Re: [R] detecting a key press
The playSudoku function in the Sudoku package reacts to keypresses using windows or Tk, you can use that as an example. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Martin Teicher Sent: Friday, August 13, 2010 8:27 PM To: r-help@r-project.org Subject: [R] detecting a key press Hi Folks, I'm relatively new to r. I'd like to have a user respond by pressing a 1 or a 2 and determining their choice and the time of response. Previous postings have indicated that keyboard responses can be processed using scan and readline but both seem to wait for the user to also press return. Is there a way to detect the initial key press without requiring them to hit return? Thank you so much. Martin H. Teicher Department of Psychiatry McLean Hospital / Harvard Medical School Belmont, MA USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] is it possible to map 2 plots(c and d) in a single output plot?
is it possible to map 2 plots(c and d) in a single output plot? library(ggplot2) c - ggplot(mtcars, aes(qsec, wt)) d - ggplot (mtcars, aes(qsec, wt)) c + stat_smooth(fill=darkgrey, colour=blue, size=2, alpha = 0.2) d + stat_smooth(fill=darkgrey, colour=red, size=2, alpha = 0.2) -- View this message in context: http://r.789695.n4.nabble.com/is-it-possible-to-map-2-plots-c-and-d-in-a-single-output-plot-tp2325023p2325023.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to perform a substitution in a loop?
Hello all: I have a data series of 500 data, and I want to limit the value of it to be less than 1. Below is my code: for (i in 1:500) +if( x[i] 1) +x[i] = 1 but the system told me it's wrong. Can anyone told me the reason? - -- Saji Ren from Shanghai China GoldenHeart Investment Group -- -- View this message in context: http://r.789695.n4.nabble.com/How-to-perform-a-substitution-in-a-loop-tp2325048p2325048.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] Learning ANOVA
how to plot 4 graphs simultaneously? Depends on what graphs were talking about: A Side-by-side boxplot containing fx 4 boxplots can be generated by a single plot() command. However often you use a primary command that activivates the plot window and plots one graph, then afterwards you use secondary commands to add new graphical objects to the same graph/window: x-c(1,2,3,4) y-c(1,2,3,4) plot(x,y) #primary graphical command activates plot window abline(-1,3) # adds line points(c(1,2),c(2,2),col=red,pch=16) #adds points Also how to save InsectSprays.aov? I think I can only save it as InsectSprays.csv. I can't find write.aov command. InsectSprays.aov is an object - in this case probably a list. If you make an assignment using the assignment operator '-' to an object not already existing in you're current work session R will create that object. Or to put it differently: InsectSprays.aov is only a name you might as well have called it HomeGrownMagic or McDonalds - the aov in the name is not to be mistaken for a file extension. The name only surves the purpose to remind you what it is a container of namely an ANOVA output of the variables Insect and Spray, that helps you not youre computer. There are different ways to save: You can store the whole worksession, thereby storing all the objects you have created during worksession and have not removed. Or you can sink particular objects I recommend that you have a look at: http://cran.r-project.org/doc/contrib/Verzani-SimpleR.pdf Very well written, want tell you all you want to know but will not drown you in computer lingo and tie you down with details. Gives you some hands on experience handling the most basic plots and objects in R. If you later want to buy Verzanis book grown from the freely available notes you go wrong there either, basically more of the same. -- View this message in context: http://r.789695.n4.nabble.com/Learning-ANOVA-tp2323660p2325057.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 ti perform a substitution in a loop?
Saji Ren wrote: ... I have a data series of 500 data, and I want to limit the value of it to be less than 1. Below is my code: for (i in 1:500) +if( x[i] 1) +x[i] = 1 but the system told me it's wrong. Can anyone told me the reason? You don't show the error message. Read the posting guide. Try ?pmin And x - pmin(x,1) will probably do what you require. /Berend -- View this message in context: http://r.789695.n4.nabble.com/How-ti-perform-a-substitution-in-a-loop-tp2325007p2325062.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 perform a substitution in a loop?
Saji Ren wrote: Hello all: I have a data series of 500 data, and I want to limit the value of it to be less than 1. Below is my code: for (i in 1:500) +if( x[i] 1) +x[i] = 1 but the system told me it's wrong. Can anyone told me the reason? Works for me, so there's something you're not telling us Possibly that x is not a vector (a data frame, maybe?). (And it's wrong is not one of the official error messages in any language. Posting guide, reproducible code, etc.) The whole thing is an inefficient way to do x - pmin(x,1), though. -- Peter Dalgaard Center for Statistics, Copenhagen Business School Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to perform a substitution in a loop?
You need {}
for (i in 1:500){ ...commands...}
Like:
x-rep(c(0.8,1.2),250)
for (i in 1:500) {
if (x[i]1) x[i]=1
}
BR
JEsper
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Re: [R] How to perform a substitution in a loop?
that's not a problem of {}.
By using {} you can put the command in one line.
Still you can put the command in several lines as well by pur enter.
That's why in the second line, the becomes +
thank you for your reply though.
saji
-
--
Saji Ren
from Shanghai China
GoldenHeart Investment Group
--
--
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Re: [R] is it possible to map 2 plots(c and d) in a single output plot?
Hi: Your example doesn't make much sense as stated, but is this what you had in mind? d - ggplot (mtcars, aes(qsec, wt) d + geom_point() + stat_smooth(aes(colour = factor(cyl)), fill=darkgrey, size=2, alpha = 0.2) + scale_colour_discrete(No. cylinders) In this case, the multiple plots were obtained by grouping on different numbers of cylinders, which was coerced to a factor in order to use it as a grouping variable. There are many ways to apply this 'grouping' trick in ggplot2 using the companion plyr package, several of which can be found in the ggplot2 list archives http://had.co.nz/ggplot2/ (look for Mailing List) and in Hadley's book: http://tinyurl.com/ggplot2-book. HTH, Dennis On Fri, Aug 13, 2010 at 11:36 PM, newbie_2010 girishb...@gmail.com wrote: is it possible to map 2 plots(c and d) in a single output plot? library(ggplot2) c - ggplot(mtcars, aes(qsec, wt)) d - ggplot (mtcars, aes(qsec, wt)) c + stat_smooth(fill=darkgrey, colour=blue, size=2, alpha = 0.2) d + stat_smooth(fill=darkgrey, colour=red, size=2, alpha = 0.2) -- View this message in context: http://r.789695.n4.nabble.com/is-it-possible-to-map-2-plots-c-and-d-in-a-single-output-plot-tp2325023p2325023.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] what is scaling in plot
code rm(list=ls()) library(vegan) library(MASS) data(varespec) print(varespec) str(varespec) #PCA vare.pca - rda(varespec) vare.pca plot(vare.pca) sum(apply(varespec, 2, var)) biplot(vare.pca, scaling = -1) Elaine On Sat, Aug 14, 2010 at 10:48 AM, Ben Bolker bbol...@gmail.com wrote: elaine kuo elaine.kuo.tw at gmail.com writes: Pls kindly advise what scaling is in plot. Sometime it could be negative but sometimes it might be positive .(I guess it is the proportion between the plot and the margin) Your question is unclear. Please give more context and/or details. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Learning ANOVA
On Aug 14, 2010, at 3:38 AM, JesperHybel wrote: JH .how to plot 4 graphs simultaneously? Depends on what graphs were talking about: A Side-by-side boxplot containing fx 4 boxplots can be generated by a single plot() command. However often you use a primary command that activivates the plot window and plots one graph, then afterwards you use secondary commands to add new graphical objects to the same graph/window: x-c(1,2,3,4) y-c(1,2,3,4) plot(x,y) #primary graphical command activates plot window abline(-1,3) # adds line points(c(1,2),c(2,2),col=red,pch=16) #adds points Apparently JesperHybel was writing in response to a question by Stephen Liu (aka satimis when viewed on Nabble) on Aug 13, 2010; 11:04pm: JH Also how to save InsectSprays.aov? I think I can only save it as InsectSprays.csv. I can't find write.aov command. If you want to save an R object such as that returned by a function like aov, there are several ways. The save() function will create a compacted form that can then be read back into an R session with load(). The dump() and dput() functions will turn an object into an ASCII representation that can be saved as a text file. If you only want to have the text that is produced when you print it (and this is what you implicitly get when to just call the function at the console), then use sink() or capture.output(). InsectSprays.aov is an object - in this case probably a list. Not probably; it is a list. Look at it with str if you have questions. If you make an assignment using the assignment operator '-' to an object not already existing in you're current work session R will create that object. Or to put it differently: InsectSprays.aov is only a name you might as well have called it HomeGrownMagic or McDonalds - the aov in the name is not to be mistaken for a file extension. The name only surves the purpose to remind you what it is a container of namely an ANOVA output of the variables Insect and Spray, that helps you not youre computer. There are different ways to save: You can store the whole worksession, thereby storing all the objects you have created during worksession and have not removed. Or you can sink particular objects No, you cannot sink objects. You sink console output. I recommend that you have a look at: http://cran.r-project.org/doc/contrib/Verzani-SimpleR.pdf Very well written, want tell you all you want to know but will not drown you in computer lingo and tie you down with details. Gives you some hands on experience handling the most basic plots and objects in R. If you later want to buy Verzanis book grown from the freely available notes you go wrong there either, basically more of the same. -- View this message in context: http://r.789695.n4.nabble.com/Learning-ANOVA-tp2323660p2325057.html Sent from the R help mailing list archive at Nabble.com. 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] How to perform a substitution in a loop?
On Aug 14, 2010, at 4:05 AM, Saji Ren wrote:
In response to JesperHybel, saji wrote:
that's not a problem of {}.
Correct, as indicated by the continuation +
By using {} you can put the command in one line.
Actually the curly brackets are more often used to put a series of
commands on successive lines. If you want to force a series of
commands after an if() else construction then { will start the
series and the interpreter will continue interpreting until it reaches
the } possibly many lines later. If you want to put multiple
commands on a single line the way to do it is to separate them by
;'s and not hit enter until complete.
Still you can put the command in several lines as well by pur enter.
That's why in the second line, the becomes +
thank you for your reply though.
You and JesperHybel would both be following recommended list behavior
if you would both start copying the questions or response to which you
are replying. It is acceptable practice to trim the questions but it
is poor form to post naked replies without any context.
View this message in context:
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[R] cca biplot (vegan) failed in matplot
Dear List, I am trying to plot the result of cca using matplot but failed. Pls kindly help and thanks. Elaine The error message was error in xy.coords(x, y, xlabel, ylabel, log = log) : (list) object cannot be coerced to type 'double' code rm(list=ls()) library(vegan) library(MASS) # input richness birdrich -read.csv(c:/migration/CCA_richness_20100814.csv,header=T, row.names=1) # input environment # GID 1 row sum = 0 therefore deleted GID1 in birdenvi birdenvi -read.csv(c:/migration/CCA_envi_20100815.csv,header=T, row.names=1) birdrich.cca - cca(birdrich ~ ., birdenvi) birdrich.cca matplot(birdrich.cca) [[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] creation package
Dear r-help, I run R CMD INSTALL : c:\RpR CMD INSTALL namepackage *installing to library'c:/PROGRA~1/R/R-210~1-1/library' *installing *source* 'namepackage'... **R **preparing package for lazy loading **help Avis:./man/namepackage-package.Rd:25:'' inatttendue **installing help indices **building package indices **MD5sums *DONEnamepackage best 2010/8/13, Uwe Ligges lig...@statistik.tu-dortmund.de: In my previous maiul I asked you to run R CMD INSTALL at first (rather than R CMD check). You could also look into the mentioned file C:/Rp/namepackage.Rcheck/00install.out But we really need that file to understand what is going on. Uwe Ligges On 13.08.2010 15:12, anderson nuel wrote: Dear r-help, I try this command R CMD INSTALL,but still there are errors. I look into the installation log file in the namepackage.Rcheck : * using log directory 'C:/Rp/namepackage.Rcheck' * using R version 2.10.1 (2009-12-14) * using session charset: ISO8859-1 * checking for file 'namepackage/DESCRIPTION' ... OK * checking extension type ... Package * this is package 'namepackage' version '1.0' * checking package dependencies ... OK * checking if this is a source package ... OK * checking for executable files ... OK * checking whether package 'namepackage' can be installed ... ERROR Installation failed. See 'C:/Rp/namepackage.Rcheck/00install.out' for details. From That , I do not know where is the error I have only one function RT, but I need several other functions within the RT function, while using package.skeleton() I put in the function list seulemnt RT or any other functions, or I shall all functions in the same page?? Best 2010/7/31, Uwe Ligges lig...@statistik.tu-dortmund.de mailto:lig...@statistik.tu-dortmund.de: On 30.07.2010 15:44, anderson nuel wrote: Dear r-help, I create a package. When I installed this package (I use this command : R CMD check namepackage),I find an error: * checking whether package 'namepackage' can be installed ... ERROR Installation failed. Could you help me to find solution for this error. Well, either take a look into the installation log file in the namepackage.Rcheck directory or start a bit more slowly by just trying to install without a check in the first step by R CMD INSTALL. Without seeing the installation log, we cannot help either. Best, Uwe Ligges Best Regards [[alternative HTML version deleted]] __ R-help@r-project.org mailto:R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html 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] cca biplot (vegan) failed in matplot
Dear List, I am trying to plot the result of cca using matplot but failed. Pls kindly help and thanks. There is no 'matplot' method for cca objects. Where did you get the idea that there was one? If you want to plot the cca object, then ?plot.cca and ?ordiplot should help. Also look at the vignettes supplied with the package as they contain examples of plotting ordination objects. HTH Gavin Elaine The error message was error in xy.coords(x, y, xlabel, ylabel, log = log) : (list) object cannot be coerced to type 'double' code rm(list=ls()) library(vegan) library(MASS) # input richness birdrich -read.csv(c:/migration/CCA_richness_20100814.csv,header=T, row.names=1) # input environment # GID 1 row sum = 0 therefore deleted GID1 in birdenvi birdenvi -read.csv(c:/migration/CCA_envi_20100815.csv,header=T, row.names=1) birdrich.cca - cca(birdrich ~ ., birdenvi) birdrich.cca matplot(birdrich.cca) [[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] Stepwise Regression + entry/exit significance level
Hi R,
Does the step function used to perform stepwise regression has the
option to specify the entry/exit significance levels for the independent
variables? (This is similar to the 'slentry' and 'slstay' option in
'Proc reg' of SAS.). Or do we have any other package which does the
above? Thanks.
Thanks and Regards,
Shubha
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[R] Re : Difference in Monte Carlo calculation between chisq.test and fisher.test
Hello, Any idea? À : r-help@r-project.org Envoyé le : Jeu 12 août 2010, 12h 17min 10s Objet : [R] Difference in Monte Carlo calculation between chisq.test and fisher.test Hello all, I would like to know what the difference is between chisq.test and fisher.test when using the Monte Carlo method with simulate.p.value=TRUE? Thank you -- View this message in context: http://r.789695.n4.nabble.com/Difference-in-Monte-Carlo-calculation-between-chisq-test-and-fisher-test-tp2322494p2322494.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] cca biplot (vegan) failed in matplot
Hello, Thanks for the response. matplot was tried for its function to distinguish assigned parts of a matrix by letters, for I wanna differentiate the dissimilar columns in a matrix. Any suggestion based on the idea above will be highly appreciated, such as points. Elaine On Sat, Aug 14, 2010 at 8:16 PM, Gavin Simpson gavin.simp...@ucl.ac.ukwrote: Dear List, I am trying to plot the result of cca using matplot but failed. Pls kindly help and thanks. There is no 'matplot' method for cca objects. Where did you get the idea that there was one? If you want to plot the cca object, then ?plot.cca and ?ordiplot should help. Also look at the vignettes supplied with the package as they contain examples of plotting ordination objects. HTH Gavin Elaine The error message was error in xy.coords(x, y, xlabel, ylabel, log = log) : (list) object cannot be coerced to type 'double' code rm(list=ls()) library(vegan) library(MASS) # input richness birdrich -read.csv(c:/migration/CCA_richness_20100814.csv,header=T, row.names=1) # input environment # GID 1 row sum = 0 therefore deleted GID1 in birdenvi birdenvi -read.csv(c:/migration/CCA_envi_20100815.csv,header=T, row.names=1) birdrich.cca - cca(birdrich ~ ., birdenvi) birdrich.cca matplot(birdrich.cca) [[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] color vector for green, red, blue
Dear list, I am plotting dots using distinct colors according to the species. Like the style of topo.colors(18), please kindly advise if any color vector exists for green, red, and blue. Thank you Elaine [[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] what is scaling in plot
In an ordination such as this, you have two sets of scores; i) one pertaining to the (dis)similarity of your samples in terms of species composition, and ii) one pertaining to the species and which species co-occur. You can't display both sets of information and retain the meaning of the scores on the one plot. Scaling controls what relationships are preserved (focussing on sites or species etc). Gabriel (forget the citation now but is was in Biometrika) suggests that scaling 3 (symmetric scaling) is optimal and to be preferred. There is a discussion of the various scalings in the Design Decision vignette that you should read. If this is too complex, then perhaps start with Jan Leps and Petr Smilauer's book Multivariate analysis of ecological data using Canoco, 2003, Cambridge University Press. HTH Gavin code rm(list=ls()) library(vegan) library(MASS) data(varespec) print(varespec) str(varespec) #PCA vare.pca - rda(varespec) vare.pca plot(vare.pca) sum(apply(varespec, 2, var)) biplot(vare.pca, scaling = -1) Elaine On Sat, Aug 14, 2010 at 10:48 AM, Ben Bolker bbol...@gmail.com wrote: elaine kuo elaine.kuo.tw at gmail.com writes: Pls kindly advise what scaling is in plot. Sometime it could be negative but sometimes it might be positive .(I guess it is the proportion between the plot and the margin) Your question is unclear. Please give more context and/or details. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] discerning species by color in cca biplot
If you only have seven species, why not draw the label for the species (the names() component of your data) at the species score, rather than a colour? plot(OBJ, display = c(sites,species), scaling = 3, type = n) points(OBJ, display = sites, scaling = 3, type = p) points(OBJ, display = species, scaling = 3, type = t) Not tested, am away at the moment but heading home. If you are still having trouble (and I see now why you wanted matplot) then email back and I'll given a longer example illustrating who to build up plots by hand. Gavin Dear List, I am running constrained correspondence analysis for abundance data of 7 birds. However, I would like to check which bird prefers which environment gradient by showing the species with different colors of the dots in cca plot (package vegan). Please kindly help and thank you Elaine [[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] color vector for green, red, blue
Take a look in RColorBrewer package. On Sat, Aug 14, 2010 at 9:13 AM, elaine kuo elaine.kuo...@gmail.com wrote: Dear list, I am plotting dots using distinct colors according to the species. Like the style of topo.colors(18), please kindly advise if any color vector exists for green, red, and blue. Thank you Elaine [[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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] discerning species by color in cca biplot
Thanks I wanna to add the species labels in a cca biplot based on species and environment conditions (maybe sites are of less relevance here). 1. Gavin's suggestion worked but the dots previously displayed using plot(birdrich.cca, scaling = -1) were gone. Pls advise how to retain the dots. 2. orditorp orditorp in vegan seems to fit my demand. However, the axis seems to be changed after running the code below. Please kindly explain why or recommend any references. Thanks again. # Now with orditorp and abbreviated species names cnam - make.cepnames(names(birdrich)) plot(birdrich.cca, dis=sp, type=n) stems - colSums(birdrich) orditorp(birdrich.cca, sp, label = cnam, priority=stems, pch=+, pcol=grey) Elaine original code rm(list=ls()) plot.new() library(vegan) library(MASS) # input richness birdrich -read.csv(c:/migration/CCA_Mig_richness_20100814.csv,header=T, row.names=1) # input environment birdenvi -read.csv(c:/migration/CCA_Mig_envi_20100815.csv,header=T, row.names=1) birdrich.cca - cca(birdrich ~ ., birdenvi) birdrich.cca plot(birdrich.cca, scaling = 3, dis=sp, type=t) On Sat, Aug 14, 2010 at 8:30 PM, Gavin Simpson gavin.simp...@ucl.ac.ukwrote: If you only have seven species, why not draw the label for the species (the names() component of your data) at the species score, rather than a colour? plot(OBJ, display = c(sites,species), scaling = 3, type = n) points(OBJ, display = sites, scaling = 3, type = p) points(OBJ, display = species, scaling = 3, type = t) Not tested, am away at the moment but heading home. If you are still having trouble (and I see now why you wanted matplot) then email back and I'll given a longer example illustrating who to build up plots by hand. Gavin [[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 perform a substitution in a loop?
Hello, Mr Dalgaard: you're right about the problem. Works for me, so there's something you're not telling us Possibly that x is not a vector (a data frame, maybe?). the mistake message is below: Error in if (track01[i] 1) track01[i] - 1 : missing value where TRUE/FALSE needed the track01 above is my data, I've used a rnorm(500) data series instead, and the code works!! One thing I do know is that my track01 data has NaN data. So I guess the system can not perform the code if (track01[i] 1) when track01[i]=NaN, or the result is missing value, thus it need a TRUE/FALSE is there any command to replace the NaN value with 1 in an efficient way? Saji - -- Saji Ren from Shanghai China GoldenHeart Investment Group -- -- View this message in context: http://r.789695.n4.nabble.com/How-to-perform-a-substitution-in-a-loop-tp2325048p2325193.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] Stepwise Regression + entry/exit significance level
The values of slentry and slstay that will avoid ruining the
statistical properties of the result are slentry=1.0 and slstay=1.0.
Frank
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt University
On Sat, 14 Aug 2010, Shubha Vishwanath Karanth wrote:
Hi R,
Does the step function used to perform stepwise regression has the
option to specify the entry/exit significance levels for the independent
variables? (This is similar to the 'slentry' and 'slstay' option in
'Proc reg' of SAS.). Or do we have any other package which does the
above? Thanks.
Thanks and Regards,
Shubha
This e-mail may contain confidential and/or privileged i...{{dropped:13}}
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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] How to add lines to lattice plot produced by rms::bplot
Once you guys figure all this out, I'm glad to modify bplot to pass
more arguments lattice if needed.
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt University
On Fri, 13 Aug 2010, David Winsemius wrote:
On Aug 13, 2010, at 11:25 PM, Duncan Mackay wrote:
Hi David
I do not know if you have done something like this.
I had tried a few efforts like that, starting with an examination of
str(bp.plot) as you demonstrate.
I tried str(bp.plot) which gave the section about the regions (for
colours) as:
$ panel.args.common:List of 8
..$ x : num [1:2500] 27 28 29 29.9 30.9 ...
..$ y : num [1:2500] 141 141 141 141 141 ...
..$ z : num [1:2500] -1.43 -1.41 -1.39 -1.36 -1.34 ...
..$ at : num [1:10] -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
..$ region : logi FALSE
..$ zlab :List of 3
.. ..$ label: chr log odds
.. ..$ rot : num 90
.. ..$ cex : num 1
..$ labels : logi TRUE
..$ contour: logi TRUE
I tried (with a bplot object named bldLT40):
bldLT40$legend$right$args$key$at - c(-0.233, seq(.50, 2.25, by=0.25),
seq(2.5, 5.0, by=0.5), 6:10)
... and then tried
bldLT40$panel.args$at - c(-0.233, seq(.50, 2.25, by=0.25), seq(2.5,
5.0, by=0.5), 6:10)
Neither of these efforts changed the boundaries beteen colors in the
plot area. The first effort changed the legend scal,e but that just
created a misalignment of the colors of plot area and the legend. I
would be interested in either a strategy that lets one alter the color
level changes of the z variable (which in bplot-created objects is
zhat, or lets one specify the values at which contour lines are
drawn in contourplot.
Thanks for your efforts.
--
David.
I added the col.region and colours from a plot levelplot that I had
done to see what would occur to the trellis parameters. No colours
were produced when plotted.
bp.plot - bplot(p, lfun=contourplot, color.key = TRUE, col.regions
=
c
(#FF
,#00,#A9E2FF,#8080FF,#FF,#FFD18F,#FF) )
$ panel.args.common:List of 10
..$ x : num [1:2500] 27 28 29 29.9 30.9 ...
..$ y : num [1:2500] 141 141 141 141 141 ...
..$ z : num [1:2500] -1.43 -1.41 -1.39 -1.36 -1.34 ...
..$ at : num [1:10] -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
..$ region : logi FALSE
..$ color.key : logi TRUE
..$ zlab :List of 3
.. ..$ label: chr log odds
.. ..$ rot : num 90
.. ..$ cex : num 1
..$ labels : logi TRUE
..$ contour: logi TRUE
..$ col.regions: chr [1:7] #FF #00 #A9E2FF
#8080FF ...
So it has been added to the panel.args.common, whether you can
access these are another matter.
I then tried
bp.plot - bplot(p, lfun=contourplot, par.settings = list(axis.text
= list(cex = 0.65)), color.key = TRUE, col.regions =
c
(#FF
,#00,#A9E2FF,#8080FF,#FF,#FFD18F,#FF) )
which changed the size of the axis text so it may be the case of
having to add the col.regions etc to the appropriate list in
par.settings
I'll leave you to amend and access the colours.
You may have to add values for the wireframe/levelplot arguments
like at etc. and col.regions (I think that is the function) to
produce an appropriate colour range of your choice
It is a while since I have delved into these sorts of plots.
Need some sustenance.
Regards
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
ARMIDALE NSW 2351
Email home: mac...@northnet.com.au
At 10:33 14/08/2010, you wrote:
I have a plot produced by function bplot (package = rms) that is
really a lattice plot (class=trellis). It is similar to this plot
produced by a very minor modification of the first example on the
bplot help page:
require(rms)
n - 1000# define sample size
set.seed(17) # so can reproduce the results
age- rnorm(n, 50, 10)
blood.pressure - rnorm(n, 120, 15)
cholesterol- rnorm(n, 200, 25)
sex- factor(sample(c('female','male'), n,TRUE))
label(age)- 'Age' # label is in Hmisc
label(cholesterol)- 'Total Cholesterol'
label(blood.pressure) - 'Systolic Blood Pressure'
label(sex)- 'Sex'
units(cholesterol)- 'mg/dl' # uses units.default in Hmisc
units(blood.pressure) - 'mmHg'
# Specify population model for log odds that Y=1
L - .4*(sex=='male') + .045*(age-50) +
(log(cholesterol - 10)-5.2)*(-2*(sex=='female') + 2*(sex=='male'))
# Simulate binary y to have Prob(y=1) = 1/[1+exp(-L)]
y - ifelse(runif(n) plogis(L), 1, 0)
ddist - datadist(age, blood.pressure, cholesterol, sex)
options(datadist='ddist')
fit - lrm(y ~ blood.pressure + sex * (age + rcs(cholesterol,4)),
x=TRUE, y=TRUE)
p - Predict(fit, age, cholesterol, sex='male', np=50) # vary sex
last
bp.plot - bplot(p, lfun=contourplot)
bp.plot
I have tried a variety of efforts at using update (which I assume
is a
lattice function although I can find no help page for it. It does
appear in some of the lattice hep pages and my
Re: [R] How to perform a substitution in a loop?
Saji Ren wrote: Hello, Mr Dalgaard: you're right about the problem. Works for me, so there's something you're not telling us Possibly that x is not a vector (a data frame, maybe?). the mistake message is below: Error in if (track01[i] 1) track01[i] - 1 : missing value where TRUE/FALSE needed the track01 above is my data, I've used a rnorm(500) data series instead, and the code works!! One thing I do know is that my track01 data has NaN data. So I guess the system can not perform the code if (track01[i] 1) when track01[i]=NaN, or the result is missing value, thus it need a TRUE/FALSE is there any command to replace the NaN value with 1 in an efficient way? Usually, you'd safeguard the if with if (is.na(x[i]) || x[i] 1)... or if (!is.na(...) ...), depending on what you want. BTW, notice also that pmin has an na.rm argument. -- Peter Dalgaard Center for Statistics, Copenhagen Business School Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Stepwise Regression + entry/exit significance level
Shubha Vishwanath Karanth wrote: Hi R, Does the step function used to perform stepwise regression has the option to specify the entry/exit significance levels for the independent variables? (This is similar to the 'slentry' and 'slstay' option in 'Proc reg' of SAS.). Or do we have any other package which does the above? Thanks. That procedure is considered unreliable (someone called Frank is likely to chime in with a reference to a book on regression modeling strategies). Cloning obsolete SAS functionality is not high on the priority list. It is probably not massively hard to modify the step function to do it, though; all the building blocks would seem to be there. -- Peter Dalgaard Center for Statistics, Copenhagen Business School Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Re : Difference in Monte Carlo calculation between chisq.test and fisher.test
Emilie Chary wrote: Hello, Any idea? Different test statistic (chi^2 vs. table probability). À : r-help@r-project.org Envoyé le : Jeu 12 août 2010, 12h 17min 10s Objet : [R] Difference in Monte Carlo calculation between chisq.test and fisher.test Hello all, I would like to know what the difference is between chisq.test and fisher.test when using the Monte Carlo method with simulate.p.value=TRUE? Thank you __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 Dalgaard Center for Statistics, Copenhagen Business School Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] confidence Intervals for predictions in GLS
Hi everyone: Is there a function in R to calculate the confidence intervals for the predictions of a GLS(Generalized Least Square) model? The function predict gives confidence intervals for the predictions of other types of models (lm, glm, etc) but not gls. Any input will be much appreciated, Best, Camilo Camilo Mora, Ph.D. Department of Biology Dalhouisie University Halifax, Canada Phone: (902) 494-7720 http://as01.ucis.dal.ca/fmap/people.php?pid=53 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 add lines to lattice plot produced by rms::bplot
On Aug 14, 2010, at 9:59 AM, Frank Harrell wrote:
Once you guys figure all this out, I'm glad to modify bplot to pass
more arguments lattice if needed.
As always, Frank, I appreciate your support. In this case I think it's
not needed. What seems to be needed is simply the correct use of the
at argument. (I thought I had tried this before but apparently
mucked it up somehow.) This gives the desired color levels separation
and labeling of the default levelplot version of bplot output:
bldLT40 - bplot(mdl.pred , perim=boundaries, at=c(-0.233, seq(.50,
2.25, by=0.25), seq(2.5, 5.0, by=0.5), 6:10) )
bldLT40
And this produces the expected output with its contouplot version:
bldLT40c - bplot(mdl.pred , perim=boundaries, lfun=contourplot,
at=c(-0.233, seq(.50, 2.25, by=0.25), seq(2.5, 5.0, by=0.5), 6:10) )
bldLT40c
My only quibble with that last one was that the labels had three
digits to the right of the decimal pt but that happily went away when
I changed the low end to 0.25. All is good here.
--
David.
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt
University
On Fri, 13 Aug 2010, David Winsemius wrote:
On Aug 13, 2010, at 11:25 PM, Duncan Mackay wrote:
Hi David
I do not know if you have done something like this.
I had tried a few efforts like that, starting with an examination of
str(bp.plot) as you demonstrate.
I tried str(bp.plot) which gave the section about the regions (for
colours) as:
$ panel.args.common:List of 8
..$ x : num [1:2500] 27 28 29 29.9 30.9 ...
..$ y : num [1:2500] 141 141 141 141 141 ...
..$ z : num [1:2500] -1.43 -1.41 -1.39 -1.36 -1.34 ...
..$ at : num [1:10] -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
..$ region : logi FALSE
..$ zlab :List of 3
.. ..$ label: chr log odds
.. ..$ rot : num 90
.. ..$ cex : num 1
..$ labels : logi TRUE
..$ contour: logi TRUE
I tried (with a bplot object named bldLT40):
bldLT40$legend$right$args$key$at - c(-0.233, seq(.50, 2.25,
by=0.25),
seq(2.5, 5.0, by=0.5), 6:10)
... and then tried
bldLT40$panel.args$at - c(-0.233, seq(.50, 2.25, by=0.25), seq(2.5,
5.0, by=0.5), 6:10)
Neither of these efforts changed the boundaries beteen colors in the
plot area. The first effort changed the legend scal,e but that just
created a misalignment of the colors of plot area and the legend. I
would be interested in either a strategy that lets one alter the
color
level changes of the z variable (which in bplot-created objects is
zhat, or lets one specify the values at which contour lines are
drawn in contourplot.
Thanks for your efforts.
--
David.
I added the col.region and colours from a plot levelplot that I had
done to see what would occur to the trellis parameters. No colours
were produced when plotted.
bp.plot - bplot(p, lfun=contourplot, color.key = TRUE, col.regions
=
c
(#FF
,#00,#A9E2FF,#8080FF,#FF,#FFD18F,#FF) )
$ panel.args.common:List of 10
..$ x : num [1:2500] 27 28 29 29.9 30.9 ...
..$ y : num [1:2500] 141 141 141 141 141 ...
..$ z : num [1:2500] -1.43 -1.41 -1.39 -1.36 -1.34 ...
..$ at : num [1:10] -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
..$ region : logi FALSE
..$ color.key : logi TRUE
..$ zlab :List of 3
.. ..$ label: chr log odds
.. ..$ rot : num 90
.. ..$ cex : num 1
..$ labels : logi TRUE
..$ contour: logi TRUE
..$ col.regions: chr [1:7] #FF #00 #A9E2FF
#8080FF ...
So it has been added to the panel.args.common, whether you can
access these are another matter.
I then tried
bp.plot - bplot(p, lfun=contourplot, par.settings = list(axis.text
= list(cex = 0.65)), color.key = TRUE, col.regions =
c
(#FF
,#00,#A9E2FF,#8080FF,#FF,#FFD18F,#FF) )
which changed the size of the axis text so it may be the case of
having to add the col.regions etc to the appropriate list in
par.settings
I'll leave you to amend and access the colours.
You may have to add values for the wireframe/levelplot arguments
like at etc. and col.regions (I think that is the function) to
produce an appropriate colour range of your choice
It is a while since I have delved into these sorts of plots.
Need some sustenance.
Regards
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
ARMIDALE NSW 2351
Email home: mac...@northnet.com.au
At 10:33 14/08/2010, you wrote:
I have a plot produced by function bplot (package = rms) that is
really a lattice plot (class=trellis). It is similar to this plot
produced by a very minor modification of the first example on the
bplot help page:
require(rms)
n - 1000# define sample size
set.seed(17) # so can reproduce the results
age- rnorm(n, 50, 10)
blood.pressure - rnorm(n, 120, 15)
cholesterol- rnorm(n, 200, 25)
sex- factor(sample(c('female','male'), n,TRUE))
label(age)- 'Age' # label is in Hmisc
label(cholesterol)-
Re: [R] confidence Intervals for predictions in GLS
install.packages('rms')
require(rms)
?Gls
?plot.Predict
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt University
On Sat, 14 Aug 2010, Camilo Mora wrote:
Hi everyone:
Is there a function in R to calculate the confidence intervals for the
predictions of a GLS(Generalized Least Square) model?
The function predict gives confidence intervals for the predictions
of other types of models (lm, glm, etc) but not gls.
Any input will be much appreciated,
Best,
Camilo
Camilo Mora, Ph.D.
Department of Biology
Dalhouisie University
Halifax, Canada
Phone: (902) 494-7720
http://as01.ucis.dal.ca/fmap/people.php?pid=53
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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 add lines to lattice plot produced by rms::bplot
Frank E Harrell Jr Professor and ChairmanSchool of Medicine Department of Biostatistics Vanderbilt University On Sat, 14 Aug 2010, David Winsemius wrote: On Aug 14, 2010, at 9:59 AM, Frank Harrell wrote: Once you guys figure all this out, I'm glad to modify bplot to pass more arguments lattice if needed. As always, Frank, I appreciate your support. In this case I think it's not needed. What seems to be needed is simply the correct use of the at argument. (I thought I had tried this before but apparently mucked it up somehow.) This gives the desired color levels separation and labeling of the default levelplot version of bplot output: bldLT40 - bplot(mdl.pred , perim=boundaries, at=c(-0.233, seq(.50, 2.25, by=0.25), seq(2.5, 5.0, by=0.5), 6:10) ) bldLT40 And this produces the expected output with its contouplot version: bldLT40c - bplot(mdl.pred , perim=boundaries, lfun=contourplot, at=c(-0.233, seq(.50, 2.25, by=0.25), seq(2.5, 5.0, by=0.5), 6:10) ) bldLT40c My only quibble with that last one was that the labels had three digits to the right of the decimal pt but that happily went away when I changed the low end to 0.25. All is good here. -- David. Excellent. Thanks David. If you think of anything that would be a good addition to the examples or other parts of the help file, shoot me a few lines and I'll paste them in. Frank Frank E Harrell Jr Professor and ChairmanSchool of Medicine Department of Biostatistics Vanderbilt University On Fri, 13 Aug 2010, David Winsemius wrote: On Aug 13, 2010, at 11:25 PM, Duncan Mackay wrote: Hi David I do not know if you have done something like this. I had tried a few efforts like that, starting with an examination of str(bp.plot) as you demonstrate. I tried str(bp.plot) which gave the section about the regions (for colours) as: $ panel.args.common:List of 8 ..$ x : num [1:2500] 27 28 29 29.9 30.9 ... ..$ y : num [1:2500] 141 141 141 141 141 ... ..$ z : num [1:2500] -1.43 -1.41 -1.39 -1.36 -1.34 ... ..$ at : num [1:10] -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 ..$ region : logi FALSE ..$ zlab :List of 3 .. ..$ label: chr log odds .. ..$ rot : num 90 .. ..$ cex : num 1 ..$ labels : logi TRUE ..$ contour: logi TRUE I tried (with a bplot object named bldLT40): bldLT40$legend$right$args$key$at - c(-0.233, seq(.50, 2.25, by=0.25), seq(2.5, 5.0, by=0.5), 6:10) ... and then tried bldLT40$panel.args$at - c(-0.233, seq(.50, 2.25, by=0.25), seq(2.5, 5.0, by=0.5), 6:10) Neither of these efforts changed the boundaries beteen colors in the plot area. The first effort changed the legend scal,e but that just created a misalignment of the colors of plot area and the legend. I would be interested in either a strategy that lets one alter the color level changes of the z variable (which in bplot-created objects is zhat, or lets one specify the values at which contour lines are drawn in contourplot. Thanks for your efforts. -- David. I added the col.region and colours from a plot levelplot that I had done to see what would occur to the trellis parameters. No colours were produced when plotted. bp.plot - bplot(p, lfun=contourplot, color.key = TRUE, col.regions = c (#FF ,#00,#A9E2FF,#8080FF,#FF,#FFD18F,#FF) ) $ panel.args.common:List of 10 ..$ x : num [1:2500] 27 28 29 29.9 30.9 ... ..$ y : num [1:2500] 141 141 141 141 141 ... ..$ z : num [1:2500] -1.43 -1.41 -1.39 -1.36 -1.34 ... ..$ at : num [1:10] -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 ..$ region : logi FALSE ..$ color.key : logi TRUE ..$ zlab :List of 3 .. ..$ label: chr log odds .. ..$ rot : num 90 .. ..$ cex : num 1 ..$ labels : logi TRUE ..$ contour: logi TRUE ..$ col.regions: chr [1:7] #FF #00 #A9E2FF #8080FF ... So it has been added to the panel.args.common, whether you can access these are another matter. I then tried bp.plot - bplot(p, lfun=contourplot, par.settings = list(axis.text = list(cex = 0.65)), color.key = TRUE, col.regions = c (#FF ,#00,#A9E2FF,#8080FF,#FF,#FFD18F,#FF) ) which changed the size of the axis text so it may be the case of having to add the col.regions etc to the appropriate list in par.settings I'll leave you to amend and access the colours. You may have to add values for the wireframe/levelplot arguments like at etc. and col.regions (I think that is the function) to produce an appropriate colour range of your choice It is a while since I have delved into these sorts of plots. Need some sustenance. Regards Duncan Duncan Mackay Department of Agronomy and Soil Science University of New England ARMIDALE NSW 2351 Email home: mac...@northnet.com.au At 10:33 14/08/2010, you wrote: I have a plot produced by function bplot (package = rms) that is really a lattice plot (class=trellis). It is similar to this plot produced by a very minor modification of the first example on the
Re: [R] Bug in t.test?
On Sat, 14 Aug 2010, ted.hard...@manchester.ac.uk wrote: Hi Thomas, I'm not too sure about your interpretation. Consider: It seems hard to interpret The formula interface is only applicable for the 2-sample tests. any other way Johannes' original query was about differences when there are NAs, corresponding to different settings of na.action. It is perhaps possible that 'na.action=na.pass' and 'na.action=na.exclude' result in different pairings in the case paired=TRUE. However, it seems to me that the differences he observed are, shall we say, obscure! No, they are perfectly straightforward. Johannes's data had two missing values, one in each group, but not in the same pair. With na.omit or na.exclude, model.frame() removes the NAs. If there are the same number of NAs in each group, this leaves the same number of observations in each group. t.test.formula() splits these according to the group variable and passes them to t.test.default. Because of the (invalid) paired=TRUE argument, t.test.default assumes these are nine pairs and gets bogus answers. On the other hand with na.pass, model.frame() does not remove NAs. t.test.formula() passes two sets of ten observations (including missing observations) to t.test.default(). Because of the paired=TRUE argument, t.test.default() assumes these are ten pairs, which happens to be true in this case, and after deleting the two pairs with missing observations it gives the right answer. Regardless of the details, however, t.test.formula() can't reliably work with paired=TRUE because the user interface provides no way to specify which observations are paired. It would be possible (though bad idea in my opinion) to specify that paired=TRUE is allowed and that the pairing is done in the order the observations appear in the data. The minimal change would be to stop doing missing-value removal in t.test.formula, although that would be undesirable if a user wanted to supply some sort of na.impute() option. I would strongly prefer having an explicit indication of pairing, eg paired=variable.name, or even better, paired=~variable.name. Relying on data frame ordering seems a really bad idea. -thomas Thomas Lumley Professor of Biostatistics University of Washington, Seattle __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Creating list from a long vector
Stupid question, but its been a long night. If I have a long vector how can I turn it into a list of the same length x-rep(seq(1,100,by=1),each=10) [[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] Creating list from a long vector
Le 14/08/10 18:22, steven mosher a écrit : Stupid question, but its been a long night. If I have a long vector how can I turn it into a list of the same length x-rep(seq(1,100,by=1),each=10) Perhaps as.list ? -- Romain Francois Professional R Enthusiast +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr |- http://bit.ly/bzoWrs : Rcpp svn revision 2000 |- http://bit.ly/b8VNE2 : Rcpp at LondonR, oct 5th `- http://bit.ly/aAyra4 : highlight 0.2-2 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 graphing impulse response functions
Dear colleagues/contributors, I'd be pleased if someone could provide insights on how to plot impulse response functions in a format that can easily be copied in a word document just as plotting time-series of variables. I had followed the outline suggested by Benhard Pfaff [see http://127.0.0.1:17693/library/vars/html/irf.html] but I am unable to get the impulse response functions in a single graphical format. I then tried this command, where each variable response to an impulse was plotted: opar -par(mfrow = c(3,2), mar=c(4.2,4.2,1,1), oma=c(0,0,0,0)) plot(svec.irfT1,main=T-G, ylab=, xlab=) plot(svec.irfT2, main=T-i,ylab=) plot(svec.irfT3,main=T-Y,ylab=) plot(svec.irfT4,main=T-T,ylab=) plot(svec.irfT5,main=T-P,ylab=) plot(svec.irfT,ylab=) par(opar) However the graphs still fail to appear in a three row, two column format that could be easily copied to word or pdf. Kind assistance will be welcome. Kind regards, Faj R: Impulse response function [[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] cacheSweave / pgfSweave driver for package vignette
Thank you everyone, I think I'll follow the technique used in Rcpp (dummy vignettes + Makefile). It would be great to have a mechanism to select the driver for vignette generation though. Also, as a side-note, the choice of a driver has its own shortcoming; I cannot use the features of highlight and, say, pgfSweave in the same document. Sincerely, baptiste On 13 August 2010 11:10, Romain Francois romain.franc...@dbmail.com wrote: Hi, I've been meaning to ask the same question before. Le 13/08/10 11:01, baptiste auguie a écrit : Dear list, I wish to use a specific driver to process an sweave document in the inst/doc directory of a package. Specifically, I would like to use either cacheSweave or pgfSweave to speed up the creation of the vignette which requires lengthy computations. The same request would also apply to the highlight driver, to provide syntax highlighting of R chunks. In writing R extensions I see that during R CMD BUILD Sweave is run first, then optionally a makefile can be used to process any other files. It doesn't seem to leave room for a choice of Sweave engine, as far as I understand. One option I am thinking of is to change the extension of the source file to something like .Rnw2 so that Sweave ignores it altogether, and then use the appropriate command in a makefile. This is what we do in Rcpp to build our 7 vignettes (since we like to use the driver from highlight). One thing to have in mind is that R needs the .Rnw file to be present in doc once the package is installed, so that the vignette function works, hence some trickery in Rcpp. A way to control which sweave (and perhaps tangle) driver is to be used for a particular vignette would be very useful. Romain I have no experience in writing makefiles, so I'm hoping someone would already have solved this issue and could provide some advice. Sincerely, baptiste -- Romain Francois Professional R Enthusiast +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr |- http://bit.ly/b8VNE2 : Rcpp at LondonR, oct 5th |- http://bit.ly/aAyra4 : highlight 0.2-2 `- http://bit.ly/94EBKx : inline 0.3.6 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Creating list from a long vector
Thx, I see my problem. more sleep required On Sat, Aug 14, 2010 at 9:25 AM, Romain Francois romain.franc...@dbmail.com wrote: Le 14/08/10 18:22, steven mosher a écrit : Stupid question, but its been a long night. If I have a long vector how can I turn it into a list of the same length x-rep(seq(1,100,by=1),each=10) Perhaps as.list ? -- Romain Francois Professional R Enthusiast +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr |- http://bit.ly/bzoWrs : Rcpp svn revision 2000 |- http://bit.ly/b8VNE2 : Rcpp at LondonR, oct 5th `- http://bit.ly/aAyra4 : highlight 0.2-2 [[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] cacheSweave / pgfSweave driver for package vignette
Le 14/08/10 19:13, baptiste auguie a écrit : Thank you everyone, I think I'll follow the technique used in Rcpp (dummy vignettes + Makefile). Cool. Let us (Rcpp team) know if you find better ways or if something is too much trickery. It would be great to have a mechanism to select the driver for vignette generation though. Also, as a side-note, the choice of a driver has its own shortcoming; I cannot use the features of highlight and, say, pgfSweave in the same document. It should be possible to create a new driver that combines both features. Let me (highlight team) know if I can do something in highlight to make this easier than it currently is. Romain Sincerely, baptiste On 13 August 2010 11:10, Romain Francoisromain.franc...@dbmail.com wrote: Hi, I've been meaning to ask the same question before. Le 13/08/10 11:01, baptiste auguie a écrit : Dear list, I wish to use a specific driver to process an sweave document in the inst/doc directory of a package. Specifically, I would like to use either cacheSweave or pgfSweave to speed up the creation of the vignette which requires lengthy computations. The same request would also apply to the highlight driver, to provide syntax highlighting of R chunks. In writing R extensions I see that during R CMD BUILD Sweave is run first, then optionally a makefile can be used to process any other files. It doesn't seem to leave room for a choice of Sweave engine, as far as I understand. One option I am thinking of is to change the extension of the source file to something like .Rnw2 so that Sweave ignores it altogether, and then use the appropriate command in a makefile. This is what we do in Rcpp to build our 7 vignettes (since we like to use the driver from highlight). One thing to have in mind is that R needs the .Rnw file to be present in doc once the package is installed, so that the vignette function works, hence some trickery in Rcpp. A way to control which sweave (and perhaps tangle) driver is to be used for a particular vignette would be very useful. Romain I have no experience in writing makefiles, so I'm hoping someone would already have solved this issue and could provide some advice. Sincerely, baptiste -- Romain Francois Professional R Enthusiast +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr |- http://bit.ly/bzoWrs : Rcpp svn revision 2000 |- http://bit.ly/b8VNE2 : Rcpp at LondonR, oct 5th `- http://bit.ly/aAyra4 : highlight 0.2-2 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 with graphing impulse response functions
On Aug 14, 2010, at 11:43 AM, Musky Dee wrote:
Dear colleagues/contributors,
I'd be pleased if someone could provide insights on how to plot
impulse response functions in a format that can easily be copied in
a word document just as plotting time-series of variables.
I had followed the outline suggested by Benhard Pfaff [see http://127.0.0.1:17693/library/vars/html/irf.html
] but I am unable to get the impulse response functions in a single
graphical format. I then tried this command, where each variable
response to an impulse was plotted:
opar -par(mfrow = c(3,2), mar=c(4.2,4.2,1,1), oma=c(0,0,0,0))
plot(svec.irfT1,main=T-G, ylab=, xlab=)
plot(svec.irfT2, main=T-i,ylab=)
plot(svec.irfT3,main=T-Y,ylab=)
plot(svec.irfT4,main=T-T,ylab=)
plot(svec.irfT5,main=T-P,ylab=)
plot(svec.irfT,ylab=)
par(opar)
However the graphs still fail to appear in a three row, two column
format that could be easily copied to word or pdf. Kind assistance
will be welcome.
That's not too surprising. Look at the function that does the plotting :
class(irf1)
[1] varirf
getAnywhere(plot.varirf)
A single object matching ‘plot.varirf’ was found
It was found in the following places
registered S3 method for plot from namespace vars
namespace:vars
with value
function (x, plot.type = c(multiple, single), names = NULL,
main = NULL, sub = NULL, lty = NULL, lwd = NULL, col = NULL,
ylim = NULL, ylab = NULL, xlab = NULL, nc, mar.multi = c(0,
4, 0, 4), oma.multi = c(6, 4, 6, 4), adj.mtext = NA,
padj.mtext = NA, col.mtext = NA, ...)
{
op - par(no.readonly = TRUE)
on.exit(par(op))
.
.
remainder of function deleted
plot.varirf does its own layout, after setting asside you par
specifications and reloading them on exit, but if you look at the
argument list, I think it is clear that there was provision made for
adjustments to mar and oma arguments. I think you need to read the
docemantion (and maybe the code as well) more closely to see how the
authors expected that function to be used.
--
David.
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] Bug in t.test?
On 14-Aug-10 16:07:14, Thomas Lumley wrote: On Sat, 14 Aug 2010, ted.hard...@manchester.ac.uk wrote: Hi Thomas, I'm not too sure about your interpretation. Consider: It seems hard to interpret The formula interface is only applicable for the 2-sample tests. any other way Johannes' original query was about differences when there are NAs, corresponding to different settings of na.action. It is perhaps possible that 'na.action=na.pass' and 'na.action=na.exclude' result in different pairings in the case paired=TRUE. However, it seems to me that the differences he observed are, shall we say, obscure! No, they are perfectly straightforward. Johannes's data had two missing values, one in each group, but not in the same pair. With na.omit or na.exclude, model.frame() removes the NAs. If there are the same number of NAs in each group, this leaves the same number of observations in each group. t.test.formula() splits these according to the group variable and passes them to t.test.default. Because of the (invalid) paired=TRUE argument, t.test.default assumes these are nine pairs and gets bogus answers. On the other hand with na.pass, model.frame() does not remove NAs. t.test.formula() passes two sets of ten observations (including missing observations) to t.test.default(). Because of the paired=TRUE argument, t.test.default() assumes these are ten pairs, which happens to be true in this case, and after deleting the two pairs with missing observations it gives the right answer. Regardless of the details, however, t.test.formula() can't reliably work with paired=TRUE because the user interface provides no way to specify which observations are paired. It would be possible (though bad idea in my opinion) to specify that paired=TRUE is allowed and that the pairing is done in the order the observations appear in the data. The minimal change would be to stop doing missing-value removal in t.test.formula, although that would be undesirable if a user wanted to supply some sort of na.impute() option. I would strongly prefer having an explicit indication of pairing, eg paired=variable.name, or even better, paired=~variable.name. Relying on data frame ordering seems a really bad idea. -thomas Thanks, Thomas, for elucidating the mechanisms of what I had suspected. Following the earlier correspondence, I did a little experimentation which helps to visualise what goes on. I had been composing a draft email about it (see below) when yours arrived. It ends with a view on how things might be arranged to work more transparently. Ted. == I previously wrote (in connection with how the parings take place): Johannes' original query was about differences when there are NAs, corresponding to different settings of na.action. It is perhaps possible that 'na.action=na.pass' and 'na.action=na.exclude' result in different pairings in the case paired=TRUE. I have now tested the latter using Johannes Dietrich's example data. Results below. # Johannes' data: testdata.A - c(1.15, -0.2, NA, 1 , -2, -0.5, 0.1, 1.2, -1.4, 0.01) testdata.B - c(1.2 , 1.1, 3 , -0.1, 3, 1.1, 0 , 1.3, 4 , NA ) # Pairwise complete data: testdata.A2 - c(1.15, -0.2, 1 , -2, -0.5, 0.1, 1.2, -1.4) testdata.B2 - c(1.2 , 1.1, -0.1, 3, 1.1, 0 , 1.3, 4 ) # Groupwise complete data: testdata.A3 - c(1.15, -0.2, 1 , -2, -0.5, 0.1, 1.2, -1.4, 0.01) testdata.B3 - c(1.2 , 1.1, 3 , -0.1, 3, 1.1, 0 , 1.3, 4) ## Johannes: t.test(testdata.A, testdata.B, paired=TRUE, alternative=two.sided, na.action=na.exclude) # Paired t-test # data: testdata.A and testdata.B # t = -1.7921, df = 7, p-value = 0.1162 # alternative hypothesis: true difference in means is not equal to 0 # 95 percent confidence interval: # -3.5517017 0.4892017 # sample estimates: mean of the differences # -1.53125 ## Johannes: testdata - c(testdata.A, testdata.B); criterion - c(0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1); ## Johannes: # Formula-based, na.pass: print(t.test(testdata ~ criterion, paired=TRUE, alternative=two.sided, na.action=na.pass)) # Paired t-test # data: testdata by criterion # t = -1.7921, df = 7, p-value = 0.1162 # alternative hypothesis: true difference in means is not equal to 0 # 95 percent confidence interval: # -3.5517017 0.4892017 # sample estimates: mean of the differences # -1.53125 ## Pairwise complete: t.test(testdata.A2,testdata.B2,paired=TRUE) # Paired t-test # data: testdata.A2 and testdata.B2 # t = -1.7921, df = 7, p-value = 0.1162 # alternative hypothesis: true difference in means is not equal to 0 # 95 percent confidence interval: # -3.5517017 0.4892017 # sample estimates: mean of the differences # -1.53125 ## Johannes: # Formula-based, na.exclude: t.test(testdata ~ criterion, paired=TRUE, alternative=two.sided, na.action=na.exclude) # Paired t-test # data: testdata by criterion # t =
Re: [R] creation package
On 14.08.2010 11:21, anderson nuel wrote: Dear r-help, I run R CMD INSTALL : c:\RpR CMD INSTALL namepackage *installing to library'c:/PROGRA~1/R/R-210~1-1/library' *installing *source* 'namepackage'... **R **preparing package for lazy loading **help Avis:./man/namepackage-package.Rd:25:'' inatttendue **installing help indices **building package indices **MD5sums *DONEnamepackage best OK, but strange that checking the package fails now in its installation checks. Can you send us the file C:/Rp/namepackage.Rcheck/00install.out please. Uwe Ligges 2010/8/13, Uwe Liggeslig...@statistik.tu-dortmund.de: In my previous maiul I asked you to run R CMD INSTALL at first (rather than R CMD check). You could also look into the mentioned file C:/Rp/namepackage.Rcheck/00install.out But we really need that file to understand what is going on. Uwe Ligges On 13.08.2010 15:12, anderson nuel wrote: Dear r-help, I try this command R CMD INSTALL,but still there are errors. I look into the installation log file in the namepackage.Rcheck : * using log directory 'C:/Rp/namepackage.Rcheck' * using R version 2.10.1 (2009-12-14) * using session charset: ISO8859-1 * checking for file 'namepackage/DESCRIPTION' ... OK * checking extension type ... Package * this is package 'namepackage' version '1.0' * checking package dependencies ... OK * checking if this is a source package ... OK * checking for executable files ... OK * checking whether package 'namepackage' can be installed ... ERROR Installation failed. See 'C:/Rp/namepackage.Rcheck/00install.out' for details. From That , I do not know where is the error I have only one function RT, but I need several other functions within the RT function, while using package.skeleton() I put in the function list seulemnt RT or any other functions, or I shall all functions in the same page?? Best 2010/7/31, Uwe Liggeslig...@statistik.tu-dortmund.de mailto:lig...@statistik.tu-dortmund.de: On 30.07.2010 15:44, anderson nuel wrote: Dear r-help, I create a package. When I installed this package (I use this command : R CMD check namepackage),I find an error: * checking whether package 'namepackage' can be installed ... ERROR Installation failed. Could you help me to find solution for this error. Well, either take a look into the installation log file in the namepackage.Rcheck directory or start a bit more slowly by just trying to install without a check in the first step by R CMD INSTALL. Without seeing the installation log, we cannot help either. Best, Uwe Ligges Best Regards [[alternative HTML version deleted]] __ R-help@r-project.orgmailto:R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html 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] Simple problem with lm/predict
Hi all, I've got an xts time series with monthly OHLC Dow Jones industrial index data from 1980 to present, the data is in stored in x. I've done an OLS fit on the data in 1982::1994 and stored it in extrapolate1 (x[,4] contains the closing value for the index). t3 - seq(1980,1994,length = length(x[1980::1994,4])) t4-t3^2 extrapolate1 - lines(lm(x[1980::1994,4]~t3+t4)$fit) extrapolate1 Call: lm(formula = x[1980::1994, 4] ~ t3 + t4) Coefficients: (Intercept) t3 t4 3.161e+07 -3.205e+048.125e+00 The plot comes up with the appropriate line fit for 1980::1994, so I can safely say the model/fit is correct. I want to plot the predicted response of extrapolate into 1995 and beyond and after much googling have got about this close -- new.t - seq(1995,len=2*12,by = 1/12) new.dat - data.frame(Time = new.t, Seas = rep(1:12,2)) predict(extrapolate1,new.dat)[1:24] 12345678 704.9639 714.5726 724.2807 734.0882 743.9951 754.0014 764.1071 774.3122 9 10 11 12 13 14 15 16 784.6168 795.0207 805.5240 816.1267 826.8288 837.6303 848.5312 859.5315 17 18 19 20 21 22 23 24 870.6313 881.8304 893.1289 904.5268 916.0241 927.6209 939.3170 951.1125 Warning message: 'newdata' had 24 rows but variable(s) found have 180 rows This is straight from the net, the trouble is with how new.dat is set up but I can't figure it out... Regards, Nick [[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] Relation 1.5*IQR/Percentile in case of a normal Distribution
Hi, can someone point me at material to understand how in http://upload.wikimedia.org/wikipedia/commons/8/89/Boxplot_vs_PDF.png the fivenum-corresponding percentages might be calculated? Thanks, Joh __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Relation 1.5*IQR/Percentile in case of a normal Distribution
Johannes Graumann wrote: Hi, can someone point me at material to understand how in http://upload.wikimedia.org/wikipedia/commons/8/89/Boxplot_vs_PDF.png the fivenum-corresponding percentages might be calculated? Looks like a pretty straightforward application of pnorm() and qnorm(). pnorm(4*qnorm(.75), lower=F) # Q3 + 1.5 IQR = 4 Q3 since IQR = 2 Q3 [1] 0.003488302 gives the tail probabilities of .35%, and the rest is by definition. -- Peter Dalgaard Center for Statistics, Copenhagen Business School Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Unequal variance ANOVA using gls function in nlme
Hi I am trying to run an ANOVA on data with unequal variance. I am new to nlme, but to my understanding I need to use the gls function. I have single response variable (distance which is continuous) and the explanatory variable is individual ID (class variable: individuals differ in the variance in their distance values hence the need to using nlme). So I would create a model Distance~Individual ID and then create a variance structure to add to the model. However the variance structure is not a linear relationship, it is simply a series of variances linked to individual ID. Can anyone guide me as to how to fit this into the model? As the model is very simple, I think that the variance is the only component I need to estimate for the model, but perhaps I am mistaken? Thanks Sam Dr Samantha Patrick EU INTERREG Post Doc Davy 618 Marine Biology Ecology Research Centre University of Plymouth Plymouth PL4 8AA T: 01752 584699 M: 07740472719 [[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] incrementing matrix elements more efficiently
I need to increment cells of a matrix (collusionM). The indexes to increment are in an index (matchIndex). This is some of the code for (j in 1:(rows-1)) matchIndex[[j]] - which(mx[j]==mx) for (j in 1:(rows-1)) collisionM[j,matchIndex[[j]]] - collisionM[j,matchIndex[[j]]] + 1 I could put them in the same loop but this is slower. The first for statement is fine. It finds the matches and creates a list of the matches. The second is too slow, it increments the sparse matrix elements. Let me know if I need to put in more complete code to make this understandable. [[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] Trouble loading saved Rdata
In the particular application I have I save test.Rdata to a sub directory dir-Example dir.create(dir) test-data.frame(a=c(1,2,3),b=c(3,4,5) full-file.path(dir,test.Rdata,fsep=.Platform$file.sep) save(test,file=full) load(full) returns NULL it works fine when the object is saved to the working directory, but fails when saved to a sub directory. The Rdata is there. Bytes are in it. but loading it doesnt work. [[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] Trouble loading saved Rdata
That worked for me once I properly quoted test.RData on sessionInfo() R version 2.11.1 (2010-05-31) x86_64-pc-mingw32 locale: [1] LC_COLLATE=English_United States.1252 [2] LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C [5] LC_TIME=English_United States.1252 If correcting the quoting does not help you, perhaps you can report the results of sessionInfo() Cheers, Josh On Sat, Aug 14, 2010 at 5:14 PM, steven mosher mosherste...@gmail.com wrote: In the particular application I have I save test.Rdata to a sub directory dir-Example dir.create(dir) test-data.frame(a=c(1,2,3),b=c(3,4,5) full-file.path(dir,test.Rdata,fsep=.Platform$file.sep) save(test,file=full) load(full) returns NULL it works fine when the object is saved to the working directory, but fails when saved to a sub directory. The Rdata is there. Bytes are in it. but loading it doesnt work. [[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 Ph.D. Student, Health Psychology University of California, Los Angeles 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.
[R] incrementing matrix elements more efficiently
I need to increment cells of a matrix (collusionM). The indexes to increment are in an index (matchIndex). This is sample code library(seqinr) library(Matrix) x - abcabcabc mx - s2c(x) collisionM - Matrix(0,nrow=10, ncol=10, sparse=TRUE) matchIndex - list() rows - length(mx) for (j in 1:(rows)) matchIndex[[j]] - which(mx[j]==mx) for (j in 1:(rows)) collisionM[j,matchIndex[[j]]] - collisionM[j,matchIndex[[j]]] + 1 Works fine, except with my data (rows=32000) it is too slow. The problem is the second for loop, where it increments the index of the sparse matrix; this needs to be rewritten so it is more efficient. Any ideas? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] infinite recursion using anonymous function
In R 2.11.1 on Mac OS 10.6.4, the following code works, but if the print
statement is removed or commented out, then an infinite recursion results. Can
someone explain what is going on? Thanks. --Steve
foo2=function(f1, f2){
print(f1)
function(x){
f1(x) + f2(x) } }
foo=function(...) Reduce(foo2, list(...))
foo(sin, cos, tan)(1)
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Re: [R] infinite recursion using anonymous function
On Aug 14, 2010, at 6:23 PM, Steven C. Bagley wrote:
In R 2.11.1 on Mac OS 10.6.4, the following code works, but if the
print statement is removed or commented out, then an infinite
recursion results. Can someone explain what is going on? Thanks. --
Steve
foo2=function(f1, f2){
print(f1)
function(x){
f1(x) + f2(x) } }
foo=function(...) Reduce(foo2, list(...))
foo(sin, cos, tan)(1)
?force
foo2=function(f1, f2){
+ force(f1)
+ function(x){
+f1(x) + f2(x) } }
foo=function(...) Reduce(foo2, list(...))
foo(sin, cos, tan)(1)
[1] 3.956286
--
David.
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Re: [R] Dealing with data
Your second fit makes no sense, as you can easily tell if you look at the regression summaries. Fitting with spray as a categorical variable gives you an overall p-value of less than 2.2e-16, while fitting with as.numeric(spray) gives an overall p-value of .2118. The fit you've done with as.numeric induces a completely invalid model, as others have tried to point out. Jonathan On Fri, Aug 13, 2010 at 1:55 PM, TGS cran.questi...@gmail.com wrote: # I wasn't trying to do ANOVA. I was simply trying to figure out how regress count on sprays (this is after I saw another poster asking an unrelated question with the InsectSprays dataset). # # Anyhow, David clarified this but also, thanks for your explanation as well. rm(list = ls()); sprays - as.numeric(InsectSprays$spray) lm(formula = count ~ 0 + spray, data = InsectSprays) lm(formula = count ~ 0 + sprays, data = InsectSprays) # besides the point, in the ANOVA problem the degrees of freedom would be 5, not 1. On Aug 13, 2010, at 12:27 PM, Greg Snow wrote: So you want 1 degree of freedom for InsectSprays? You believe that the difference between A and B is exactly the same as between B and C which is exactly the same as between D and E (etc.)? that seems an odd assumption, but you can get that by using as.numeric (as I and others have already stated). If on the other hand you want InsectSprays to be treated correctly with the correct number of degrees of freedom, but have the output on a single line testing the overall effect, then you want to use the aov function rather than lm (internally they do the same thing, but the default summary output for aov is 1 line per term). Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: TGS [mailto:cran.questi...@gmail.com] Sent: Friday, August 13, 2010 11:51 AM To: Greg Snow Cc: r-help@r-project.org Subject: Re: [R] Dealing with data # Greg, if R automatically does that then I don't know why it's treating each indicator # as a different regressor. In other words, I am interested in treating 'spray' as one # independent variable. # # Erik, which book do you suggest I read? Thanks. data(InsectSprays) lm(InsectSprays$count ~ 0 + InsectSprays$spray) On Aug 13, 2010, at 10:34 AM, Greg Snow wrote: R/S does all of that automatically for you, you do not need to manually create the indicator variables. If you do something like: fit - lm( Sepal.Width ~ Species, data=iris, x=TRUE) Then look at the matrix actually used: fit$x Or the output: summary(fit) You will see that Species was automatically converted into indicator variables and those were used in the regression. If you really need the indicator variables yourself, look at the model.matrix function, e.g.: model.matrix( ~Species, data=iris ) Or model.matrix( ~Species - 1, data=iris ) If you really want 1 for A, 2 for B, etc. then look at as.numeric on a factor variable (e.g. as.numeric(iris$Species) ). Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of TGS Sent: Friday, August 13, 2010 11:22 AM To: David Winsemius Cc: r-help@r-project.org Subject: Re: [R] Dealing with data To clarify, I'd like to create a column of indicators for the respective letters so that I could maybe do regression on indicators, etc. For instance, A gets 1, B gets 2, and so on. On Aug 13, 2010, at 10:19 AM, David Winsemius wrote: On Aug 13, 2010, at 1:03 PM, TGS wrote: # how would I code in R to look at the letter of the alphabet # in the second column and create a indicator column for the # corresponding letter? data(InsectSprays) InsectSprays$spray It's already what most people mean when they say indicator column, i.e., a factor variable (and not a character vector) so, what do _you_ mean? -- 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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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
[R] band pass filter
Hello list, Is there any way to bandpass filter in R thanks nuncio -- Nuncio.M Research Scientist National Center for Antarctic and Ocean research Head land Sada Vasco da Gamma Goa-403804 [[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] band pass filter
Try this:
library(sos)
findFn('bandpass')
If necessary, install the package first. I got 24 hits in eight packages.
Hopefully one or more of them will be suitable.
HTH,
Dennis
On Sat, Aug 14, 2010 at 8:52 PM, nuncio m nunci...@gmail.com wrote:
Hello list,
Is there any way to bandpass filter in R
thanks
nuncio
--
Nuncio.M
Research Scientist
National Center for Antarctic and Ocean research
Head land Sada
Vasco da Gamma
Goa-403804
[[alternative HTML version deleted]]
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Re: [R] 64 bit RSQLite
Hi Stephen, On 8/12/10 7:10 PM, Stephen Liu wrote: Hi folks, Ubuntu 10.04 64 bit Where can I find 64 bit RSQLite? It seems not there; RSQLite: SQLite interface for R http://cran.r-project.org/web/packages/RSQLite/index.html You should be able to install the RSQLite source package to obtain a 64 bit package. install.packages(RSQLite, type=source) + seth __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
