # $ d: num [1:10] 2.109 1.243 0.627 -2.343 -6.071 ... #... par(mfcol=c(5,2)) lapply(df1, plot) This plots each element of the list, but the label on the vertical axis is X[[0L]] (as expected from the documentation in ?lapply). I'd like the heading for each plot to be the name of that item in the list. This can be achieved by using a for-loop: for (i in names(df1)) plot(df1[[i]], ylab=i) but can it somehow be achieved bu using lapply? I would be hoping for something like lapply(df1, function(x) plot(x, ylab=parent.index())) or some way to parse the index number out of the call, using match.call() or something like that. Thanks in advance for any comments, Bjarke Christensen -- Romain Francois Professional R Enthusiast +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr |- http://tr.im/EAD5 : LondonR slides |- http://tr.im/BcPw : celebrating R commit #5 `- http://tr.im/ztCu : RGG #158:161: examples of package IDPmisc __ 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] Accessing list names in lapply
You can try this: par(mfcol=c(5,2)) lapply(df1, function(x){ nm - names(eval(as.list(sys.call(-1))[[2]]))[as.numeric(gsub([^0-9], , deparse(substitute(x] plot(x, main = nm) }) On Thu, Nov 19, 2009 at 10:27 AM, Bjarke Christensen bjarke.christen...@sydbank.dk wrote: Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? A trivial example: df1 - split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num [1:10] -1.14 -1.773 2.512 -2.072 -0.904 ... # $ d: num [1:10] 2.109 1.243 0.627 -2.343 -6.071 ... #... par(mfcol=c(5,2)) lapply(df1, plot) This plots each element of the list, but the label on the vertical axis is X[[0L]] (as expected from the documentation in ?lapply). I'd like the heading for each plot to be the name of that item in the list. This can be achieved by using a for-loop: for (i in names(df1)) plot(df1[[i]], ylab=i) but can it somehow be achieved bu using lapply? I would be hoping for something like lapply(df1, function(x) plot(x, ylab=parent.index())) or some way to parse the index number out of the call, using match.call() or something like that. Thanks in advance for any comments, Bjarke Christensen __ 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] re placing the dates format in R for exporting the data set...
On Nov 18, 2009, at 11:00 PM, ychu066 wrote: hey Jim , I have solve the column name problems now. But i am still unable to read the date in R ... toms_dat- replace(toms_dat, toms_dat ==2009-08-24, 6) replace needs its first argument to be a vector, while you have given it a dataframe. Look at these examples: toms - data.frame(a=letters[1:10], b=Sys.Date() + 1:10) toms a b 1 a 2009-11-20 2 b 2009-11-21 3 c 2009-11-22 4 d 2009-11-23 5 e 2009-11-24 6 f 2009-11-25 7 g 2009-11-26 8 h 2009-11-27 9 i 2009-11-28 10 j 2009-11-29 replace(toms$b, toms$b==2009-11-23, 6) Error in as.Date.numeric(value) : 'origin' must be supplied Notice that this did not replace(toms$b, toms$b==2009-11-23, 2008-01-01) [1] 2009-11-20 2009-11-21 2009-11-22 2008-01-01 2009-11-24 2009-11-25 2009-11-26 [8] 2009-11-27 2009-11-28 2009-11-29 toms a b 1 a 2009-11-20 2 b 2009-11-21 3 c 2009-11-22 4 d 2009-11-23 5 e 2009-11-24 6 f 2009-11-25 7 g 2009-11-26 8 h 2009-11-27 9 i 2009-11-28 10 j 2009-11-29 Notice that the replace() operation did not do anything to toms. If you had wanted it to, you would have needed to do: toms$b - replace(toms$b, toms$b==2009-11-23, 2008-01-01) Now, if you want further assistance you need to provide a working excaple that has the same features as your problem. Use str(toms_dat) to see what type your columns are ant then perhaps: dput(head(toms_dat)) or: dump(toms_dat, file=stdout() ) or if toms_dat is big, then: smalltoms - head(toms_dat) dump(smalltoms, stdout() ) the toms_dat is a data frame , and I want to replace the date to be a single number eg:1,2,3, regards, Tom. jholtman wrote: First of all '2009-08-06' is 1995; this is probably not what you were expecting. What do you what your expression to do? Is 'toms_dat' a dataframe? if so, your expression 'toms_dat ==2009-08-06' seem strange. So tell us what you want to do, not how you want to do it. On Tue, Nov 17, 2009 at 4:54 PM, ychu066 ychu...@aucklanduni.ac.nz wrote: hi everyone, i am having difficulties with replacing the dates format in R for exporting the data set... eg: the code that i used was toms_dat- replace(toms_dat, toms_dat ==2009-08-06, 2) toms_dat- replace(toms_dat, toms_dat ==2009-08-04, 1) but when i export the data as into txt file or excel file the dates come up with very large numbers .:drunk: please help me ...=) -- View this message in context: http://old.nabble.com/replacing-the-dates-format-in-R-for-exporting-the-data-set...-tp26396492p26396492.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 Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://old.nabble.com/replacing-the-dates-format-in-R-for-exporting-the-data-set...-tp26396492p26420068.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Acceptance sampling
Hi I'm trying to learn how to use R when it comes to acceptance sampling. What I want to do is to type in a lot size, an AQL level and a LQ level and get the required sample size in accordance with inspection level II (ISO 2859-2). I have tried find.plan (c(0.01,0.95), c(0.07,0.05), type=hypergeom, 100)$n, but the value did not correspond to the value of the standard. I then tried x-OC2c(500,10,type=hypergeom,N=10001,pd=0.0315) and came to the conclusion that R is using inspection level III. Can I use any command to get inspection level II? Best regards Bodil Tufvesson [[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 do I change the colour and format for the trelli plot ?
Add col=i in the histogram call. -- View this message in context: http://old.nabble.com/How-do-I-change-the-colour-and-format-for-the-trelli-plot---tp26418382p26421302.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] [Follow-up] Re: PLoS, Arial, R linux
Also see of the journal will let you use NimbusSan: pdf('my.pdf', onefile=FALSE, pointsize=18, family=NimbusSan,height=6,width=8,paper=special) ... dev.off() embedFonts('my.pdf') Frank Dirk Eddelbuettel wrote: On 19 November 2009 at 13:00, (Ted Harding) wrote: | However, | (a) I do not have the 'ttf2afm' program; | (b) Synaptic cannot find it in the repositories. | | So, I'm off on another hunt! i) You should try to get over the 'Synaptic is the only interface to package' syndrome. It limits your ability to use your system to the fullest. ii) E.g. consider http://packages.debian.org/ which has the handy shortcut http://packages.debian.org/file:path for the search for paths ending in the keyword. iii) This was another question for r-sig-debian, not r-help iv) You want the texlive-font-utils package as per ii) and http://packages.debian.org/search?searchon=contentskeywords=ttf2afmmode=filenamesuite=unstablearch=any Dirk -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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] name of a name
On Nov 19, 2009, at 4:39 AM, baptiste auguie wrote: Hi, Try this, d - data.frame(a=1:4, b=3:6) var - a mean(d[var]) I tried get() on Simpson's pasted argument without success. Assuming the OP still wanted to use only text arguments, then this approach works dfn - d get(dfn)[var] So get() works only on complete objects and not on expressions that include extraction operations. ## or, if you are not aware of ## fortune(parse) xx - paste(d$,var, sep=) mean(eval(parse(text=xx))) HTH, baptiste 2009/11/19 William Simpson william.a.simp...@gmail.com: I have quite a complicated problem that's hard to describe. Suppose I have a dataframe d. I want to access the vector d$var, where var is one of the variables in d. Just take for granted that there's a good reason I want to do this as follows. var-DeOxyA xx-paste(d$,var, sep=) mean(xx) [1] NA Warning message: In mean.default(xx) : argument is not numeric or logical: returning NA mean(d$DeOxyA) [1] 21.98904 How can I convert xx so I can do mean(xx), for example? Currently xx is a character, not a variable. as.numeric doesn't do it. I want R to see mean(DeOxyA), not mean(DeOxyA) I'm stumped. Thanks very much for any help. Bill __ 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. 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] ddply function nesting problems
While putting my R code into functions, I've encountered a ddply function nesting issue and need a bit of advice on the proper way to fix it. I've tried several approahces, but neither worked and I need to have the ability to include the cut, range, and fullseq methods within ddply. (For a bit of that explanation refer to http://finzi.psych.upenn.edu/Rhelp08/2009-February/187331.html) Thus, in order to preserve that functionality, and put my code within functions, I needed to have an architecture similar to the following implemented, where you end up running: function_nesting() Unfortunately this produced errors within the ddply where it does not appear to be recognizing or allowing variables or functions to be processed within side its function. Thank you for any advice about how to proceed forward. determine_counts-function() { min_range-1 max_range-30 bin_range_size-5 Me_df-data.frame(Data = c(1:15), Person = Me) You_df-data.frame(Data = c(10:20), Person = You) Them_df-data.frame(Data = c(15:25), Person = Them) Group_df_tmp-rbind(Me_df,You_df) Group_df-rbind(Group_df_tmp,Them_df) Group_df$Person - factor(Group_df$Person, levels = c(Them, You, Me)) #counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range(Data), 5)), Person), nrow) # Approach 1 counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range(c(Group_df$Data, min_range, max_range)), bin_range_size)), Person), nrow) # Approach 2 range_tmp-range(c(Group_df$Data, min_range, max_range)) counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range_tmp, bin_range_size)), Person), nrow) names(counts) - c(Bin, Person, Frequency) qplot(Person, Frequency, data = counts, fill = Person, geom=bar, stat=identity, width = 0.9, xlab=Person) + facet_grid(. ~ Bin) } function_nesting-function() { determine_counts() } However, if the code is just run straight through without being nested it works fine: min_range-1 max_range-30 bin_range_size-5 Me_df-data.frame(Data = c(1:15), Person = Me) You_df-data.frame(Data = c(10:20), Person = You) Them_df-data.frame(Data = c(15:25), Person = Them) Group_df_tmp-rbind(Me_df,You_df) Group_df-rbind(Group_df_tmp,Them_df) Group_df$Person - factor(Group_df$Person, levels = c(Them, You, Me)) #counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range(Data), 5)), Person), nrow) counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range(c(Group_df$Data, min_range, max_range)), bin_range_size)), Person), nrow) Unfortunately this is not within a function, so thanks again for any advice on how to approach this issue. __ 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] t-criterion calculation using means and SE
Dear all, I need to create n*n table with sums of all possible pair combinations of numbers from n-row column. What function allows it? Best regards, A.Morkovin __ 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] Presentation of data in Graphical format
Hello Sir Thanx even I will try to work out on your suggestions, will keep you updated on the progress. Thanx a lot Regards Our Thoughts have the Power to Change our Destiny. Sunita On Thu, Nov 19, 2009 at 3:50 PM, Jim Lemon j...@bitwrit.com.au wrote: On 11/19/2009 03:13 AM, Sunita Patil wrote: Hello Sir I have got 150 observations, got 10 posts/ 6 departments/ tasks vary from 5 to 10, A few of the variables are crossed specially in case of Office boy, where the tasks are like open the door, put on the lights, Yes time variable I have used Chron package, so that it works well My aim for this study is to check the amount of time and its variability for groups of tasks Its my project work so need to work this out myself if it doesnt work then I will have to consult a statistician Thanks for guiding me to put up the question in more clearer way, I will sure take care next time Hi Sunita, You seem to have two aims, one to display the tasks, and the other to summarize the times. I have been looking at the plot.dendrite function and it might perform the first task with a bit of rewriting (which it needs anyway). The second task might be handled by the hierobarp function. I'll try to work out whether these will do the job in the next day or two. Jim [[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] Presentation of data in Graphical format
Hello Sir Thanx a lot, will try Pareto chart for my data Regards Our Thoughts have the Power to Change our Destiny. Sunita On Thu, Nov 19, 2009 at 12:29 PM, Petr PIKAL petr.pi...@precheza.cz wrote: Well, from what you say it seems to me that you could also use Pareto charts together with some aggregation of data. But it depends on what you want to show to your audience. Below is some code which I slightly adapted form original author. Regards Petr #-- # pareto. Produces a Pareto plot of effects. # # Parameters: # effects - vector or matrix of effects to plot. # names - vector of names to label the effects. # xlab - String to display as the x axis label. # ylab - String to display as the y axis label. # perlab - Label for the cumulative percentage label. # heading - Vector of names for plot heading. # pareto - function(effects, names=NULL, xlab=NULL, ylab=Magnitude of Effect, indicate.percent=TRUE, perlab=Cumulative Percentage, heading=NULL, trunc.perc=.95, long.names=FALSE,...) { # set up graphics parameters, note: set las=2 for perpendicular axis. oldpar - par( mar=c(6, 4, 2, 4) + 0.1 , las=3) on.exit(par(oldpar)) if( ! is.matrix(effects)) effects-as.matrix( effects ) for( i in 1:ncol(effects) ) { if( i==2 ) oldpar$ask-par(ask=TRUE)$ask # draw bar plot eff.ord - rev(order(abs(effects[,i]))) ef - abs(effects[eff.ord,i]) names-as.character(names)[eff.ord] # plot barplot # get cumulative sum of effects sumeff - cumsum(ef) m-max(ef) sm-sum(ef) sumeff - sumeff/sm vyber-sumefftrunc.perc suma.ef-sum(ef[vyber]) sumeff-c(sumeff[!vyber],1)*m ef-c(ef[!vyber],suma.ef) names-c(as.character(names[!vyber]),Dalsi) ylimit-max(ef) + max(ef)*0.19 ylimit-c(0,ylimit) par( mar=c(6, 4, 2, 4) + 0.1 , las=3) if (long.names) { x- barplot(ef, names.arg=names, ylim=ylimit, xlab=xlab, ylab=ylab, main=heading[i], plot=F, ...) x- barplot(ef, ylim=ylimit, xlab=xlab, ylab=ylab, main=heading[i], ...) text(x,ylimit[2]/10, names, srt=90, adj=0, cex=.7)} else { x-barplot(ef, names.arg=names, ylim=ylimit, xlab=xlab, ylab=ylab, main=heading[i], ...) } if( indicate.percent == TRUE ){ # draws curve. lines(x, sumeff, lty=solid, lwd=2, col=purple) # draw 80% line lines( c(0,max(x)), rep(0.8*m,2) ) # draw axis labling percentage. at - c(0:5)* m/5 axis(4, at=at, labels=c(0,20,40,60,80,100), pos=max(x)+.6) # add axis lables par(las=0) mtext(perlab, 4, line=2) } } # end for each col } #Don Wingate r-help-boun...@r-project.org napsal dne 18.11.2009 16:17:32: yes in my data the 1st column is the main category say suppose Secretary the second column is the sub category HR Dept the 3rd column is the list of duties performed by the Secretary from HR dept and 4th column is time required to perform the duty so there are many such posts and dept with varied duties and times resp. Regards Our Thoughts have the Power to Change our Destiny. Sunita On Wed, Nov 18, 2009 at 8:42 PM, Petr PIKAL petr.pi...@precheza.cz wrote: Hi r-help-boun...@r-project.org napsal dne 18.11.2009 16:01:27: Yes I tried all the basic ones like box plot, pie chart, etc but the data representation isnt that clear. I agree with Tal. But it partly depends on your data. If you have many levels and only few time values in each boxplot would not look well. Maybe you could check also ?xtabs or ?table and/or R graph gallery http://addictedtor.free.fr/graphiques/ if you find suitable graph. Regards Petr Regards Our Thoughts have the Power to Change our Destiny. Sunita On Wed, Nov 18, 2009 at 7:20 PM, Tal Galili tal.gal...@gmail.com wrote: I would start with ?boxplot -- My contact information: Tal Galili E-mail: tal.gal...@gmail.com Phone number: 972-52-7275845 FaceBook: Tal Galili My Blogs: http://www.talgalili.com (Web and general, Hebrew) http://www.biostatistics.co.il (Statistics, Hebrew) http://www.r-statistics.com/ (Statistics,R, English)
Re: [R] Accessing list names in lapply
Thanks to everybody who replied - I got three distinct, very useful suggestions. Bjarke Christensen Romain Francois romain.francois@ dbmail.com Til Bjarke Christensen 19-11-2009 14:33 bjarke.christen...@sydbank.dk cc r-help@r-project.org Emne Re: [R] Accessing list names in lapply Maybe this : http://tolstoy.newcastle.edu.au/R/e4/help/08/04/8720.html Romain On 11/19/2009 01:27 PM, Bjarke Christensen wrote: Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? A trivial example: df1- split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num [1:10] -1.14 -1.773 2.512 -2.072 -0.904 ... # $ d: num [1:10] 2.109 1.243 0.627 -2.343 -6.071 ... #... par(mfcol=c(5,2)) lapply(df1, plot) This plots each element of the list, but the label on the vertical axis is X[[0L]] (as expected from the documentation in ?lapply). I'd like the heading for each plot to be the name of that item in the list. This can be achieved by using a for-loop: for (i in names(df1)) plot(df1[[i]], ylab=i) but can it somehow be achieved bu using lapply? I would be hoping for something like lapply(df1, function(x) plot(x, ylab=parent.index())) or some way to parse the index number out of the call, using match.call() or something like that. Thanks in advance for any comments, Bjarke Christensen -- Romain Francois Professional R Enthusiast +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr |- http://tr.im/EAD5 : LondonR slides |- http://tr.im/BcPw : celebrating R commit #5 `- http://tr.im/ztCu : RGG #158:161: examples of package IDPmisc __ 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] t-criterion calculation using means and SE
On Nov 19, 2009, at 9:40 AM, Антон Морковин wrote: Dear all, I need to create n*n table with sums of all possible pair combinations of numbers from n-row column. What function allows it? ?expand.grid Best regards, A.Morkovin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 3D plot, rotatable and with adjustable symbols
Look at the ggobi program and the rggobi package for interactions between R and ggobi. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Joel Fürstenberg-Hägg Sent: Thursday, November 19, 2009 2:27 AM To: r-help@r-project.org Subject: [R] 3D plot, rotatable and with adjustable symbols Hi all, I've tried to make a 3D plot, but have run into some problems. I'd like to have a plot that I can rotate interactively using the mouse, which is possible using the plots3d {R.basic}. However, I would like to change the symbols used as the points, but there's no pch in plot3d(). If I use the Scatterplot3d package, I'm able to change this, but not able to rotate the plot interactively. Does anyone know a solution to this? Maybe another package is better? Best regards, Joel _ Nya Windows 7 gör allt lite enklare. Hitta en dator som passar dig! http://windows.microsoft.com/shop [[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] ddply function nesting problems
Hi, I think your ddply call with a calculation inside .( ) is the problem. Are you sure you need to do this? Performing the cut outside ddply seems to work fine, determine_counts-function() { min_range-1 max_range-30 bin_range_size-5 Me_df-data.frame(Data = c(1:15), Person = Me) You_df-data.frame(Data = c(10:20), Person = You) Them_df-data.frame(Data = c(15:25), Person = Them) Group_df_tmp-rbind(Me_df,You_df) Group_df-rbind(Group_df_tmp,Them_df) Group_df$Person - factor(Group_df$Person, levels = c(Them, You, Me)) Group_df - transform(Group_df, cut=cut(Data, breaks=fullseq(range(c(Data, min_range, max_range)), bin_range_size))) counts - ddply(Group_df, .(cut, Person), nrow) names(counts) - c(Bin, Person, Frequency) qplot(Person, Frequency, data = counts, fill = Person, geom=bar, stat=identity, width = 0.9, xlab=Person) + facet_grid(. ~ Bin) } function_nesting() HTH, baptiste 2009/11/19 Jason Rupert jasonkrup...@yahoo.com: While putting my R code into functions, I've encountered a ddply function nesting issue and need a bit of advice on the proper way to fix it. I've tried several approahces, but neither worked and I need to have the ability to include the cut, range, and fullseq methods within ddply. (For a bit of that explanation refer to http://finzi.psych.upenn.edu/Rhelp08/2009-February/187331.html) Thus, in order to preserve that functionality, and put my code within functions, I needed to have an architecture similar to the following implemented, where you end up running: function_nesting() Unfortunately this produced errors within the ddply where it does not appear to be recognizing or allowing variables or functions to be processed within side its function. Thank you for any advice about how to proceed forward. determine_counts-function() { min_range-1 max_range-30 bin_range_size-5 Me_df-data.frame(Data = c(1:15), Person = Me) You_df-data.frame(Data = c(10:20), Person = You) Them_df-data.frame(Data = c(15:25), Person = Them) Group_df_tmp-rbind(Me_df,You_df) Group_df-rbind(Group_df_tmp,Them_df) Group_df$Person - factor(Group_df$Person, levels = c(Them, You, Me)) #counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range(Data), 5)), Person), nrow) # Approach 1 counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range(c(Group_df$Data, min_range, max_range)), bin_range_size)), Person), nrow) # Approach 2 range_tmp-range(c(Group_df$Data, min_range, max_range)) counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range_tmp, bin_range_size)), Person), nrow) names(counts) - c(Bin, Person, Frequency) qplot(Person, Frequency, data = counts, fill = Person, geom=bar, stat=identity, width = 0.9, xlab=Person) + facet_grid(. ~ Bin) } function_nesting-function() { determine_counts() } However, if the code is just run straight through without being nested it works fine: min_range-1 max_range-30 bin_range_size-5 Me_df-data.frame(Data = c(1:15), Person = Me) You_df-data.frame(Data = c(10:20), Person = You) Them_df-data.frame(Data = c(15:25), Person = Them) Group_df_tmp-rbind(Me_df,You_df) Group_df-rbind(Group_df_tmp,Them_df) Group_df$Person - factor(Group_df$Person, levels = c(Them, You, Me)) #counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range(Data), 5)), Person), nrow) counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range(c(Group_df$Data, min_range, max_range)), bin_range_size)), Person), nrow) Unfortunately this is not within a function, so thanks again for any advice on how to approach this issue. __ 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] advice about R for windows speed
Dear All, I appreciate any advice or hints you could provide about the following. We are running R code in a server (running Windows XP and QuadCore Xeon processors, see details below) and we would like to use the server efficiently. Our code takes a bit more than 6 seconds per 25 iterations in the server using a default R 2.10.0 installation. We tested our code in two other computers, a Dell Latitute and a MacBook Pro, and from the details that i include below you will notice that the code needs almost twice the time when we used R for Windows compared against the time the code needs when we use Linux or MacOSX 10.6.2 in each of these computers. I'm sorry I don't provide details on the code we are using. The code consists of all sort of operations (matrix inverses, random number generation, vectorized functions, a few loops, and so on). I hope I can get some advice from you despite the lack of specific code details. Is there any important R feature we should configure manually in the windows server to speed the code up? Is there an optimized BLAS available somewhere for this type of machine? Is these something else apart of an optimized BLAS that we could do to improve the timing? Best regards, Carlos **Server running WinXP (QuadCore Xeon 2.6GHz 8G Ram) Time per 25 Iterations 6.17 **Dell Latitude running Linux (R 2.9.2, Intel Core 2 Duo P9500 @ 2.53GHz, 4GB ram) Time per 25 iterations 2.88 **Dell Latitude running Win Vista (R 2.10.0, Intel Core 2 Duo P9500 @ 2.53GHz, 4GB ram) with New DLL in terminal Time per 25 iterations 5.53 --- **Macbook pro (2.16GHz Intel Core 2 Duo 2GB ram) Time per 25 Iterations 4.58 **Macbook pro running WinXp (2.16GHz Intel Core 2 Duo 2GB ram) Time per 25 Iterations 8.23 note: for the Dell and MacBook Pro we replaced the Rblas.dll file of R for Windows with the file available here http://cran.r-project.org/bin/windows/contrib/ATLAS/C2D/ [[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] Efficient cbind of elements from two lists
Hi! I have a data.frame data and splitted it. data - split(data, data[,1]) This is a quite slow procedure; and I do not want to do it again. So, any unsplit and resplit is no option for me. But: I have to cbind variables to the splitted data from another list, that contains of vectors with matching sizes, so for (i in 1:length(data)) { data[[i]] - cbind(data[[i]], l[[i]])) } works well; but very, very slowly. The lapply solution: data - lapply(1:k, function(i) cbind(data[[i]], l[[i]])) does not improve the situation, but allows for mclapply from the multicore package... Is there a more efficient way to combine elements from two lists? Thank you very much! Greetings, Stephan __ 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] Efficient cbind of elements from two lists
Dear Stephan, Here is a suggestion using do.call(): res - do.call(cbind, yourlist) res HTH, Jorge On Thu, Nov 19, 2009 at 10:03 AM, Stephan Dlugosz wrote: Hi! I have a data.frame data and splitted it. data - split(data, data[,1]) This is a quite slow procedure; and I do not want to do it again. So, any unsplit and resplit is no option for me. But: I have to cbind variables to the splitted data from another list, that contains of vectors with matching sizes, so for (i in 1:length(data)) { data[[i]] - cbind(data[[i]], l[[i]])) } works well; but very, very slowly. The lapply solution: data - lapply(1:k, function(i) cbind(data[[i]], l[[i]])) does not improve the situation, but allows for mclapply from the multicore package... Is there a more efficient way to combine elements from two lists? Thank you very much! Greetings, Stephan __ 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] advice about R for windows speed
On Nov 19, 2009, at 9:25 AM, Carlos Hernandez wrote: Dear All, I appreciate any advice or hints you could provide about the following. We are running R code in a server (running Windows XP and QuadCore Xeon processors, see details below) and we would like to use the server efficiently. Our code takes a bit more than 6 seconds per 25 iterations in the server using a default R 2.10.0 installation. We tested our code in two other computers, a Dell Latitute and a MacBook Pro, and from the details that i include below you will notice that the code needs almost twice the time when we used R for Windows compared against the time the code needs when we use Linux or MacOSX 10.6.2 in each of these computers. I'm sorry I don't provide details on the code we are using. The code consists of all sort of operations (matrix inverses, random number generation, vectorized functions, a few loops, and so on). I hope I can get some advice from you despite the lack of specific code details. Is there any important R feature we should configure manually in the windows server to speed the code up? Is there an optimized BLAS available somewhere for this type of machine? Is these something else apart of an optimized BLAS that we could do to improve the timing? Best regards, Carlos **Server running WinXP (QuadCore Xeon 2.6GHz 8G Ram) Time per 25 Iterations 6.17 **Dell Latitude running Linux (R 2.9.2, Intel Core 2 Duo P9500 @ 2.53GHz, 4GB ram) Time per 25 iterations 2.88 **Dell Latitude running Win Vista (R 2.10.0, Intel Core 2 Duo P9500 @ 2.53GHz, 4GB ram) with New DLL in terminal Time per 25 iterations 5.53 --- **Macbook pro (2.16GHz Intel Core 2 Duo 2GB ram) Time per 25 Iterations 4.58 **Macbook pro running WinXp (2.16GHz Intel Core 2 Duo 2GB ram) Time per 25 Iterations 8.23 note: for the Dell and MacBook Pro we replaced the Rblas.dll file of R for Windows with the file available here http://cran.r-project.org/bin/windows/contrib/ATLAS/C2D/ Are you running 32 bit R on each platform or are you using 64 bit R on Linux and OSX? On the Dell, you are running two different versions of R and you don't indicate the R versions on the MacBook. The RAM configuration on each computer is different, which will impact the timings to some extent, depending upon how much RAM you may require for your R code, given other processes that are running and before any disk swapping kicks in. You might want to review R Windows FAQ 2.9, if you have not already: http://cran.r-project.org/bin/windows/base/rw-FAQ.html#There-seems-to-be-a-limit-on-the-memory-it-uses_0021 For Windows on the MacBook, are you using Boot Camp to run Windows natively or are you using virtualization (eg. Parallels, VMWare, VirtualBox) to run Windows under OSX? If the latter, some of the time increase will be due to the virtualization overhead. You should be using the same version of R across each platform for a fair comparison, as there is also the potential, if not the likelihood, that some code has been improved between versions, which may yield some performance differences. 32 bit versus 64 bit will also yield some differences. Differences in tuned BLAS libraries across each OS can also account for performance differences. You should look into using the one provided by R across each to enable more balanaced comparisons. I am also not sure of what differences across each Windows test is attributable to WinXP versus Vista. There are others here with more insight into that aspect of things. While there is a consistent increase for Windows timing as you have above, some of the differences may be due to not really having a (pardon the pun) Apples to Apples comparison across each platform. HTH, Marc Schwartz __ 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] Efficient cbind of elements from two lists
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Stephan Dlugosz Sent: Thursday, November 19, 2009 7:03 AM To: r-help@r-project.org Subject: [R] Efficient cbind of elements from two lists Hi! I have a data.frame data and splitted it. data - split(data, data[,1]) This is a quite slow procedure; and I do not want to do it again. So, any unsplit and resplit is no option for me. But: I have to cbind variables to the splitted data from another list, that contains of vectors with matching sizes, so for (i in 1:length(data)) { data[[i]] - cbind(data[[i]], l[[i]])) } works well; but very, very slowly. The lapply solution: data - lapply(1:k, function(i) cbind(data[[i]], l[[i]])) does not improve the situation, but allows for mclapply from the multicore package... Is there a more efficient way to combine elements from two lists? Can you restructure your analysis so you don't need to split the data.frame itself? I'm assuming the split was slow because there are a lot of groups. Splitting a data.frame into lots of pieces is considerably slower than splitting a few numeric or character columns in it. df - data.frame(group=rep(1:1e5, each=2), score=1:2e5) system.time(split(df, df$group)) # split entire data.frame into 1e5 parts user system elapsed 117.32 38.42 154.34 system.time(split(df$score, df$group)) # split 2nd column into 1e5 parts user system elapsed 0.430.030.46 If R does things the way S+ does this is because splitting simple vectors is done in C code but splitting data.frames invokes the S-language [.data.frame function, which is relatively slow when selecting rows from a data.frame. I'd suggest using ave() (or a function from the plyr package), working on columns from your data.frame and adding ave's output as a column in your big data.frame. E.g., to compute the average score in each group system.time(df$meanScore - ave(df$score, df$group, FUN=mean)) user system elapsed 3.370.003.50 df[1:6,] group score meanScore 1 1 1 1.5 2 1 2 1.5 3 2 3 3.5 4 2 4 3.5 5 3 5 5.5 6 3 6 5.5 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com Thank you very much! Greetings, Stephan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problem on using Rexcel in the vba code
Hello everybody, I started to use Rexcel and I am getting an error with the following code: Sub AutoForma1_Clique() Call RInterface.StartRServer Call RInterface.RRun(setwd(C:/Program Files/R/R-2.10.0/Working Directory)) Call RInterface.RRun(getwd()) End Sub The error is the following when I run the macro: error in time execution '424', the object is obligatory Any idea? - Anna Lippel new in R so be careful I should be asking a lt of questions!:teeth: -- View this message in context: http://old.nabble.com/Problem-on-using-Rexcel-in-the-vba-code-tp26421377p26421377.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] Printing labeled summary to text file ?
Dear List, I am trying to run a mixed model which, on the R console, prints output as follows: [1] Marker [1] perm no. [1] NA Linear mixed model fit by REML Formula: peg.no.prm ~ 1 + (1 | family/f) Data: modeldf AIC BIC logLik deviance REMLdev 3119 3134 -1555 31123111 Random effects: Groups NameVariance Std.Dev. f:family (Intercept) 0.0 0.000 family (Intercept) 0.0 0.000 Residual 178.513.360 Number of obs: 388, groups: f:family, 73; family, 60 Here 'Marker' is one of 426 markers, and 'perm.no' is one of 10,000 permutations of variance for each marker. Each time R begins with a new marker column, it prints 'Marker', and for each permutation it prints a number from 1 to 1 (it hasn't here, but even 'NA' tells me where one data block begins and another ends-- MUCH easier for eyeballing). Of course, since at the end, I will have 426*1 blocks of data, the console doesn't handle it and I tried printing out to a text file (.txt) using capture.output(). The problem with this is that in the text file, my labels ('Marker' and 'Perm no.') aren't printed (since the capture command only uses output from summary(*model*). This leaves me with about 4 million blocks of output (hypothetically) and with no way of telling where one marker ends and the next begins. Eyeballing this will take years. Is there a way to therefore produce labeled output while writing to a text file? Thanks a lot for any suggestions that might help, Aditi - Code for a test file with 10 markers: model-read.table(...) modeldf-data.frame(model) modeldf[2:13]-lapply(modeldf[2:13],factor) colms-(modeldf)[4:13] ## ten marker columns se-c(1:1) peg.no-(modeldf)[,14] library(lme4) for(f in colms) { print(Marker) { for( i in 1:1) { print(perm no.) print(se[i]) { peg.no.prm-sample(peg.no, length(peg.no)) try(fit5-lmer(data=modeldf, peg.no.prm~1 + (1|family/f))) print(summary(fit5)) capture.output(fit5, file=testperm5.txt, append=T) }}} } The data files are at: http://www.4shared.com/file/131980362/460bdafe/Testvcomp10.ht ml (excel) http://www.4shared.com/file/131980512/dc7308b/Testvcomp10.html (txt) -- A Singh aditi.si...@bristol.ac.uk School of Biological Sciences University of Bristol __ 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] Splitting massive output into multiple text files
Dear List, I thought it would be much easier to put a second query into a second mail. I need to print 426*1 blocks of variance components data, where 426 is the number of columns of data that have 1 permutations of variance generated for each of them. I have tried printing out a smaller number of permutations for a smaller number of markers and that has worked. However, since a text file will not handle 4 million blocks in a single file (which is what I ultimately need to do), is there a way to tell R to create a new file for every 10 or so columns? I tried to use some suggested code that looked like: for (j in 1:426) { write(cbind(modeldf[,j:(j+9)]),file=as.character(j)) j - j+10 } ..but can't figure out how to put it into my own code and make it work. I did find one example of code for split files, each successive file being labeled as a series of numbers, but I couldn't figure out how to even adapt that to my model. I cannot figure out what other way there is to conveniently view 4 m. items of data without losing some of it somewhere.. Any help will be much much appreciated.. Aditi -- Code for shorter sample file: model-read.table(...) modeldf-data.frame(model) modeldf[2:13-lapply(modeldf[2:13],factor) colms-(modeldf)[4:13] ## 10 markers only in this file se-c(1:1000) for(f in colms) { print(Marker) { for( i in 1:1000) { print(perm no.) print(se[i]) { peg.no.prm-sample(peg.no, length(peg.no)) try(fit5-lmer(data=modeldf, peg.no.prm~1 + (1|family/f))) print(summary(fit5)) capture.output(fit5, file=testperm5.txt, append=T) }}} } The data files are at: http://www.4shared.com/file/131980362/460bdafe/Testvcomp10.ht ml (excel) http://www.4shared.com/file/131980512/dc7308b/Testvcomp10.html (txt) -- -- A Singh aditi.si...@bristol.ac.uk School of Biological Sciences University of Bristol __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problem with zoo and BootPR packages
Hi, I'm trying to plot the forecasts I generated using the Plot.Fore function of the BootPR package. But I got an error from zoo: My data: Time Series: Start = 1 End = 18 Frequency = 1 [1] 38731 38628 39117 92809 71984 31226 58613 72360 107956 92066 [11] 95208 99098 95848 120383 110717 105680 98469 101916 Script: y1-ts(y1); forey1-BootBC(y1,p=2,h=3,nboot=5000,type=const+trend,correct=ssf) Plot.Fore(y1,forey1$forecast,start=1966,end=1984,frequency=1) The Error: Error in zooreg(x, start, end, frequency) : data : attempt to define illegal zoo object Any Help? Thanks Rick [[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] person-level to person-period xfm
Dear R experts, I have a so-called person-level data frame that I need to transform into a person-period data frame. If the lingo is unclear, the data have one row for each subject, with repeated measures data each in a separate column. I need to transform these data so that each subject has multiple rows, one for each repeated measure value. Is there a quick-and-dirty way to do this transformation? Many thanks, -- Robert Terwilliger Biomedical Physicist Laboratory of Neurocognitive Development Western Psychiatric Institute and Clinic University of Pittsburgh Medical Center Loeffler Building 121 Meyran Avenue #114 Pittsburgh, PA 15213 412.383.8174 - Office 412.383.8179 - Fax em: rater...@gmail.com http://www.wpic.pitt.edu/research/lncd/ __ 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] Problem with zoo and BootPR packages
Contact the BootPR maintainer regarding a bug in this line of Plot.Fore: y1 - zooreg(x, start, end, frequency) where x is a ts object but that may not be used in that context. as.zooreg is available for converting ts series (and certain other objects) to zooreg objects. 2009/11/19 Ricardo Gonçalves Silva ricard...@terra.com.br: Hi, I'm trying to plot the forecasts I generated using the Plot.Fore function of the BootPR package. But I got an error from zoo: My data: Time Series: Start = 1 End = 18 Frequency = 1 [1] 38731 38628 39117 92809 71984 31226 58613 72360 107956 92066 [11] 95208 99098 95848 120383 110717 105680 98469 101916 Script: y1-ts(y1); forey1-BootBC(y1,p=2,h=3,nboot=5000,type=const+trend,correct=ssf) Plot.Fore(y1,forey1$forecast,start=1966,end=1984,frequency=1) The Error: Error in zooreg(x, start, end, frequency) : data : attempt to define illegal zoo object Any Help? Thanks Rick [[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] advice about R for windows speed
Thanks for your reply! I just added some more details below. Our code needs around 1GB of RAM and all machines and R configurations have its default maximum above this number. Our suspicion is that the windows server could run the code in half of its current time (given the apparent factor of 2 between windows and other OS timing). There may be something very important either in the R configuration or in our code that we should take care of? I appreciate a lot any further advice or hints, specially about speeding up the code in the windows xp server with QuadCore Xeon processors. Best regards, Carlos **Server running WinXP 64bit (R 2.10.0 32bit , QuadCore Xeon 2.6GHz 8G Ram) Time per 25 Iterations 6.17 **Dell Latitude running Linux 32bit (R 2.9.2, Intel Core 2 Duo P9500 @ 2.53GHz, 4GB ram) Time per 25 iterations 2.88 **Dell Latitude running Win Vista 32bit (R 2.10.0, Intel Core 2 Duo P9500 @ 2.53GHz, 4GB ram) with New DLL in terminal Time per 25 iterations 5.53 --- **Macbook pro running Snow Leopard (R 2.10.0, 2.16GHz Intel Core 2 Duo 2GB ram) Time per 25 Iterations 4.58 (both R 2.10.0 32bit and 64bit produce almost identical timings) **Macbook pro running WinXp natively (R 2.10.0 32bit, 2.16GHz Intel Core 2 Duo 2GB ram) Time per 25 Iterations 8.23 note: for the Dell and MacBook Pro we replaced the Rblas.dll file of R for Windows with the file available here http://cran.r-project.org/bin/windows/contrib/ATLAS/C2D/ == On Thu, Nov 19, 2009 at 5:06 PM, Marc Schwartz marc_schwa...@me.com wrote: On Nov 19, 2009, at 9:25 AM, Carlos Hernandez wrote: Dear All, I appreciate any advice or hints you could provide about the following. We are running R code in a server (running Windows XP and QuadCore Xeon processors, see details below) and we would like to use the server efficiently. Our code takes a bit more than 6 seconds per 25 iterations in the server using a default R 2.10.0 installation. We tested our code in two other computers, a Dell Latitute and a MacBook Pro, and from the details that i include below you will notice that the code needs almost twice the time when we used R for Windows compared against the time the code needs when we use Linux or MacOSX 10.6.2 in each of these computers. I'm sorry I don't provide details on the code we are using. The code consists of all sort of operations (matrix inverses, random number generation, vectorized functions, a few loops, and so on). I hope I can get some advice from you despite the lack of specific code details. Is there any important R feature we should configure manually in the windows server to speed the code up? Is there an optimized BLAS available somewhere for this type of machine? Is these something else apart of an optimized BLAS that we could do to improve the timing? Best regards, Carlos **Server running WinXP (QuadCore Xeon 2.6GHz 8G Ram) Time per 25 Iterations 6.17 **Dell Latitude running Linux (R 2.9.2, Intel Core 2 Duo P9500 @ 2.53GHz, 4GB ram) Time per 25 iterations 2.88 **Dell Latitude running Win Vista (R 2.10.0, Intel Core 2 Duo P9500 @ 2.53GHz, 4GB ram) with New DLL in terminal Time per 25 iterations 5.53 --- **Macbook pro (2.16GHz Intel Core 2 Duo 2GB ram) Time per 25 Iterations 4.58 **Macbook pro running WinXp (2.16GHz Intel Core 2 Duo 2GB ram) Time per 25 Iterations 8.23 note: for the Dell and MacBook Pro we replaced the Rblas.dll file of R for Windows with the file available here http://cran.r-project.org/bin/windows/contrib/ATLAS/C2D/ Are you running 32 bit R on each platform or are you using 64 bit R on Linux and OSX? On the Dell, you are running two different versions of R and you don't indicate the R versions on the MacBook. The RAM configuration on each computer is different, which will impact the timings to some extent, depending upon how much RAM you may require for your R code, given other processes that are running and before any disk swapping kicks in. You might want to review R Windows FAQ 2.9, if you have not already: http://cran.r-project.org/bin/windows/base/rw-FAQ.html#There-seems-to-be-a-limit-on-the-memory-it-uses_0021 For Windows on the MacBook, are you using Boot Camp to run Windows natively or are you using virtualization (eg. Parallels, VMWare, VirtualBox) to run Windows under OSX? If the latter, some of the time increase will be due to the virtualization overhead. You should be using the same version of R across each platform for a fair comparison, as there is also the potential, if not the likelihood, that some code has been improved between versions, which may yield some performance differences. 32 bit versus 64 bit will also yield some differences. Differences in tuned BLAS libraries across each OS can also account for performance differences. You should look into using the one provided by R across each to enable more balanaced comparisons. I
Re: [R] Problem with zoo and BootPR packages
On Thu, 19 Nov 2009, Ricardo Gonçalves Silva wrote: Hi, I'm trying to plot the forecasts I generated using the Plot.Fore function of the BootPR package. But I got an error from zoo: My data: Time Series: Start = 1 End = 18 Frequency = 1 [1] 38731 38628 39117 92809 71984 31226 58613 72360 107956 92066 [11] 95208 99098 95848 120383 110717 105680 98469 101916 Script: y1-ts(y1); forey1-BootBC(y1,p=2,h=3,nboot=5000,type=const+trend,correct=ssf) Plot.Fore(y1,forey1$forecast,start=1966,end=1984,frequency=1) The Error: Error in zooreg(x, start, end, frequency) : data : attempt to define illegal zoo object Any Help? This is a bug in Plot.Fore() which does not use the zoo functions correctly. It should be reported to the package maintainer. To avoid it, you can do Plot.Fore(as.vector(y1), ...) instead of Plot.Fore(y1, ...) Note to the maintainer (Jae Kim, Cc now): Plot.Fore() calls zooreg(x) where x is a ts object. This isn't the appropriate use of zooreg() which expects a numeric vector/matrix (or a factor). In the case above as.zoo(y1) would already be enough (and preserve all time information). Or you can manually call zooreg(coredata(y1), start = ..., end = ..., ). See the zoo vignettes/examples for more details. Best, Z Thanks Rick [[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] Problem with zoo and BootPR packages
Ok, Thanks all. Rick. -- From: Achim Zeileis achim.zeil...@wu-wien.ac.at Sent: Thursday, November 19, 2009 3:06 PM To: Ricardo Gonçalves Silva ricard...@terra.com.br Cc: R-Help r-help@r-project.org; j@latrobe.edu.au Subject: Re: [R] Problem with zoo and BootPR packages On Thu, 19 Nov 2009, Ricardo Gonçalves Silva wrote: Hi, I'm trying to plot the forecasts I generated using the Plot.Fore function of the BootPR package. But I got an error from zoo: My data: Time Series: Start = 1 End = 18 Frequency = 1 [1] 38731 38628 39117 92809 71984 31226 58613 72360 107956 92066 [11] 95208 99098 95848 120383 110717 105680 98469 101916 Script: y1-ts(y1); forey1-BootBC(y1,p=2,h=3,nboot=5000,type=const+trend,correct=ssf) Plot.Fore(y1,forey1$forecast,start=1966,end=1984,frequency=1) The Error: Error in zooreg(x, start, end, frequency) : data : attempt to define illegal zoo object Any Help? This is a bug in Plot.Fore() which does not use the zoo functions correctly. It should be reported to the package maintainer. To avoid it, you can do Plot.Fore(as.vector(y1), ...) instead of Plot.Fore(y1, ...) Note to the maintainer (Jae Kim, Cc now): Plot.Fore() calls zooreg(x) where x is a ts object. This isn't the appropriate use of zooreg() which expects a numeric vector/matrix (or a factor). In the case above as.zoo(y1) would already be enough (and preserve all time information). Or you can manually call zooreg(coredata(y1), start = ..., end = ..., ). See the zoo vignettes/examples for more details. Best, Z Thanks Rick [[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. No virus found in this incoming message. Checked by AVG - www.avg.com Version: 9.0.707 / Virus Database: 270.14.73/2513 - Release Date: 11/19/09 05:51:00 __ 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] person-level to person-period xfm
Robert Terwilliger wrote: Dear R experts, I have a so-called person-level data frame that I need to transform into a person-period data frame. If the lingo is unclear, the data have one row for each subject, with repeated measures data each in a separate column. I need to transform these data so that each subject has multiple rows, one for each repeated measure value. Is there a quick-and-dirty way to do this transformation? Many thanks, There are several ways. This is one. tmp - data.frame(id=letters[1:3], x1=1:3, x2=4:6, x3=7:9, x4=10:12) tmp id x1 x2 x3 x4 1 a 1 4 7 10 2 b 2 5 8 11 3 c 3 6 9 12 reshape(tmp, direction=long, varying=list(names(tmp)[-1]), ids=id) id time x1 a.1 a1 1 b.1 b1 2 c.1 c1 3 a.2 a2 4 b.2 b2 5 c.2 c2 6 a.3 a3 7 b.3 b3 8 c.3 c3 9 a.4 a4 10 b.4 b4 11 c.4 c4 12 __ 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] plot filled.contour over continent map
Dear all, As a newbie in R I would like to do the following (simple?) thing: to plot a filled.contour plot over a map showing country boundaries (e.g. for Europe) What i do is: map('worldHires',xlim=c(-10,40),ylim=c(35,70),boundary = TRUE,border=0.1) map.axes() filled.contour(mslp, zlim=c(1000,1020),color.palette = colorRampPalette(c(blue, white, red)),main=Avegared MLSP (hPa) ERA40 JJA [1996-2002], xlab=Longitude,ylab=Latitude) in which the mslp file is a netcdf file, with mean sea level pressure for a range of lat/lon values. If I run the above-mentioned, I just get the map of Europe, without the contourplot. When commenting the map statements I can see the contourplot. So I am doing something wrong, but I really have no idea what? Anybody could help me out here? Thanks in advance, Matthias - Department of Earth Environmental Sciences Physical and Regional Geography Research Group Regional climate studies Celestijnenlaan 200E 3001 Heverlee (Leuven) BELGIUM Tel: + 32 16 326424 Fax: + 32 16 322980 http://geo.kuleuven.be/aow/ www.kuleuven.be/geography - [[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 system is computationally singular by using function dmvnorm
Alla Bulashevska wrote: Dear R users, i try to use function dmvnorm(x, mean, sigma, log=FALSE) from R package mvtnorm to calculate the probability of x under the multivariate normal distribution with mean equal to mean and covariance matrix sigma. I become the following Error in solve.default(cov, ...) : system is computationally singular: reciprocal condition number = 1.81093e-19 Probably your sigma is almost singular and can't be inverted in the calculation of the Mahalanobis distance. Uwe Ligges What could be the reason of it? Thank you in Advance, Alla. __ 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] converting a vector of bytes to a PNG/JPEG image
Can't be too hard to convert it in pixmap representation, but you need to tell us hoe the representation looks like, probbaly with a short code that makes it reproducible (e.g. generate some data that we can use in R). Uwe Ligges Rajarshi Guha wrote: Hi, I have some code that uses rJava. One of the Java side methods returns a byte[] representing the bytes from a PNG image. What I'd like to do is to be able to bring up the PNG on the R side (I can bring up a Swing window to show the PNG but I want to avoid that). I have looked at the pixmap and rimage packages but don't seem to be able to work out how I'd go about this (or if it's at all possible). Does anybody have any pointers? Thanks, __ 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] ddply function nesting problems
Awesome! Thanks a ton! I guess I had overlooked how it was really working. I will still have to reflect on why it was working running it straight through, but not being nested. That is kind of a mystery. Oh well... Thanks again. - Original Message From: baptiste auguie baptiste.aug...@googlemail.com To: Jason Rupert jasonkrup...@yahoo.com Cc: R-help@r-project.org Sent: Thu, November 19, 2009 9:24:29 AM Subject: Re: [R] ddply function nesting problems Hi, I think your ddply call with a calculation inside .( ) is the problem. Are you sure you need to do this? Performing the cut outside ddply seems to work fine, determine_counts-function() { min_range-1 max_range-30 bin_range_size-5 Me_df-data.frame(Data = c(1:15), Person = Me) You_df-data.frame(Data = c(10:20), Person = You) Them_df-data.frame(Data = c(15:25), Person = Them) Group_df_tmp-rbind(Me_df,You_df) Group_df-rbind(Group_df_tmp,Them_df) Group_df$Person - factor(Group_df$Person, levels = c(Them, You, Me)) Group_df - transform(Group_df, cut=cut(Data, breaks=fullseq(range(c(Data, min_range, max_range)), bin_range_size))) counts - ddply(Group_df, .(cut, Person), nrow) names(counts) - c(Bin, Person, Frequency) qplot(Person, Frequency, data = counts, fill = Person, geom=bar, stat=identity, width = 0.9, xlab=Person) + facet_grid(. ~ Bin) } function_nesting() HTH, baptiste 2009/11/19 Jason Rupert jasonkrup...@yahoo.com: While putting my R code into functions, I've encountered a ddply function nesting issue and need a bit of advice on the proper way to fix it. I've tried several approahces, but neither worked and I need to have the ability to include the cut, range, and fullseq methods within ddply. (For a bit of that explanation refer to http://finzi.psych.upenn.edu/Rhelp08/2009-February/187331.html) Thus, in order to preserve that functionality, and put my code within functions, I needed to have an architecture similar to the following implemented, where you end up running: function_nesting() Unfortunately this produced errors within the ddply where it does not appear to be recognizing or allowing variables or functions to be processed within side its function. Thank you for any advice about how to proceed forward. determine_counts-function() { min_range-1 max_range-30 bin_range_size-5 Me_df-data.frame(Data = c(1:15), Person = Me) You_df-data.frame(Data = c(10:20), Person = You) Them_df-data.frame(Data = c(15:25), Person = Them) Group_df_tmp-rbind(Me_df,You_df) Group_df-rbind(Group_df_tmp,Them_df) Group_df$Person - factor(Group_df$Person, levels = c(Them, You, Me)) #counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range(Data), 5)), Person), nrow) # Approach 1 counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range(c(Group_df$Data, min_range, max_range)), bin_range_size)), Person), nrow) # Approach 2 range_tmp-range(c(Group_df$Data, min_range, max_range)) counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range_tmp, bin_range_size)), Person), nrow) names(counts) - c(Bin, Person, Frequency) qplot(Person, Frequency, data = counts, fill = Person, geom=bar, stat=identity, width = 0.9, xlab=Person) + facet_grid(. ~ Bin) } function_nesting-function() { determine_counts() } However, if the code is just run straight through without being nested it works fine: min_range-1 max_range-30 bin_range_size-5 Me_df-data.frame(Data = c(1:15), Person = Me) You_df-data.frame(Data = c(10:20), Person = You) Them_df-data.frame(Data = c(15:25), Person = Them) Group_df_tmp-rbind(Me_df,You_df) Group_df-rbind(Group_df_tmp,Them_df) Group_df$Person - factor(Group_df$Person, levels = c(Them, You, Me)) #counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range(Data), 5)), Person), nrow) counts - ddply(Group_df, .(cut(Data, breaks=fullseq(range(c(Group_df$Data, min_range, max_range)), bin_range_size)), Person), nrow) Unfortunately this is not within a function, so thanks again for any advice on how to approach this issue. __ 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
Re: [R] person-level to person-period xfm
On Nov 19, 2009, at 12:23 PM, Richard M. Heiberger wrote: Robert Terwilliger wrote: Dear R experts, I have a so-called person-level data frame that I need to transform into a person-period data frame. If the lingo is unclear, the data have one row for each subject, with repeated measures data each in a separate column. I need to transform these data so that each subject has multiple rows, one for each repeated measure value. Is there a quick-and-dirty way to do this transformation? Many thanks, There are several ways. This is one. Here's another: data.frame(id=tmp$id, stack(tmp, select=-id)) id values ind 1 a 1 x1 2 b 2 x1 3 c 3 x1 4 a 4 x2 5 b 5 x2 6 c 6 x2 7 a 7 x3 8 b 8 x3 9 c 9 x3 10 a 10 x4 11 b 11 x4 12 c 12 x4 tmp - data.frame(id=letters[1:3], x1=1:3, x2=4:6, x3=7:9, x4=10:12) tmp id x1 x2 x3 x4 1 a 1 4 7 10 2 b 2 5 8 11 3 c 3 6 9 12 reshape(tmp, direction=long, varying=list(names(tmp)[-1]), ids=id) id time x1 a.1 a1 1 b.1 b1 2 c.1 c1 3 a.2 a2 4 b.2 b2 5 c.2 c2 6 a.3 a3 7 b.3 b3 8 c.3 c3 9 a.4 a4 10 b.4 b4 11 c.4 c4 12 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with zoo and BootPR packages
Actually it may be that the documentation is at fault more than the code. The help page for Plot.Fore says that the first argument is a time series data set and although that seems to suggest that it should be a ts object the code seems to be written assuming a plain numeric vector; therefore, you could do this: library(zoo) Plot.Fore(coredata(y1), forey1$forecast, start = 1966, end = 1984, frequency = 1) 2009/11/19 Gabor Grothendieck ggrothendi...@gmail.com: Contact the BootPR maintainer regarding a bug in this line of Plot.Fore: y1 - zooreg(x, start, end, frequency) where x is a ts object but that may not be used in that context. as.zooreg is available for converting ts series (and certain other objects) to zooreg objects. 2009/11/19 Ricardo Gonçalves Silva ricard...@terra.com.br: Hi, I'm trying to plot the forecasts I generated using the Plot.Fore function of the BootPR package. But I got an error from zoo: My data: Time Series: Start = 1 End = 18 Frequency = 1 [1] 38731 38628 39117 92809 71984 31226 58613 72360 107956 92066 [11] 95208 99098 95848 120383 110717 105680 98469 101916 Script: y1-ts(y1); forey1-BootBC(y1,p=2,h=3,nboot=5000,type=const+trend,correct=ssf) Plot.Fore(y1,forey1$forecast,start=1966,end=1984,frequency=1) The Error: Error in zooreg(x, start, end, frequency) : data : attempt to define illegal zoo object Any Help? Thanks Rick [[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] How do I specify a partially completed survival analysis model?
Hello, I just started using R to do epidemiologic simulation research using the Cox proportional hazard model. I have 2 covariates X1 and X2 which I want to model as h(t,X)=h0(t)*exp(b1*X1+b2*X2). I assume independence of X from t. After I simulate Time and Censor data vectors denoting the censoring time and status respectively, I can call the following function to fit the data into the Cox model (a is a data.frame containing 4 columns X1, X2, Time and Censor): b = coxph (Surv (Time, Censor) ~ X1 + X2, data = a, method = breslow); Now the purpose of me doing simulation is that I have another mechanism to generate the number b2. From the given b2 (say it's 4.3), Cox model can be fit to generate b1 and check how feasible the new model is. Thus, my question is, how do I specify such a model that is partially completed (as in b2 is known). I tried things like Surv(Time,Censor)~X1+4.3*X2, but it's not working. Thanks very much. -- View this message in context: http://old.nabble.com/How-do-I-specify-a-partially-completed-survival-analysis-model--tp26421391p26421391.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] Reading in a series of files using a for loop
Dear R Users, I am trying to read in a series of csv files which vary by the letter on the end of he file name. When I input what seems to be a logical for loop I get an error message that doesn't make sense to me. for(i in 1:12){ paste(GP, LETTERS[i],sep='') -read.csv(paste(/Users/thomasjackson/Data/GEPCO/GEPCO,LETTERS[i],/HPLC_,LETTERS[i],12.csv,sep=''), header=T, sep=',')} Error in paste(GP, LETTERS[i], sep = ) - read.csv(paste(/Users/thomasjackson/Data/GEPCO/GEPCO, : target of assignment expands to non-language object For example the first file name is HPLC_A12.csv in the folder GEPCOA Thanks for any help, Thomas Jackson __ 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 install older version of R?
Dear Paul, Thank you for your response. I am investigating a differentially expressed gene detected method that cannot be applied on R 2.9.1 that currently I used. The reason about why it cannot work due to the changes of new version of R. And, recently, I study a lots of methods that developed by some statisticians, they kindly provided me source code, but some of codes only implements on the older version of R or packages. I have met many cases of this situation. That's why I am trying to get an older version. Jia-Chiun Pan 2009/11/19 Paul Hiemstra p.hiems...@geo.uu.nl Dear Jia-Chiun Pan, First of all, why install an older version of R? In my experience the backward compatibility of R is quite good. So using R 2.10.0 should be ok. But if you really want this you can build R from source. The source can be downloaded from CRAN: http://cran.r-project.org/src/base/R-2/R-2.7.2.tar.gz Extract, ./configure, make, make install and you have R 2.7.2 installed on your system. cheers, Paul Pan, Jia-chiun/¼ï®a¸s wrote: Dear list This is much like a linux problem, but I can't find any reference for it. My OS is ubuntu 9.04 and a version of 2.9.2 of R has been already installed in. Now, I need to install the version of 2.7.1. I google a lot of websites and it seems like without a painless way provided me to do it. If any one offers me some suggestions/reference, I will appreciate. Jia-Chiun Pan __ 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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Drs. Paul Hiemstra Department of Physical Geography Faculty of Geosciences University of Utrecht Heidelberglaan 2 P.O. Box 80.115 3508 TC Utrecht Phone: +3130 274 3113 Mon-Tue Phone: +3130 253 5773 Wed-Fri http://intamap.geo.uu.nl/~paul [[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] Reading in a series of files using a for loop
Hi Thomas, Here are two suggestions which storage all the files in a list: # Parameters route - /Users/thomasjackson/Data/GEPCO/GEPCO lts - LETTERS[1:12] toread - paste(route,lts,/HPLC_,lts,12.csv,sep='') # Reading in the data -- option 1 myfiles - list() for(i in 1:12) myfiles[i] - read.csv(toread, header=T, sep=',') names(myfiles) - paste(GP, lts, sep='') myfiles # Option 2 -- no for() loop myfiles2 - lapply(toread, read.csv, header=T, sep=',') myfiles2 To access the first file just type either myfiles[[1]] or myfiles2[[1]] For further analysis using each file, the lapply (see ?lapply) function might be useful. HTH, Jorge On Thu, Nov 19, 2009 at 12:29 PM, Thomas Jackson wrote: Dear R Users, I am trying to read in a series of csv files which vary by the letter on the end of he file name. When I input what seems to be a logical for loop I get an error message that doesn't make sense to me. for(i in 1:12){ paste(GP, LETTERS[i],sep='') -read.csv(paste(/Users/thomasjackson/Data/GEPCO/GEPCO,LETTERS[i],/HPLC_,LETTERS[i],12.csv,sep=''), header=T, sep=',')} Error in paste(GP, LETTERS[i], sep = ) - read.csv(paste(/Users/thomasjackson/Data/GEPCO/GEPCO, : target of assignment expands to non-language object For example the first file name is HPLC_A12.csv in the folder GEPCOA Thanks for any help, Thomas Jackson __ 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] Reading in a series of files using a for loop
Try something like: gp - lapply( paste((/Users/thomasjackson/Data/GEPCO/GEPCO,LETTERS[1:12],/HPLC_,LETTERS[1:12],12.csv,sep=''), read.csv, header=TRUE, sep=',' ) names(gp) - paste(GandP, LETTERS[1:12], sep='') Now gp (or whatever you want to call it) will be a list with your 12 data files as elements. You can analyze individual elements, or use lapply to perform the same analysis on each of them. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Thomas Jackson Sent: Thursday, November 19, 2009 10:29 AM To: r-help@r-project.org Subject: [R] Reading in a series of files using a for loop Dear R Users, I am trying to read in a series of csv files which vary by the letter on the end of he file name. When I input what seems to be a logical for loop I get an error message that doesn't make sense to me. for(i in 1:12){ paste(GP, LETTERS[i],sep='') - read.csv(paste(/Users/thomasjackson/Data/GEPCO/GEPCO,LETTERS[i],/H PLC_,LETTERS[i],12.csv,sep=''), header=T, sep=',')} Error in paste(GP, LETTERS[i], sep = ) - read.csv(paste(/Users/thomasjackson/Data/GEPCO/GEPCO, : target of assignment expands to non-language object For example the first file name is HPLC_A12.csv in the folder GEPCOA Thanks for any help, Thomas Jackson __ 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] plot filled.contour over continent map
On Fri, 20 Nov 2009, Matthias Demuzere wrote: Dear all, As a newbie in R I would like to do the following (simple?) thing: to plot a filled.contour plot over a map showing country boundaries (e.g. for Europe) What i do is: map('worldHires',xlim=c(-10,40),ylim=c(35,70),boundary = TRUE,border=0.1) map.axes() filled.contour(mslp, zlim=c(1000,1020),color.palette = colorRampPalette(c(blue, white, red)),main=Avegared MLSP (hPa) ERA40 JJA [1996-2002], xlab=Longitude,ylab=Latitude) in which the mslp file is a netcdf file, with mean sea level pressure for a range of lat/lon values. If I run the above-mentioned, I just get the map of Europe, without the contourplot. When commenting the map statements I can see the contourplot. So I am doing something wrong, but I really have no idea what? Does the filled.contour() call work on its own? It should, right? What I would expect to see is just the contour plot, since that should overwrite the map() plot. Once you get filled.contour() working, try adding the map using: map('worldHires',xlim=c(-10,40),ylim=c(35,70),boundary = TRUE, border=0.1, add = TRUE) You may have to put the xlim= and ylim= into the filled.contour() call to get what you want. HTH Ray Brownrigg Anybody could help me out here? Thanks in advance, Matthias - Department of Earth Environmental Sciences Physical and Regional Geography Research Group Regional climate studies Celestijnenlaan 200E 3001 Heverlee (Leuven) BELGIUM Tel: + 32 16 326424 Fax: + 32 16 322980 http://geo.kuleuven.be/aow/ www.kuleuven.be/geography - [[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] Reading in a series of files using a for loop
You need to look at the assign function: http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-turn-a-string-into-a-variable_003f (... and in addition to R FAQ 7.21 also perhaps read the rest of the R FAQ.) There are a ton of very similar questions in the r-help archives, so you could also try RSiteSearch(read.table paste) -- David On Nov 19, 2009, at 12:29 PM, Thomas Jackson wrote: Dear R Users, I am trying to read in a series of csv files which vary by the letter on the end of he file name. When I input what seems to be a logical for loop I get an error message that doesn't make sense to me. for(i in 1:12){ paste(GP, LETTERS[i],sep='') -read.csv(paste(/ Users/thomasjackson/Data/GEPCO/GEPCO,LETTERS[i],/ HPLC_,LETTERS[i],12.csv,sep=''), header=T, sep=',')} Error in paste(GP, LETTERS[i], sep = ) - read.csv(paste(/ Users/thomasjackson/Data/GEPCO/GEPCO, : target of assignment expands to non-language object For example the first file name is HPLC_A12.csv in the folder GEPCOA 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] plot filled.contour over continent map
You probably want to put the map boundaries on top of the contour plot rather than the other way around. Here is one example of doing that (find the correct asp may be the hardest part): library(maps) x - seq( -124.7, -67, length.out=25 ) y - seq( 25, 49, length.out=25 ) z - outer(x,y,'+') filled.contour(x,y,z, asp=1.5, plot.axes=map('state',add=TRUE) ) hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Ray Brownrigg Sent: Thursday, November 19, 2009 11:50 AM To: r-help@r-project.org Cc: Matthias Demuzere Subject: Re: [R] plot filled.contour over continent map On Fri, 20 Nov 2009, Matthias Demuzere wrote: Dear all, As a newbie in R I would like to do the following (simple?) thing: to plot a filled.contour plot over a map showing country boundaries (e.g. for Europe) What i do is: map('worldHires',xlim=c(-10,40),ylim=c(35,70),boundary = TRUE,border=0.1) map.axes() filled.contour(mslp, zlim=c(1000,1020),color.palette = colorRampPalette(c(blue, white, red)),main=Avegared MLSP (hPa) ERA40 JJA [1996-2002], xlab=Longitude,ylab=Latitude) in which the mslp file is a netcdf file, with mean sea level pressure for a range of lat/lon values. If I run the above-mentioned, I just get the map of Europe, without the contourplot. When commenting the map statements I can see the contourplot. So I am doing something wrong, but I really have no idea what? Does the filled.contour() call work on its own? It should, right? What I would expect to see is just the contour plot, since that should overwrite the map() plot. Once you get filled.contour() working, try adding the map using: map('worldHires',xlim=c(-10,40),ylim=c(35,70),boundary = TRUE, border=0.1, add = TRUE) You may have to put the xlim= and ylim= into the filled.contour() call to get what you want. HTH Ray Brownrigg Anybody could help me out here? Thanks in advance, Matthias - Department of Earth Environmental Sciences Physical and Regional Geography Research Group Regional climate studies Celestijnenlaan 200E 3001 Heverlee (Leuven) BELGIUM Tel: + 32 16 326424 Fax: + 32 16 322980 http://geo.kuleuven.be/aow/ www.kuleuven.be/geography - [[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-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] Lme
Hi In my experiment I have 2 Groups (control patient) and 4 blocks of trials per subject, n=12 for Control and n=11 for patient The variances of the patients is very much higher than that of the Control. 1- I tried an ANOVA with Groups as between factor and Blocks as within factor (repeated measures). aov(Y ~ gr*block+Error(subj/block), data=d) works OK but without the significant effect I was looking for, which may be caused by the variances. So I though I¹d use lme 2- lme.m = lme(Y ~ gr*block, data=d, random=~1 + block | subj); anova(lme.m) I¹d like to make sure that my lme model is correctly psecified. Thanks in advance for your help! __ Patrick Bédard Ph.D. Dept. of Neuroscience Brown University [[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] Calling R (GAMM) from Fortran
Hello, I am currently working on a modeling project using Fortran to run large repetitive loops (many DO loops). As part of this process I would like to use a model fit in R and currently stored as an R object. This is a rather complex model, a GAMM, and I am curious whether there is a way to call this model from Fortran. I am not sure call is correct terminology, but I would basically like to use this GAMM to make a prediction as part of each DO loop. Is this possible? I have found instructions, etc. for calling Fortran from R but not vice versa. Thank you for any information or advice. General advice to address this situation is also welcome. Thanks again! -Paul Simonin [[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] Partial derivatives of the multivariate cumulative distribution
I'm currently using the mvtnorm package to model unobserved heterogeneity in a structural model and using optim to estimate the model. I have got good clues that convergence is not really a problem but the hessian matrix estimate is very bad. To overcome this problem, I'm constructing an OPG estimator of the information matrix and I was wondering if there were an easy way to obtain partial derivatives of say for instance: P1 - pmvnorm(lower=c(-Inf,-Inf,-Inf,-Inf),upper=c(theta1,theta2,theta3,theta4),corr=ssigma) with respect to the mean parameters theta1, theta2, theta3, theta4 and the non-diagonal parameters in sigma, hence $\partial P_1 / \partial \theta_1$, etc... I can deal with numerical or analytical partial derivatives - a gradient would be fine since all observations share the same partial derivative. Stephane __ 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] Calling R (GAMM) from Fortran
On Thu, Nov 19, 2009 at 12:14 PM, Paul Warren Simonin paul.simo...@uvm.edu wrote: Hello, I am currently working on a modeling project using Fortran to run large repetitive loops (many DO loops). As part of this process I would like to use a model fit in R and currently stored as an R object. This is a rather complex model, a GAMM, and I am curious whether there is a way to call this model from Fortran. I am not sure call is correct terminology, but I would basically like to use this GAMM to make a prediction as part of each DO loop. Is this possible? I have found instructions, etc. for calling Fortran from R but not vice versa. Thank you for any information or advice. General advice to address this situation is also welcome. Thanks again! -Paul Simonin The R library and interpreter are written in C and available using a C interface. Therefore, it would be theoretically possible to call this interface from Fortran-- however there are details that probably make a direct call impossible in practice. This is because interfacing with R requires passing representations of R objects (SEXPs) which are far removed from the basic C variable types supported by standardized C--Fortran interfaces. It may be that the only plausible way to approach this problem may involve writing a set of bindings in C that interface with R and extract the information you are interested in and then express it using basing C variables that Fortran can understand. With such a system in place your Fortran program would call a set of routines written in C that called routines from the R libraries and then reformatted the results and returned them to Fortran. This is a conclusion based on my limited experience-- not any formal expertise. There are definitely others on this list that could provide a better qualified answer. Good luck! -Charlie __ 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 wait for a user response in Rscript?
Why I don't see 'par(ask=T)' in /usr/lib/R/library/graphics/R-ex/barplot.R? Is 'par(ask=T)' implicitly called by example()? On Mon, Sep 7, 2009 at 10:55 AM, RIOS,ALFREDO ARTURO ar...@ufl.edu wrote: Hi Peng I think this is what you are looking for par(ask=T) Alfredo On Sun Sep 06 12:52:31 EDT 2009, Peng Yu pengyu...@gmail.com wrote: Hi, In 'example(barplot)' running in R, I see 'Hit Return to see next plot:', then R waits for my input. I am wondering how to wait for a user response in Rscript. Regards, Peng __ 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. -- RIOS,ALFREDO ARTURO __ 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] Performance of 'by' and 'ddply' on a large data frame
I've only recently started using R. One of the problems I come up against is after having extracted a large dataset (5M rows) out of database, I realize I need another variable. In this case I have data frame with dates. I want to find the minimum date for each value of x1 and add that minimum date to my data.frame. randomdf - function(p) { data.frame(x1=sample(1:10^4, 10^p, replace=T), x2=sample(seq.Date(Sys.Date() - 356*3,Sys.Date(), by=day), 10^p, replace=T), y1=sample(1:100, 10^p, replace=T)) } testby - function(p) { df - randomdf(p) system.time(by(df, df$x1, function(dfi) { min(dfi$x2) })) } lapply(c(1,2,3,4,5), testby) [[1]] user system elapsed 0.006 0.000 0.006 [[2]] user system elapsed 0.024 0.000 0.025 [[3]] user system elapsed 0.233 0.000 0.234 [[4]] user system elapsed 1.996 0.026 2.022 [[5]] user system elapsed 11.030 0.000 11.032 Strangely enough, not sure why this is, the result of by with the min function is not date objects but instead integers representing days from an origin. Is there a min function that would return me a date instead of an integer? Or is this a result of using by? I also wanted to see how ddply compares. testddply - function(p) { pdf - randomdf(p); system.time(ddply(pdf, .(x1), function(df) { return (data.frame(min(df$x2))) })) } lapply(c(1,2,3,4,5), testddply) [[1]] user system elapsed 0.020 0.000 0.021 [[2]] user system elapsed 0.119 0.000 0.119 [[3]] user system elapsed 1.008 0.000 1.008 [[4]] user system elapsed 8.425 0.001 8.428 [[5]] user system elapsed 23.070 0.000 23.075 Once the data frame gets above 1M rows, the timings are a bit too long (on a previous run it went up to 8000s user time). This seems quite a bit slower than I expected. Maybe there's a better and faster way to add such variables to a data frame that are derived using some aggregation. Also, ddply seems to take twice as long as by. Are these two operations not equivalent? Thanks, Tahir __ 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] Is there an variant of apply() that does not return anything?
There are a few version of apply() (e.g., lapply(), sapply()). I'm wondering if there is one that does not return anything but just silently apply a function to the list argument. For example, the plot function is applied to each element in 'alist'. It is redundant to return anything from apply. apply(alist,function(x){ plot each element of alist}) __ 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] Is there an variant of apply() that does not return anything?
On Nov 20, 2009, at 10:21 AM, Peng Yu wrote: There are a few version of apply() (e.g., lapply(), sapply()). I'm wondering if there is one that does not return anything but just silently apply a function to the list argument. For example, the plot function is applied to each element in 'alist'. It is redundant to return anything from apply. apply(alist,function(x){ plot each element of alist}) Just use a for() loop. If you are plotting things, the performance bottleneck is not going to be in the loop. Sometimes, we get too anal about avoiding for() loops. HTH, Marc Schwartz __ 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] Is there an variant of apply() that does not return anything?
On Thu, Nov 19, 2009 at 2:21 PM, Peng Yu pengyu...@gmail.com wrote: There are a few version of apply() (e.g., lapply(), sapply()). I'm wondering if there is one that does not return anything but just silently apply a function to the list argument. For example, the plot function is applied to each element in 'alist'. It is redundant to return anything from apply. apply(alist,function(x){ plot each element of alist}) Take a look at the l_ply(), a_ply() and d_ply() functions from Hadley's plyr package. They are a refinement and extension to the apply family of functions and the underscore, _, in the function names indicates that they have no return value. -Charlie __ 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] Is there an variant of apply() that does not return anything?
invisible(apply(...)) On Thu, Nov 19, 2009 at 5:21 PM, Peng Yu pengyu...@gmail.com wrote: There are a few version of apply() (e.g., lapply(), sapply()). I'm wondering if there is one that does not return anything but just silently apply a function to the list argument. For example, the plot function is applied to each element in 'alist'. It is redundant to return anything from apply. apply(alist,function(x){ plot each element of alist}) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there an variant of apply() that does not return anything?
On Thu, Nov 19, 2009 at 4:27 PM, Marc Schwartz marc_schwa...@me.com wrote: On Nov 20, 2009, at 10:21 AM, Peng Yu wrote: There are a few version of apply() (e.g., lapply(), sapply()). I'm wondering if there is one that does not return anything but just silently apply a function to the list argument. For example, the plot function is applied to each element in 'alist'. It is redundant to return anything from apply. apply(alist,function(x){ plot each element of alist}) Just use a for() loop. If you are plotting things, the performance bottleneck is not going to be in the loop. Sometimes, we get too anal about avoiding for() loops. Is there a way to get the name of the list in the loop body? List=list(a='c',b='x',e='q') for(x in List) { print(x) } [1] c [1] x [1] q __ 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] Is there an variant of apply() that does not return anything?
I want to use both the name and the content. Although, I could do the following thing. for(x in names(List)) { do some thing with x do some thing with List[[x]] } However, I'd prefer something like the following if R offers such functionality. But it seems not. for(x in List) { do something with the name of x do something with x } On Fri, Nov 20, 2009 at 4:35 PM, Jorge Ivan Velez jorgeivanve...@gmail.com wrote: If I understood correctly for(x in names(List)) print(x) should do what you asked. HTH, Jorge On Thu, Nov 19, 2009 at 5:31 PM, Peng Yu wrote: On Thu, Nov 19, 2009 at 4:27 PM, Marc Schwartz marc_schwa...@me.com wrote: On Nov 20, 2009, at 10:21 AM, Peng Yu wrote: There are a few version of apply() (e.g., lapply(), sapply()). I'm wondering if there is one that does not return anything but just silently apply a function to the list argument. For example, the plot function is applied to each element in 'alist'. It is redundant to return anything from apply. apply(alist,function(x){ plot each element of alist}) Just use a for() loop. If you are plotting things, the performance bottleneck is not going to be in the loop. Sometimes, we get too anal about avoiding for() loops. Is there a way to get the name of the list in the loop body? List=list(a='c',b='x',e='q') for(x in List) { print(x) } [1] c [1] x [1] q __ 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] Is there an variant of apply() that does not return anything?
On Thu, Nov 19, 2009 at 5:31 PM, Peng Yu pengyu...@gmail.com wrote: Is there a way to get the name of the list in the loop body? This was just discussed! See this thread: https://stat.ethz.ch/pipermail/r-help/2009-November/218919.html __ 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] Change color of single panel in xyplot
Hi there, I have created a single page, multi-panel, xyplot using lattice. Each panel is a trial (total of 8) from an experiment with 3 variables on the x axis and the observed value on the y axis. I have added the mean of all trials so the entire plot looks like this (where M=mean panel and 1-8 = trial panels): M 1 2 3 4 5 6 7 8 I want to change the colour of the mean panel (M) so that it stands out from the trials. I have tried using: col=c(red,rep(black,8)), but this changes the first point to red on all panels. Is there a way to specify the colour for a single panel without changing the rest? Thanks, James -- View this message in context: http://old.nabble.com/Change-color-of-single-panel-in-xyplot-tp26421512p26421512.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] Is there an variant of apply() that does not return anything?
On Nov 19, 2009, at 4:31 PM, Peng Yu wrote: On Thu, Nov 19, 2009 at 4:27 PM, Marc Schwartz marc_schwa...@me.com wrote: On Nov 20, 2009, at 10:21 AM, Peng Yu wrote: There are a few version of apply() (e.g., lapply(), sapply()). I'm wondering if there is one that does not return anything but just silently apply a function to the list argument. For example, the plot function is applied to each element in 'alist'. It is redundant to return anything from apply. apply(alist,function(x){ plot each element of alist}) Just use a for() loop. If you are plotting things, the performance bottleneck is not going to be in the loop. Sometimes, we get too anal about avoiding for() loops. Is there a way to get the name of the list in the loop body? List=list(a='c',b='x',e='q') for(x in List) { print(x) } [1] c [1] x [1] q Here is one approach to give you some insight into how to get the name (as opposed to the object itself) of an argument passed to a function: MyList - list(a = 1, b = 2, e = 3) PlotFn - function(x) { # Get the name of the object passed as 'x' (eg. MyList) Main - deparse(substitute(x)) for (i in seq(along = x)) { plot(x[[i]], main = Main, ylab = names(x[i])) } } # Set to pause between each graphic par(ask = TRUE) PlotFn(MyList) Take note that the main title for each graphic will be MyList and the y axis label for each, will be the name of each of the three components in MyList (eg. a, b and e) HTH, Marc Schwartz __ 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 wait for a user response in Rscript?
You could try ?example and see what the help page says. -Peter Ehlers Peng Yu wrote: Why I don't see 'par(ask=T)' in /usr/lib/R/library/graphics/R-ex/barplot.R? Is 'par(ask=T)' implicitly called by example()? On Mon, Sep 7, 2009 at 10:55 AM, RIOS,ALFREDO ARTURO ar...@ufl.edu wrote: Hi Peng I think this is what you are looking for par(ask=T) Alfredo On Sun Sep 06 12:52:31 EDT 2009, Peng Yu pengyu...@gmail.com wrote: Hi, In 'example(barplot)' running in R, I see 'Hit Return to see next plot:', then R waits for my input. I am wondering how to wait for a user response in Rscript. Regards, Peng __ 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. -- RIOS,ALFREDO ARTURO __ 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] loess smoothing
Hello, In reading the loess description I see: span: the parameter alpha which controls the degree of smoothing. The default seems to be 0.75. Would it be possible to expand on this decription so I can avoid trail and error? Can I increase this pass 'span' 1? Qualitatively to what degree changing this value affects the smoothing of the data? Thank you. Kevin __ 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] optim(.. ,SANN,..)
I have a problem using optim, so I am hoping someone can help me out with it: Suppose I have the list: list(D,R,P) [[1]] V1 V2 V3 V4 1 0 1 0 1 2 1 1 0 0 3 1 0 1 0 4 0 0 1 1 5 0 1 0 1 6 1 1 0 0 7 1 0 1 0 8 0 0 1 1 9 1 0 1 0 10 0 0 1 1 [[2]] [1] 23 85 12 73 23 24 25 56 78 1200 [[3]] V1 1 25 2 80 3 15 4 60 5 56 6 15 7 20 8 70 9 45 10 12000 and the strange function fr which I want to maximise: fr-function(x){y-x*D C-apply(t(apply(y,1,function(c){c[c0]})),1,prod) R2-R*C w-R2[R2P] g-sum(w) return(g)} v-c(1,1,1,1,1,1,1,1,1,1) res - optim(v, fr, method=SANN, control=list(maxit=200, temp=20,v)) I get the following error: Error in prod(..., na.rm = na.rm) : invalid 'type' (list) of argument I'm not sure if its because this function is too disjoint to be opimised using SANN or I am not using optim properly. Any help or suggestions would be appreciated. Thanks __ 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] Change color of single panel in xyplot
This was just answered by Deepayan earlier today: This code did not produce the plot you have linked to. The answer to your question depends on how you created the plot, so you have to tell us that. Changing the color in all panels is easy: histogram(rnorm(100), col = goldenrod) Different colors in different panels is a little more work: histogram(~rnorm(100) | gl(3, 1, 100), mycolors = sample(colors(), 3), panel = function(..., col, mycolors) { panel.histogram(..., col = mycolors[panel.number()]) }) -Deepayan -- David On Nov 19, 2009, at 5:43 PM, jimdare wrote: Hi there, I have created a single page, multi-panel, xyplot using lattice. Each panel is a trial (total of 8) from an experiment with 3 variables on the x axis and the observed value on the y axis. I have added the mean of all trials so the entire plot looks like this (where M=mean panel and 1-8 = trial panels): M 1 2 3 4 5 6 7 8 I want to change the colour of the mean panel (M) so that it stands out from the trials. I have tried using: col=c(red,rep(black,8)), but this changes the first point to red on all panels. Is there a way to specify the colour for a single panel without changing the rest? Thanks, James -- View this message in context: http://old.nabble.com/Change-color-of-single-panel-in-xyplot-tp26421512p26421512.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Passing filenames to the getopt package
Hi all, I've finally started to use Rscript for my statistical scripting needs, and find I'm being blocked by what must be a very simple problem. Specifically, the command lines for my scripts usually contain: (1) the script name, (2) one or more options and their arguments, and finally, (3) one or more filenames to be processed. While the getopt package seems to provide me with excellent option reading functionality, it however, also wants to read the trailing filenames as options as well -- and then fails. Am I missing something obvious? Thanks, -Jim version _ platform i386-apple-darwin8.11.1 arch i386 os darwin8.11.1 system i386, darwin8.11.1 status major 2 minor 9.2 year 2009 month 08 day24 svn rev49384 language R version.string R version 2.9.2 (2009-08-24) -- = Jim Nikelski, Ph.D. Postdoctoral Research Fellow Bloomfield Centre for Research in Aging Lady Davis Institute for Medical Research Sir Mortimer B. Davis - Jewish General Hospital McGill University __ 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 do I change the colour and format for the trelli plot ?
what is that ? Hrishi Mittal wrote: Add col=i in the histogram call. -- View this message in context: http://old.nabble.com/How-do-I-change-the-colour-and-format-for-the-trelli-plot---tp26418382p26421453.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] sqldf
Hello, I would like some help with sqldf syntax. Suppose I have table 1 and table 2. What do I have to do to generate a table with columns 2,5,6 from table 1 (for example), and columns 3,4,5,9 from table 2, but only when values in column 2 from table 1 are equal to values in column 5 from table 2. tnks -- View this message in context: http://old.nabble.com/sqldf-tp26421449p26421449.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 do I change the colour and format for the trelli plot ?
To have a different colour for each histogram. Wasn't that your question? Sorry if I misunderstood. Did you try it? ychu066 wrote: what is that ? Hrishi Mittal wrote: Add col=i in the histogram call. -- View this message in context: http://old.nabble.com/How-do-I-change-the-colour-and-format-for-the-trelli-plot---tp26418382p26421456.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] Tinn-R related problem
As of last night I have this exact same problem, Enter, Backspace and the arrow keys do not work for me in TinnR! I am using windows vista prof on a 64bit laptop. I have uninstalled, re-installed many times and even cleaned the registry etc. Does anyone know of a reason and a fix for this problem? thanks Alison davidseres wrote: David Hewitt wrote: A few weeks ago all of a sudden the backspace, enter and direction keys were not working. I updated Tinn-R to the newest version but still no sollution. After this I tried reinstalling it (prior to that I removed Tinn-R and deleted all the leftovers manually) and still no change. In every other execution (e.g. when I save a file) every key works fine. I've used Tinn-R with R on Win XP ever since I started with R, and I've never had this problem. The only immediate thing that comes to mind is that you should be installing R in SDI mode to get it working with Tinn-R. At least that's what they say, and I've never tried it the other way (MDI). Maybe just uninstall R and Tinn-R, then reload R, use Custom installation and pick SDI, then reinstall Tinn-R. Worth a shot. From what I have read in the other forums I believe this issue is not necessarily R or Tinn-R related but might be some hidden Windows settings (I'm using XP) but of this I'm not sure. If that's the case, I can't help. What occurred a few weeks ago that might have been related? Did you upgrade R? I am using R in the SDI mode. I can't really recall what happened a few weeks ago. There is too much going on on my laptop. But yes, I've upgraded R to 2.6.2 but I don't remember this causing the problem. -- View this message in context: http://old.nabble.com/Tinn-R-related-problem-tp15950714p26421486.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 do I change the colour and format for the trelli plot ?
Fore example my code is histogram(~data[,8]|data[,2], ylab = Frequency, xlab = Score, xlim = c(1,5), ylim = c(0,100),layout = c(3,1),col=data[,2] ) and i want the colour to be depended on the level of the factor in data[,2]. how do i do that ? ychu066 wrote: what is that ? Hrishi Mittal wrote: Add col=i in the histogram call. -- View this message in context: http://old.nabble.com/How-do-I-change-the-colour-and-format-for-the-trelli-plot---tp26418382p26421461.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 do I change the colour and format for the trelli plot ?
I want to change the colour of the histogram in each panel by the levels of the conoditonal factor. I have 3 levels in the factor levels(data[,2])[1] levels(data[,2])[2] levels(data[,2])[3] what can i do , in order to chnage the colour ?? ychu066 wrote: Fore example my code is histogram(~data[,8]|data[,2], ylab = Frequency, xlab = Score, xlim = c(1,5), ylim = c(0,100),layout = c(3,1),col=data[,2] ) and i want the colour to be depended on the level of the factor in data[,2]. how do i do that ? ychu066 wrote: what is that ? Hrishi Mittal wrote: Add col=i in the histogram call. -- View this message in context: http://old.nabble.com/How-do-I-change-the-colour-and-format-for-the-trelli-plot---tp26418382p26421465.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] RE shaping large dataset
I am doing a project which involve reshaping a large dataset, can any of you please sugguest me some good reading/websites/ examples can be in R and SAS Thanks everyone !!! -- View this message in context: http://old.nabble.com/REshaping-large-dataset-tp26421513p26421513.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] sqldf
Google for sql join and see the examples in Example 4 on the sqldf home page: http://code.google.com/p/sqldf/#Example_4._Join On Thu, Nov 19, 2009 at 2:30 PM, JoK LoQ jok...@gmail.com wrote: Hello, I would like some help with sqldf syntax. Suppose I have table 1 and table 2. What do I have to do to generate a table with columns 2,5,6 from table 1 (for example), and columns 3,4,5,9 from table 2, but only when values in column 2 from table 1 are equal to values in column 5 from table 2. tnks -- View this message in context: http://old.nabble.com/sqldf-tp26421449p26421449.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] How do I change the colour and format for the trelli plot ?
On Nov 19, 2009, at 3:10 PM, ychu066 wrote: I want to change the colour of the histogram in each panel by the levels of the conoditonal factor. I have 3 levels in the factor levels(data[,2])[1] levels(data[,2])[2] levels(data[,2])[3] what can i do , in order to chnage the colour ?? Why do you suppose there is a FAQ? Assuming that the underlying levels are coercible to numeric type then: col = as.numeric(as.character(data[ , 2] ) ) and of course: library(fortunes) fortune(dog) -- David ychu066 wrote: Fore example my code is histogram(~data[,8]|data[,2], ylab = Frequency, xlab = Score, xlim = c(1,5), ylim = c(0,100),layout = c(3,1),col=data[,2] ) and i want the colour to be depended on the level of the factor in data[,2]. how do i do that ? ychu066 wrote: what is that ? Hrishi Mittal wrote: Add col=i in the histogram call. -- View this message in context: http://old.nabble.com/How-do-I-change-the-colour-and-format-for-the-trelli-plot---tp26418382p26421465.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to plot mean+/-SD from externally calculated values
Hi, I am trying to plot a set of means+/-SD calculated by an external program (an RDBMS) since the data set is too large to happily fit in R (740M x 100 values - which are summarised to 100 means/SD by the DB). I want to have a mean with whiskers at +/-1SD. Can anyone suggest a way to do this? thanks Dan __ 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 plot mean+/-SD from externally calculated values
On Nov 19, 2009, at 6:33 PM, Dan Kortschak wrote: Hi, I am trying to plot a set of means+/-SD calculated by an external program (an RDBMS) since the data set is too large to happily fit in R (740M x 100 values - which are summarised to 100 means/SD by the DB). I want to have a mean with whiskers at +/-1SD. Can anyone suggest a way to do this? install.packages(plotrix) library(plotrix) ?plotCI -- 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] Finding replacing non-ASCII characters
Hi guys, Are there any feasible methods in searching finding non-ASCII characters in R? For example, from the following object, x - mia. SzaÌmitaÌó The desired output is, x.out - mia. SzaImitaIA Your help in resolving this would be greatly appreciated. [[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.