[R] : Followup Error in library(gplots) : there is no package called 'gplots' : 64bits
Dear James and everybody interested, James'idea was correct : the problem was related to the 64bits version of R. Trying on the same computer a 32bits version of R did not produce this 'gplots' error (but many other conflicts linked to 64 / 32 bits) when installing RODBC package. I will install a 32 Linux version in parallel ... Valère -Ursprüngliche Nachricht- Von: Martin Valere Gesendet: Freitag, 16. April 2010 12:13 An: 'r-help@R-project.org' Betreff: AW: [SPAM] Re: [R] Error in library(gplots) : there is no package called 'gplots' Dear James, i have tried to install the package RODBC1-3.1 with R version 2.9.0 and 2.10.1 (using Opensuse 11.1, 64bits). I have tried install.packages locally (package downloaded and stored locally on the computer) or directly from Internet (using different mirrors !). Same results each time ... Same outcome either if I try to install other packages like for instance e1071. So it appears this is not linked to the package itself but rather with R-base or R-devel (both are installed) ... Regards, Valère -Ursprüngliche Nachricht- Von: james [mailto:ja...@ipec.co.uk] Gesendet: Freitag, 16. April 2010 10:47 An: Martin Valere Cc: R Help List Betreff: [SPAM] Re: [R] Error in library(gplots) : there is no package called 'gplots' Wichtigkeit: Niedrig Hi Vava, What version of R are you using? I'm not sure but I think that R will refuse to install a package in this way if the version of gplots is incompatiable with the version of R you're using. You can check the depends of packages on CRAN. Regards, James Vava wrote: Thanks for your suggestion Tal. Unfortunately, still no luck with me ... still get the usual error message: Error in library(gplots) : there is no package called 'gplots' , whatever I try to install. This is a mystery to me with respect to why /how. I am really stuck with that problem. Best, Valère -Ursprüngliche Nachricht- Von:Martin Valere Gesendet: Donnerstag, 25. März 2010 10:58 An: 'r-help@R-project.org' Betreff:Error in library(gplots) : there is no package called 'gplots' Dear all, I have an issue trying to install new packages (have tried with RODBC_1.3-1, gplots_2.6.1, gtools_2.7.4 packages) and get the same error message : Error in library(gplots) : there is no package called 'gplots' Only clue I have found so far on the Web is related to Perl (Perl modules are installed on my computer, but which one is related to gplots if any ?); no gplots in usr/lib or /usr/lib64 at least ... I am somewhat lost here, having no idea about Perl (if Perl is really the issue ?). I am using OpenSuse 11.1 (64bits); and R version 2.9.0. Installation of package is performed offline as Root. Valère, Switzerland __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Words appear to be bolded in the PDF output
Hi all, I have written a note near each of my graphs using mtext. mtext(text,side=1,line=4,cex=0.5,adj=0) Then I have exported the graphs as a PDF file. pdf(file=name,paper='a4',width=7.27,height=10.69) The mtext appears OK in R. But it looks like it is bolded in the PDF file. http://n4.nabble.com/file/n2016971/graph.png I am not sure if this is actually my monitor/computer's problem. But I want to see if it can be fixed in R. Many thanks, Chris -- View this message in context: http://n4.nabble.com/Words-appear-to-be-bolded-in-the-PDF-output-tp2016971p2016971.html Sent from the R help mailing list archive at Nabble.com. [[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] Exporting an rgl graph
Thanks a lot Both Asymptote and MeshLab work fine. Thanks for this article, Luke. On my particular case, I need to export lines (1D object in a 3D space) and not surfaces (2D objects). Is it possible to draw lines with misc3d ? Christophe Genolini Luke Tierney a écrit : The current issue of JCGS (Vol 18 No 1, http://pubs.amstat.org/toc/jcgs/19/1) has an editorial on including animations, 3D visualizations, and movies in on-line PDF files supporting JCGS articles. The online supplements to the editorial include examples. The 3D examples related to the misc3d packages are also available in http://www.stat.uiowa.edu/~luke/R/misc3d/misc3d-pdf/ http://www.stat.uiowa.edu/%7Eluke/R/misc3d/misc3d-pdf/. At some point the code there will be added to misc3d. It should be possible to adapt these ideas to other objects rendered with rgl. luke __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] glmer with non integer weights
hello, it's the Morisita Horn Index, which is an ecological index for similarity between two multivariate objects (vegetation samples with species and its abundance) where a value of one indicates completely same relative importance of species in both samples and 0 denotes total absence of any same species. it can be expressed as a probability: (prob. that an individual drawn from sample j and one from sample k belong to the same species) - = MH-INdex (prob. that two ind. drawn from either sample will belong to same species) it is also covered in library(vegan);?vegdist here it is its complement: 1-MH, which then is a dissimilarity measure best regards, kay -- View this message in context: http://n4.nabble.com/glmer-with-non-integer-weights-tp1837179p2017027.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] apropos and find
Hello, I'm starting working myself in the use of R, reading M. J. Crawley, The R Book. The problem I do encounter is concerning the commands apropos and find: apropos(edit) Fehler: is.character(what) is not TRUE find(edit) Fehler: is.character(what) is not TRUE I get the same error message typing anything else instead of edit. The command ?edit seems to work well. What could be the reason for that I can't use apopos and find? Thanks for your help! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] local and global variables
Hi all,
I have a question about global and local variables.First of all, a variable
defined in a for loop is it a local or global variable??
Second, I'm trying to build a loop in the following way:
I have these 3 data frames
bilanci_2005-bilanci1[ANNO==2005,]
bilanci_2006-bilanci1[ANNO==2006,]
bilanci_2007 --bilanci1[ANNO==2007,]
LOOP:
v=list(2005,2006,2007)
for (a in v){
anno1=a
anno2=a+1
for (i in 1:length(bilanci_a)) {
assign(paste(bilanci,anno1,anno2,sep=), rbind(paste(bilanci,anno1,anno2,
sep=),
bilanci_anno2[bilanci_anno2$CFISCALE==bilanci_anno1
[i,]$CFISCALE,],
bilanci_anno1[i,])
}
}
what I'm trying to do is for example to give the contents of a (so the
values which are in the list named v) to the bilanci variable. In Stata this
means using a particular kind of .
How can I do this in R?
__
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and provide commented, minimal, self-contained, reproducible code.
[R] how to select the last non-'NA' observation in a row
I have a matrix of the following form: time id0 2 4 6 9 12 14 3 9 8 NA NA NA NA NA 7 3 NA 3 NA 3 NA 4 13 11 6 7 NA 5 NA 6 . I hope for each row to select the last observation which is not 'NA'. For example, for the first row, id=3, the value I want to select is 8 for the second row, id=7, the value I want to select is 4 for the third row, id=13, the value I want to select is 6 if it would be easier for you to demonstrate for me, here I include the code to generate the above three-row matrix: a1=c(9,8,rep(NA,5)) a2=c(3,NA,3,NA,3,NA,4) a3=c(11,6,7,NA,5,NA,6) matrix=rbind(a1,a2,a3) [[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] apropos and find
?apropos says that it expects a character string. Therefore use quotes: apropos(edit) [1] edit file.edit xedit Christian On Tue, 2010-04-20 at 10:24 +0200, Karsten Rincke wrote: Hello, I'm starting working myself in the use of R, reading M. J. Crawley, The R Book. The problem I do encounter is concerning the commands apropos and find: apropos(edit) Fehler: is.character(what) is not TRUE find(edit) Fehler: is.character(what) is not TRUE I get the same error message typing anything else instead of edit. The command ?edit seems to work well. What could be the reason for that I can't use apopos and find? Thanks for your help! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to select the last non-'NA' observation in a row
one way is the following: a1 - c(9,8,rep(NA,5)) a2 - c(3,NA,3,NA,3,NA,4) a3 - c(11,6,7,NA,5,NA,6) M - rbind(a1,a2,a3) ind - !is.na(M) tapply(M[ind], row(M)[ind], tail, 1) I hope it helps. Best, Dimitris On 4/20/2010 10:33 AM, gallon li wrote: I have a matrix of the following form: time id0 2 4 6 9 12 14 3 9 8 NA NA NA NA NA 7 3 NA 3 NA 3 NA 4 13 11 6 7 NA 5 NA 6 . I hope for each row to select the last observation which is not 'NA'. For example, for the first row, id=3, the value I want to select is 8 for the second row, id=7, the value I want to select is 4 for the third row, id=13, the value I want to select is 6 if it would be easier for you to demonstrate for me, here I include the code to generate the above three-row matrix: a1=c(9,8,rep(NA,5)) a2=c(3,NA,3,NA,3,NA,4) a3=c(11,6,7,NA,5,NA,6) matrix=rbind(a1,a2,a3) [[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. -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] apropos and find
On Tue, 2010-04-20 at 10:24 +0200, Karsten Rincke wrote: Hello, I'm starting working myself in the use of R, reading M. J. Crawley, The R Book. The problem I do encounter is concerning the commands apropos and find: apropos(edit) Fehler: is.character(what) is not TRUE find(edit) Fehler: is.character(what) is not TRUE I get the same error message typing anything else instead of edit. From ?apropos Usage: apropos(what, where = FALSE, ignore.case = TRUE, mode = any) find(what, mode = any, numeric = FALSE, simple.words = TRUE) Arguments: what: character string with name of an object, or more generally a regular expression to match against. And edit is not a character string but edit is. Hence the very clear error. The command ?edit seems to work well. From ?? Usage: ?topic type?topic Arguments: topic: Usually, a name or character string specifying the topic for which help is sought. Alternatively, a function call to ask for documentation on a corresponding S4 method: see the section on S4 method documentation. The calls ‘pkg::topic’ and ‘pkg:::topic’ are treated specially, and look for help on ‘topic’ in package ‘pkg’. Hence `?` allows a wider range of object types for it's argument topic. What could be the reason for that I can't use apopos and find? You aren't providing a character a string. Not sure where you got the usage you are using from, but it is wrong. The docs are quite clear here. HTH G Thanks for your help! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] local and global variables
On 20/04/2010 4:27 AM, n.via...@libero.it wrote:
Hi all,
I have a question about global and local variables.First of all, a variable
defined in a for loop is it a local or global variable??
R doesn't create a local scope for for loops. So it will have the
same scope as other variables created in the same scope.
Second, I'm trying to build a loop in the following way:
I have these 3 data frames
bilanci_2005-bilanci1[ANNO==2005,]
bilanci_2006-bilanci1[ANNO==2006,]
bilanci_2007 --bilanci1[ANNO==2007,]
LOOP:
v=list(2005,2006,2007)
I would recommend using c(2005,2006, 2007) rather than list(), but I
don't think that has much effect.
for (a in v){
anno1=a
anno2=a+1
for (i in 1:length(bilanci_a)) {
Using the name of one variable as part of another name won't cause it to
be magically expanded. You should have set things up differently from
the beginning:
bilanci_subsets - list(2005 = bilanci1[ANNO==2005,],
2006 = bilanci1[ANNO==2006,],
2007 = bilanci1[ANNO==2007,] )
Then you can use
for (i in names(bilanci_subsets)) {
bilanci_subsets[[i]]
Duncan Murdoch
assign(paste(bilanci,anno1,anno2,sep=), rbind(paste(bilanci,anno1,anno2,
sep=),
bilanci_anno2[bilanci_anno2$CFISCALE==bilanci_anno1
[i,]$CFISCALE,],
bilanci_anno1[i,])
}
}
what I'm trying to do is for example to give the contents of a (so the
values which are in the list named v) to the bilanci variable. In Stata this
means using a particular kind of .
How can I do this in R?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Words appear to be bolded in the PDF output
On Mon, 19-Apr-2010 at 10:54PM -0800, chrisli1223 wrote:
|
| Hi all,
|
| I have written a note near each of my graphs using mtext.
| mtext(text,side=1,line=4,cex=0.5,adj=0)
|
| Then I have exported the graphs as a PDF file.
| pdf(file=name,paper='a4',width=7.27,height=10.69)
That seems backwards to me. I'd begin with the pdf() call, then do the
plot, then dev.off(). Maybe that's what you meant: if not, I've no
idea what you did.
|
| The mtext appears OK in R. But it looks like it is bolded in the PDF file.
| http://n4.nabble.com/file/n2016971/graph.png
That's a fairly low res png file, not a pdf file, but even so, it
doesn't look like font = 2 to me. It's a considerably larger cex
value than the other text, but it's still not bold.
|
| I am not sure if this is actually my monitor/computer's
| problem. But I want to see if it can be fixed in R.
Decide after seeing what it looks like printed.
|
| Many thanks,
| Chris
|
| --
| View this message in context:
http://n4.nabble.com/Words-appear-to-be-bolded-in-the-PDF-output-tp2016971p2016971.html
| Sent from the R help mailing list archive at Nabble.com.
|
| [[alternative HTML version deleted]]
|
| __
| R-help@r-project.org mailing list
| https://stat.ethz.ch/mailman/listinfo/r-help
| PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
| and provide commented, minimal, self-contained, reproducible code.
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_ Average minds discuss events
(:_~*~_:) Small minds discuss people
(_)-(_) . Eleanor Roosevelt
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
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Re: [R] Words appear to be bolded in the PDF output
Hi, Taking a wild guess, it looks to me that you might have overlaid several times the same text, plot.new() text(0.5,0.5,rep(test,10)) HTH, baptiste On 20 April 2010 08:54, chrisli1223 chri...@austwaterenv.com.au wrote: Hi all, I have written a note near each of my graphs using mtext. mtext(text,side=1,line=4,cex=0.5,adj=0) Then I have exported the graphs as a PDF file. pdf(file=name,paper='a4',width=7.27,height=10.69) The mtext appears OK in R. But it looks like it is bolded in the PDF file. http://n4.nabble.com/file/n2016971/graph.png I am not sure if this is actually my monitor/computer's problem. But I want to see if it can be fixed in R. Many thanks, Chris -- View this message in context: http://n4.nabble.com/Words-appear-to-be-bolded-in-the-PDF-output-tp2016971p2016971.html Sent from the R help mailing list archive at Nabble.com. [[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] problem with retreaving column from a data.frame
Hi all, I have a problem with retreaving column from a data.frame that is I have a data.frame named temp in that dataframe some column are there whose names are IDENTIFIERUNIQUEID TRIALGROUPSID GRPNUMBER GRPDESC SEXDIDSTDID STUDYTYPE SDID DOSDIDSOPDIDSPECIES i can retreave any column with using command like temp$STDID but my problem is the column name i have to get from another xml its comming like STDID that one i assigned to a variable namedkk like kk-STDID but now while i am trying to retreave this column from dataframe temp like temp$kk or with using command paste like paste(temp,kk,sep$) both the times its giving null result but its having values in table can any one help me how to get column values from dataframe like these cases thanks in advance kiran [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to set proxy settings for R
Dear Pete, Thanks, it works now! I did as you suggested: --internet flag to the target line (right click, properties) e.g. C:\Program Files\R\R-2.8.1\bin\Rgui.exe --internet2 Strangely enough, I can now easly download packages but I still get these messages: chooseCRANmirror() Warning message: In open.connection(con, r) : cannot open: HTTP status was '407 Proxy Authentication Required' utils:::menuInstallPkgs() Warning: unable to access index for repository http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.10 ..but eventually it does install what I want...any clue? Alessandra -- View this message in context: http://n4.nabble.com/How-to-set-proxy-settings-for-R-tp2016158p2017172.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to draw multiple vertical bands
On 04/15/2010 12:36 AM, senne wrote: hi R gurus I saw some graphs with vertical band like this one: http://pragcap.com/wp-content/uploads/2010/04/GS.png how to draw the blue band in R, can't find any clue to do this,any ideas? See the source code for nber.xy and nberShade.ggplot in the tis package by Jeff Hallman. They both do what you want. I'm working on something similar. Besides drawing the bands, my version will also vary heights optionally (rather than have bands stretching the entire height of the graph) and use ggplot2. Marsh Feldman __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] log-linear regression question
I am trying to estimate a demand function: Y=K * X1^b1 * X2^b2 * X3 ^(-1-b1) in log form: ln Y = ln K + b1 ln X1 + b2 ln X2 + (-1-b1) ln X3 As the regression coefficients are related for 2 of the regressors, I am not sure of the appropriate methodology or function in R to handle this. Any hints? thx, Tarun [[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] Effective distance measures for Text Clustering
Hi useRs, Disclaimer: My question is more statistical than pertaining specifically to the R system) I am using the tm package in R to create a Document-Term Matrix, with Tf-Idf measures. A) Once done, I create a distance matrix using euclidean distance measure. B) After this, I use hierarchical clustering to find an appropriate separation in the data using ward measure For A above, what are generally the best practices for distance measures on TfIdf. I used the cosine similarity measure, but that creates NaN/Inf values which have to be converted to zero. For B above, I used ward since the Details alluded to it being the most used measure which provides better results. I understand that such a question requires extensive research since the underlying data (emails in my case) may have a great influence on the results. I have used a Part of Speech tagger to extract nouns as features to use as the dictionary in order to weed out trivial words. Any feedback/link to online knowledge resources/your experience would be greatly appreciated. Thank you for your time. Regards, Harsh Singhal __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] multiple plots problem
hello, i try to plot 3 graphs which have the same x.axis underneath each other. i'd like the plots to be aligned without margings between the boxes and draw a single x axis beneath the lowest plot. i managed to get the alignment by setting par(mar), but the middle box is stretched and i cant't figure out how to get around this. par(pin) was my guess, but this doesn't do the job because it seems to be overridden by par(mar). sorry for the long example data, but this was quick at hand.. mydata: ### ad-data.frame( list(structure(list(pos = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), .Label = c(Gaps, Surroundings), class = factor), stage = structure(c(4L, 2L, 3L, 1L, 4L, 2L, 3L, 1L), .Label = c(Alpine_Grassl., Early, Late, Pioneer), class = factor), mw = c(12.06293707, 26.09265735, 37.12287713, 81.53846154, 28.88005449, 68.44004261, 97.61124961, 100)), .Names = c(pos, stage, mw), class = data.frame, row.names = c(NA, -8L seedl-data.frame( list(structure(list(pos = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), .Label = c(Gaps, Surroundings), class = factor), stage = structure(c(4L, 2L, 3L, 1L, 4L, 2L, 3L, 1L), .Label = c(Alpine_Grassl., Early, Late, Pioneer), class = factor), mw = c(24.52797203, 17.20571096, 41.68401043, 34.03846154, 13.92056379, 9.955734802, 23.54039945, 26.11543325)), .Names = c(pos, stage, mw ), class = data.frame, row.names = c(NA, -8L rich-data.frame( list(structure(list(pos = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), .Label = c(Gaps, Surroundings), class = factor), stage = structure(c(4L, 2L, 3L, 1L, 4L, 2L, 3L, 1L), .Label = c(Alp._Grassl., Early, Late, Pioneer), class = factor), n = c(44L, 44L, 45L, 10L, 44L, 44L, 45L, 10L), mw = c(1.93, 3.59, 4.24, 6.8, 1.84, 5.61, 7.2, 10.6)), .Names = c(pos, stage, n, mw), class = data.frame, row.names = c(NA, -8L ### windows(5,15) par(ps=8,mgp=c(2.25,1,0),mfrow=c(3,1)) ### par(mar=c(0,4,4,2)) attach(seedl) plot(c(1:4),mw[1:4],type=o,xaxt=n,xlab=, ylab=Abundance (%),ylim=c(0,100),xlim=c(0.7,4.3),pch=22,bg=white,cex=0.75,las=1) par(new=T) plot(c(1:4),mw[5:8],type=o,xaxt=n,xlab=,yaxt=n, ylab=,ylim=c(0,100),xlim=c(0.7,4.3),pch=15,cex=0.75) text(0.7,90,adj=0,Seedlings,font=2) legend(topright,c(Surr.,Gaps),bty=n,pch=c(15,0)) ### par(ps=8,mgp=c(2.25,1,0),mar=c(0,4,0,2)) attach(ad) plot(c(1:4),mw[1:4],type=o,xaxt=n,xlab=, ylab=Abundance (%),ylim=c(0,100),xlim=c(0.7,4.3),pch=22,bg=white,cex=0.75,las=1) par(new=T) plot(c(1:4),mw[5:8],type=o,xaxt=n,xlab=,yaxt=n, ylab=,ylim=c(0,100),xlim=c(0.7,4.3),pch=15,cex=0.75) text(0.7,90,adj=0,Adults,font=2) ### par(ps=8,mgp=c(2.25,1,0),mar=c(4,4,0,2)) attach(rich) plot(c(1:4),mw[1:4],type=o,xaxt=n,xlab=, ylab=Abundance (%),ylim=c(0,12),xlim=c(0.7,4.3),pch=22,bg=white,cex=0.75,las=1) par(new=T) plot(c(1:4),mw[5:8],type=o,xaxt=n,xlab=,yaxt=n, ylab=,ylim=c(0,12),xlim=c(0.7,4.3),pch=15,cex=0.75) axis(1,at=c(1:4),labels=c(rep(,4))) axis(1,at=c(1:4),line=1.2,lwd=0,adj=0,hadj=0.5,padj=0.5,las=2, labels=c(Pioneer,Early,Late,Alpine\nGrassland)) text(0.7,11,adj=0,Richness,font=2) ### dev.off() -- View this message in context: http://n4.nabble.com/multiple-plots-problem-tp2017326p2017326.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to set proxy settings for R
Hi Alessandra Did you restart R? You should be able to download packages using the Packages Install Package(s) ... from the menu bar in the GUI. Once the package has been downloaded select Packages Load Package ... you should see what you just downloaded in a list of available packages Pete From: danda [via R] [mailto:ml-node+2017172-614719349-63...@n4.nabble.com] Sent: Tuesday, April 20, 2010 5:02 AM To: Brecknock, Peter Subject: Re: How to set proxy settings for R Dear Pete, Thanks, it works now! I did as you suggested: --internet flag to the target line (right click, properties) e.g. C:\Program Files\R\R-2.8.1\bin\Rgui.exe --internet2 Strangely enough, I can now easly download packages but I still get these messages: chooseCRANmirror() Warning message: In open.connection(con, r) : cannot open: HTTP status was '407 Proxy Authentication Required' utils:::menuInstallPkgs() Warning: unable to access index for repository http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.10 ..but eventually it does install what I want...any clue? Alessandra View message @ http://n4.nabble.com/How-to-set-proxy-settings-for-R-tp2016158p2017172.h tml To unsubscribe from Re: How to set proxy settings for R, click here (link removed) ub2NrQGJwLmNvbXwyMDE2NTExfC0xNDg0OTk1MTQx . -- View this message in context: http://n4.nabble.com/How-to-set-proxy-settings-for-R-tp2016158p2017327.html Sent from the R help mailing list archive at Nabble.com. [[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] multiple variables pointing to single dataframe?
That's great, that will work perfectly..
Thanks All!
-Alex
-Original Message-
From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Sent: Monday, April 19, 2010 10:34 AM
To: Alex Bryant
Cc: r-help@r-project.org
Subject: Re: [R] multiple variables pointing to single dataframe?
If you only need to retrieve x by referring to x2 and you don`t have
to modify x via x2 then this works:
x - Orange
makeActiveBinding(x2, function() x, .GlobalEnv)
x$age - 50
head(x2)
Tree age circumference
11 5030
21 5058
31 5087
41 50 115
51 50 120
61 50 142
On Mon, Apr 19, 2010 at 10:15 AM, Alex Bryant abry...@i-review.com wrote:
Hi, for example:
x - Orange
x2 - x
x[1,]$age - 50
x2[1,]
Tree age circumference
1 1 118 30
I would like a way for x2 to also reference the modified x data frame without
having to reassign x2x each time x is modified.
Thanks,
Alex
-Original Message-
From: Petr PIKAL [mailto:petr.pi...@precheza.cz]
Sent: Monday, April 19, 2010 3:18 AM
To: Alex Bryant
Cc: r-help@r-project.org
Subject: Odp: [R] multiple variables pointing to single dataframe?
Hi
r-help-boun...@r-project.org napsal dne 16.04.2010 16:15:40:
Hi, I have a need to have 2 variables point to the same dataframe (d1),
I
What does it mean to point to data frame? Seems to me that it is something
from C+.
You can reference data frame by $ or by square brackets with as many
variables as you want.
see
?[
regards
Petr
don't want to simply copy the dataframe ( d2-d1 ) as my understanding
is that
this will create a second dataframe. Any suggestions on best practice
here?
Thank You,
//
// Alex Bryant
// Software Developer
// Integrated Clinical Systems, Inc.
// 908-996-7208
Confidentiality Note: This e-mail, and any attachment
to...{{dropped:13}}
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Re: [R] how to draw multiple vertical bands
Hi you can also use abline function x-rnorm(10) plot(1:10, x) abline(v=seq(2,10,2), lwd=50, col=lightblue) lines(1:10, x, type=b) Regards Petr r-help-boun...@r-project.org napsal dne 19.04.2010 20:59:15: On 04/15/2010 12:36 AM, senne wrote: hi R gurus I saw some graphs with vertical band like this one: http://pragcap.com/wp-content/uploads/2010/04/GS.png how to draw the blue band in R, can't find any clue to do this,any ideas? See the source code for nber.xy and nberShade.ggplot in the tis package by Jeff Hallman. They both do what you want. I'm working on something similar. Besides drawing the bands, my version will also vary heights optionally (rather than have bands stretching the entire height of the graph) and use ggplot2. Marsh Feldman __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] log-linear regression question
open up the brackets (-1-b1) and you would get an equation in which you would have to estimate only b1 and b2... then you can proceed by the normal method.. On Tue, Apr 20, 2010 at 4:44 PM, rajiv guha dabr...@gmail.com wrote: I am trying to estimate a demand function: Y=K * X1^b1 * X2^b2 * X3 ^(-1-b1) in log form: ln Y = ln K + b1 ln X1 + b2 ln X2 + (-1-b1) ln X3 As the regression coefficients are related for 2 of the regressors, I am not sure of the appropriate methodology or function in R to handle this. Any hints? thx, Tarun [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help: coxph() in {survival} package
The best answer to your question is help(coxph.object) This is a manual page describing the contents of a coxph fit. Terry Therneau __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 retreaving column from a data.frame
Maybe you want something like this. temp-data.frame(id=rbinom(10,5,.5),grp=rbinom(10,3,.3),stdid=rnorm(10)) kk-stdid temp[,kk] -tgs On Tue, Apr 20, 2010 at 7:32 AM, venkata kirankumar kiran4u2...@gmail.comwrote: Hi all, I have a problem with retreaving column from a data.frame that is I have a data.frame named temp in that dataframe some column are there whose names are IDENTIFIERUNIQUEID TRIALGROUPSID GRPNUMBER GRPDESC SEXDIDSTDID STUDYTYPE SDID DOSDIDSOPDIDSPECIES i can retreave any column with using command like temp$STDID but my problem is the column name i have to get from another xml its comming like STDID that one i assigned to a variable namedkk like kk-STDID but now while i am trying to retreave this column from dataframe temp like temp$kk or with using command paste like paste(temp,kk,sep$) both the times its giving null result but its having values in table can any one help me how to get column values from dataframe like these cases thanks in advance kiran [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] the bar width of barchart plot in lattice package
Dear R users,
I am trying to use the following code to make a barchar plot. The bars in
the plot turn out to be a little narrow. Is there any way to modify the
width of the bars? Thank you!
library(lattice)
scores = gl(2, 5, label=c('Sensitivity', 'PPV'), length = 100)
sequences = gl(5, 1, label=c('Lemna minor', 'Dugesia japonica A',
'Gymnosporangium sabinae', 'Hymeniacidon sanguinea', 'Streptomyces
griseus'), length = 100)
levels = gl(10, 10, label = c('Cycle 1', 'Cycle 2', 'Cycle 3', 'Cycle 4',
'Order 1', 'Order 2', 'Order 3', 'MaxPairs = 20', 'MaxPairs = Average
Length', 'MaxPairs = 500'))
factors = c(rep('Cycles', 40), rep('Order', 30), rep('MaxPairs', 30))
values = rnorm(100) # this is toy data
a = data.frame(values, scores, sequences, levels, factors)
bc.factors =
barchart(values ~ sequences | scores * factors , data = a,
groups = levels,
layout = c(2,3),
between = list(y=0.5),
clip = list(strip = 'off'),
par.strip.text = list(cex=0.7),
par.settings = list(fontsize=list(text=8)),
auto.key = list(rectangles = TRUE, space = 'right', columns = 1),
draw.key = TRUE,
scales = list(x = list(rot = 45)))
--
Best,
Zhenjiang
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Re: [R] multiple plots problem
The stretching of plot two occurs because you are allotting more space for plot two. You set Plot 1: mar=c(0,4,4,2) Plot 2: mar=c(0,4,0,2) Plot 3: mar=c(4,4,0,2) In plot one your are dedicating 4 lines to the top margin, in plot three you are dedicating 4 lines to the bottom margin. In plot two, you are not dedicating any lines to either the top or bottom margin. It follows that plot 2 is 4 lines taller than the other plots. The cheapest solution is: Before the plots: par(oma=c(4,0,4,0)) Plot 1: mar=c(0,4,0,2) Plot 2: mar=c(0,4,0,2) Plot 3: mar=c(0,4,0,2) -tgs On Tue, Apr 20, 2010 at 8:58 AM, Kay Cichini kay.cich...@uibk.ac.at wrote: hello, i try to plot 3 graphs which have the same x.axis underneath each other. i'd like the plots to be aligned without margings between the boxes and draw a single x axis beneath the lowest plot. i managed to get the alignment by setting par(mar), but the middle box is stretched and i cant't figure out how to get around this. par(pin) was my guess, but this doesn't do the job because it seems to be overridden by par(mar). sorry for the long example data, but this was quick at hand.. mydata: ### ad-data.frame( list(structure(list(pos = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), .Label = c(Gaps, Surroundings), class = factor), stage = structure(c(4L, 2L, 3L, 1L, 4L, 2L, 3L, 1L), .Label = c(Alpine_Grassl., Early, Late, Pioneer), class = factor), mw = c(12.06293707, 26.09265735, 37.12287713, 81.53846154, 28.88005449, 68.44004261, 97.61124961, 100)), .Names = c(pos, stage, mw), class = data.frame, row.names = c(NA, -8L seedl-data.frame( list(structure(list(pos = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), .Label = c(Gaps, Surroundings), class = factor), stage = structure(c(4L, 2L, 3L, 1L, 4L, 2L, 3L, 1L), .Label = c(Alpine_Grassl., Early, Late, Pioneer), class = factor), mw = c(24.52797203, 17.20571096, 41.68401043, 34.03846154, 13.92056379, 9.955734802, 23.54039945, 26.11543325)), .Names = c(pos, stage, mw ), class = data.frame, row.names = c(NA, -8L rich-data.frame( list(structure(list(pos = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), .Label = c(Gaps, Surroundings), class = factor), stage = structure(c(4L, 2L, 3L, 1L, 4L, 2L, 3L, 1L), .Label = c(Alp._Grassl., Early, Late, Pioneer), class = factor), n = c(44L, 44L, 45L, 10L, 44L, 44L, 45L, 10L), mw = c(1.93, 3.59, 4.24, 6.8, 1.84, 5.61, 7.2, 10.6)), .Names = c(pos, stage, n, mw), class = data.frame, row.names = c(NA, -8L ### windows(5,15) par(ps=8,mgp=c(2.25,1,0),mfrow=c(3,1)) ### par(mar=c(0,4,4,2)) attach(seedl) plot(c(1:4),mw[1:4],type=o,xaxt=n,xlab=, ylab=Abundance (%),ylim=c(0,100),xlim=c(0.7,4.3),pch=22,bg=white,cex=0.75,las=1) par(new=T) plot(c(1:4),mw[5:8],type=o,xaxt=n,xlab=,yaxt=n, ylab=,ylim=c(0,100),xlim=c(0.7,4.3),pch=15,cex=0.75) text(0.7,90,adj=0,Seedlings,font=2) legend(topright,c(Surr.,Gaps),bty=n,pch=c(15,0)) ### par(ps=8,mgp=c(2.25,1,0),mar=c(0,4,0,2)) attach(ad) plot(c(1:4),mw[1:4],type=o,xaxt=n,xlab=, ylab=Abundance (%),ylim=c(0,100),xlim=c(0.7,4.3),pch=22,bg=white,cex=0.75,las=1) par(new=T) plot(c(1:4),mw[5:8],type=o,xaxt=n,xlab=,yaxt=n, ylab=,ylim=c(0,100),xlim=c(0.7,4.3),pch=15,cex=0.75) text(0.7,90,adj=0,Adults,font=2) ### par(ps=8,mgp=c(2.25,1,0),mar=c(4,4,0,2)) attach(rich) plot(c(1:4),mw[1:4],type=o,xaxt=n,xlab=, ylab=Abundance (%),ylim=c(0,12),xlim=c(0.7,4.3),pch=22,bg=white,cex=0.75,las=1) par(new=T) plot(c(1:4),mw[5:8],type=o,xaxt=n,xlab=,yaxt=n, ylab=,ylim=c(0,12),xlim=c(0.7,4.3),pch=15,cex=0.75) axis(1,at=c(1:4),labels=c(rep(,4))) axis(1,at=c(1:4),line=1.2,lwd=0,adj=0,hadj=0.5,padj=0.5,las=2, labels=c(Pioneer,Early,Late,Alpine\nGrassland)) text(0.7,11,adj=0,Richness,font=2) ### dev.off() -- View this message in context: http://n4.nabble.com/multiple-plots-problem-tp2017326p2017326.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. [[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] problem with retreaving column from a data.frame
On Apr 20, 2010, at 7:32 AM, venkata kirankumar wrote: Hi all, I have a problem with retreaving column from a data.frame that is I have a data.frame named temp in that dataframe some column are there whose names are IDENTIFIERUNIQUEID TRIALGROUPSID GRPNUMBER GRPDESC SEXDIDSTDID STUDYTYPE SDID DOSDIDSOPDIDSPECIES i can retreave any column with using command like temp$STDID but my problem is the column name i have to get from another xml its comming like STDID that one i assigned to a variable namedkk like kk-STDID but now while i am trying to retreave this column from dataframe temp like temp$kk or with using command paste like paste(temp,kk,sep$) both the times its giving null result but its having values in table Don't use the $ operator for such tasks. Instead use the indexing operator [: temp[ , kk] testdf - data.frame(aa=letters[1:10], bb=LETTERS[1:10], cc=1:10) colname - aa testdf[ , colname] [1] a b c d e f g h i j Levels: a b c d e f g h i j -- David. can any one help me how to get column values from dataframe like these cases thanks in advance kiran [[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. David Winsemius, MD 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] log-linear regression question
I think you'll need the offset option. ln( Y ) = ln( K ) + b1 ln( X1 ) + b2 ln( X2 / X3 ) + 1 ln( 1 / X3 ) as in: glm( log(Y) ~ log( X1 ) + I(log( X2 / X3 )), offset=I(log( 1 / X3 ))) -tgs On Tue, Apr 20, 2010 at 7:14 AM, rajiv guha dabr...@gmail.com wrote: I am trying to estimate a demand function: Y=K * X1^b1 * X2^b2 * X3 ^(-1-b1) in log form: ln Y = ln K + b1 ln X1 + b2 ln X2 + (-1-b1) ln X3 As the regression coefficients are related for 2 of the regressors, I am not sure of the appropriate methodology or function in R to handle this. Any hints? thx, Tarun [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] the bar width of barchart plot in lattice package
On Apr 20, 2010, at 9:46 AM, zhenjiang xu wrote:
Dear R users,
I am trying to use the following code to make a barchar plot. The
bars in
the plot turn out to be a little narrow. Is there any way to modify
the
width of the bars? Thank you!
library(lattice)
scores = gl(2, 5, label=c('Sensitivity', 'PPV'), length = 100)
sequences = gl(5, 1, label=c('Lemna minor', 'Dugesia japonica A',
'Gymnosporangium sabinae', 'Hymeniacidon sanguinea', 'Streptomyces
griseus'), length = 100)
levels = gl(10, 10, label = c('Cycle 1', 'Cycle 2', 'Cycle 3',
'Cycle 4',
'Order 1', 'Order 2', 'Order 3', 'MaxPairs = 20', 'MaxPairs = Average
Length', 'MaxPairs = 500'))
factors = c(rep('Cycles', 40), rep('Order', 30), rep('MaxPairs', 30))
values = rnorm(100) # this is toy data
a = data.frame(values, scores, sequences, levels, factors)
bc.factors =
barchart(values ~ sequences | scores * factors , data = a,
groups = levels,
layout = c(2,3),
between = list(y=0.5),
clip = list(strip = 'off'),
par.strip.text = list(cex=0.7),
par.settings = list(fontsize=list(text=8)),
auto.key = list(rectangles = TRUE, space = 'right',
columns = 1),
draw.key = TRUE,
scales = list(x = list(rot = 45)))
?barchart
Looking at the arguments to barchart in the help page I would have
guessed that box.ratio would do what you want. Since that is clearly
not reproducible code , (in the absence of test dataset of the
appropriate structure) I suppose guessing will remain the level of my
knowledge in this instance.
--
Best,
Zhenjiang
David Winsemius, MD
West Hartford, CT
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Re: [R] Drawing a line with misc3d
misc3d only knows about triangular mesh scenes. For our JSS paper we added pointsTetrahedra that represents data points as small terahedra. You could do something analogous by creating a triangular mesh representation of a long thin bar for your lines. luke On Mon, 19 Apr 2010, Christophe Genolini wrote: Hi the list, I would like to draw some lines with misc3d. I find a lot of tools to draw surfaces, but nothing for simple line... Is it possible? Note that I know that it is possible to draw lines with rgl (using lines3d), but I need to do it with misc3d to export the drawing in .asy format. Any solution? Christophe __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Luke Tierney Chair, Statistics and Actuarial Science Ralph E. Wareham Professor of Mathematical Sciences University of Iowa Phone: 319-335-3386 Department of Statistics andFax: 319-335-3017 Actuarial Science 241 Schaeffer Hall email: l...@stat.uiowa.edu Iowa City, IA 52242 WWW: http://www.stat.uiowa.edu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Exporting an rgl graph
On Fri, 16 Apr 2010, Michael Friendly wrote: l...@stat.uiowa.edu wrote: The current issue of JCGS (Vol 18 No 1, http://pubs.amstat.org/toc/jcgs/19/1) has an editorial on including animations, 3D visualizations, and movies in on-line PDF files supporting JCGS articles. The online supplements to the editorial include examples. The 3D examples related to the misc3d packages are also available in http://www.stat.uiowa.edu/~luke/R/misc3d/misc3d-pdf/. At some point the code there will be added to misc3d. It should be possible to adapt these ideas to other objects rendered with rgl. luke Luke, Your misc3d-pdf example is very instructive and the .tex file shows how to embed in LaTeX. Thanks! (JCGS 19(1) is actually one of the nicest issues in a long time.) Of the two approaches you describe, the Asymptote route seems easier and more capable than the MeshLab one. The Asymptote/PRC route was the only one I could find (with a limited amoutn of time and effort I could put in) that would support both color and transparency. The downside is that PRC suport requires very new Adobe readers and seems to result in huge files. I know the U3D format support color but MeshLab doesn't seem to put color into its U3D exports. I forget whether U3D supports transparency. Someone with the energy and motivation to do so can read the binary file format specs and write these file formats directly usign alltheir cababilities, but I wasn't up to doing that at the time. It would be particularly useful to have this capability available for rgl. Any plans for this? Not on my part. misc3d scenes are very simple -- just triangular mesh objects with optional color or transparency. rgl handles much richer scenes so figuring out how to translate such scenes to one of the binary formats would be a lot more work. On the other hand it may already have been done in the OpenGL community. One note: With Adobe Acrobat Pro 9.3.1, the U3D and PRC images display on screen, but do not print (replaced by the filename). Is this your experience too? I believe so. There may well be a way of including a static image in the LaTeX that would be used by printing and readers that don't understand the embedded formats, but I haven't had the chance to check the movie15 documentation for that. luke -Michael -- Luke Tierney Chair, Statistics and Actuarial Science Ralph E. Wareham Professor of Mathematical Sciences University of Iowa Phone: 319-335-3386 Department of Statistics andFax: 319-335-3017 Actuarial Science 241 Schaeffer Hall email: l...@stat.uiowa.edu Iowa City, IA 52242 WWW: http://www.stat.uiowa.edu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] QCA3 segfault
Hi I have just dwonloaded QCA3 onto this machine (ubuntu, karmic, amd64) and a mac. The examples run fine on teh mac, but crashed R on ubuntu. Any help much apprecia\ted. Thanks Richard -- Terminal log: conditions - c(Developed.FZ,Urban.FZ,Literate.FZ,Industrial.FZ, Stable.FZ) reduce(mydata=Lipset_fs,Survived.FZ,conditions,explain=positive,remaind=exclude,prepro=fs,consistency=0.7) *** caught segfault *** address 0x4, cause 'memory not mapped' Traceback: 1: .C(lpslink, direction = as.integer(direction), x.count = as.integer(x.count), objective = as.double(objective), const.count = as.integer(const.count), constraints = as.double(constraints), int.count = as.integer(int.count), int.vec = as.integer(int.vec), bin.count = as.integer(bin.count), binary.vec = as.integer(binary.vec), num.bin.solns = as.integer(num.bin.solns), objval = as.double(objval), solution = as.double(solution), presolve = as.integer(presolve), compute.sens = as.integer(compute.sens), sens.coef.from = as.double(sens.coef.from), sens.coef.to = as.double(sens.coef.to), duals = as.double(duals), duals.from = as.double(duals.from), duals.to = as.double(duals.to), scale = as.integer(scale), use.dense = as.integer(use.dense), dense.col = as.integer(dense.col), dense.val = as.double(dense.val), dense.const.nrow = as.integer(dense.const.nrow), dense.ctr = as.integer(dense.ctr), use.rw = as.integer(use.rw), tmp = as.character(tmp), status = as.integer(status), PACKAGE = lpSolve) 2: lpSolve:::lp(direction = min, objective.in = rep(1, nrow(PIChart)), const.mat = t(PIChart), const.dir = =, 1, all.bin = TRUE) 3: solvePIChart(PIChart) 4: reduce.default(mydata = Lipset_fs, Survived.FZ, conditions, explain = positive, remaind = exclude, prepro = fs, consistency = 0.7) 5: reduce(mydata = Lipset_fs, Survived.FZ, conditions, explain = positive, remaind = exclude, prepro = fs, consistency = 0.7) Possible actions: 1: abort (with core dump, if enabled) 2: normal R exit 3: exit R without saving workspace 4: exit R saving workspace Selection: -- Sys.getlocale() [1] LC_CTYPE=en_GB.UTF-8;LC_NUMERIC=C;LC_TIME=en_GB.UTF-8;LC_COLLATE=en_GB.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_GB.UTF-8;LC_PAPER=en_GB.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_GB.UTF-8;LC_IDENTIFICATION=C -- View this message in context: http://n4.nabble.com/QCA3-segfault-tp2017493p2017493.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] Exporting an rgl graph
On 20/04/2010 10:43 AM, l...@stat.uiowa.edu wrote: On Fri, 16 Apr 2010, Michael Friendly wrote: l...@stat.uiowa.edu wrote: The current issue of JCGS (Vol 18 No 1, http://pubs.amstat.org/toc/jcgs/19/1) has an editorial on including animations, 3D visualizations, and movies in on-line PDF files supporting JCGS articles. The online supplements to the editorial include examples. The 3D examples related to the misc3d packages are also available in http://www.stat.uiowa.edu/~luke/R/misc3d/misc3d-pdf/. At some point the code there will be added to misc3d. It should be possible to adapt these ideas to other objects rendered with rgl. luke Luke, Your misc3d-pdf example is very instructive and the .tex file shows how to embed in LaTeX. Thanks! (JCGS 19(1) is actually one of the nicest issues in a long time.) Of the two approaches you describe, the Asymptote route seems easier and more capable than the MeshLab one. The Asymptote/PRC route was the only one I could find (with a limited amoutn of time and effort I could put in) that would support both color and transparency. The downside is that PRC suport requires very new Adobe readers and seems to result in huge files. I know the U3D format support color but MeshLab doesn't seem to put color into its U3D exports. I forget whether U3D supports transparency. Someone with the energy and motivation to do so can read the binary file format specs and write these file formats directly usign alltheir cababilities, but I wasn't up to doing that at the time. It would be particularly useful to have this capability available for rgl. Any plans for this? Not on my part. misc3d scenes are very simple -- just triangular mesh objects with optional color or transparency. rgl handles much richer scenes so figuring out how to translate such scenes to one of the binary formats would be a lot more work. On the other hand it may already have been done in the OpenGL community. It's on my wish list, but I'm spending too much time fighting to get my email working to actually work on anything. The way the rgl.postcript conversion works (using the gl2ps library) is to redirect OpenGL calls into calls to generate Postscript. Working at this level seems like the right approach, but I don't know if anyone has done it for these newer formats. Duncan One note: With Adobe Acrobat Pro 9.3.1, the U3D and PRC images display on screen, but do not print (replaced by the filename). Is this your experience too? I believe so. There may well be a way of including a static image in the LaTeX that would be used by printing and readers that don't understand the embedded formats, but I haven't had the chance to check the movie15 documentation for that. luke -Michael __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] multiple plots problem
hi thomas, thanks a lot- of course that's it. i knew why the middle plot is stretcht, but didn't figure out that settting par(oma) is the key.. greetings, kay -- View this message in context: http://n4.nabble.com/multiple-plots-problem-tp2017326p2017551.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] label the bars by the percentage values in the conditional histogram?
HI, Dear R Community, Does anyone know how to label the values in the conditional histogram? Thanks so much!!! h- sample(1:14, 319, rep=T) c- sample(1:14, 608, rep=T) n- sample(1:14, 1140, rep=T) vt-c(h, c, n) ta-rep(c(h, c, n), c(319, 608, 1140)) to-data.frame(vt,ta) library(lattice) histogram(~ vt|ta, data=to, layout=c(1,3), labels=TRUE, main=Histograms by target, col=skyblue) -- Sincerely, Changbin -- [[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] stupid regexp question
Both methods solve my problem. Thanks a lot to Gabor and David! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Weights in binomial glm
Thomas, Thierry, Thank your for your answers and my appogies for my late reply. Thomas, from your reply it seems that dividing the weights by their average would also make finding a suitable starting value more robust. This indeed seems to be case from test I've ran. However, your comments about me misusing glm() make me wonder at how the weights are used in glm(). I was under the impression that weights in glm are usually replicae/frequency weights. When I look at how weights are usually taken into account in the likelihood, namely L = Product_i L_i^w_i each record i contributes w_i times a factor L_i to the total likelihood, which is the same as saying that record i represents w_i observations. This is exactly what I want to express with my weights. Writing out the likelihood for binomial regression also shows that in that case it is possible to have Y's that are not only 0 or 1, but can also have values between 0 and 1 as in R. However, looking at this I do not see any problems with large weights and Y's equal to 0 or one. Is it just the fact that R has problems generating the starting values when the Y's are all between 0 and 1, or are there also some other reasons for this? Regards, Jan On Fri, Apr 16, 2010 at 6:28 PM, Thomas Lumley tlum...@u.washington.edu wrote: Jan, Thierry is correct in saying that you are misusing glm(), but there is also a numerical problem. You are misusing glm() because your model specification claims to have Binomial(n,p) observations with w in the vicinity of 100, where there is a single common p but the observed binomial proportion is either 1 or 0, never anything in between. These data are a very poor fit to a binomial model. The correct specification if you have what you call replicate weights and I call frequency weights is to produce a single data record for each covariate pattern that has both the 1 and 0 observations. This can either be two columns for successes and failures, or one column of proportions and one column of weights. As your quote from MASS says weights are used to give the number of trials when the response is the proportion of successes. In your data the response is *not* the proportion of successes. However, the MLE should still be equal to the weighted mean even with this misuse. The reason it is not is because of the starting values. R has to find some starting values for the iterative maximization of the likelihood, and for binomial data with y successes out of n it uses starting values for the fitted means of (y+0.5)/(n+1). Starting the iteration at the data in this way usually makes the Fisher scoring algorithm very reliable -- it is correctly scaled to the data, in some sense. Unfortunately, if you separate out the successes and failures, you have some points starting with values very close to 0. When I used your code the starting value for the point with the largest weight was 0.5/199. At iteration 2, the estimated mean ends up very small for all observations, and then the iteration diverges. However, if you provide a starting value then the fitting works, even if you start the iteration at, say beta=1, corresponding to a fitted mean of over 70%. So, the result is wrong in the sense that it is not the mle, because of a failure of convergence, which happens because specifying the weights the way you did rather than the documented way leads to bad default starting values for the iteration. You need either to specify the data as recommended or supply starting values. =thomas On Fri, 16 Apr 2010, Jan van der Laan wrote: I have some questions about the use of weights in binomial glm as I am not getting the results I would expect. In my case the weights I have can be seen as 'replicate weights'; one respondent i in my dataset corresponds to w[i] persons in the population. From the documentation of the glm method, I understand that the weights can indeed be used for this: For a binomial GLM prior weights are used to give the number of trials when the response is the proportion of successes. From Modern applied statistics with S-Plus 3rd ed. I understand the same. However, I am getting some strange results. I generated an example: Generate some data which is simular to my dataset Z - rbinom(1000, 1, 0.1) W - round(rnorm(1000, 100, 40)) W[W 1] - 1 Probability of success can either be estimated using: sum(Z*W)/sum(W) [1] 0.09642109 Or using glm: model - glm(Z ~ 1, weights=W, family=binomial()) Warning message: In glm.fit(x = X, y = Y, weights = weights, start = start, etastart = etastart, : fitted probabilities numerically 0 or 1 occurred predict(model, type=response)[1] 1 2.220446e-16 These two results are obviously not the same. The strange thing is that when I scale the weights, such that the total equals one, the probability is correctly estimated: model - glm(Z ~ 1, weights=W/sum(W), family=binomial()) Warning
[R] [R-pkgs] cudaBayesreg update
Dear all, Package cudaBayesreg, version 0.3-3, is now online. Apart some minor corrections, the package now includes a brief report of the tests used to assess the quality of the random number generator. Best regards, Adelino Ferreira da Silva http://cran.at.r-project.org/src/contrib/cudaBayesreg_0.3-3.tar.gz [[alternative HTML version deleted]] ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] lattice visualization - superpose multiple columns over another (fixed) column
Hi, I've been struggling with a lattice visualiation. I have a
data.frame with 4 columns. What I'd like to have is a set of 3 panels.
Ecah panel will have the first column plotted against serial number
and then will superimpose the relevant column. My non-lattice version
is as follows:
x - data.frame( ... )
par(mfrow=c(3,1))
for (i in 2:4) {
plot(x[,1])
points(x[,i])
}
Any suggestions as to how I could convert this to a lattice version
would be much appreciated
Thanks,
--
Rajarshi Guha
NIH Chemical Genomics Center
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[R] Visualize a fitted line in a log plot of a power law distribution
Hi, I am trying to fit a line in the log plot of my networks degree distribution to show that it is a power-law distribution. I am using the following commands. However, I am not able to see the fitted line. Any comments to help? I am using following packages: igraph, splines,base,VGAM, netmodels. g is my network, d is the degree of nodes in the network, and dd is the degree distribution d -degree(g) dd - degree.distribution(g) fitpower.law.fit(d,20) plot(d,log=xy) abline(fit) Thanks in advance. Regards, *Narcissus* [[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] sum() returns 0 not NA
Dear all, just a stupid R question, since the results puzzle me a bit: sum(c(NA,NA), na.rm=TRUE) [1] 0 NA + NA [1] NA NA + 1 [1] NA Why does sum(c(NA,NA), na.rm=TRUE) return 0 and not NA? Thanks in advance, Will __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] fit a line to power law distribution
Hi, I am trying to fit a line in the log plot of my networks degree distribution to show that it is a power-law distribution. I am using the following commands. However, I am not able to see the fitted line. Any comments to help? I am using following packages: igraph, splines,base,VGAM, netmodels. g is my network, d is the degree of nodes in the network, and dd is the degree distribution d -degree(g) dd - degree.distribution(g) fitpower.law.fit(d,20) plot(dd,log=xy) abline(fit) Thanks in advance. Regards, *Narcissus* [[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] sum() returns 0 not NA
So that the two lines below give the same answer: xx - c(); yy - 1:3 sum(xx) + sum(yy) sum(c(xx, yy)) On Tue, Apr 20, 2010 at 12:42 PM, will.ea...@gmx.net wrote: Dear all, just a stupid R question, since the results puzzle me a bit: sum(c(NA,NA), na.rm=TRUE) [1] 0 NA + NA [1] NA NA + 1 [1] NA Why does sum(c(NA,NA), na.rm=TRUE) return 0 and not NA? Thanks in advance, Will __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] QCA3 segfault
this should have gone to the package manager apologies to all rg ded0rg wrote: Hi I have just dwonloaded QCA3 onto this machine (ubuntu, karmic, amd64) and a mac. The examples run fine on teh mac, but crashed R on ubuntu. Any help much apprecia\ted. Thanks Richard -- Terminal log: conditions - c(Developed.FZ,Urban.FZ,Literate.FZ,Industrial.FZ, Stable.FZ) reduce(mydata=Lipset_fs,Survived.FZ,conditions,explain=positive,remaind=exclude,prepro=fs,consistency=0.7) *** caught segfault *** address 0x4, cause 'memory not mapped' Traceback: 1: .C(lpslink, direction = as.integer(direction), x.count = as.integer(x.count), objective = as.double(objective), const.count = as.integer(const.count), constraints = as.double(constraints), int.count = as.integer(int.count), int.vec = as.integer(int.vec), bin.count = as.integer(bin.count), binary.vec = as.integer(binary.vec), num.bin.solns = as.integer(num.bin.solns), objval = as.double(objval), solution = as.double(solution), presolve = as.integer(presolve), compute.sens = as.integer(compute.sens), sens.coef.from = as.double(sens.coef.from), sens.coef.to = as.double(sens.coef.to), duals = as.double(duals), duals.from = as.double(duals.from), duals.to = as.double(duals.to), scale = as.integer(scale), use.dense = as.integer(use.dense), dense.col = as.integer(dense.col), dense.val = as.double(dense.val), dense.const.nrow = as.integer(dense.const.nrow), dense.ctr = as.integer(dense.ctr), use.rw = as.integer(use.rw), tmp = as.character(tmp), status = as.integer(status), PACKAGE = lpSolve) 2: lpSolve:::lp(direction = min, objective.in = rep(1, nrow(PIChart)), const.mat = t(PIChart), const.dir = =, 1, all.bin = TRUE) 3: solvePIChart(PIChart) 4: reduce.default(mydata = Lipset_fs, Survived.FZ, conditions, explain = positive, remaind = exclude, prepro = fs, consistency = 0.7) 5: reduce(mydata = Lipset_fs, Survived.FZ, conditions, explain = positive, remaind = exclude, prepro = fs, consistency = 0.7) Possible actions: 1: abort (with core dump, if enabled) 2: normal R exit 3: exit R without saving workspace 4: exit R saving workspace Selection: -- Sys.getlocale() [1] LC_CTYPE=en_GB.UTF-8;LC_NUMERIC=C;LC_TIME=en_GB.UTF-8;LC_COLLATE=en_GB.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_GB.UTF-8;LC_PAPER=en_GB.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_GB.UTF-8;LC_IDENTIFICATION=C -- Richard Gott 1 Willow Court Finghall Leyburn N Yorks DL8 5NL 01677 450 974 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Huge data sets and RAM problems
Stella, A few brief words of advice: 1. Work through your code a line at a time, making sure that each is what you would expect. I think some of your later problems are a result of something early not being as expected. For example, if the read.delim() is in fact not giving you what you expect, stop there before moving onwards. I suspect some funny character(s) or character encodings might be a problem. 2. 32-bit Windows can be limiting. With 2 GB of RAM, you're probably not going to be able to work effectively in native R with objects over 200-300 MB, and the error indicates that something (you or a package you're using) simply have run out of memory. So... 3. Consider more RAM (and preferably with 64-bit R). Other solutions might be possible, such as using a database to hand the data transition into R. 2.5 million rows by 18 columns is apt to be around 360 MB. Although you can afford 1 (or a few) copies of this, it doesn't leave you much room for the memory overhead of working with such an object. Part of the oringal message below. Jay - Message: 80 Date: Mon, 19 Apr 2010 22:07:03 +0200 From: Stella Pachidi stella.pach...@gmail.com To: r-h...@stat.math.ethz.ch Subject: [R] Huge data sets and RAM problems Message-ID: g2j133363581004191307t2a48c1bfqd9d57cf0d6c62...@mail.gmail.com Content-Type: text/plain; charset=ISO-8859-1 Dear all, I am using R 2.10.1 in a laptop with Windows 7 - 32bit system, 2GB RAM and CPU Intel Core Duo 2GHz. . Finally, another problem I have is when I perform association mining on the data set using the package arules: I turn the data frame into transactions table and then run the apriori algorithm. When I put too low support in order to manage to find the rules I need, the vector of rules becomes too big and I get problems with the memory such as: Error: cannot allocate vector of size 923.1 Mb In addition: Warning messages: 1: In items(x) : Reached total allocation of 153Mb: see help(memory.size) Could you please help me with how I could allocate more RAM? Or, do you think there is a way to process the data by loading them into a document instead of loading all into RAM? Do you know how I could manage to read all my data set? I would really appreciate your help. Kind regards, Stella Pachidi -- John W. Emerson (Jay) Associate Professor of Statistics Department of Statistics Yale University http://www.stat.yale.edu/~jay [[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] sum() returns 0 not NA
As a disclaimer, I cannot say that this is why sum() was designed as it was.
0 is the sum of a set with no elements, the empty set {}. When
na.rm=TRUE, NA values are removed. When the only values are NA (as in
your example c(NA, NA) ), and you remove them all, you are taking the
sum of no elements, which is 0.
Also note the behavior of
sum() # returns 0
sum is one of the few functions that you can simply call that will not
return an error.
HTH,
Josh
On Tue, Apr 20, 2010 at 9:42 AM, will.ea...@gmx.net wrote:
Dear all,
just a stupid R question, since the results puzzle me a bit:
sum(c(NA,NA), na.rm=TRUE)
[1] 0
NA + NA
[1] NA
NA + 1
[1] NA
Why does sum(c(NA,NA), na.rm=TRUE) return 0 and not NA?
Thanks in advance,
Will
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--
Joshua Wiley
Senior in Psychology
University of California, Riverside
http://www.joshuawiley.com/
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Re: [R] sum() returns 0 not NA
On 20-Apr-10 17:07:31, Joshua Wiley wrote:
As a disclaimer, I cannot say that this is why sum() was designed as it
was.
0 is the sum of a set with no elements, the empty set {}. When
na.rm=TRUE, NA values are removed. When the only values are NA (as in
your example c(NA, NA) ), and you remove them all, you are taking the
sum of no elements, which is 0.
Also note the behavior of
sum() # returns 0
sum is one of the few functions that you can simply call that will not
return an error.
HTH,
Josh
On Tue, Apr 20, 2010 at 9:42 AM, will.ea...@gmx.net wrote:
Dear all,
just a stupid R question, since the results puzzle me a bit:
sum(c(NA,NA), na.rm=TRUE)
[1] 0
_NA + NA
[1] NA
NA + 1
[1] NA
Why does sum(c(NA,NA), na.rm=TRUE) return 0 and not NA?
Thanks in advance,
Will
Exactly the same question was asked, in exactly the same words,
5 days ago:
https://stat.ethz.ch/pipermail/r-help/2010-April/235337.html
and, I suspect, by exactly the same person (thoug using a
different email address). And, also, it was answered 5 days ago.
Was the second sending an accident?
Ted.
E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 20-Apr-10 Time: 18:24:38
-- XFMail --
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Re: [R] What is mclust up to? Different clusters found if x and y interchanged
Thanks Prof. Fraley. This is very helpful. I had looked at the way mclust gets its start and that looked like a possible source of the differences, but had not had time to get up to speed and figure it out. I was thinking it was more of a set.seed sort of thing. Now I know more. Thanks so much. Bryan On 4/20/10 11:41 AM, Chris Fraley fra...@u.washington.edu wrote: This is a really interesting case, but it is probably not that unusual. mod23 - Mclust(tur[,2:3]) mod32 - Mclust(tur[,3:2]) mod23[c(modelName,G)] mod32[c(modelName,G)] The 2:3 case is choosing the VVV model (more parameters per cluster) with 3 clusters. The 3:2 case is choosing the EEV model (more parsimonious), but with 7 clusters. If you look at the BIC curves, you can see that mclust is indeed choosing the model corresponding to the maximum BIC in each case. plot( mclustBIC( tur[,2:3], modelNames = c(EEV, VVV)) plot( mclustBIC( tur[,3:2], modelNames = c(EEV, VVV)) The default initialization for Mclust is via hierarchical clustering, and that's why the results differ hc23 - hcVVV( tur[,2:3]) # default initialization for mod23 hc32 - hcVVV( tur[,3:2]) # default initialization for mod32 all.equal(hc32,hc23) # they differ! The hierarchical clustering merges two groups at each stage. If there is more than one choice at any particular stage for which the merge criteria are equal or within roundoff error, a different pair could be chosen for merging if columns (or rows) are permuted. This will affect later merges, and could ultimately affect the mclust results. If you initialize both col permutations in the same way, you'll get the same results. mod23mod - Mclust(tur[,2:3], initialization = list( hcPairs = hc32)) # uses mod32 initialization mod32mod - Mclust(tur[,3:2], initialization = list( hcPairs = hc23)) # uses mod23 initialization mod23mod[c(modelName,G)] # same as mod32 mod32mod[c(modelName,G)] # same as mod23 Hope this helps, Chris Fraley On Tue, 20 Apr 2010, Bryan Hanson wrote: Prof. Fraley, I wonder if you would have a couple of moments to respond to this question I put to the R help list. Thanks in advance for your time. Bryan * Bryan Hanson Acting Chair Professor of Chemistry Biochemistry DePauw University, Greencastle IN USA -- Forwarded Message From: Bryan Hanson han...@depauw.edu Date: Mon, 19 Apr 2010 19:31:03 -0400 To: han...@gapps.depauw.edu, R Help r-h...@stat.math.ethz.ch Subject: [R] What is mclust up to? Different clusters found if x and y interchanged Hello All... I gave a task to my students that involved using mclust to look for clusters in some bivariate data of isotopes vs various mining locations. They discovered something I didn¹t expect; the data (called tur) is appended below. p - qplot(x = dD, y = dCu65, data = tur, color = mine) print(p) # simple bivariate plot of the data; looks fine mod1 - Mclust(tur[,2:3]) mod1$G mod2 - Mclust(tur[,3:2]) mod2$G One can use coordProj to see the clusters found, but the basic result is that mclust found 3 clusters in mod1, but if you interchange the x and y columns it finds 7 clusters (mod2). Since this is bivariate data, I find this result pretty strange. The only thing I can think of is that it has something to do with how the algorithm is seeded, but that isn't too convincing. Since I couldn't believe my eyes, I made up some bivariate data with known clusters (not given here) and interchanged the x and y columns and got the same clusters. I'm at a loss. Hopefully someone can set my understanding on the right path. Thanks for any insight! Bryan sessionInfo() R version 2.10.1 (2009-12-14) x86_64-apple-darwin9.8.0 locale: [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] datasets tools grid graphics grDevices utils stats methods base other attached packages: [1] faraway_1.0.4 GGally_0.1 xtable_1.5-6 mvbutils_2.5.1 ggplot2_0.8.7 [6] digest_0.4.2 reshape_0.8.3 proto_0.3-8ChemoSpec_1.43 R.utils_1.4.0 [11] R.oo_1.7.1 R.methodsS3_1.2.0 rgl_0.91 lattice_0.18-3 mvoutlier_1.4 [16] plyr_0.1.9 RColorBrewer_1.0-2 chemometrics_0.8 som_0.3-5 robustbase_0.5-0-1 [21] rpart_3.1-46 pls_2.1-0 pcaPP_1.8 mvtnorm_0.9-9 nnet_7.3-1 [26] mclust_3.4.4 MASS_7.3-5 lars_0.9-7 e1071_1.5-23 class_7.3-2 And the data: dput(tur) structure(list(mine = structure(c(13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 6L, 6L, 6L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 14L, 14L, 14L, 14L, 14L, 14L, 14L, 14L, 9L, 9L, 9L, 9L, 12L, 12L, 12L, 12L, 12L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 10L, 10L, 10L,
Re: [R] Unwanted boxes in legend
On 2010-04-19 8:11, Thomas Stewart wrote: Try border=c(0,0,1,0). -tgs If the 'border' argument is not recognized, then this won't work. Steve: What version of R are you using? I have no problems with the suggestions I gave you in R 2.10.1 or R 2.11.0 alpha. -Peter Ehlers On Mon, Apr 19, 2010 at 4:21 AM, Steve Murraysmurray...@hotmail.comwrote: Dear all, Thanks for the response, however I'm getting the following error message when I execute the legend command using the 'border' argument: Error in legend(10, par(usr)[4], c(A, B, : unused argument(s) (border = FALSE) Is anyone aware of any alternative means of switching off boxes around all but one of the elements in a legend? Many thanks for any input, Steve Date: Thu, 15 Apr 2010 12:13:40 -0600 From: ehl...@ucalgary.ca To: smurray...@hotmail.com CC: r-help@r-project.org Subject: Re: [R] Unwanted boxes in legend On 2010-04-15 11:10, Steve Murray wrote: Dear all, I am using the following code to generate a legend in my plot (consisting of both bars and points), but end up with boxes around my points: legend(10, par(usr)[4], c(A, B, C, D), fill=c(NA,NA, grey28, NA), pch=c(16,4,NA,18), col=c(red,blue,grey28,yellow), lty=FALSE, bty=n, horiz=FALSE) I want a box around the third element of the legend (to represent the bar 'fill' colour), but not for the others, where points are shown instead. What am I doing wrong above and how do I correct it? Add the 'border' argument: either border = FALSE # in which case no box is drawn for any element or border = c(NA, NA, black, NA) -Peter Ehlers Many thanks, Steve __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Peter Ehlers University of Calgary __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Peter Ehlers University of Calgary __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] shift and pop equivalent in R
Dear All, I am wondering is there any shift (or pop or push or unshift) equivalent in R? For example, shift(x) # should return x[1], and x becomes x[-1] Thanks a lot. Best Regards, Xie Chao __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] comparing attitudes of 2 groups / likert scales?
On 2010-04-19 9:15, Dieter Menne wrote:
David Winsemius wrote:
If you are thinking of using that quote, you might want to check the
spelling of his name. My memory is van Belle.
Sorry, I thought I had corrected that before mailing.
@BOOK{vanBelle2002,
title = {Statistical rules of thumb},
publisher = {Wiley series in probability and statistics},
year = {2002},
author = {Gerald van Belle}
}
And a very fine book it is, too. Highly recommended.
--
Peter Ehlers
University of Calgary
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Re: [R] shift and pop equivalent in R
On 20/04/2010 2:03 PM, Xie Chao wrote: Dear All, I am wondering is there any shift (or pop or push or unshift) equivalent in R? For example, shift(x) # should return x[1], and x becomes x[-1] Not in base R. Generally speaking, functions with side effects are frowned upon. It is possible to write it yourself, but you need to watch out for weird cases like shift(x + y) where it's not at all clear what the side effect should be. Duncan Murdoch Thanks a lot. Best Regards, Xie Chao __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] shift and pop equivalent in R
On Tue, Apr 20, 2010 at 7:03 PM, Xie Chao xiech...@gmail.com wrote: Dear All, I am wondering is there any shift (or pop or push or unshift) equivalent in R? For example, shift(x) # should return x[1], and x becomes x[-1] I seem to have implemented a FIFO stack in 2003: http://finzi.psych.upenn.edu/R/Rhelp02/archive/15541.html but note the reply commenting on the existence of the stack function in R which stacks data on top of other data. Here's sample usage once you've loaded the code from that old posting: s=stack() push(s,1) s [[1]] [1] 1 push(s,3) push(s,4) pop(s) [1] 4 pop(s) [1] 3 s [[1]] [1] 1 If you want to implement a 'shift' method, just look at the existing pop and push methods and write something similar. Note how the functions are created on the object, and then called from S3 methods. Note the use of - within the functions on the object. There's probably better ways to do this, and with S4 methods. Barry -- blog: http://geospaced.blogspot.com/ web: http://www.maths.lancs.ac.uk/~rowlings web: http://www.rowlingson.com/ twitter: http://twitter.com/geospacedman pics: http://www.flickr.com/photos/spacedman __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] shift and pop equivalent in R
Thanks Barry for the code sample, and Thanks Duncan for the clarification. Both replies help a lot! Best Regards, Xie Chao On Wed, Apr 21, 2010 at 2:50 AM, Barry Rowlingson b.rowling...@lancaster.ac.uk wrote: On Tue, Apr 20, 2010 at 7:03 PM, Xie Chao xiech...@gmail.com wrote: Dear All, I am wondering is there any shift (or pop or push or unshift) equivalent in R? For example, shift(x) # should return x[1], and x becomes x[-1] I seem to have implemented a FIFO stack in 2003: http://finzi.psych.upenn.edu/R/Rhelp02/archive/15541.html but note the reply commenting on the existence of the stack function in R which stacks data on top of other data. Here's sample usage once you've loaded the code from that old posting: s=stack() push(s,1) s [[1]] [1] 1 push(s,3) push(s,4) pop(s) [1] 4 pop(s) [1] 3 s [[1]] [1] 1 If you want to implement a 'shift' method, just look at the existing pop and push methods and write something similar. Note how the functions are created on the object, and then called from S3 methods. Note the use of - within the functions on the object. There's probably better ways to do this, and with S4 methods. Barry -- blog: http://geospaced.blogspot.com/ web: http://www.maths.lancs.ac.uk/~rowlings web: http://www.rowlingson.com/ twitter: http://twitter.com/geospacedman pics: http://www.flickr.com/photos/spacedman __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 with Partial dependence bar graph
Hello, I need to draw a partial dependence bar graph. My the my predictor vectors are continous and so is the response variable. Iam using the partialPlot function of the randomForest package. I get a line graph. How can I edit it to get a bar graph instead? (partialPlot(randomForest object ,data-matrix, number of predictor vectors, Temp)) -- Daudi Jjingo [[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] Exporting PDF
I run the script and it exports a PDF called version 1.
I want it to check if version 1 already exists. If so,
then I want the new graphs to be exported as
version 2, and so on.
Is it possible to do it in R?
Someone may know a way. However its certainly possible to execute a command in
linus from R. So you can ls the file (in windows try dir) and see if it
exists, and build a loop round that.
Just bew3are it'll be quick if there is only 1 file. If you have files it
may slow things down a bit!
Here's my example...
#Example script to check if a file exists (linux system)
filename = somefile
fileno = 0
extension= .pdf
path=/home/user/
repeat
{
fullfile= paste(path,filename,fileno,extension, sep=)
if (length (system(paste(ls,fullfile), intern = T)) == 0) break
fileno-fileno+1
}
#your command to save the file... using 'fullfile' as the filename and path
This message may contain confidential information. If yo...{{dropped:21}}
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[R] having more than one plot in one figure
Hi All, I have been trying to plot multiple line plots with different colors on one figure. in my example below I was able to plot cat vs num1 as a dot plot connected with lines but was not able to do that for cat vs num2 and I do not know how to add the third plot cat vs num3. below is my code df - data.frame(cat=1:10, num1=rnorm(10), num2=rnorm(10), num3=rnorm(10)) plot(df$num1, type=b, col=red, xlab=categories, ylab=random numbers) points(df$num1, type=p, pch=21, col=red,bg=red)#plot num1 points(df$num2, type=p, pch=21, col=green) #plot num2 legend(x=topright, legend=c(num1, num2), pch=c(16,16), col=c(red,green))#add a legend how to make the second plot of type=b and how to add a third plot cat vs num3? I appreciate your help -- View this message in context: http://n4.nabble.com/having-more-than-one-plot-in-one-figure-tp2017471p2017471.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] Building Rblas.dll failing on Windows 7 (64-bit)
I'm currently attempting to build R-2.10.1 from the source on Windows 7 (64-bit version). I've downloaded the latest Rtools (Rtools211.exe) from the Murdoch-Sutherland site and have done a full install. I've also downloaded the HTML Help Workshop, MikTeX, and the InnoSetup installer. I am not using any 64-bit versions of the tools, such as the MinGW-w64, or a 64-bit version of TCL as I'm just trying to build the 32-bit version of R first. The only change necessary in Mkrules was the location of the Inno Setup directory. The make process gets a considerable way through before exiting with the following error: . . . gcc -std=gnu99 -shared -s -mwindows -o R.dll R.def console.o dataentry.o dynload.o edit.o editor.o embeddedR.o extra.o opt.o pager.o preferences.o psignal.o rhome.o rt_complete.o rui.o run.o shext.o sys-win32.o system.o dos_wglob.o e_pow.o malloc.o ../main/libmain.a ../appl/libappl.a ../nmath/libnmath.a getline/gl.a ../extra/xdr/libxdr.a ../extra/pcre/libpcre.a ../extra/bzip2/libbz2.a ../extra/intl/libintl.a ../extra/trio/libtrio.a ../extra/tzone/libtz.a ../extra/tre/libtre.a ../extra/xz/liblzma.a dllversion.o -L. -lgfortran -lRblas -L../../bin -lRzlib -lRgraphapp -lRiconv -lcomctl32 -lversion cp R.dll ../../bin/ Building ../../../bin/Rblas.dll gcc -std=gnu99 -shared -o ../../../bin/Rblas.dll blas.o cmplxblas.o ../../gnuwin32/dllversion.o Rblas.def -L../../../bin -lR -lgfortran c:/factory/rtools/mingw/bin/../lib/gcc/mingw32/4.2.1-sjlj/../../../../mingw32/bin/ld.exe: cannot find -lR collect2: ld returned 1 exit status make[3]: *** [../../../bin/Rblas.dll] Error 1 make[2]: *** [Rblas] Error 2 make[1]: *** [rbuild] Error 2 make: *** [all] Error 2 Thus we see that the prior step of building the R DLL has suceeded, but it fails on the Rblas DLL. Interestingly, I did notice that if I repeat make all recommended, then it fails on the copy command cp R.dll ../../bin/ as it tries to overwrite the file created on the proir run. I think this is a permission issue. For some reason the R.dll file is read only. I'm wondering whether this is somehow related to the failure in creating the Rblas.dll file. Any thoughts would be appreciated. Thank you. ubk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] having more than one plot in one figure
On Apr 20, 2010, at 10:34 AM, kayj wrote: Hi All, I have been trying to plot multiple line plots with different colors on one figure. in my example below I was able to plot cat vs num1 as a dot plot connected with lines but was not able to do that for cat vs num2 and I do not know how to add the third plot cat vs num3. below is my code df - data.frame(cat=1:10, num1=rnorm(10), num2=rnorm(10), num3=rnorm(10)) plot(df$num1, type=b, col=red, xlab=categories, ylab=random numbers) points(df$num1, type=p, pch=21, col=red,bg=red)#plot num1 points(df$num2, type=p, pch=21, col=green) #plot num2 legend(x=topright, legend=c(num1, num2), pch=c(16,16), col=c(red,green))#add a legend Add the x values since points does not have exactly the same default behavior when given just one vector for plotting: df - data.frame(cat=1:10, num1=rnorm(10), num2=rnorm(10), num3=rnorm(10)) plot(df$num1, type=b, col=red, xlab=categories, ylab=random numbers) points(1:10, df$num2, type=b, pch=21, col=red,bg=red)#plot num1 points(1:10, df$num3, type=b, pch=21, col=green) #plot num2 legend(x=topright, legend=c(num1, num2), pch=c(16,16), col=c(red,green))#add a legend If you want to have all of the random point visible than use range(c(df $num1, df$num2, df$num3) as your argument to ylim. how to make the second plot of type=b and how to add a third plot cat vs num3? I don't understand why you did not look at the help page for points and realize that type=b would work. I appreciate your help David Winsemius, MD 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] lattice code to plot columns over another variable
Hi, I've been struggling with a lattice visualiation. I have a
data.frame with 4 columns. What I'd like to have is a set of 3 panels.
Ecah panel will have the first column plotted against serial number
and then will superimpose the relevant column. My non-lattice version
is as follows:
x - data.frame( ... )
par(mfrow=c(3,1))
for (i in 2:4) {
plot(x[,1])
points(x[,i])
}
Any suggestions as to how I could convert this to a lattice version
would be much appreciated
Thanks,
--
Rajarshi Guha
NIH Chemical Genomics Center
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[R] 3D surface plot with wireframe or persp?
Hello Dear,
I have a function, like z=f(x,y), and try a surface plot with this function.
But, on the reference of wireframe requires data option, so I generated x
and y, and computed z with them. But, still I have a problem to draw a
surface plot. The code and errors are
##
mle_beta0=64.43707;
mle_beta1=-24365.16;
# generating for the requirement of wireframe
data_for_c = runif(1000,30,85); # range is (30,50)
data_for_s = sort(1/(273+data_for_c));
data_for_time = sort(runif(1000,0,10)) # range is (0,10)
data_for_R = exp((-exp(mle_beta0+mle_beta1*data_for_s))*data_for_time)
data_all = cbind(data_for_s,data_for_time,data_for_R)
# function: plot_R_i_3d = f(data_for_time,data_for_s)
plot_R_i_3d = function(data_for_time,data_for_s)
{
R_i = exp((-exp(mle_beta0+mle_beta1*data_for_s))*data_for_time);
return(R_i)
}
# tried 1) persp or 2) wireframe
persp(data_for_time,data_for_s,plot_R_i_3d)
== Error in min(x, na.rm = na.rm) : invalid 'type' (list) of argument
wireframe(formula=data_for_R ~ data_for_s*data_for_time,data=data_all)
== Error in nrow(x) :
element 1 is empty;
the part of the args list of 'dim' being evaluated was:
(x)
###
I have two questions:
1) which one is better to plot in this case - persp and wireframe, Is
there another suggestion?
2) Is it possible to plot with only a function?
3) If not, what is wrong in the code above?
Many thank you in advance,
Jin
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[R] Error in from:to : NA/NaN argument
Hello R gurus,
I am having difficulties running a chunk of code that I otherwise thought
was correct..
if (lower max(length(IC_peaks),length(IC_valleys))) {
+ valley_index - IC_valleys[lower+1]
+ for (i in seq(peak_index,valley_index-1)) {
+ IC_peaks_and_valleys - c(IC_peaks_and_valleys, v)
+ }
+ }
Error in from:to : NA/NaN argument
I can not pin point the issue and if you have any suggestions I would
greatly appreciate them. If more code is needed to figure out the problem, I
will post it right away.
Thank you all.
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Re: [R] Error in from:to : NA/NaN argument
On Apr 20, 2010, at 3:54 PM, cobbler_squad wrote:
Hello R gurus,
I am having difficulties running a chunk of code that I otherwise
thought
was correct..
if (lower max(length(IC_peaks),length(IC_valleys))) {
+ valley_index - IC_valleys[lower+1]
+ for (i in seq(peak_index,valley_index-1)) {
+ IC_peaks_and_valleys - c(IC_peaks_and_valleys, v)
+ }
+ }
Error in from:to : NA/NaN argument
Perhaps an error in you seq() call which has from and to arguments.
What do you get with this:
str(peak_index)
str(valley_index)
I can not pin point the issue and if you have any suggestions I would
greatly appreciate them. If more code is needed to figure out the
problem, I
will post it right away.
Thank you all.
--
David Winsemius, MD
West Hartford, CT
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[R] Basic doubt on ylim
Hi all, I'm a newbie with R and with a very basic question. Can I define the minor unit for ylim? For example, I have a y scale ranging from 1 to 14, jumping automatically every 2 units, but I want it to jump 1 unit at a time...is it possible? I tried something like boxplot(bla[,1], xlim=c(1,15,1) and it didn't obbey... :( any suggestion?? Cheers from Portugal! PatrÃcia -- View this message in context: http://n4.nabble.com/Basic-doubt-on-ylim-tp2017891p2017891.html Sent from the R help mailing list archive at Nabble.com. [[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] Exporting PDF
Since R provides the file.exists() function for exactly this
purpose, I would recommend avoiding using shell scripts for
such a simple task. file.exists() is a multiplatform solution
to the problem.
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Tue, 20 Apr 2010, Polwart Calum (County Durham and Darlington NHS Foundation
Trust) wrote:
I run the script and it exports a PDF called version 1.
I want it to check if version 1 already exists. If so,
then I want the new graphs to be exported as
version 2, and so on.
Is it possible to do it in R?
Someone may know a way. However its certainly possible to execute a command in
linus from R. So you can ls the file (in windows try dir) and see if it
exists, and build a loop round that.
Just bew3are it'll be quick if there is only 1 file. If you have files it
may slow things down a bit!
Here's my example...
#Example script to check if a file exists (linux system)
filename = somefile
fileno = 0
extension= .pdf
path=/home/user/
repeat
{
fullfile= paste(path,filename,fileno,extension, sep=)
if (length (system(paste(ls,fullfile), intern = T)) == 0) break
fileno-fileno+1
}
#your command to save the file... using 'fullfile' as the filename and path
This message may contain confidential information. If yo...{{dropped:21}}
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Re: [R] Error in from:to : NA/NaN argument
Hi,
The problem, I think, comes from seq(peak_index,valley_index-1)
Since I don't know what is peak_index and valley_index, it is difficult
to further help you.
Ivan
Le 20-Apr-10 21:54, cobbler_squad a écrit :
Hello R gurus,
I am having difficulties running a chunk of code that I otherwise thought
was correct..
if (lower max(length(IC_peaks),length(IC_valleys))) {
+ valley_index- IC_valleys[lower+1]
+ for (i in seq(peak_index,valley_index-1)) {
+ IC_peaks_and_valleys- c(IC_peaks_and_valleys, v)
+ }
+ }
Error in from:to : NA/NaN argument
I can not pin point the issue and if you have any suggestions I would
greatly appreciate them. If more code is needed to figure out the problem, I
will post it right away.
Thank you all.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Basic doubt on ylim
Hi, Take a look at ?par and especially the xaxp argument. I'm not quite sure, but if it's not this one, it's probably another one in par() HTH, Ivan Le 20-Apr-10 21:21, poliveira a écrit : Hi all, I'm a newbie with R and with a very basic question. Can I define the minor unit for ylim? For example, I have a y scale ranging from 1 to 14, jumping automatically every 2 units, but I want it to jump 1 unit at a time...is it possible? I tried something like boxplot(bla[,1], xlim=c(1,15,1) and it didn't obbey... :( any suggestion?? Cheers from Portugal! PatrÃcia __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Basic doubt on ylim
There's a worked example of using axis() at the bottom of ?plot.default On Apr 20, 2010, at 4:48 PM, Ivan Calandra wrote: Hi, Take a look at ?par and especially the xaxp argument. I'm not quite sure, but if it's not this one, it's probably another one in par() HTH, Ivan Le 20-Apr-10 21:21, poliveira a écrit : Hi all, I'm a newbie with R and with a very basic question. Can I define the minor unit for ylim? For example, I have a y scale ranging from 1 to 14, jumping automatically every 2 units, but I want it to jump 1 unit at a time...is it possible? I tried something like boxplot(bla[,1], xlim=c(1,15,1) and it didn't obbey... :( any suggestion?? Cheers from Portugal! Patrà cia David Winsemius, MD 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] some problems
Dear R community! I am a mathematician listener, and I have to rewrite some source codes from Matlab to R. I would like to ask a solution of people who are skilled at him better hereby, because it is very new for me R and I do not receive it final result than in Matlab. So the problem: --- Let us generate 2005 standard normal random numbers, we depict it these the histogram of density together with the standard normal density function! We do this so that on [0,1] we write even random numbers the normal distribution function into inverse one's! --- I have a problem with it mainly when who I want to have it drawn. The drawing the arrangement of the histogram and the density function slips compared to each other. But the same situation then, if barplot() I use it. I apologise for my ridiculous problem, and my bad composition (translator I use a program). Thank you very much: Barjak Tamas [[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] Results from clogit out of range?
Hi,
I'm calculating a conditional logit on some data stratified by group.
My understanding was that a conditional logit by definition returns a
value between 0 and 1 a a probability. Can anyone suggest why I'm
seeing results outside of the {0,1} range??
The call in R is:
m - clogit(score ~ val_1 + val_2 + strata(group), data=data)
Then
prediction - predict(m,newdata)
A sample of the data with resulting predictive values is:
group score val_1 val_2prediction
1 2009-01-04_1 1 0.5913962 -1.121589 1.62455210
2 2009-01-04_1 1 0.6175472 -3.249820 -0.20093346
3 2009-01-04_1 1 0.5439640 -2.424501 0.46651849
4 2009-01-04_1 0 0.3745209 -2.477424 0.31263855
5 2009-01-04_1 0 0.6329855 -3.424174 -0.34200448
6 2009-01-04_1 0 0.4571999 -2.770247 0.11190788
7 2009-01-04_1 0 0.3822623 -2.259422 0.50627534
8 2009-01-04_1 0 0.2605742 -4.424806 -1.44566070
9 2009-01-04_1 0 0.4449604 -2.357060 0.46174993
10 2009-01-04_1 0 0.6595178 -2.246518 0.69427842
11 2009-01-04_1 0 0.5260032 -2.977887 -0.02393871
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Re: [R] Using all variables in a linear model
What about using x[,-predictor]. For instance: x - matrix(1:9, nrow=3) x[,-2] # all but the second column Perhaps using your code it would be something like... lm=lm(x[,dim(x)[2]] ~ x[,-dim(x)[2]], data=x) Best regards, Josh On Tue, Apr 20, 2010 at 3:12 PM, Walter Yund IV wyun...@gmail.com wrote: Hello, I am trying to automate linear regression for many different datasets, each with the same rough format (the last variable is the response). I've been doing something like this: lm=lm(x[,dim(x)[2]] ~ ., data=x) where the dot denotes all variables. However, this means that the response is included as a predictor, which is obviously what I don't want. How do I request that all the columns in my dataset be used as predictors, except for the response? Thanks, Walter [[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. -- Joshua Wiley Senior in Psychology University of California, Riverside http://www.joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] some problems
On 20-Apr-10 21:25:28, tamas barjak wrote: Dear R community! I am a mathematician listener, and I have to rewrite some source codes from Matlab to R. I would like to ask a solution of people who are skilled at him better hereby, because it is very new for me R and I do not receive it final result than in Matlab. So the problem: --- Let us generate 2005 standard normal random numbers, we depict it these the histogram of density together with the standard normal density function! We do this so that on [0,1] we write even random numbers the normal distribution function into inverse one's! --- I have a problem with it mainly when who I want to have it drawn. The drawing the arrangement of the histogram and the density function slips compared to each other. But the same situation then, if barplot() I use it. I apologise for my ridiculous problem, and my bad composition (translator I use a program). Thank you very much: Barjak Tamas Here is an example of making a histogram of 2005 standard Normal random numbers, drawing a histogram, and plotting the curve of the standard Normal density on top of it. Explanations are added as comments (#). # set the RNG seed (for reproducibility of this example) set.seed(54321) # Generate 2005 standard Normal numbers X - rnorm(2005) # Draw a histogram (arbitrary break-points) # breAkpoints at -4.0, -3.6, -3.2, ... -0.4, 0, 0.4, ... 3.6, 4.0 hist(X, breaks = 0.4*(-10:10)) # Draw the curve of the Normal distributiom # using 10 points per interval of the histogram # and multiplying by the width 0.4 of the interval # in order to match the probabilities of the intervals # and also by N=2005 to scale the curve up to give counts x0 - 0.04*(-100:100) y0 - 2005*0.4*dnorm(x0) lines(x0,y0) There is no perceptible slip between the histigram and the Normal curve Ithe slight differences are due to random variation in the positions of the X values). To see it better, use a much larger random sample: set.seed(54321) # Generate 20 standard Normal numbers X - rnorm(20) hist(X, breaks = 0.4*(-12:12)) x0 - 0.04*(-120:120) y0 - 20*0.4*dnorm(x0) lines(x0,y0) (For this, the range of the histogram has been extended, to include all the points) I hope this helps to make it clearer how to do this in R. Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 20-Apr-10 Time: 23:28:07 -- 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] Results from clogit out of range?
On Apr 20, 2010, at 5:59 PM, Noah Silverman wrote:
Hi,
I'm calculating a conditional logit on some data stratified by group.
My understanding was that a conditional logit by definition returns a
value between 0 and 1 a a probability. Can anyone suggest why I'm
seeing results outside of the {0,1} range??
Probably because you did not read the help page for the function
predict.coxph. Pay special attention to the type argument.
type=c(lp, risk, expected, terms)
The call in R is:
m - clogit(score ~ val_1 + val_2 + strata(group), data=data)
Then
prediction - predict(m,newdata)
I do not see that you have defined any newdata.
A sample of the data with resulting predictive values is:
group score val_1 val_2prediction
1 2009-01-04_1 1 0.5913962 -1.121589 1.62455210
2 2009-01-04_1 1 0.6175472 -3.249820 -0.20093346
3 2009-01-04_1 1 0.5439640 -2.424501 0.46651849
4 2009-01-04_1 0 0.3745209 -2.477424 0.31263855
5 2009-01-04_1 0 0.6329855 -3.424174 -0.34200448
6 2009-01-04_1 0 0.4571999 -2.770247 0.11190788
7 2009-01-04_1 0 0.3822623 -2.259422 0.50627534
8 2009-01-04_1 0 0.2605742 -4.424806 -1.44566070
9 2009-01-04_1 0 0.4449604 -2.357060 0.46174993
10 2009-01-04_1 0 0.6595178 -2.246518 0.69427842
11 2009-01-04_1 0 0.5260032 -2.977887 -0.02393871
By default (which is what you have implicitly chosen) you are
requesting lp = linear predictors rather than risk.
--
David.
__
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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
West Hartford, CT
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using all variables in a linear model
lm can take a data frame whose first column is the response so: lm(rev(x)) On Tue, Apr 20, 2010 at 6:12 PM, Walter Yund IV wyun...@gmail.com wrote: Hello, I am trying to automate linear regression for many different datasets, each with the same rough format (the last variable is the response). I've been doing something like this: lm=lm(x[,dim(x)[2]] ~ ., data=x) where the dot denotes all variables. However, this means that the response is included as a predictor, which is obviously what I don't want. How do I request that all the columns in my dataset be used as predictors, except for the response? Thanks, Walter [[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] Results from clogit out of range?
Thanks David,
That explains a lot. I appreciate it.
--
Noah
On 4/20/10 3:48 PM, David Winsemius wrote:
On Apr 20, 2010, at 5:59 PM, Noah Silverman wrote:
Hi,
I'm calculating a conditional logit on some data stratified by group.
My understanding was that a conditional logit by definition returns a
value between 0 and 1 a a probability. Can anyone suggest why I'm
seeing results outside of the {0,1} range??
Probably because you did not read the help page for the function
predict.coxph. Pay special attention to the type argument.
type=c(lp, risk, expected, terms)
The call in R is:
m - clogit(score ~ val_1 + val_2 + strata(group), data=data)
Then
prediction - predict(m,newdata)
I do not see that you have defined any newdata.
A sample of the data with resulting predictive values is:
group score val_1 val_2prediction
1 2009-01-04_1 1 0.5913962 -1.121589 1.62455210
2 2009-01-04_1 1 0.6175472 -3.249820 -0.20093346
3 2009-01-04_1 1 0.5439640 -2.424501 0.46651849
4 2009-01-04_1 0 0.3745209 -2.477424 0.31263855
5 2009-01-04_1 0 0.6329855 -3.424174 -0.34200448
6 2009-01-04_1 0 0.4571999 -2.770247 0.11190788
7 2009-01-04_1 0 0.3822623 -2.259422 0.50627534
8 2009-01-04_1 0 0.2605742 -4.424806 -1.44566070
9 2009-01-04_1 0 0.4449604 -2.357060 0.46174993
10 2009-01-04_1 0 0.6595178 -2.246518 0.69427842
11 2009-01-04_1 0 0.5260032 -2.977887 -0.02393871
By default (which is what you have implicitly chosen) you are
requesting lp = linear predictors rather than risk.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Results from clogit out of range?
I just read the help page for predict.coxph.
It indicates that the risk score is just exp(lp)
What I'm trying to find, and have seen with some other implementations
is the conditional probability within group. Neither the lp or the
risk options seem to deliver this.
What am I missing?
-N
On 4/20/10 4:22 PM, Noah Silverman wrote:
Thanks David,
That explains a lot. I appreciate it.
--
Noah
On 4/20/10 3:48 PM, David Winsemius wrote:
On Apr 20, 2010, at 5:59 PM, Noah Silverman wrote:
Hi,
I'm calculating a conditional logit on some data stratified by group.
My understanding was that a conditional logit by definition returns a
value between 0 and 1 a a probability. Can anyone suggest why I'm
seeing results outside of the {0,1} range??
Probably because you did not read the help page for the function
predict.coxph. Pay special attention to the type argument.
type=c(lp, risk, expected, terms)
The call in R is:
m - clogit(score ~ val_1 + val_2 + strata(group), data=data)
Then
prediction - predict(m,newdata)
I do not see that you have defined any newdata.
A sample of the data with resulting predictive values is:
group score val_1 val_2prediction
1 2009-01-04_1 1 0.5913962 -1.121589 1.62455210
2 2009-01-04_1 1 0.6175472 -3.249820 -0.20093346
3 2009-01-04_1 1 0.5439640 -2.424501 0.46651849
4 2009-01-04_1 0 0.3745209 -2.477424 0.31263855
5 2009-01-04_1 0 0.6329855 -3.424174 -0.34200448
6 2009-01-04_1 0 0.4571999 -2.770247 0.11190788
7 2009-01-04_1 0 0.3822623 -2.259422 0.50627534
8 2009-01-04_1 0 0.2605742 -4.424806 -1.44566070
9 2009-01-04_1 0 0.4449604 -2.357060 0.46174993
10 2009-01-04_1 0 0.6595178 -2.246518 0.69427842
11 2009-01-04_1 0 0.5260032 -2.977887 -0.02393871
By default (which is what you have implicitly chosen) you are
requesting lp = linear predictors rather than risk.
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Changing the position of label title and rotating the title
Dear all, I want to plot a simple xy-plot but I have some problems with the label titles. In my plot, y-axis title should be at the top of the y-axis and it should be perpendicular to the axis. Similarly, x-axis title should be just next to the x-axis not at the bottom. By using mgp, I can change the position of the title vertically for y-axis and horizontally for x-axis but I want to change the positions in other direction too.Can I do this in R? If you can help me, I appreciate a lot. Thanks a lot, Ceylan [[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] Words appear to be bolded in the PDF output
Problem solved. Yes, I found that the text is overlaid several times! Thank you very much for your help, greatly appreciated. :) -- View this message in context: http://n4.nabble.com/Words-appear-to-be-bolded-in-the-PDF-output-tp2016971p2018219.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] sum specific rows in a data frame
Or try data.table 1.4 on r-forge, its grouping is faster than aggregate : agg datatable X100.012 0.008 X100 0.020 0.008 X1000 0.172 0.020 X1 1.164 0.144 X1e.05 9.397 1.180 install.packages(data.table, repos=http://R-Forge.R-project.org;) require(data.table) dt = as.data.table(df) t3 - system.time(zz3 - dt[, list(sumflt=sum(fltval), sumint=sum (intval)), by=id]) Matthew On Thu, 15 Apr 2010 13:09:17 +, hadley wickham wrote: On Thu, Apr 15, 2010 at 1:16 AM, Chuck vijay.n...@gmail.com wrote: Depending on the size of the dataframe and the operations you are trying to perform, aggregate or ddply may be better. In the function below, df has the same structure as your dataframe. Current version of plyr: agg ddply X100.005 0.007 X100 0.007 0.026 X1000 0.086 0.248 X1 0.577 3.136 X1e.05 4.493 44.147 Development version of plyr: agg ddply X100.003 0.005 X100 0.007 0.007 X1000 0.042 0.044 X1 0.410 0.443 X1e.05 4.479 4.237 So there are some big speed improvements in the works. Hadley __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] User inputs
Hi everyone, I have been searching for answers for the following questions but I don't have much success. The following questions may actually be quite simple. Any help would be greatly appreciated. (1) I have written a script which requires user input. I am using the readline() command.However, everytime when I run the script, R does not wait for the user input and proceed to the next line. Is there something like par(ask=T) to solve this problem? (2) In my script, I want it to stop running when a certain condition is met. I have tried using the stop() function, but apparently R only stops reading that line and start reading the following lines. I have also thought about quit() but it is not quite what I want. May someone please lead me to the right function please? (3) When a minor error happens, I would like to get the user permission by pressing the return key before the script continues to run. What function should I be looking at? Many thanks, Chris -- View this message in context: http://n4.nabble.com/User-inputs-tp2018251p2018251.html Sent from the R help mailing list archive at Nabble.com. [[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] Best subset of models for glm.nb()
Dear List, I am looking for a function that will find the best subset of negative binomial models. I have a large data set with 15 variables that I am interested in. I want an easy way to run all possible models and find a subset of the best models that I can then look at in more detail. I have found two functions that seem to provide what I am looking for, but am not sure which one (if either) are appropriate. glmulti() in package glmulti does an exhaustive search of all models and gives a number of candidate models to choose from based on your choice of Information Criterion. This seems to be exactly what I am after, but I found nothing about it on this list which makes me think there is some reason no one is using it. gl1ce() in package lasso2 uses the least absolute shrinkage and selection operator (lasso) to do something. I found it at another thread: http://tolstoy.newcastle.edu.au/R/help/05/03/0121.html I did not understand the paper it was based on, and want to know if it even does what I am interested in before investing a lot of time in trying to understand it. Yes, I have read about the problems with stepwise algorithms and am looking for a valid alternative to narrowing down models when you have a lot of data and a large number of variables your interested in. Any thoughts on either of these methods? Or should I be doing something else? Thanks for your help, Tim Tim Clark Department of Zoology University of Hawaii __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ggplot and scale_x_date
Hi all,
I have a question about setting arbitrary breaks/labels when using GGPLOT
and date/time data on the x-axis. I want to specify the breaks/limits
arbitrarily rather than using scale_x_date(major = 'blah'), much like when
arbitrarily defining breaks/labels using scale_x_discrete(breaks = blah,
labels = blah)
I have tried variants such as:
scale_x_date(major = seq(from = as.Date('2000-01-01'), to =
as.Date('2009-01-01'), by = 'months'));
etc but can't get this to work. I want to arbitrarily define the
breaks/labels because I want to be able to control when the ticks/labels
start (i.e. I want to be able to set where the first and last ticks/labels
occur - you do not appear to be able to do this using the major = 'blah'
command under scale_x_date).
R-version: 2.9.2
ggplotversion: 0.8.5
OS: Windows 7
Many thanks,
Liam
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Re: [R] Results from clogit out of range?
On Apr 20, 2010, at 7:31 PM, Noah Silverman wrote:
I just read the help page for predict.coxph.
It indicates that the risk score is just exp(lp)
What I'm trying to find, and have seen with some other implementations
is the conditional probability within group. Neither the lp or the
risk options seem to deliver this.
What am I missing?
I'm not sure that clogit lets you estimate that quantity:
http://tolstoy.newcastle.edu.au/R/help/06/02/21443.html
Do you have citations that back up your original assumption?
--
David,
-N
On 4/20/10 4:22 PM, Noah Silverman wrote:
Thanks David,
That explains a lot. I appreciate it.
--
Noah
On 4/20/10 3:48 PM, David Winsemius wrote:
On Apr 20, 2010, at 5:59 PM, Noah Silverman wrote:
Hi,
I'm calculating a conditional logit on some data stratified by
group.
My understanding was that a conditional logit by definition
returns a
value between 0 and 1 a a probability. Can anyone suggest why I'm
seeing results outside of the {0,1} range??
Probably because you did not read the help page for the function
predict.coxph. Pay special attention to the type argument.
type=c(lp, risk, expected, terms)
The call in R is:
m - clogit(score ~ val_1 + val_2 + strata(group), data=data)
Then
prediction - predict(m,newdata)
I do not see that you have defined any newdata.
A sample of the data with resulting predictive values is:
group score val_1 val_2
prediction
1 2009-01-04_1 1 0.5913962 -1.121589 1.62455210
2 2009-01-04_1 1 0.6175472 -3.249820 -0.20093346
3 2009-01-04_1 1 0.5439640 -2.424501 0.46651849
4 2009-01-04_1 0 0.3745209 -2.477424 0.31263855
5 2009-01-04_1 0 0.6329855 -3.424174 -0.34200448
6 2009-01-04_1 0 0.4571999 -2.770247 0.11190788
7 2009-01-04_1 0 0.3822623 -2.259422 0.50627534
8 2009-01-04_1 0 0.2605742 -4.424806 -1.44566070
9 2009-01-04_1 0 0.4449604 -2.357060 0.46174993
10 2009-01-04_1 0 0.6595178 -2.246518 0.69427842
11 2009-01-04_1 0 0.5260032 -2.977887 -0.02393871
By default (which is what you have implicitly chosen) you are
requesting lp = linear predictors rather than risk.
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David Winsemius, MD
West Hartford, CT
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Re: [R] Changing the position of label title and rotating the title
On Apr 20, 2010, at 5:41 PM, Ceylan Yozgatligil wrote: Dear all, I want to plot a simple xy-plot but I have some problems with the label titles. In my plot, y-axis title should be at the top of the y-axis and it should be perpendicular to the axis. Similarly, x-axis title should be just next to the x-axis not at the bottom. By using mgp, I can change the position of the title vertically for y-axis and horizontally for x- axis but I want to change the positions in other direction too.Can I do this in R? If you can help me, I appreciate a lot. ?mtext Thanks a lot, Ceylan [[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. David Winsemius, MD 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] 'cuts' option in levelplot()
Hi All, Does anyone knows what exactly cuts=some number (say 15) does in lattice's levelplot(). I doubt this but does it uses cut() internally, something like cut(z, 15)? Thanks, Gurmeet [[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] 'cuts' option in levelplot()
On Apr 20, 2010, at 10:43 PM, Gurmeet wrote: Hi All, Does anyone knows what exactly cuts=some number (say 15) does in lattice's levelplot(). I doubt this but does it uses cut() internally, something like cut(z, 15)? C'mon, man, look at the code: seq(zrng[1], zrng[2], length.out = cuts + 2) Thanks, Gurmeet David Winsemius, MD 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] Converting daily data series to monthly series
Hi Users, I have daily series of data from 1962 - 2000, with the data for February 29th in leap years excluded, leaving 365 daily values for each year. I wish to convert the daily series to monthly series. How can I do this using the zoo package or any other package? Thanks ZABLONE OWITI GRADUATE STUDENT Nanjing University of Information, Science and Technology College of International Education Add: 219 Ning Liu Rd, Nanjing, Jiangsu, 21004, P.R. China Tel: +86-25-58731402 Fax: +86-25-58731456 Mob. 15077895632 Website: www.nuist.edu.cn [[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] 'cuts' option in levelplot()
Thanks David! :P On Tue, Apr 20, 2010 at 11:02 PM, David Winsemius dwinsem...@comcast.netwrote: On Apr 20, 2010, at 10:43 PM, Gurmeet wrote: Hi All, Does anyone knows what exactly cuts=some number (say 15) does in lattice's levelplot(). I doubt this but does it uses cut() internally, something like cut(z, 15)? C'mon, man, look at the code: seq(zrng[1], zrng[2], length.out = cuts + 2) Thanks, Gurmeet David Winsemius, MD West Hartford, CT [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Converting daily data series to monthly series
There is a monthly aggregation of daily data example in the zoo-quickref vignette: vignette(zoo-quickref) and there are also relevant examples using aggregate.zoo or duplicated in the example section of ?aggregate.zoo On Tue, Apr 20, 2010 at 11:16 PM, zow...@ncst.go.ke zow...@ncst.go.ke wrote: Hi Users, I have daily series of data from 1962 - 2000, with the data for February 29th in leap years excluded, leaving 365 daily values for each year. I wish to convert the daily series to monthly series. How can I do this using the zoo package or any other package? Thanks ZABLONE OWITI GRADUATE STUDENT Nanjing University of Information, Science and Technology College of International Education Add: 219 Ning Liu Rd, Nanjing, Jiangsu, 21004, P.R. China Tel: +86-25-58731402 Fax: +86-25-58731456 Mob. 15077895632 Website: www.nuist.edu.cn [[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] Converting daily data series to monthly series
-Mensaje original- De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] En nombre de zow...@ncst.go.ke Enviado el: miércoles, 21 de abril de 2010 5:16 Para: r-help@r-project.org Asunto: [R] Converting daily data series to monthly series Hi Users, I have daily series of data from 1962 - 2000, with the data for February 29th in leap years excluded, leaving 365 daily values for each year. I wish to convert the daily series to monthly series. How can I do this using the zoo package or any other package? Thanks ZABLONE OWITI GRADUATE STUDENT Nanjing University of Information, Science and Technology College of International Education Let df be your dataframe, #In case you have to format your data as date before setting the montth df$date - as.Date(df$date, %d/%m/%Y) #Getting year, month and week from your correctly formatted date df$Year - as.numeric(format(df$date, %Y))#Year df$Month - as.numeric(format(df$date, %m))#Month df$Week - as.numeric(format(df$date, %W)) +1 #Week Dr. Rubén Roa-Ureta AZTI - Tecnalia / Marine Research Unit Txatxarramendi Ugartea z/g 48395 Sukarrieta (Bizkaia) SPAIN __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
