Re: [R] How to shadow 'power' area?
Look at the following link: http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=88 This should be pretty close to what you want. HTH -Christos > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Andrej Kastrin > Sent: Monday, June 25, 2007 4:51 PM > To: R-help > Subject: [R] How to shadow 'power' area? > > Dear all, > > Suppose I plot two normal distributions (A and B) side by > side and add vertical line which hipotheticaly represent > alpha value; e.g.: > > x <- seq(-3.5,5, length=1000) > y <- dnorm(x) > # Plot distribution A > plot(y~x, type='l',axes=F,xlab="",ylab="",lwd=2) > # Plot distribution B > y2 <- dnorm(x-1.5) > lines(y2~x,lwd=2) > # Plot vertical line for alpha value > abline(h=0) > segments(qnorm(.5)+1.5,0,qnorm(.5)+1.5,dnorm(qnorm(.5))) > text(2,0.2,"Power") > > Now I want to shadow area labeled as "Power". Any suggestion > how to do that using 'polygon' function? > > Thanks in advance for any suggestion. > > Andrej > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] vectorize a function
How about:
sum(sapply(unique(a), function(x) {b <- which(a==x); sum(M[b, b])}))
HTH
-Christos
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Robin Hankin
> Sent: Friday, June 22, 2007 9:28 AM
> To: RHelp help
> Subject: [R] vectorize a function
>
> Hello everyone
>
> suppose I have an integer vector "a" of length "n" and a
> symmetric matrix "M" of size n-by-n.
>
> Vector "a" describes a partition of a set of "n" elements and
> matrix M describes a penalty function: row i column j
> represents the penalty if element i and element j are in the
> same partition.
>
> Toy example follows; the real case is much larger and I need
> to evaluate my penalty function many times.
>
> If a <- c(1,1,2,1,3) then elements 1,2,4 are in the same
> partition; element 3 is in a partition on its own and element
> 5 is in a partition on its own.
>
> The total penalty can be described by the following (ugly)
> function:
>
> f <- function(a,M){
>out <- 0
>for(i in unique(a)){
> out <- out + sum(M[which(a==i),which(a==i)])
>}
>return(out)
> }
>
>
> so with
>
> M <- matrix(rpois(25,3),5,5)
> M <- M+t(M)
> diag(M) <- 0
> a <- c(1,2,1,1,3)
>
> f(a,M) gives the total penalty.
>
>
> QUESTION: how to rewrite f() so that it has no loop?
>
>
>
>
>
>
> --
> Robin Hankin
> Uncertainty Analyst
> National Oceanography Centre, Southampton European Way,
> Southampton SO14 3ZH, UK
> tel 023-8059-7743
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
__
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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] Removing vertical line in Tinn R editor
Go to Options/Main/Editor Click on EdgeColor and select white as the color (or to whatever color you have set your page's background). This will make the edge line disappear. -Christos > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Judith Flores > Sent: Wednesday, June 06, 2007 1:50 PM > To: RHelp > Subject: [R] Removing vertical line in Tinn R editor > > Hi, > >I haven't been able to figure out how to remove a vertical > line that appears when I open an r file. How can I do this? > > Thank you, > > Judith > > > > __ > __ > Looking for a deal? Find great prices on flights and hotels > with Yahoo! FareChase. > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] pie chart in lattice - trellis class
In case you haven't seen this, there is an example in Paul Murrell's book that plots temperatures on a map using 'thermometer' charts. I would imagine it should be relatively straight forward to combine the floating.pie function with Paul's grid-base code (but I have not tried it myself). http://www.stat.auckland.ac.nz/~paul/RGraphics/chapter7.html See Figure 7.18 and code -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Ben Bolker > Sent: Monday, May 28, 2007 8:51 AM > To: r-help@stat.math.ethz.ch > Subject: Re: [R] pie chart in lattice - trellis class > > P > > Yes indeed. Thats' likely what I am going to do. Anyway, to > plot axes, > > labels of sophisticated graphs on maps may be interesting > anyway. For > > instance, we are monitoring fox and hare populations in > tens of game > > areas. Drawing observations (panel.xyplot) over time and > representing > > the trend variations (panel.loess) at the very place on the > map where > > the observations were done gives an absolutely interesting > view where > > spatial relationships between trends can be visualized. > > > > Patrick > > There is a floating.pie in the plotrix package, and a > hidden floating.pie.asp function in the ape package. I agree > that grid objects would be a more elegant way to implement these ... > (The standard argument is that "thermometers" or > mini-barplots would be a better way to view this information, > but I agree that pie charts seem familiar to people.) I have > the feeling that I've seen pie-charts-on-maps somewhere ... > searching the R Graphics Gallery for "pie" also produces the > "hexbin pie" plot (which doesn't use grid either ...) > > Ben Bolker > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] names of objects in .rda
You're right. I didn't realize that what happens when attaching is that "a new environment is created on the search path and the elements of a list (including columns of a data frame) or objects in a save file or an environment are *** copied *** into the new environment" Thanks. -Christos > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Seth Falcon > Sent: Friday, May 11, 2007 1:32 PM > To: r-help@stat.math.ethz.ch > Subject: Re: [R] names of objects in .rda > > "Christos Hatzis" <[EMAIL PROTECTED]> writes: > > > An approach would be to attach it and then use ls() > > No, that really is not an approach. If you load it, then > there is no problem to read the names. The point is not to > load it. This is important when dealing with large objects > or large collections of objects. > > + seth > > -- > Seth Falcon | Computational Biology | Fred Hutchinson Cancer > Research Center http://bioconductor.org > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] names of objects in .rda
An approach would be to attach it and then use ls() attach(myarchive.rda) ls(pos=2) detach(myarchive.rda) -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of > Benilton Carvalho > Sent: Friday, May 11, 2007 11:54 AM > To: [EMAIL PROTECTED] server posting > Subject: [R] names of objects in .rda > > Hi everyone, > > sorry if this was discussed before (and in this situation, > could you please point me to the discussion in the archive? > My search didn't seem to be effective). > > Is there a way of getting the names of objects in a .rda file > without having to load it? > > Thank you very much, > > benilton > > -- > PhD Candidate > Department of Biostatistics > Bloomberg School of Public Health > Johns Hopkins University > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] voronoi.mosaic chokes?
Have you tried 'debug'?
With one of the datasets that crashes your system, run the first
voronoi.mosaic, then the torus function and then
debug(voronoi.mosaic)
and run through the second call of voronoi.mosaic. You can step though the
code and at least will you find the point where it bombs.
-Christos
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Andrew Pierce
> Sent: Wednesday, May 09, 2007 12:21 AM
> To: r-help@stat.math.ethz.ch
> Subject: [R] voronoi.mosaic chokes?
>
> Hi all,
>
> I am running R 2.5.0 under Windows XP Media Center Edition.
> Here's a problem that's been stumping me for a few days now,
> and I can't find anything useful in the archives. I am using
> voronoi.mosaic (tripack
> package) to create proximity polygons for a study of
> vegetation competition and dynamics. The points lists are
> read in from a file for each plot, then 8 duplicates are
> translated around the edges of the plot (Toroidal edge
> correction). This is completed using the torus(...) function
> that I wrote myself.
>
> VMuncorrected is the voronoi mosaic that is not toroidally
> edge corrected VMcorrected is the voronoi mosaic that is
> toroidally edge corrected
>
> >treemap = read.table('af1.txt', sep = "\t", header = 1)
> >VMuncorrected = voronoi.mosaic(treemap$X, treemap$Y)
>
> ###Use the torus function to create 8 copies around the study
> region >toroid = torus(treemap$X, treemap$Y, 25, 25)
>
> >VMcorrected = voronoi.mosaic(toroid[,1], toroid[,2],
> duplicate = "remove")
>
> Here's the problem. When I read in the points for the data
> file listed above ('af1.txt'), both calls to voronoi.mosaic()
> work fine. (The second one takes about 1.5 seconds because
> there are 1147 points in the toroidally corrected set).
>
> However, when I read in the points from the next file
> ('af2.txt'), the first call to voronoi.mosaic() works. The
> next call (to torus()) also works fine. But the second call
> to voronoi.mosaic() causes R to freeze completely requiring
> Ctrl-Alt-Del.
>
> I have 10 sets of points and this problem happens for about 5 of them.
>
> Factors I have ruled out:
> -too many points in the call (one set had 1147 and worked
> fine while the next set had 801 and froze R) -duplicate
> points (taken care of by voronoi.mosaic(..., duplicate =
> "remove") and also independently verified by exporting the
> data. no duplicates in either the original or the toroidally
> corrected set) -points too close together in space (minimum
> distance between two points in 'af1.txt' is 0.1414 and works
> fine. minimum distance in the second set, 'af2.txt', is
> 0.2236, and this set causes R to freeze) -not enough memory
> (each data set is run in a new R session-i.e. R was quit
> between each attempt) -'flukiness' (the problem happens the
> same way every time for the problem data sets, and the code
> runs fine every time for the non-problem data sets) -file
> formats (each text file has the same number of columns, all
> the labels for the columns are identical, and the columns are
> always in the same order) -outdated versions (I am using R
> 2.5.0 and updated the tripack package within the last week.
> also, I update packages about once a month)
>
> This is a very frustrating problem because I get no errors
> indicating any problem with the data, and I have checked the
> data over and over for any type of error and found none. If
> anyone has ANY helpful suggestions, I would love to hear
> about any and all of them.
>
> Andrew
>
> p.s. - for those of you who are really intrigued, I can email
> you the .txt files and the code for the torus() function.
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] A function for raising a matrix to a power?
Here is a recursive version of the same function:
"%^%" <- function(A, n) if(n == 1) A else A %*% (A %^% (n-1))
> a <- matrix(1:4, 2)
> a %^% 1
[,1] [,2]
[1,]13
[2,]24
> a %^% 2
[,1] [,2]
[1,]7 15
[2,] 10 22
> a %^% 3
[,1] [,2]
[1,] 37 81
[2,] 54 118
-Christos
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Ron E.
> VanNimwegen
> Sent: Sunday, May 06, 2007 10:48 AM
> To: r-help@stat.math.ethz.ch; [EMAIL PROTECTED]
> Subject: Re: [R] A function for raising a matrix to a power?
>
> > Hi,
> >
> > Is there a function for raising a matrix to a power? For example if
> > you like to compute A%*%A%*%A, is there an abbreviation
> similar to A3?
> >
> > Atte Tenkanen
> Hi Atte,
>
> I was looking for a similar operator, because R uses scalar products
> when raising a matrix to a power with "^". There might be something
> more elegant, but this little loop function will do what you
> need for a
> matrix "mat" raised to a power "pow":
>
> mp <- function(mat,pow){
> ans <- mat
> for ( i in 1:(pow-1)){
> ans <- mat%*%ans
> }
> return(ans)
> }
>
> Then, for your example:
> > > A=rbind(c(1,1),c(-1,-2))
> > > mp(A,3)
> > [,1] [,2]
> > [1,]12
> > [2,] -2 -5
> Cheers,
> Ron
>
> ---
> Ron E. VanNimwegen
> Ph.D. Candidate, Division of Biology (EEB)
> Kansas Cooperative Fish & Wildlife Research Unit
> 205 Leasure Hall
> Kansas State University
> Manhattan, KS 66506-3501
> [EMAIL PROTECTED]
> ---
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
__
R-help@stat.math.ethz.ch mailing list
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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] R package development in windows
After cleaning up your workspace to keep only the objects that you will need to have in your "package", you can save it as MyStuff.RData save(list=ls(), file=file.path(my.package.dir, "MyStuff.RData")) To use this "package" in future sessions all you need is attach(file.path(my.package.dir, "MyStuff.RData")) This will make your objects available in your current session without cluttering your workspace. Of course, there is a price to pay for simplicity (i.e. no error checks, no documentation etc). -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Lucke, Joseph F > Sent: Friday, May 04, 2007 10:03 AM > To: Doran, Harold; Duncan Murdoch > Cc: r-help@stat.math.ethz.ch > Subject: Re: [R] R package development in windows > > Might there be an (semi-)automated procedure to create a > minimal, personal package, for my eyes only, that I can load > with a "libray(MyStuff)" command? This would be preferable > to having to > source() the files. Is there already such a procedure? > Joe > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Doran, Harold > Sent: Thursday, May 03, 2007 2:33 PM > To: Duncan Murdoch > Cc: r-help@stat.math.ethz.ch > Subject: Re: [R] [SPAM] - Re: R package development in > windows - BayesianFilter detected spam > > Thanks, Duncan. I'll look into that. Is there an > authoritative document that codifies the new package > development procedures for 2.5.0 (windows-specific), or is > that Writing R Extensions? In this thread alone I've received > multiple emails pointing to multiple web sites with > instructions for windows. Inasmuch as its appreciated, I'm a > bit confused as to which I should consider authoritative. > > I do hope I can resolve this and appreciate the help I've received. > However, I feel a bit compelled to note how very difficult > this process is. > > Harold > > > > -Original Message- > > From: Duncan Murdoch [mailto:[EMAIL PROTECTED] > > Sent: Thursday, May 03, 2007 3:24 PM > > To: Doran, Harold > > Cc: Gabor Grothendieck; r-help@stat.math.ethz.ch > > Subject: [SPAM] - Re: [R] R package development in windows > - Bayesian > > Filter detected spam > > > > On 5/3/2007 3:04 PM, Doran, Harold wrote: > > > Thanks Gabor, Sundar, and Tony. Indeed, Rtools was > missing from the > > > path. With that resolved, and another 10 minute windows > > restart, I get > > > the following below. The log suggests that hhc is not > installed. It > > > is, and, according to the directions I am following, I have > > placed it > > > in the c:\cygwin directory. > > > > I think the problem is that you are following a real mix of > > instructions, and they don't make sense. > > > > It would be nice if folks would submit patches to the R > Admin manual > > (or to the Rtools web site) rather than putting together web sites > > with advice that is bad from day one, and quickly gets > worse when it > > is not updated. > > > > > BTW, package.skeleton() doesn't seem to create the correct > > DESCRIPTION > > > template. I had to add the DEPENDS line. Without this, I > > get another > > > error. > > > > > > > > > C:\Program Files\R\R-2.4.1\bin>Rcmd build --force --binary g:\foo > > > > R 2.4.1 is no longer current; the package building > instructions in R > > 2.5.0 have been simplified a bit. You might want to try those. > > > > Duncan Murdoch > > > > > * checking for file 'g:\foo/DESCRIPTION' ... OK > > > * preparing 'g:\foo': > > > * checking DESCRIPTION meta-information ... OK > > > * removing junk files > > > * checking for LF line-endings in source files > > > * checking for empty or unneeded directories > > > * building binary distribution > > > WARNING > > > * some HTML links may not be found > > > installing R.css in c:/TEMP/Rinst40061099 > > > > > > Using auto-selected zip options '' > > > latex: not found > > > latex: not found > > > latex: not found > > > > > > -- Making package foo > > > latex: not found > > > adding build stamp to DESCRIPTION > > > latex: not found > > > latex: not found &
Re: [R] the Surv function
Try using the following (without naming the arguments) - it should work. Surv(fup, status) I think what is happening is that if you use one or two arguments (unnamed), Surv assumes that you have right-censored data, with the first argument being time and the second being status. By specifying 'event' as you did, Surv assumes that you have 3 arguments of "type"=counting and it is looking for the ending time required by the counting data format and this is why you got the error. If you are not afraid of R code, type Surv (without the parentheses) to see the code behind the function. -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Jennifer Dillon > Sent: Tuesday, May 01, 2007 11:16 PM > To: r-help@stat.math.ethz.ch > Subject: [R] the Surv function > > Hi, > > I'm trying to do a simple survival analysis on some data, and > I'm having the following problem (here's my code and the > error message): > > out <- Surv(fup,event=status) > Error in Surv(fup, event = status) : argument "time2" is > missing, with no default > > >From reading the documentation, it seems that I should be able to > >simply > write: Surv(time1, event) if my data is right-censored, which it is. > Help! > > Thanks a million, > > Jen > > > > -- > Jennifer Dillon > Doctoral Student > Harvard Biostatistics > Room 414B, Building 1 > > [[alternative HTML version deleted]] > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to code the censor variable for "survfit"
The Surv object contains the information on the type of censoring. Look at ?Surv for an explanation of how censored events are represented. -Christos > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Lu, Jiang > Sent: Saturday, April 28, 2007 11:10 PM > To: r-help@stat.math.ethz.ch > Subject: [R] how to code the censor variable for "survfit" > > Dear r-helpers, > > This is my first time to run survival analysis. Currently, I > have a data set which contains two variables, the variable of > time to event (or time to censoring) and the variable of > censor indicator. For the indicator variable, it was coded as > 0 and 1. 0 represents right censor, 1 means event of > interest. Now I try to use "survfit" in the package of > "survival". I wrote the following code: > > rptsurv <- survfit(surv(time,censor)~1,data=x) > > Before I run the code, I am concerned with my 0/1 coding to > the censor indicator because I did not see any argument in > the syntax of "survfit", which may tell the program that > value 1 means event. I checked the documentations and R-help > archive, but ended in vain. > > Would you please kindly tell me how "survfit" treats censor variables? > In 0/1 coding, is it the default that 1 means event and 0 > means right censor? What if the censor was coded as 2 or 3 > instead of 0 or 1? I means how the "survfit" knows the > difference. In SAS, if a "lifetest" > procedure (similar to survfit) is performed, there is an > argument specifying which value in the censor variable is > treated as event. > > I know I could just compare the results from R and from SAS > to see the difference. However, I really want to know exactly > how "survfit" deals with this problem. Thank you very much in advance. > > sincerely, > > Jiang Lu > University of Pittsburgh > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extract p-value from survdiff function
If sdf <- survdiff(...) is your survdiff object, the p-value can be computed as follows: p.val <- 1 - pchisq(sdf$chisq, length(sdf$n) - 1) and then use it in your K-M plot. -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Armin Goralczyk > Sent: Thursday, April 26, 2007 3:03 PM > To: R-help@stat.math.ethz.ch > Subject: [R] Extract p-value from survdiff function > > Hi list, > I want to use the p-value from the survdiff function (package > survival) to reuse within a function in a Kaplan-Meier plot. > The p-value is somehow not a component of the value list ?! > > Thanks in advance > -- > A. Goralczyk > Göttingen, Ger. > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] graphs superimposed on pictures?
Take a look at Figure 3.26 from Paul Murrell's book and associated code posted here: http://www.stat.auckland.ac.nz/~paul/RGraphics/chapter3.html -Christos > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Robert Biddle > Sent: Wednesday, April 11, 2007 9:40 AM > To: r-help@stat.math.ethz.ch > Subject: [R] graphs superimposed on pictures? > > Hi: > > I am doing some work that involves plotting points of > interest superimposed on photographs and maps. I can produce > the plots fine in R, but so far I have had to do the > superimposition externally, which makes it tedious to do > exploratory work. > I have looked to see if there is some capability to put a > background picture on a plot window, but I have not found anything. > Advice, anyone? > > Cheers > Robert Biddle > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ISwR library
Take a look at the "R Installation and Administration" manual - Section 6.1 Installing packages. It is platform-dependent and you didn't say what platform you're on. -Christos > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of slomascolo > Sent: Monday, April 09, 2007 3:04 PM > To: r-help@stat.math.ethz.ch > Subject: [R] ISwR library > > > I am learning to use R using a book that uses examples from > the ISwR library so I need to have that library. When I type > library (ISwR), R says: Error in library(ISwR) : there is no > package called 'ISwR' > > Do I need to install ISwR separately? How do I do this? > > Thanks! > -- > View this message in context: > http://www.nabble.com/ISwR-library-tf3548959.html#a9907582 > Sent from the R help mailing list archive at Nabble.com. > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] transition matrices
I think what you are looking for is also called seriation or ordination, i.e. find the "best" enumeration order of a set of objects. Take a look at the following link for more information and on criteria for defining "best" http://www.lirmm.fr/~caraux/PermutMatrix/EN/Seriation.htm The link is for a windows application (PermutMatrix) that can perform seriation on distance matrices used in hierarchical clustering, but might be useful in your context as well. -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Richard Rowe > Sent: Wednesday, April 04, 2007 2:35 AM > To: r-help@stat.math.ethz.ch > Subject: [R] transition matrices > > I am working with transition matrices of sequences of animal > behaviours. What I would like to do is parse the original > matrices, adjusting row/column order so that the matrix has > its main values in blocks surrounding the diagonal. This > would cause behaviours involved in functional groupings (e.g. > grooming, resting, foraging etc) to appear as blocks. > This can be done manually by applying subjective 'prior > knowledge' of sequences, however I would like to have an > algorithmic/objective method to generate at least a first cut ... > Any suggestions or hints (even just thoughts) would be much > appreciated, > > Richard Rowe > > Dr Richard Rowe > Zoology & Tropical Ecology > School of Tropical Biology > James Cook University > Townsville 4811 > AUSTRALIA > > ph +61 7 47 81 4851 > fax +61 7 47 25 1570 > JCU has CRICOS Provider Code 00117J > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Create a new var reflecting the order of subjects in existingvar
Try this: y <- rle(dat$ID) unlist(sapply(y$lengths, FUN=function(x) seq(1,x))) Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Nguyen > Dinh Nguyen > Sent: Sunday, April 01, 2007 10:00 PM > To: r-help@stat.math.ethz.ch > Subject: [R] Create a new var reflecting the order of > subjects in existingvar > > Dear R helpers > I have a data set sth like this: > set.seed(123);dat <- data.frame(ID= c(rep(1,2),rep(2,3), > rep(3,3), rep(4,4), rep(5,5)), > var1 =rnorm(17, 35,2), > var2=runif(17,0,1)) > dat >ID var1 var2 > 1 1 33.87905 0.02461368 > 2 1 34.53965 0.47779597 > 3 2 38.11742 0.75845954 > 4 2 35.14102 0.21640794 > 5 2 35.25858 0.31818101 > 6 3 38.43013 0.23162579 > 7 3 35.92183 0.14280002 > 8 3 32.46988 0.41454634 > 9 4 33.62629 0.41372433 > 10 4 34.10868 0.36884545 > 11 4 37.44816 0.15244475 > 12 4 35.71963 0.13880606 > 13 5 35.80154 0.23303410 > 14 5 35.22137 0.46596245 > 15 5 33.88832 0.26597264 > 16 5 38.57383 0.85782772 > 17 5 35.99570 0.04583117 > I would like to create a new var in dat which reflects the > order of each subject (ID), like this >ID var1 var2 IDorder > 1 1 33.87905 0.02461368 1 > 2 1 34.53965 0.47779597 2 > 3 2 38.11742 0.75845954 1 > 4 2 35.14102 0.21640794 2 > 5 2 35.25858 0.31818101 3 > 6 3 38.43013 0.23162579 1 > 7 3 35.92183 0.14280002 2 > 8 3 32.46988 0.41454634 3 > 9 4 33.62629 0.41372433 1 > 10 4 34.10868 0.36884545 2 > 11 4 37.44816 0.15244475 3 > 12 4 35.71963 0.13880606 4 > 13 5 35.80154 0.23303410 1 > 14 5 35.22137 0.46596245 2 > 15 5 33.88832 0.26597264 3 > 16 5 38.57383 0.85782772 4 > 17 5 35.99570 0.04583117 5 > > Thank you very much for your help > Regards > Nguyen > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] multi-level modeling & R?
Also, take a look at Andrew Gelman's recent book with Jennifer Hill on
multilevel modeling.
It is written as a practical how-to text book using R as the primary
platform. I have found it very easy to read, extremely useful and very good
value for its price (I am not related to the authors or the publisher in any
way).
http://www.stat.columbia.edu/~gelman/arm/
-Christos
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of John Kane
> Sent: Wednesday, March 28, 2007 2:25 PM
> To: Chuck Cleland
> Cc: R R-help
> Subject: Re: [R] multi-level modeling & R?
>
> Thanks to everyone for the information and to Chuck for the
> links. I must have misspelt something because I missed most
> of them in the RsiteSearch.
>
> Anyway back to selling R :)
>
> --- Chuck Cleland <[EMAIL PROTECTED]> wrote:
>
> > John Kane wrote:
> > > A colleague was asking me if R does multi-level modelling
> as opposed
> > > to multiple regression.
> > Since I
> > > have no knowledge of multi-level modelling (except
> > 5
> > > minutes googling ) I thought that I would as
> > here.
> > >
> > > Does are offer any multi-level modeling packages?
> > It
> > > looked like arm might be one but I was not sure.
> >
> > RSiteSearch("multilevel model") and a search for "multilevel" on
> > CRAN point to a number of other relevant packages and docs,
> including:
> >
> >
> http://finzi.psych.upenn.edu/R/library/nlme/html/lme.html
> >
> >
> http://cran.r-project.org/src/contrib/Descriptions/lme4.html
> >
> > http://cran.r-project.org/doc/packages/mlmRev.pdf
> >
> >
> http://cran.r-project.org/src/contrib/Descriptions/multilevel.html
> >
> >
> http://cran.r-project.org/doc/contrib/Bliese_Multilevel.pdf
> >
> > > Thanks
> > >
> > > __
> > > R-help@stat.math.ethz.ch mailing list
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained,
> > reproducible code.
> >
> > --
> > Chuck Cleland, Ph.D.
> > NDRI, Inc.
> > 71 West 23rd Street, 8th floor
> > New York, NY 10010
> > tel: (212) 845-4495 (Tu, Th)
> > tel: (732) 512-0171 (M, W, F)
> > fax: (917) 438-0894
> >
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] what is the difference between survival analysis and (...)
On the same point, transforming time-to-event data to binary outcomes so
that contingency-table analysis (odds ratios etc) or logistic regression can
be applied will result in loss of information that could lead to misleading
conclusions.
For example, assuming that there is a good-prognosis group (low risk) and a
poor-prognosis group (high risk) that need to be compared. By definition,
patients in the good prognosis group are those that have been followed up
for a longer time in the study, whereas patients with poor prognosis will
tend to die earlier. Therefore censoring will occur later in the good
prognosis group and thus the two groups will not have a homogeneous
censorship structure. In this case, naïve analysis could be misleading.
For more details and a simulation example take a look at
http://jnci.oxfordjournals.org/cgi/data/99/2/147/DC1/3
HTH
-Christos
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Lucke, Joseph F
> Sent: Wednesday, March 28, 2007 12:10 PM
> To: Eric Elguero; R-help@stat.math.ethz.ch
> Subject: Re: [R] what is the difference between survival
> analysis and (...)
>
> You can (and I have) fit survival data with logistic
> regression. Agresti (1990, pp 189--196) has an introductory
> discussion.
>
> The issue is whether the occurrence of the event is of
> interest or whether the time-to-event is of interest. If the
> study lasts 180 days (as in my case) logistic regression
> treats an event at 1 day the same as an event at 179 days.
> Similarly, non-occurrence censored at 5 days is treated the
> same as non-occurrence censored at 180 days. These
> assumptions only make sense if the hazard rate is constant
> and (therefore) the time-to-failure distribution is exponential.
>
> One can include exposure time as a offset (non-estimated
> covariate) to handle non-constant hazard rates. One can also
> model the hazard rate directly as a log-linear model.
>
> Based on what he said (number events/sample size, using
> cumulative times), the hostile medical epidemiologist was
> implicitly assuming the survival time followed an exponential
> distribution. This assumption is often incorrect. His
> arrogance was exceeded only by his ignorance.
>
> Joe
>
> @BOOK{Agresti1990,
> author = {Agresti, Alan},
> title = {Categorical data analysis},
> year = {1990},
> publisher = {John Wiley \& Sons},
> address = {New York, NY},
> series = {Wiley Series in Probability and Mathematical Statistics},
> keywords = {loglinear; logistic}
> }
>
>
>
>
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Eric Elguero
> Sent: Wednesday, March 28, 2007 8:40 AM
> To: R-help@stat.math.ethz.ch
> Subject: Re: [R] what is the difference between survival
> analysis and (...)
>
> Hi everybody,
>
> recently I had to teach a course on Cox model, of which I am
> not a specialist, to an audience of medical epidemiologists.
> Not a good idea you might say.. anyway, someone in the
> audience was very hostile. At some point, he sayed that Cox
> model was useless, since all you have to do is count who dies
> and who survives, divide by the sample sizes and compute a
> relative risk, and if there was significant censoring, use
> cumulated follow-up instead of sample sizes and that's it!
> I began arguing that in Cox model you could introduce several
> variables, interactions, etc, then I remembered of logistic
> models ;-) The only (and poor) argument I could think of was
> that if mr Cox took pains to devise his model, there should
> be some reason...
>
> but the story doesn't end here. When I came back to my
> office, I tried these two methods on a couple of data sets,
> and true, crude RRs are very close to those coming from Cox model.
>
> hence this question: could someone provide me with a dataset
> (preferably real) where there is a striking difference
> between estimated RRs and/or between P-values? and of course
> I am interested in theoretical arguments and references.
>
> sorry that this question has nothing to do with R and thank
> you in advance for your leniency.
>
> Eric Elguero
> GEMI-UMR 2724 IRD-CNRS,
> Équipe "Évolution des Systèmes Symbiotiques"
> 911 avenue Agropolis, BP 64501,
> 34394 Montpellier cedex 5 FRANCE
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-projec
Re: [R] Replacement in an expression - can't use parse()
A way to do this is through substitute:
e1 <- substitute(expression(u1 + u2 + u3), list(u2=quote(x), u3=1))
But I am not sure whether you will run into similar limitations regarding
the length of the expression to be substituted.
-Christos
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Daniel Berg
> Sent: Tuesday, March 27, 2007 9:57 AM
> To: r-help@stat.math.ethz.ch
> Subject: [R] Replacement in an expression - can't use parse()
>
> Dear all,
>
> Suppose I have a very long expression e. Lets assume, for
> simplicity, that it is
>
> e = expression(u1+u2+u3)
>
> Now I wish to replace u2 with x and u3 with 1. I.e. the 'new'
> expression, after replacement, should be:
>
> > e
> expression(u1+x+1)
>
> My question is how to do the replacement?
>
> I have tried using:
>
> > e = parse(text=gsub("u2","x",e))
> > e = parse(text=gsub("u3",1,e))
>
> Even though this works fine in this simple example, the use
> of parse when e is very long will fail since parse has a
> maximum line length and will cut my expressions. I need to
> keep mode(e)=expression since I will use e further in
> symbolic derivation and division.
>
> Any suggestions are most welcome.
>
> Thank you.
>
> Regards,
> Daniel Berg
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Select the last two rows by id group
You can try the following: n.len <- nrow(iris) n.sel <- 3 iris[(n.len-n.sel+1):n.len, ] -Christos > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Lauri Nikkinen > Sent: Tuesday, March 20, 2007 10:33 AM > To: r-help@stat.math.ethz.ch > Subject: [R] Select the last two rows by id group > > Hi R-users, > > Following this post > http://tolstoy.newcastle.edu.au/R/help/06/06/28965.html , how > do I get last two rows (or six or ten) by id group out of the > data frame? Here the example gives just the last row. > > Sincere thanks, > Lauri > > [[alternative HTML version deleted]] > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Highlight overlapping area between two curves
See http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=7 and code therein. -Christos > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Nguyen > Dinh Nguyen > Sent: Tuesday, March 13, 2007 12:20 AM > To: r-help@stat.math.ethz.ch > Subject: [R] Highlight overlapping area between two curves > > Dear R helpers, > I have a graph as following; I would like to highlight the > overlapping area between the two curves. Do you know how to do this? > Thank you in advance for your help. > Nguyen > > ###START > x1 <- rnorm(1, 0.70,0.12) > x2 <- rnorm(1, 0.90,0.12) > > d1 <- density(x1) > d2 <- density(x2) > > plot(range(d1$x,d2$x), range(d1$y, d2$y), type = "n", > xlab = "X value", ylab = "Probability Density" ) > > lines(d1, col = "red",lwd=4, lty=2) > lines(d2, col = "blue",lwd=4) > > ##END CODE > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Table Construction from calculations
Your "data table" basis is actually a dataframe, whose first column is
non-numeric. That's what is causing the problem.
Try removing the first column of the dataframe before adding the row to your
matrix:
test <- latpoints + basis[2, -1]
-Christos
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Seth Imhoff
> Sent: Friday, March 09, 2007 9:23 PM
> To: r-help@stat.math.ethz.ch
> Subject: [R] Table Construction from calculations
>
> Hi-
>
> I am trying to create a table of values by adding pairs of
> vectors, but am running into some problems. The problem is
> best expressed by a simple example.
>
> Starting with a data table "basis":
> atom x y z
> 1 Cu 0.0 0.0 0.0
> 2 Cu 0.5 0.5 0.5
>
> I want to add 0.5 0.5 0.5 (and also the 0 0 0 but it
> wouldn't change the values below so I won't refer to it in
> the rest of the example) to a list of vectors in the form of:
> > latpoints
>V1 V2 V3
> 1 0 0 0
> 2 0 0 1
> 3 0 0 2
> 4 0 0 3
> 5 0 1 1
>
> so that I end up with a table such as:
> V1 V2 V3
> 0.5 0.5 0.5
> 0.5 0.5 1.5
> 0.5 0.5 2.5
> 0.5 0.5 3.5
> 0.5 1.5 1.5
>
> I've tried many variations on the following: (not just cat,
> but most of the data/data.table options)
>
> test = for(i in 1:5) {cat(basis[1,2:4] +
> latticemultipliers[i,], append=TRUE)}
>
> However, I either end up with an error telling me that cat
> doesn't handle "type 'list' " or with a table with length of
> 1 such as:
> xyz
> 2 0.5 1.5 1.5
>
> Which is simply the last value that the loop calculates.
>
> Does anyone know what function handles lists of the form I am
> using, or have a better suggestion on how to get the form that I want.
>
> Thanks in advance,
> Seth Imhoff
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
__
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Re: [R] Duplicate rows of matrix
Try this: a <- matrix(c(8, 4.2, 9.4, 1.1),2) b <- c(3,1) a[rep(1:nrow(a), b), ] -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Bruno C. > Sent: Friday, March 09, 2007 12:17 PM > To: r-help > Subject: [R] Duplicate rows of matrix > > Hello my problem is the following: > > I have a matrix A and a vector B which contains as many rows as A. > > I need to build a matrix C which contains B[i]-times the row > A[i,] and this for each line of A. > > if for example A is > > [1][2] > [1] 8 9.4 > [2] 4.21.1 > > and B is (3,1). Then C will be: > [1][2] > [1] 8 9.4 > [2] 8 9.4 > [3] 8 9.4 > [4] 4.21.1 > > > I have some working code which go through all the lines of A > and for each line does a rbind(C, A[i,]) B[i]-times But this > is quite time consuming given that each rbind rebuild a new > matrix ... is there any faster way? > I can think of some minor improvements like building a matrix > C of zeros, containing as many columns as A and as many > columns as the sum of elements of B ... and the filing it. > > But I was more looking for some already implemented > function/package, is there any? > > Thanx > > > > -- > Leggi GRATIS le tue mail con il telefonino i-modeT di Wind > http://i-mode.wind.it > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix conversion question
Try split(x, row(x)) -Christos > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of > Johannes Graumann > Sent: Friday, March 09, 2007 10:30 AM > To: r-help@stat.math.ethz.ch > Subject: [R] Matrix conversion question > > Hello, > > Please help - I'm blanking on this ... > > I have a matrix like this: > > [,1] [,2] > [1,]12 > [2,]13 > [3,]23 > > and would like to have a list of vectors, where a vector > contains the entries in a matrix row ... > > Can somebody nudge me to the place I need to go? > > Thanks, Joh > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] understanding print.summary.lm and perhaps print/show in general
Paul, Usually summary methods perform some computations if needed and then change the class of the original object so that a print method can be called for the new summary object. In this case, this is done at the end of the summary.lm method: ... if (!is.null(z$na.action)) ans$na.action <- z$na.action class(ans) <- "summary.lm" ^^ ans } So then print.summary.lm does all the job displaying the summary.lm object. To see that function do getAnywhere(print.summary.lm) Then you can then modify that function as needed. -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Paul Bailey > Sent: Friday, March 09, 2007 9:34 AM > To: r-help@stat.math.ethz.ch > Subject: [R] understanding print.summary.lm and perhaps > print/show in general > > I'm trying to understand how R prints summary.lm objects and > trying to change it slightly for a summary function that > calculates standard errors using an alternative method. > > I've found that I can modify a summary.lm object and then it > prints the modified way but I want to change a few things in > the print method that I think I might just be able to do. One > is that I want the coefficients table to print a different > header (other than "Std. Error"). I've tried changing the > column name of the summary$coef matrix and this works for > calls to printCoefmat but it still prints out "Std. Error" > when I pass the summary.lm to the command line by itself. I > don't understand this behavior. When I do this (enter an > object on the command line by itself), does it then calls the > print / show method associated with that objects class, in > this case, summary.lm? Below is some sample code to reproduce > the behavior I don't understand and a comment regarding the > result I don't understand. > > Cheers, > Paul > > # > lma <- lm(dist ~ speed, data=cars) > suma <- summary(lma) > colnames(suma$coef) <- c(LETTERS[1:4]) > printCoefmat(suma$coef) # prints what I expect suma # the > above is the print behavior question regards, # why does the > coefficients matrix have in its header # the usual "Estimate > Std. Error t value Pr(>|t|)" > # I expect "A B C D" as above in the call to printCoefmat > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Memory Limits in Ubuntu Linux
Take a look at Windows FAQ 2.9. Following the instructions there, I was able to make WinXP use at least 3GB of RAM (physical RAM installed) with Rgui.exe. -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of > [EMAIL PROTECTED] > Sent: Tuesday, March 06, 2007 3:44 PM > To: r-help@stat.math.ethz.ch > Subject: [R] Memory Limits in Ubuntu Linux > > I am an R user trying to get around the 2Gig memory limit in > Windows, so here I am days later with a working Ubuntu, and R > under Ubuntu. But - the memory problems seem worse than ever. > R code that worked under windows fails, unable to allocate memory. > > Searching around the web, it appears that the problem may be > the ability to find contguous memory for my big vectors, but > a fresh boot of Ubuntu does not help either. > > Which way to go? > > 1) Try to install 64-bit version for bigger address space. > Would this help? Is this workable for my Athlon 64 Dual-core? > (the live cd seems to work but I never got it to boot after a > disk install, but then the 386 version was no better until I > learned more about Grub...I could try again if this might solve the > problem) > > 2) Recompile R to get bigger memory capability? (I'll have to > cross-post to some R forums too) This will be a challenge for > a Linux newbie...like me. > > 3) Any other suggestions? My goal is to create a bigger > neural network than fits in my Windows R version. > -- > David Katz > www.davidkatzconsulting.com >541 482-1137 > > [[alternative HTML version deleted]] > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with paste()
What do you want to do with rbind?
paste produces a single vector of 24 character-valued elements.
If you want a column vector, you could do that by
x <- paste('txt.est',1:24, sep = '')
matrix(x, ncol=1)
-Christos
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Michael Kubovy
> Sent: Saturday, March 03, 2007 2:30 PM
> To: r-help@stat.math.ethz.ch list
> Subject: [R] Help with paste()
>
> Dear r-helpers,
>
> Could you please tell me what's missing:
> rbind(paste('txt.est',1:24, sep = ''))
> txt.est1, ... txt.est24 are vectors that I wish to rbind.
> _
> Professor Michael Kubovy
> University of Virginia
> Department of Psychology
> USPS: P.O.Box 400400Charlottesville, VA 22904-4400
> Parcels:Room 102Gilmer Hall
> McCormick RoadCharlottesville, VA 22903
> Office:B011+1-434-982-4729
> Lab:B019+1-434-982-4751
> Fax:+1-434-982-4766
> WWW:http://www.people.virginia.edu/~mk9y/
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
__
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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] Can a data.frame column contain lists/arrays?
Why do you need to use a data frame? A list will give you the flexibility you want: d <- list( x=list( c(1,2), c(5,2), c(9,1) ), y=c( 1, -1, -1) ) Then you can access the individual elements > d$x [[1]] [1] 1 2 [[2]] [1] 5 2 [[3]] [1] 9 1 > d$y [1] 1 -1 -1 > d$x[[1]] [1] 1 2 -Christos > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of > Christian Convey > Sent: Monday, February 12, 2007 11:29 PM > To: r-help@stat.math.ethz.ch > Subject: [R] Can a data.frame column contain lists/arrays? > > I'd like to have a data.frame structured something like the following: > > d <- data.frame ( >x=list( c(1,2), c(5,2), c(9,1) ), >y=c( 1, -1, -1) > ) > > The reason is this: 'd' is the training data for a machine > learning algorithm. d$x is the independent data, and d$y is > the dependent data. > > In general my machine learning code will work where each > element of d$x is a vector of one or more real numbers. So > for instance, the same code should work when d$x[1] = 42, or > when d$x[1] = (42, 3, 5). > All that matters is that all element within d$x are > lists/vectors of the same length. > > Does anyone know if/how I can get a data.frame set up like that? > > Thanks, > Christian > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] blank upper or lower triangle of cor-matrix
And if you want to know how it is done, take a look at stats:::print.dist -Christos > -Original Message- > From: Christos Hatzis [mailto:[EMAIL PROTECTED] > Sent: Wednesday, February 07, 2007 2:16 PM > To: 'Leo Gürtler'; 'r-help@stat.math.ethz.ch' > Subject: RE: [R] blank upper or lower triangle of cor-matrix > > You can try > > as.dist(d) > > > Christos Hatzis, Ph.D. > Nuvera Biosciences, Inc. > 400 West Cummings Park > Suite 5350 > Woburn, MA 01801 > Tel: 781-938-3830 > www.nuverabio.com > > > > > -Original Message- > > From: [EMAIL PROTECTED] > > [mailto:[EMAIL PROTECTED] On Behalf Of Leo Gürtler > > Sent: Wednesday, February 07, 2007 1:36 PM > > To: r-help@stat.math.ethz.ch > > Subject: [R] blank upper or lower triangle of cor-matrix > > > > Dear altogether, > > > > I want to blank the lower (or upper) part of a correlation > matrix as > > it is done by dist() > > > > example: > > > > ( d <- cor(matrix(runif(12),nrow=4)) ) > > > > If I do the following > > > > d[lower.tri(d)] <- "" > > > > of course everything is changed to character - that's not > what should > > be. > > Additionally, it does not work to assign "0" or anything else. The > > same is true for assigning "NA". > > > > However, what I want is like the following: > > > > ( dist(matrix(runif(12),nrow=4)) ) > > > > Looking into dist(), it seems that the calculation and the > matrix are > > done in C and not in plain R. > > > > How can I realize it? > > > > thanks! > > > > best, > > > > leo > > > > __ > > R-help@stat.math.ethz.ch mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide > > http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. > > > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] blank upper or lower triangle of cor-matrix
You can try as.dist(d) Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Leo Gürtler > Sent: Wednesday, February 07, 2007 1:36 PM > To: r-help@stat.math.ethz.ch > Subject: [R] blank upper or lower triangle of cor-matrix > > Dear altogether, > > I want to blank the lower (or upper) part of a correlation > matrix as it is done by dist() > > example: > > ( d <- cor(matrix(runif(12),nrow=4)) ) > > If I do the following > > d[lower.tri(d)] <- "" > > of course everything is changed to character - that's not > what should be. > Additionally, it does not work to assign "0" or anything > else. The same is true for assigning "NA". > > However, what I want is like the following: > > ( dist(matrix(runif(12),nrow=4)) ) > > Looking into dist(), it seems that the calculation and the > matrix are done in C and not in plain R. > > How can I realize it? > > thanks! > > best, > > leo > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding Histograms to Leaves of Rpart Tree or other Dendrogram
Hi Jon, Take a look at this graph http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=85 I think it is very close to what you need (source code provided at the site). -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Jonathan Zelner > Sent: Friday, February 02, 2007 5:07 PM > To: r-help@stat.math.ethz.ch > Subject: [R] Adding Histograms to Leaves of Rpart Tree or > other Dendrogram > > Hi - I'm trying to append simple density histograms of a > continuous variable to the leaves of an rpart tree. > > The splits in the tree are all levels of a factor and I'm > hoping to make the histograms out of the subsets of the > dataframe corresponding to the splits and for them to be > attached to the appropriate leaf of the final tree. > > Any help would be much appreciated, > thanks, > Jon Zelner > > University of Michigan > Gerald R. Ford School of Public Policy > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lining up x-y datasets based on values of x
Marc,
The data structure is a list of data frames generated from read.table:
> class(nmr.spectra.serum)
[1] "list"
> class(nmr.spectra.serum[[1]])
[1] "data.frame"
> dim(nmr.spectra.serum[[1]])
[1] 32768 2
Converting the data.frames to matrices does not have much of an effect on
timing.
-Christos
-Original Message-
From: Marc Schwartz [mailto:[EMAIL PROTECTED]
Sent: Thursday, February 01, 2007 11:06 PM
To: [EMAIL PROTECTED]
Cc: 'Prof Brian Ripley'; r-help@stat.math.ethz.ch
Subject: Re: [R] Lining up x-y datasets based on values of x
On Thu, 2007-02-01 at 22:46 -0500, Christos Hatzis wrote:
> Marc,
>
> I don't think the issue is duplicates in the matching columns. The
> data were generated by an instrument (NMR spectrometer), processed by
> the instrument's software through an FFT transform and other
> transformations and finally reported as a sequence of chemical shift (x)
vs intensity (y) pairs.
> So all x values are unique. For the example that I reported earlier:
>
> > length(nmr.spectra.serum[[1]]$V1)
> [1] 32768
> > length(unique(nmr.spectra.serum[[1]]$V1))
> [1] 32768
> > length(nmr.spectra.serum[[2]]$V1)
> [1] 32768
> > length(unique(nmr.spectra.serum[[2]]$V1))
> [1] 32768
>
> And most of the x-values are common
> > sum(nmr.spectra.serum[[1]]$V1 %in% nmr.spectra.serum[[2]]$V1)
> [1] 32625
>
> For this reason, merge is probably an overkill for this problem and my
> initial thought was to align the datasets through some simple
> index-shifting operation.
>
> Profiling of the merge code in my case shows that most of the time is
> spent on data frame subsetting operations and on internal merge and
> rbind calls secondarily (if I read the summary output correctly). So
> even if most of the time in the internal merge function is spent on
> sorting (haven't checked the source code), this is in the worst case a
> rather minor effect, as suggested by Prof. Ripley.
>
> > Rprof("merge.out")
> > zz <- merge(nmr.spectra.serum[[1]], nmr.spectra.serum[[2]], by="V1",
> all=T, sort=T)
> > Rprof(NULL)
> > summaryRprof("merge.out")
>
> $by.self
>self.time self.pct total.time total.pct
> merge.data.frame6.56 50.0 11.84 90.2
> [.data.frame2.42 18.4 3.68 28.0
> merge 1.28 9.8 13.12 100.0
> rbind 1.24 9.5 1.36 10.4
> names<-.default 1.16 8.8 1.16 8.8
> row.names<-.data.frame 0.12 0.9 0.18 1.4
> duplicated.default 0.12 0.9 0.12 0.9
> make.unique 0.10 0.8 0.10 0.8
> data.frame 0.02 0.2 0.04 0.3
> * 0.02 0.2 0.02 0.2
> is.na 0.02 0.2 0.02 0.2
> match 0.02 0.2 0.02 0.2
> order 0.02 0.2 0.02 0.2
> unclass 0.02 0.2 0.02 0.2
> [ 0.00 0.0 3.68 28.0
> do.call 0.00 0.0 1.18 9.0
> names<- 0.00 0.0 1.16 8.8
> row.names<- 0.00 0.0 0.18 1.4
> any 0.00 0.0 0.14 1.1
> duplicated 0.00 0.0 0.12 0.9
> cbind 0.00 0.0 0.04 0.3
> as.vector 0.00 0.0 0.02 0.2
> seq 0.00 0.0 0.02 0.2
> seq.default 0.00 0.0 0.02 0.2
>
> $by.total
>total.time total.pct self.time self.pct
> merge 13.12 100.0 1.28 9.8
> merge.data.frame11.84 90.2 6.56 50.0
> [.data.frame 3.68 28.0 2.42 18.4
> [3.68 28.0 0.00 0.0
> rbind1.36 10.4 1.24 9.5
> do.call 1.18 9.0 0.00 0.0
> names<-.default 1.16 8.8 1.16 8.8
> names<- 1.16 8.8 0.00 0.0
> row.names<-.data.frame 0.18 1.4 0.12 0.9
> row.names<- 0.18 1.4 0.00 0.0
> any 0.14 1.1 0.00 0.0
> duplicated.default 0.12 0.9 0.12 0.9
> duplicated
Re: [R] Lining up x-y datasets based on values of x
Thanks Gabor.
This is along the lines of what I was looking for. In fact the merge
function for zoo objects (ordered) turns out to be almost an order of
magnitude faster than the generic merge function for my problem:
> system.time(
+ zz <- merge( spec.1 = zoo(nmr.spectra.serum[[1]]$V2,
nmr.spectra.serum[[1]]$V1),
+spec.2 = zoo(nmr.spectra.serum[[2]]$V2, nmr.spectra.serum[[2]]$V1),
fill=NA )
+ )
[1] 0.74 0.07 0.82 NA NA
> system.time(
+ ww <- merge(nmr.spectra.serum[[1]], nmr.spectra.serum[[2]], by="V1",
all=T, sort=T)
+ )
[1] 6.85 0.05 6.94 NA NA
> head(zz)
spec.1 spec.2
-1322.2 -0.651 NA
-1321.9 -0.266 NA
-1321.7 -0.962 NA
-1321.4 -0.602 NA
-1321.2 0.753 NA
-1320.9 1.212 NA
> head(ww)
V1 V2.x V2.y
1 -1322.2 -0.651 NA
2 -1321.9 -0.266 NA
3 -1321.7 -0.962 NA
4 -1321.4 -0.602 NA
5 -1321.2 0.753 NA
6 -1320.9 1.212 NA
>
Thanks again.
-Christos
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Thursday, February 01, 2007 7:25 PM
To: [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Lining up x-y datasets based on values of x
The zoo package has a multiway merge with optional zero fill.
Here are two ways:
library(zoo)
merge(x = zoo(x[,2], x[,1]),
y = zoo(y[,2], y[,1]),
z = zoo(z[,2], z[,1]),
fill = 0)
# or
library(zoo)
X <- list(x = x, y = y, z = z)
merge0 <- function(..., fill = 0) merge(..., fill = fill) do.call("merge0",
lapply(X, function(x) zoo(x[,2], x[,1])))
To get more info on zoo try:
vignette("zoo")
On 2/1/07, Christos Hatzis <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I was wondering if there is a direct approach for lining up 2-column
> matrices according to the values of the first column. An example and
> a brute-force approach is given below:
>
> x <- cbind(1:10, runif(10))
> y <- cbind(5:14, runif(10))
> z <- cbind((-4):5, runif(10))
>
> xx <- seq( min(c(x[,1],y[,1],z[,1])), max(c(x[,1],y[,1],z[,1])), 1) w
> <- cbind(xx, matrix(rep(0, 3*length(xx)), ncol=3))
>
> w[ xx >= x[1,1] & xx <= x[10,1], 2 ] <- x[,2] w[ xx >= y[1,1] & xx <=
> y[10,1], 3 ] <- y[,2] w[ xx >= z[1,1] & xx <= z[10,1], 4 ] <- z[,2]
>
> w
>
> I appreciate any pointers.
>
> Thanks.
>
> Christos Hatzis, Ph.D.
> Nuvera Biosciences, Inc.
> 400 West Cummings Park
> Suite 5350
> Woburn, MA 01801
> Tel: 781-938-3830
> www.nuverabio.com
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lining up x-y datasets based on values of x
Marc,
I don't think the issue is duplicates in the matching columns. The data
were generated by an instrument (NMR spectrometer), processed by the
instrument's software through an FFT transform and other transformations and
finally reported as a sequence of chemical shift (x) vs intensity (y) pairs.
So all x values are unique. For the example that I reported earlier:
> length(nmr.spectra.serum[[1]]$V1)
[1] 32768
> length(unique(nmr.spectra.serum[[1]]$V1))
[1] 32768
> length(nmr.spectra.serum[[2]]$V1)
[1] 32768
> length(unique(nmr.spectra.serum[[2]]$V1))
[1] 32768
And most of the x-values are common
> sum(nmr.spectra.serum[[1]]$V1 %in% nmr.spectra.serum[[2]]$V1)
[1] 32625
For this reason, merge is probably an overkill for this problem and my
initial thought was to align the datasets through some simple index-shifting
operation.
Profiling of the merge code in my case shows that most of the time is spent
on data frame subsetting operations and on internal merge and rbind calls
secondarily (if I read the summary output correctly). So even if most of
the time in the internal merge function is spent on sorting (haven't checked
the source code), this is in the worst case a rather minor effect, as
suggested by Prof. Ripley.
> Rprof("merge.out")
> zz <- merge(nmr.spectra.serum[[1]], nmr.spectra.serum[[2]], by="V1",
all=T, sort=T)
> Rprof(NULL)
> summaryRprof("merge.out")
$by.self
self.time self.pct total.time total.pct
merge.data.frame6.56 50.0 11.84 90.2
[.data.frame2.42 18.4 3.68 28.0
merge 1.28 9.8 13.12 100.0
rbind 1.24 9.5 1.36 10.4
names<-.default 1.16 8.8 1.16 8.8
row.names<-.data.frame 0.12 0.9 0.18 1.4
duplicated.default 0.12 0.9 0.12 0.9
make.unique 0.10 0.8 0.10 0.8
data.frame 0.02 0.2 0.04 0.3
* 0.02 0.2 0.02 0.2
is.na 0.02 0.2 0.02 0.2
match 0.02 0.2 0.02 0.2
order 0.02 0.2 0.02 0.2
unclass 0.02 0.2 0.02 0.2
[ 0.00 0.0 3.68 28.0
do.call 0.00 0.0 1.18 9.0
names<- 0.00 0.0 1.16 8.8
row.names<- 0.00 0.0 0.18 1.4
any 0.00 0.0 0.14 1.1
duplicated 0.00 0.0 0.12 0.9
cbind 0.00 0.0 0.04 0.3
as.vector 0.00 0.0 0.02 0.2
seq 0.00 0.0 0.02 0.2
seq.default 0.00 0.0 0.02 0.2
$by.total
total.time total.pct self.time self.pct
merge 13.12 100.0 1.28 9.8
merge.data.frame11.84 90.2 6.56 50.0
[.data.frame 3.68 28.0 2.42 18.4
[3.68 28.0 0.00 0.0
rbind1.36 10.4 1.24 9.5
do.call 1.18 9.0 0.00 0.0
names<-.default 1.16 8.8 1.16 8.8
names<- 1.16 8.8 0.00 0.0
row.names<-.data.frame 0.18 1.4 0.12 0.9
row.names<- 0.18 1.4 0.00 0.0
any 0.14 1.1 0.00 0.0
duplicated.default 0.12 0.9 0.12 0.9
duplicated 0.12 0.9 0.00 0.0
make.unique 0.10 0.8 0.10 0.8
data.frame 0.04 0.3 0.02 0.2
cbind0.04 0.3 0.00 0.0
*0.02 0.2 0.02 0.2
is.na0.02 0.2 0.02 0.2
match0.02 0.2 0.02 0.2
order0.02 0.2 0.02 0.2
unclass 0.02 0.2 0.02 0.2
as.vector0.02 0.2 0.00 0.0
seq 0.02 0.2 0.00 0.0
seq.default 0.02 0.2 0.00 0.0
$sampling.time
[1] 13.12
Thanks again for your time in looking into this.
-Christos
-Original Message-
From: Marc Schwartz [mailto:[EMAIL PROTECTED]
Sent: Thursday, February 01, 2007 9:59 PM
To: Prof Brian
Ripley
Cc: r-help@stat.math.ethz.ch; [EMAIL PROTECTED]
Subject: Re: [R] Lining up x-y datasets based on values of x
On Thu, 2007-02-01 at 23:34 +, Prof Brian Ripley wrote:
> On Thu, 1 Feb 2007, Marc Schwar
Re: [R] Lining up x-y datasets based on values of x
[Sorry I meant to reply to the list] Thanks, Marc. That's what I have done. However, there seems to be a penalty from using merge repeatedly as it appears to internally re-sort the datasets. In my case the datasets are long (~35K rows) and already sorted so this step adds considerable and unnecessary overhead. There doesn't seem to be an option for disabling sorting. Setting 'sort=F' only affects sorting of the final data.frame. > system.time(merge(nmr.spectra.serum[[1]], nmr.spectra.serum[[2]], > by="V1", all=T, sort=T)) [1] 6.96 0.00 7.24 NA NA > system.time(merge(nmr.spectra.serum[[1]], nmr.spectra.serum[[2]], > by="V1", all=T, sort=F)) [1] 6.82 0.00 7.14 NA NA > I was wondering if perhaps there is a parallel between this problem and methods for linining up time-series data, since such data are also usually sorted on the time dimension. -Christos -Original Message- From: Marc Schwartz [mailto:[EMAIL PROTECTED] Sent: Thursday, February 01, 2007 4:21 PM To: [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Lining up x-y datasets based on values of x On Thu, 2007-02-01 at 15:45 -0500, Christos Hatzis wrote: > Thanks Marc and Phil. > > My dataset actually consists of 50+ individual files, so I will have > to do this one column at a time in a loop... > I might look into SQL and outer joints as an alternative to avoid looping. > > Thanks again. > -Christos If the files conform to some naming convention and/or are all located in a common sub-directory, you can use list.files() to get the file names into a vector. If not, you could use file.choose() interactively. Then use either a for() loop or sapply() to loop over the filenames, read them in to data frames using read.table() and merge them together in the same loop. When it comes to basic data manipulation like this, loops are not a bad thing. The overhead of a loop is typically outweighed by the file I/O and related considerations. HTH, Marc __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lining up x-y datasets based on values of x
Thanks Marc and Phil.
My dataset actually consists of 50+ individual files, so I will have to do
this one column at a time in a loop...
I might look into SQL and outer joints as an alternative to avoid looping.
Thanks again.
-Christos
-Original Message-
From: Marc Schwartz [mailto:[EMAIL PROTECTED]
Sent: Thursday, February 01, 2007 3:29 PM
To: [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Lining up x-y datasets based on values of x
On Thu, 2007-02-01 at 15:05 -0500, Christos Hatzis wrote:
> Hi,
>
> I was wondering if there is a direct approach for lining up 2-column
> matrices according to the values of the first column. An example and
> a brute-force approach is given below:
>
> x <- cbind(1:10, runif(10))
> y <- cbind(5:14, runif(10))
> z <- cbind((-4):5, runif(10))
>
> xx <- seq( min(c(x[,1],y[,1],z[,1])), max(c(x[,1],y[,1],z[,1])), 1) w
> <- cbind(xx, matrix(rep(0, 3*length(xx)), ncol=3))
>
> w[ xx >= x[1,1] & xx <= x[10,1], 2 ] <- x[,2] w[ xx >= y[1,1] & xx <=
> y[10,1], 3 ] <- y[,2] w[ xx >= z[1,1] & xx <= z[10,1], 4 ] <- z[,2]
>
> w
>
> I appreciate any pointers.
>
> Thanks.
How about this:
x <- cbind(1:10, runif(10))
y <- cbind(5:14, runif(10))
z <- cbind((-4):5, runif(10))
colnames(x) <- c("X", "Y")
colnames(y) <- c("X", "Y")
colnames(z) <- c("X", "Y")
xy <- merge(x, y, by = "X", all = TRUE)
xyz <- merge(xy, z, by = "X", all = TRUE)
xyz[is.na(xyz)] <- 0
> xyz
X Y.x Y.y Y
1 -4 0.000 0.000 0.3969099
2 -3 0.000 0.000 0.8943127
3 -2 0.000 0.000 0.4882819
4 -1 0.000 0.000 0.0275787
5 0 0.000 0.000 0.7562341
6 1 0.6873130 0.000 0.6185218
7 2 0.1930880 0.000 0.2318025
8 3 0.1164783 0.000 0.7336057
9 4 0.7408532 0.000 0.3006347
10 5 0.7112887 0.6383823 0.8515126
11 6 0.2719079 0.5952721 0.000
12 7 0.2067017 0.8178048 0.000
13 8 0.2085043 0.5714917 0.000
14 9 0.2251435 0.4032660 0.000
15 10 0.3471888 0.5247478 0.000
16 11 0.000 0.6899197 0.000
17 12 0.000 0.7188912 0.000
18 13 0.000 0.9133252 0.000
19 14 0.000 0.9186001 0.000
Note that 'xyz' will be a data frame, so just use as.matrix(xyz) to coerce
back to a numeric matrix if needed.
See ?merge
HTH,
Marc Schwartz
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[R] Lining up x-y datasets based on values of x
Hi, I was wondering if there is a direct approach for lining up 2-column matrices according to the values of the first column. An example and a brute-force approach is given below: x <- cbind(1:10, runif(10)) y <- cbind(5:14, runif(10)) z <- cbind((-4):5, runif(10)) xx <- seq( min(c(x[,1],y[,1],z[,1])), max(c(x[,1],y[,1],z[,1])), 1) w <- cbind(xx, matrix(rep(0, 3*length(xx)), ncol=3)) w[ xx >= x[1,1] & xx <= x[10,1], 2 ] <- x[,2] w[ xx >= y[1,1] & xx <= y[10,1], 3 ] <- y[,2] w[ xx >= z[1,1] & xx <= z[10,1], 4 ] <- z[,2] w I appreciate any pointers. Thanks. Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Loading functions in R
The recommended approach is to make a package for your functions that will include documentation, error checks etc. Another way to accomplish what you want is to start a new R session and 'source' your .R files and then to save the workspace in a .RData file, e.g. myFunctions.RData. Finally attach(myFunctions.RData) should do the trick without cluttering your workspace. -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Forest Floor Sent: Wednesday, January 31, 2007 10:41 PM To: r-help@stat.math.ethz.ch Subject: [R] Loading functions in R Hi all, This information must be out there, but I can't seem to find it. What I want to do is to store functions I've created (as .R files or in whatever form) and then load them when I need them (or on startup) so that I can access without cluttering my program with the function code. This seems like it should be easy, but Thanks! Jeff __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix operations in a list
Try,
mapply('%*%', a, b, SIMPLIFY=FALSE)
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Doran, Harold
Sent: Tuesday, January 23, 2007 10:22 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Matrix operations in a list
I have matrices stored within a list like something as follows:
a <- list(matrix(rnorm(50), ncol=5), matrix(rnorm(50), ncol=5)) b <-
list(matrix(rnorm(50), nrow=5), matrix(rnorm(50), nrow=5))
I don't recall how to perform matrix multiplication on each list element
such that the result is a new list
result <- list(a[[1]]%*%b[[1]], a[[2]]%*%b[[2]])
I think I'm close with mapply(), but I'm doing something silly
mapply('%*%', a,b)
Thanks.
Harold
[[alternative HTML version deleted]]
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Re: [R] comparing two matrices
Here is a slightly more compact version of your function which might run
faster (I did not test timings) since it does not use the sum:
apply(mat2, 1, function(x) which(apply(mat1, 1, function(y) all(x == y)) ==
TRUE))
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Adrian Dusa
Sent: Saturday, January 20, 2007 5:15 PM
To: r-help@stat.math.ethz.ch
Subject: [R] comparing two matrices
Dear helpeRs,
I have two matrices:
mat1 <- expand.grid(0:2, 0:2, 0:2)
mat2 <- aa[c(19, 16, 13, 24, 8), ]
where mat2 is always a subset of mat1
I need to find the corersponding row numbers in mat1 for each row in mat2.
For this I have the following code:
apply(mat2, 1, function(x) {
which(apply(mat1, 1, function(y) {
sum(x == y)
}) == ncol(mat1))
})
The code is vectorized, but I wonder if there is a simpler (hence faster)
matrix computation that I miss.
Thank you,
Adrian
--
Adrian Dusa
Romanian Social Data Archive
1, Schitu Magureanu Bd
050025 Bucharest sector 5
Romania
Tel./Fax: +40 21 3126618 \
+40 21 3120210 / int.101
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Re: [R] simple q: returning a logical vector of substring matches
You can try the following:
a == grep("ear", a, value=T)
-Christos
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED]
Sent: Saturday, January 20, 2007 1:31 PM
To: r-help@stat.math.ethz.ch
Subject: [R] simple q: returning a logical vector of substring matches
I'm a relative R novice, and sometimes the simple things trip me up.
Suppose I have
a <- c("apple", "pear")
and I want a logical vector of whether each of these strings contains "ear"
(in this case, F T). What is the idiom?
Quizzically,
Mark Lindeman
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Re: [R] how to get the index of entry with max value in an array?
Or which.max(a) -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of talepanda Sent: Wednesday, January 17, 2007 11:45 PM To: Feng Qiu Cc: r-help@stat.math.ethz.ch Subject: Re: [R] how to get the index of entry with max value in an array? In R language, one solution is: a<-c(3,4,6,2,3) which(a==max(a)) On 1/18/07, Feng Qiu <[EMAIL PROTECTED]> wrote: > Hi all: > A short question: > For example, a=[3,4,6,2,3], obviously the 3rd entry of the > array has the maxium value, what I want is index of the maxium value: > 3. is there a neat expression to get this index? > > Thank you! > > Best, > > Feng > > > [[alternative HTML version deleted]] > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A vectorization question
That's true. Just need to negate the difference. Actually, straight diff can be used after reversing the vector: apply(mat, 1, function(x) diff(sort(x, decreasing = TRUE)[2:1])) I only have 3 columns in my matrix so sorting should not add much overhead, but I will time both versions. Thanks again. -Christos -Original Message- From: Marc Schwartz [mailto:[EMAIL PROTECTED] Sent: Tuesday, January 09, 2007 5:07 PM To: [EMAIL PROTECTED] Cc: 'R-help' Subject: RE: [R] A vectorization question Welcome Christos. Note that my first example can actually be simplified to: apply(mat, 1, function(x) -diff(sort(x, decreasing = TRUE)[1:2])) Since we really just need to negate the difference, rather than take the abs(). The advantage of this approach is that the two max values will always be the first and second values, so will be independent of the length of 'x' (number of columns in the matrix). Using the second example more generally, you would have to use something like: apply(mat, 1, function(x) diff(sort(x)[-c(1:(length(x) - 2))])) in the subsetting of the sort() results or precalcuate the indices (ie. ncol(mat) and ncol(mat) - 1). Might add a bit more overhead, but testing would give you more empiric timing data. That might have to be balanced by whether the rows tend to be random in order or closer to being sorted in increasing/decreasing order, which would affect the sort time. Worst case scenario is generally having to reverse the sort order. Of course, if the matrices are "relatively" small, sorting time would likely be a non-issue. HTH, Marc On Tue, 2007-01-09 at 16:39 -0500, Christos Hatzis wrote: > Thanks, Marc. > This is what I was trying to do but could not get it to work. > > -Christos __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A vectorization question
Thanks, Marc. This is what I was trying to do but could not get it to work. -Christos -Original Message- From: Marc Schwartz [mailto:[EMAIL PROTECTED] Sent: Tuesday, January 09, 2007 4:32 PM To: [EMAIL PROTECTED] Cc: 'R-help' Subject: Re: [R] A vectorization question On Tue, 2007-01-09 at 16:10 -0500, Christos Hatzis wrote: > Hi, > > A function calculates the absolute difference between the two largest > values of each row of a matrix, as shown in the following example code: > > cx <- matrix(runif(15),5) > cy <- t( apply(cx, 1, order, decreasing=TRUE) ) > > cz <- rep(0, nrow(cx)) > for( i in 1:nrow(cx) ) cz[i] <- abs(diff(cx[i, cy[i,1:2]])) > > Anybody has any ideas on how the last loop can be vectorized? > > Thanks. How about this: > mat <- matrix(sample(1:50, 12), ncol = 4) > mat [,1] [,2] [,3] [,4] [1,] 391 22 11 [2,] 34 28 13 48 [3,] 25 40 383 > apply(mat, 1, function(x) abs(diff(sort(x, decreasing = TRUE)[1:2]))) [1] 17 14 2 Or > apply(mat, 1, function(x) diff(sort(x)[3:4])) [1] 17 14 2 HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] A vectorization question
Hi, A function calculates the absolute difference between the two largest values of each row of a matrix, as shown in the following example code: cx <- matrix(runif(15),5) cy <- t( apply(cx, 1, order, decreasing=TRUE) ) cz <- rep(0, nrow(cx)) for( i in 1:nrow(cx) ) cz[i] <- abs(diff(cx[i, cy[i,1:2]])) Anybody has any ideas on how the last loop can be vectorized? Thanks. Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com <http://www.nuverabio.com/> __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] vectorizing an iterative process.
In your case, the recurrence relationship for x can be solved easily:
Notice that
sum{i=1,n}(x[i]-x[i-1]) = x[n] - x[0]
and therefore
x[n] = x[0] + sum{i=1,n}(y[i-1] for n=1, N, with the appropriate initial
condition for i=0, (x[0],y[0]).
Thus cumsum on y will give you a direct answer.
-Christos
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Geoffrey Zhu
Sent: Tuesday, December 26, 2006 11:06 AM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] vectorizing an iterative process.
I meant x[i] <- x[i-1] + y[i-1] and Y[i] <- y[i-1] + x[i] below.
The mailing list software just keep adding 3D's. Sorry.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Geoffrey Zhu
Sent: Tuesday, December 26, 2006 10:03 AM
To: Richard M. Heiberger; r-help@stat.math.ethz.ch
Subject: Re: [R] vectorizing an iterative process.
Hi Richard,
3D is automatically generated by the mailing list software, probably because
I had ] followed by =3D3D without a space in the original post.
What I meant was to calculate x[i] =3D3D x[i-1] + y[i-1]
For example, if X <- 1:10
Then I want the vector Y to be 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, or in
other words Y[i] =3D3D y[i-1] + x[i].
Yes, cumsum does the trick for this. This is what I need. Thanks.
Just curious, do you know how to calculate the more generic x[i] <- f(
x[i-1], y[i-1] )?
Thanks,
Geoffrey
-Original Message-
From: Richard M. Heiberger [mailto:[EMAIL PROTECTED]
Sent: Tuesday, December 26, 2006 9:56 AM
To: Geoffrey Zhu; r-help@stat.math.ethz.ch
Subject: Re: [R] vectorizing an iterative process.
> x[i]=3D3D3Dx[i-1]+y[i-1] for all i, how can I do this without a loop?
It looks like
x <- cumsum(y)
What does 3D mean?
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Re: [R] how to 'get' an object that is part of a list
True. Thanks again.
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Monday, December 25, 2006 8:34 AM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: [R] how to 'get' an object that is part of a list
my.length.2 also has the advantage of eliminating the eval.
On 12/25/06, Christos Hatzis <[EMAIL PROTECTED]> wrote:
> Thank you Gabor. Very interesting solution.
> If I get it right, the first argument in function f is just a
> placeholder to help extract the right element out of the list(...)
> that is passed to length. Very smart trick.
>
> Jim's solution appears a bit simpler, at least along the lines that I
> was
> thinking:
>
> my.length <- function(...) {
>names <- as.character(substitute(list(...)))[-1]
>sapply(names, function(x){y <- eval(parse(text=x)); length(y)})
> }
>
> -Christos
>
> -Original Message-
> From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
> Sent: Sunday, December 24, 2006 7:41 AM
> To: jim holtman
> Cc: [EMAIL PROTECTED];
> Subject: Re: [R] how to 'get' an object that is part of a list
>
> Is this what you are looking for:
>
> > my.length.2 <- function(...) {
> +f <- function(nm, val) length(val)
> +mapply(f, make.names(as.list(match.call()[-1])), list(...)) }
> > my.length.2(xx, xx$b)
> xx xx.b
> 25
>
> On 12/24/06, jim holtman <[EMAIL PROTECTED]> wrote:
> > use 'eval' and 'parse'
> >
> >
> > > xx
> > $a
> > [1] 1 2 3 4 5
> >
> > $b
> > [1] "a" "b" "c" "d" "e"
> >
> > > eval(parse(text='xx$a'))
> > [1] 1 2 3 4 5
> > >
> >
> > So that way you can pass in the character string and then 'parse' it.
> >
> >
> >
> > On 12/24/06, Christos Hatzis <[EMAIL PROTECTED]> wrote:
> > >
> > > This might be an trivial thing but I am stuck.
> > >
> > > Consider:
> > >
> > > xx <- list(a=1:5, b=letters[1:5])
> > >
> > > Although object xx is accessible through its name, how can object
> > > xx$b be accessed similarly through its name?
> > >
> > > > get("xx")
> > > $a
> > > [1] 1 2 3 4 5
> > >
> > > $b
> > > [1] "a" "b" "c" "d" "e"
> > >
> > > > get("xx$b")
> > > Error in get(x, envir, mode, inherits) : variable "xx$b" was not
> > > found
> > >
> > > get("xx")$b will not work in my case because it will probably
> > > require parsing to make it work within a function. E.g.
> > >
> > > my.length <- function(...) {
> > >names <- as.character(substitute(list(...)))[-1]
> > >sapply(names, FUN=function(x){y <- get(x); length(y)}) }
> > > > my.length(xx)
> > > xx
> > > 2
> > > > my.length(xx$a)
> > > Error in get(x, envir, mode, inherits) : variable "xx$a" was not
> > > found
> > > > my.length(xx$a, xx$b)
> > > Error in get(x, envir, mode, inherits) : variable "xx$a" was not
> > > found
> > >
> > > Thank you.
> > >
> > > Christos Hatzis, Ph.D.
> > > Nuvera Biosciences, Inc.
> > > 400 West Cummings Park
> > > Suite 5350
> > > Woburn, MA 01801
> > > Tel: 781-938-3830
> > > www.nuverabio.com
> > >
> > > __
> > > R-help@stat.math.ethz.ch mailing list
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > PLEASE do read the posting guide
> > > http://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible code.
> > >
> >
> >
> >
> > --
> > Jim Holtman
> > Cincinnati, OH
> > +1 513 646 9390
> >
> > What is the problem you are trying to solve?
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to 'get' an object that is part of a list
Thank you Gabor. Very interesting solution.
If I get it right, the first argument in function f is just a placeholder to
help extract the right element out of the list(...) that is passed to
length. Very smart trick.
Jim's solution appears a bit simpler, at least along the lines that I was
thinking:
my.length <- function(...) {
names <- as.character(substitute(list(...)))[-1]
sapply(names, function(x){y <- eval(parse(text=x)); length(y)})
}
-Christos
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Sunday, December 24, 2006 7:41 AM
To: jim holtman
Cc: [EMAIL PROTECTED];
Subject: Re: [R] how to 'get' an object that is part of a list
Is this what you are looking for:
> my.length.2 <- function(...) {
+f <- function(nm, val) length(val)
+mapply(f, make.names(as.list(match.call()[-1])), list(...)) }
> my.length.2(xx, xx$b)
xx xx.b
25
On 12/24/06, jim holtman <[EMAIL PROTECTED]> wrote:
> use 'eval' and 'parse'
>
>
> > xx
> $a
> [1] 1 2 3 4 5
>
> $b
> [1] "a" "b" "c" "d" "e"
>
> > eval(parse(text='xx$a'))
> [1] 1 2 3 4 5
> >
>
> So that way you can pass in the character string and then 'parse' it.
>
>
>
> On 12/24/06, Christos Hatzis <[EMAIL PROTECTED]> wrote:
> >
> > This might be an trivial thing but I am stuck.
> >
> > Consider:
> >
> > xx <- list(a=1:5, b=letters[1:5])
> >
> > Although object xx is accessible through its name, how can object
> > xx$b be accessed similarly through its name?
> >
> > > get("xx")
> > $a
> > [1] 1 2 3 4 5
> >
> > $b
> > [1] "a" "b" "c" "d" "e"
> >
> > > get("xx$b")
> > Error in get(x, envir, mode, inherits) : variable "xx$b" was not
> > found
> >
> > get("xx")$b will not work in my case because it will probably
> > require parsing to make it work within a function. E.g.
> >
> > my.length <- function(...) {
> >names <- as.character(substitute(list(...)))[-1]
> >sapply(names, FUN=function(x){y <- get(x); length(y)}) }
> > > my.length(xx)
> > xx
> > 2
> > > my.length(xx$a)
> > Error in get(x, envir, mode, inherits) : variable "xx$a" was not
> > found
> > > my.length(xx$a, xx$b)
> > Error in get(x, envir, mode, inherits) : variable "xx$a" was not
> > found
> >
> > Thank you.
> >
> > Christos Hatzis, Ph.D.
> > Nuvera Biosciences, Inc.
> > 400 West Cummings Park
> > Suite 5350
> > Woburn, MA 01801
> > Tel: 781-938-3830
> > www.nuverabio.com
> >
> > __
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem you are trying to solve?
>
>[[alternative HTML version deleted]]
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
__
R-help@stat.math.ethz.ch mailing list
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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 'get' an object that is part of a list
Great. Thanks, Jim.
_
From: jim holtman [mailto:[EMAIL PROTECTED]
Sent: Sunday, December 24, 2006 1:11 AM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: [R] how to 'get' an object that is part of a list
use 'eval' and 'parse'
> xx
$a
[1] 1 2 3 4 5
$b
[1] "a" "b" "c" "d" "e"
> eval(parse(text='xx$a'))
[1] 1 2 3 4 5
>
So that way you can pass in the character string and then 'parse' it.
On 12/24/06, Christos Hatzis <[EMAIL PROTECTED]> wrote:
This might be an trivial thing but I am stuck.
Consider:
xx <- list(a=1:5, b=letters[1:5])
Although object xx is accessible through its name,
how can object xx$b be accessed similarly through its name?
> get("xx")
$a
[1] 1 2 3 4 5
$b
[1] "a" "b" "c" "d" "e"
> get("xx$b")
Error in get(x, envir, mode, inherits) : variable "xx$b" was not found
get("xx")$b will not work in my case because it will probably require
parsing to make it work within a function. E.g.
my.length <- function(...) {
names <- as.character(substitute(list(...)))[-1]
sapply(names, FUN=function(x){y <- get(x); length(y)})
}
> my.length(xx)
xx
2
> my.length (xx$a)
Error in get(x, envir, mode, inherits) : variable "xx$a" was not found
> my.length(xx$a, xx$b)
Error in get(x, envir, mode, inherits) : variable "xx$a" was not found
Thank you.
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] how to 'get' an object that is part of a list
This might be an trivial thing but I am stuck.
Consider:
xx <- list(a=1:5, b=letters[1:5])
Although object xx is accessible through its name,
how can object xx$b be accessed similarly through its name?
> get("xx")
$a
[1] 1 2 3 4 5
$b
[1] "a" "b" "c" "d" "e"
> get("xx$b")
Error in get(x, envir, mode, inherits) : variable "xx$b" was not found
get("xx")$b will not work in my case because it will probably require
parsing to make it work within a function. E.g.
my.length <- function(...) {
names <- as.character(substitute(list(...)))[-1]
sapply(names, FUN=function(x){y <- get(x); length(y)})
}
> my.length(xx)
xx
2
> my.length(xx$a)
Error in get(x, envir, mode, inherits) : variable "xx$a" was not found
> my.length(xx$a, xx$b)
Error in get(x, envir, mode, inherits) : variable "xx$a" was not found
Thank you.
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Speeding up small functions
Wee-Jin, The other option that you have is to set up your function as an expression and then evaluate the expression for each new value of x. This might be faster in some cases. sigmoid.fun <- function(x) 1/(1+exp(-x)) sigmoid.expr <- expression( 1/(1+exp(-x)) ) x <- runif(10^6) # non-vectorized function system.time( for(i in seq(along=x)) sigmoid.fun(x[i]) ) [1] 6.76 0.01 7.13 NA NA # vectorized function system.time( sigmoid.fun(x) ) [1] 0.56 0.00 0.59 NA NA # vectorized expression system.time( eval(sigmoid.expr) ) [1] 0.37 0.01 0.39 NA NA -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Wee-Jin Goh Sent: Sunday, November 19, 2006 4:26 PM To: r-help@stat.math.ethz.ch Subject: [R] Speeding up small functions Greetings list, In my code, I have a few small functions that are called very very frequently. An example of one such function is the following : sigmoid<-function(x) 1/(1+exp(-x)) Now, is there anyway to make this go faster? For example, in C++ we could make it inline. Is there a corresponding feature in R? Cheers, Wee-Jin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] X-fold cross validation function for discriminant analysis
Apologies for the typos there. I meant to say see the Bioconductor web site for details. -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Christos Hatzis Sent: Thursday, November 16, 2006 11:53 AM To: 'Wensui Liu'; 'Wade Wall' Cc: r-help@stat.math.ethz.ch Subject: Re: [R] X-fold cross validation function for discriminant analysis One option is the Bioconductor package MLInterfaces that provides a unified interface for several machine learning alrogirithms and methods for cross-validation etc. See the algorithms web site for details. -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Wensui Liu Sent: Thursday, November 16, 2006 11:37 AM To: Wade Wall Cc: r-help@stat.math.ethz.ch Subject: Re: [R] X-fold cross validation function for discriminant analysis how hard is it to write one though? On 11/16/06, Wade Wall <[EMAIL PROTECTED]> wrote: > Hi all, > > I ran a discriminant analysis with some data and want to get a general > idea of prediction error rate. Some have suggested using X-fold cross > validation procedure. Anyone know if there is a function for this in R? > > Thanks, > > Wade > > [[alternative HTML version deleted]] > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- WenSui Liu (http://spaces.msn.com/statcompute/blog) Senior Decision Support Analyst Cincinnati Children Hospital Medical Center __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] X-fold cross validation function for discriminant analysis
One option is the Bioconductor package MLInterfaces that provides a unified interface for several machine learning alrogirithms and methods for cross-validation etc. See the algorithms web site for details. -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Wensui Liu Sent: Thursday, November 16, 2006 11:37 AM To: Wade Wall Cc: r-help@stat.math.ethz.ch Subject: Re: [R] X-fold cross validation function for discriminant analysis how hard is it to write one though? On 11/16/06, Wade Wall <[EMAIL PROTECTED]> wrote: > Hi all, > > I ran a discriminant analysis with some data and want to get a general > idea of prediction error rate. Some have suggested using X-fold cross > validation procedure. Anyone know if there is a function for this in R? > > Thanks, > > Wade > > [[alternative HTML version deleted]] > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- WenSui Liu (http://spaces.msn.com/statcompute/blog) Senior Decision Support Analyst Cincinnati Children Hospital Medical Center __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix-vector multiplication without loops
Ravi,
Here is another version, somewhat similar to what Dimitris proposed but
without using 'rep'. For large n, K 'rep' tries allocating two vectors,
each of length n*(K+1)^2, which can be a problem. In this version 'outer'
buys you some efficiency and compactness, but your looping version is
faster.
n <- 1000
K <- 1000
p <- rep(0, n)
cf <- runif(K+1)
U <- matrix(runif(n*(2*K+1)), n, 2*K+1)
# original code
system.time(
{
for (i in 0:K)
for (j in 0:K)
p <- p + cf[i+1]* cf[j+1] * U[,i+j+1]
p.1 <- p
} )
# 'vectorized'
system.time(
{
ind <- sapply(1:(K+1), seq, length = K+1)
cc <- outer(cf,cf)
p.2 <- apply(U, 1, FUN=function(u) sum(cc * u[ind]))
} )
all.equal(p.1, p.2)
rm(n,K,p,U,cf,cc,ind,p.1,p.2)
-Christos
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Dimitris Rizopoulos
Sent: Tuesday, November 14, 2006 11:20 AM
To: Ravi Varadhan
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Matrix-vector multiplication without loops
I think you need something along these lines:
ind <- c(sapply(1:(K+1), seq, length = K + 1))
cf1 <- rep(rep(coef, each = n), K + 1)
cf2 <- rep(rep(coef, each = n), each = K + 1)
rowSums(cf1 * cf2 * U[, ind])
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: "Ravi Varadhan" <[EMAIL PROTECTED]>
To:
Sent: Tuesday, November 14, 2006 4:45 PM
Subject: [R] Matrix-vector multiplication without loops
> Hi,
>
>
>
> I am trying to do the following computation:
>
>
>
> p <- rep(0, n)
>
> coef <- runif(K+1)
>
> U <- matrix(runif(n*(2*K+1)), n, 2*K+1)
>
> for (i in 0:K){
>
> for (j in 0:K){
>
> p <- p + coef[i+1]* coef[j+1] * U[,i+j+1]
>
> } }
>
>
>
> I would appreciate any suggestions on how to perform this
> computation
> efficiently without the "for" loops?
>
>
>
> Thank you,
>
> Ravi.
>
>
> ---
>
> Ravi Varadhan, Ph.D.
>
> Assistant Professor, The Center on Aging and Health
>
> Division of Geriatric Medicine and Gerontology
>
> Johns Hopkins University
>
> Ph: (410) 502-2619
>
> Fax: (410) 614-9625
>
> Email: [EMAIL PROTECTED]
>
> Webpage:
> http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html
>
>
>
>
>
>
>
>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] R and Fortran 9x -- advice
Mike, Can you recommend any good books on Fortran 90/95? I had been an old user of Fortran 77 but haven't followed the developments in the last 15 years or so... Thanks. Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Mike Prager Sent: Monday, November 13, 2006 1:00 PM To: r-help@stat.math.ethz.ch Subject: Re: [R] R and Fortran 9x -- advice Tamas K Papp <[EMAIL PROTECTED]> wrote: > I found some bottlenecks in my R code with Rprof. First I wanted to > rewrite them in C, but a colleague keeps suggesting that I learn > Fortran, so maybe this is the time to do it... > > 1) I hear bad things about Fortran. Sure, F77 looks archaic, but > F90/95 seems nicer. Is it worth learning, especially when I plan to > use it mainly from R? Dusting off my C knowledge would take a bit of > time too, and I never liked C that much. I'll answer this from the perspective of someone who uses Fortran 95 regularly. It is a modern language, far more reliable and flexible than Fortran 77 and quite well suited to most scientific problems. I do think it's worth learning, particularly if C is not to your taste. Two free compilers for Fortran 95 are available. It seems that g95 is complete, while gfortran is nearing completion. There are also several high-quality commercial compilers, some of which are free under certain operating systems and/or conditions and others of which (Lahey) are quite inexpensive if one is willing to work from the command line or one's own editor. I can't address questions of R interoperability -- not something I've done. -- Mike Prager, NOAA, Beaufort, NC * Opinions expressed are personal and not represented otherwise. * Any use of tradenames does not constitute a NOAA endorsement. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] CPU or memory
Great. I will try it.
Thank you.
-Christos
-Original Message-
From: Prof Brian Ripley [mailto:[EMAIL PROTECTED]
Sent: Wednesday, November 08, 2006 1:21 PM
To: Christos Hatzis
Cc: 'Stefan Grosse'; r-help@stat.math.ethz.ch; 'Taka Matzmoto'
Subject: RE: [R] CPU or memory
On Wed, 8 Nov 2006, Christos Hatzis wrote:
> Prof. Ripley,
>
> Do you mind providing some pointers on how "coarse-grained parallelism"
> could be implemented on a Windows environment? Would it be as simple
> as running two R-console sessions and then (manually) combining the
> results of these simulations. Or it would be better to run them as batch
processes.
That is what I would do in any environment (I don't do such things under
Windows since all my fast machines run Linux/Unix).
Suppose you want to do 1 simulations. Set up two batch scripts that
each run 5000, and save() the results as a list or matrix under different
names, and set a different seed at the top. Then run each via R CMD BATCH
simultaneously. When both have finished, use an interactive session to
load() both sets of results and merge them.
> RSiteSearch('coarse grained') did not produce any hits so this topic
> might have not been discussed on this list.
>
> I am not really familiar with running R in any mode other than the
> default (R-console in Windows) so I might be missing something really
> obvious. I am interested in running Monte-Carlo cross-validation in
> some sort of a parallel mode on a dual core (Pentium D) Windows XP
machine.
>
> Thank you.
> -Christos
>
> Christos Hatzis, Ph.D.
> Nuvera Biosciences, Inc.
> 400 West Cummings Park
> Suite 5350
> Woburn, MA 01801
> Tel: 781-938-3830
> www.nuverabio.com
>
>
>
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Prof Brian
> Ripley
> Sent: Wednesday, November 08, 2006 5:29 AM
> To: Stefan Grosse
> Cc: r-help@stat.math.ethz.ch; Taka Matzmoto
> Subject: Re: [R] CPU or memory
>
> On Wed, 8 Nov 2006, Stefan Grosse wrote:
>
>> 64bit does not make anything faster. It is only of use if you want to
>> use more then 4 GB of RAM of if you need a higher precision of your
>> variables
>>
>> The dual core question: dual core is faster if programs are able to
>> use that. What is sure that R cannot make (until now) use of the two
>> cores if you are stuck on Windows. It works excellent if you use
>> Linux. So if you want dual core you should work with linux (and then
>> its faster of course).
>
> Not necessarily. We have seen several examples in which using a
> multithreaded BLAS (the only easy way to make use of multiple CPUs
> under Linux for a single R process) makes things many times slower.
> For tasks that are do not make heavy use of linear algebra, the
> advantage of a multithreaded BLAS is small, and even from those which
> do the speed-up is rarely close to double for a dual-CPU system.
>
> John mentioned simulations. Often by far the most effective way to
> use a multi-CPU platform (and I have had one as my desktop for over a
> decade) is to use coarse-grained parallelism: run two or more
> processes each doing some of the simulation runs.
>
>> The Core 2 duo is the fastest processor at the moment however.
>>
>> (the E6600 has a good price/performance ration)
>>
>> What I already told Taka is that it is probably always a good idea to
>> improve your code for which purpose you could ask in this mailing
>> list... (And I am very sure that you have there a lot of potential).
>> Another speeding up possibility is e.g. using the atlas library...
>> (where I am not sure if you already use it)
>>
>> Stefan
>>
>> John C Frain schrieb:
>>> *Can I extend Taka's question?*
>>> **
>>> *Many of my programs in (mainly simulations in R which are cpu
>>> bound) on a year old PC ( Intel(R) Pentium(R) M processor 1.73GHz or
>>> Dell GX380 with 2.8Gh Pentium) are taking hours and perhaps days to
>>> complete on a one year old PC. I am looking at an upgrade but the
>>> variety of cpu's available is
>>> confusing at least. Does any one know of comparisons of the Pentium
>>> 9x0, Pentium(r)
>>> Extreme/Core 2 Duo, AMD(r) Athlon(r) 64 , AMD(r) Athlon(r) 64
>>> FX/Dual Core AM2 and
>>> similar chips when used for this kind of work. Does anyone have any
>>> advice on (1) the use of a single core or dual core cpu or (2) on
>>> the use of 32 bit and 64 bit cpu. This question is now much more
>>> difficult as the numbers on the various chips do not necessarily
>&
Re: [R] CPU or memory
Prof. Ripley,
Do you mind providing some pointers on how "coarse-grained parallelism"
could be implemented on a Windows environment? Would it be as simple as
running two R-console sessions and then (manually) combining the results of
these simulations. Or it would be better to run them as batch processes.
RSiteSearch('coarse grained') did not produce any hits so this topic might
have not been discussed on this list.
I am not really familiar with running R in any mode other than the default
(R-console in Windows) so I might be missing something really obvious. I am
interested in running Monte-Carlo cross-validation in some sort of a
parallel mode on a dual core (Pentium D) Windows XP machine.
Thank you.
-Christos
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Prof Brian Ripley
Sent: Wednesday, November 08, 2006 5:29 AM
To: Stefan Grosse
Cc: r-help@stat.math.ethz.ch; Taka Matzmoto
Subject: Re: [R] CPU or memory
On Wed, 8 Nov 2006, Stefan Grosse wrote:
> 64bit does not make anything faster. It is only of use if you want to
> use more then 4 GB of RAM of if you need a higher precision of your
> variables
>
> The dual core question: dual core is faster if programs are able to
> use that. What is sure that R cannot make (until now) use of the two
> cores if you are stuck on Windows. It works excellent if you use
> Linux. So if you want dual core you should work with linux (and then
> its faster of course).
Not necessarily. We have seen several examples in which using a
multithreaded BLAS (the only easy way to make use of multiple CPUs under
Linux for a single R process) makes things many times slower. For tasks
that are do not make heavy use of linear algebra, the advantage of a
multithreaded BLAS is small, and even from those which do the speed-up is
rarely close to double for a dual-CPU system.
John mentioned simulations. Often by far the most effective way to use a
multi-CPU platform (and I have had one as my desktop for over a decade) is
to use coarse-grained parallelism: run two or more processes each doing some
of the simulation runs.
> The Core 2 duo is the fastest processor at the moment however.
>
> (the E6600 has a good price/performance ration)
>
> What I already told Taka is that it is probably always a good idea to
> improve your code for which purpose you could ask in this mailing
> list... (And I am very sure that you have there a lot of potential).
> Another speeding up possibility is e.g. using the atlas library...
> (where I am not sure if you already use it)
>
> Stefan
>
> John C Frain schrieb:
>> *Can I extend Taka's question?*
>> **
>> *Many of my programs in (mainly simulations in R which are cpu bound)
>> on a year old PC ( Intel(R) Pentium(R) M processor 1.73GHz or Dell
>> GX380 with 2.8Gh Pentium) are taking hours and perhaps days to
>> complete on a one year old PC. I am looking at an upgrade but the
>> variety of cpu's available is
>> confusing at least. Does any one know of comparisons of the Pentium
>> 9x0, Pentium(r)
>> Extreme/Core 2 Duo, AMD(r) Athlon(r) 64 , AMD(r) Athlon(r) 64
>> FX/Dual Core AM2 and
>> similar chips when used for this kind of work. Does anyone have any
>> advice on (1) the use of a single core or dual core cpu or (2) on
>> the use of 32 bit and 64 bit cpu. This question is now much more
>> difficult as the numbers on the various chips do not necessarily
>> refer to the relative speed of the chips.
>> *
>> *John
>>
>> * On 06/11/06, Taka Matzmoto <[EMAIL PROTECTED]> wrote:
>>
>>
>>> Hi R users
>>>
>>> Having both a faster CPU and more memory will boost computing power.
>>> I was wondering if only adding more memory (1GB -> 2GB) will
>>> significantly reduce R computation time?
>>>
>>> Taka,
>>>
>>> _
>>> Get FREE company branded e-mail accounts and business Web site from
>>> Microsoft Office Live
>>>
>>> __
>>> R-help@stat.math.ethz.ch mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>>
>>
>>
>>
>>
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.
Re: [R] subsetting a matrix and filling other
Use a list to store the partial matrices:
z <- vector("list", 189)
for(i in 1:189)
z[[i]] <- subset(...)
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of antonio rodriguez
Sent: Tuesday, November 07, 2006 2:01 PM
To: R-Help
Subject: [R] subsetting a matrix and filling other
Hi,
Having a matrix F(189,6575) I want to do this:
z1<-subset(F[,1], F[,1] >= 5 & F[,1] <= 10) .
.
.
z189<-subset(F[,189], F[,189] >= 5 & F[,189] <= 10)
I would prefer to have an empty matrix, say 'z' in order to fill its columns
with the output of subsetting F. But each of the subsets can differ in
length. It's to say:
length(z1)
1189
length(z2)
1238
Don't know how to set up this with a for loop or the use of apply:
z<-189
for (i in 1:189)
{z[i]<-subset(F.zoo[,i], F.zoo[,i] >= 5 & z1 <= 10)}
Error in z[i] <- subset(F.zoo[, i], F.zoo[, i] >= 5 & z1 <= 10) :
nothing to replace with
or
z<-matrix(0,6575,189)
for (i in 1:189)
{z[i]<-subset(F.zoo[,i], F.zoo[,i] >= 5 & z1 <= 10)}
new error
Thanks,
Antonio
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Re: [R] alist()
You need to start by first allocating the list. The problem is that you
cannot reference a list member that does not exist yet (as in x$two):
x <- vector("list", 10)
x
List x has 10 NULL elements by default. You can then assign values to these
as needed.
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Glen Herrmannsfeldt
Sent: Monday, November 06, 2006 3:37 PM
To: r-help@stat.math.ethz.ch
Subject: [R] alist()
In trying to get NULL members into a list, I found out about alist().
x<-alist()
x$one<-1
x$two<-NULL
but x$two doesn't exist.
It seems, though, that an alist is just a list.
How can one put NULL members into a list?
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Re: [R] Subset and levels
Hi Florent, A simple example with the expected output would help. If I understood correctly what you want, perhaps the following would work: X[ X$code %in% levels(Y$code), ] assuming Y$code is a factor. If not, you can used instead unique(Y$code) in the above. -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Florent Bresson Sent: Monday, November 06, 2006 12:15 PM To: R-help Subject: [R] Subset and levels Hi, I've got a very simple problem but cannot find the solution. I'm using two data frames (say X and Y) and I want to get a subset of one according to the different levels of a variable "code" of the other data frame. I tried something like Z<-subset(X, code==levels(Y$code))(1) but it does not work. I do not want to do something like Z<-subset(X,code==level1 | code==level2...) because length(levels(Y$code)) is 130. So what's wrong with (1)? Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to obtain the estimate of baseline survival function?
See ?basehaz in 'survival' package. -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Zheng Yuan Sent: Friday, November 03, 2006 6:20 PM To: r-help@stat.math.ethz.ch Subject: [R] How to obtain the estimate of baseline survival function? Hi, If I fit a Cox model using "coxph", is there a R function so that I could obtain the estimate of baseline survival function? Thank you. Zheng -- Zheng Yuan Ph.D student Department of Biostatistics University of Michigan Ann Arbor, MI 48109 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] multiple plots in the same graph
See ?points and ?lines for adding new groups of points or lines to an existing graph created by plot. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Li Zhang Sent: Friday, November 03, 2006 9:55 PM To: R-help@stat.math.ethz.ch Subject: [R] multiple plots in the same graph I'd like to plot y vs x according to the third variable "group" which has three levels. I am wondering how can I put the three plots in one graph? Thank you (http://groups.yahoo.com) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] psigamma derivative
Try
?Special
For information on the special functions included in the base R
distribution.
Also,
RSiteSearch("digamma")
gives several hits on additional packages that might be useful.
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED]
Sent: Monday, October 30, 2006 3:26 PM
To: r-help@stat.math.ethz.ch
Subject: [R] psigamma derivative
Hello,
I am trying to find a hessian matrix that involves log(gamma(1/p)) second
derivative, p being one of the parameters of the function. I am using a
function "deriv" with the hessian=TRUE option, but psigamma is not on the
list of derivative functions.
I know that it is possible to use 'psigamma(p,deriv)', but it doesn't work
with 1/p. Does anybody can help with this?
Thanks,
Ana Maria
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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] help fo finding R package "utils"
Please reply to the list, not to me personally.
When I want to import an Excel file, I usually save it as a tab-delimited
text file and use read.delim to import it:
my.dataframe <- read.delim("file.name.txt", header=TRUE, sep="\t")
See ?read.delim for further details. There are many other options, I'm
sure, but this is simple and flexible and does not require any additional
packages.
Good luck.
-Christos
From: amna khan [mailto:[EMAIL PROTECTED]
Sent: Thursday, October 26, 2006 2:41 PM
To: [EMAIL PROTECTED]
Subject: Re: [R] help fo finding R package "utils"
sir i have downloaded poptools for excel to import data but still feeling
problems. I am not understanding what I should do?
On 10/26/06, Christos Hatzis <[EMAIL PROTECTED]> wrote:
Amina,
The best way for entering data in R is by importing an existing data
file (text, Excel etc). This will reduce time and potential data entry
errors.
Please read Section 7 of the "Introduction to R" manual for a
detailed discussion of available options. The manuals are available on your
system (Help menu in the R Console for Windows) or through the R web site.
Regards,
-Christos
From: amna khan [mailto:[EMAIL PROTECTED]
Sent: Thursday, October 26, 2006 1:21 PM
To: [EMAIL PROTECTED]
Subject: Re: [R] help fo finding R package "utils"
Respected sir
I have a great problem for data entry in R. using sessionInfo()
utils has appeared. but while entring data there is a message of no object
found. Sir is there any option for spreadsheet for data entry.
I shall be thankful to you.
Amina
On 10/25/06, Christos Hatzis <[EMAIL PROTECTED]
<mailto:[EMAIL PROTECTED]> > wrote:
Amina,
utils is a base package and should be already installed in a
functional R
system.
You can try
sessionInfo()
to verify that it is loaded.
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto: [EMAIL PROTECTED] On Behalf Of amna
khan
Sent: Thursday, October 26, 2006 12:11 AM
To: R-help@stat.math.ethz.ch
<mailto:R-help@stat.math.ethz.ch>
Subject: [R] help fo finding R package "utils"
Sir I have a problem that from which country i can assess R
package "utils"
for data entry for loading it in R
Please help me in this regard.
--
AMINA SHAHZADI
Department of Statistics
GC University Lahore, Pakistan.
Email:
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[[alternative HTML version deleted]]
__
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mailing list
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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.
--
AMINA SHAHZADI
Department of Statistics
GC University Lahore, Pakistan.
Email:
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED]
--
AMINA SHAHZADI
Department of Statistics
GC University Lahore, Pakistan.
Email:
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] help fo finding R package "utils"
Amina, The best way for entering data in R is by importing an existing data file (text, Excel etc). This will reduce time and potential data entry errors. Please read Section 7 of the "Introduction to R" manual for a detailed discussion of available options. The manuals are available on your system (Help menu in the R Console for Windows) or through the R web site. Regards, -Christos _ From: amna khan [mailto:[EMAIL PROTECTED] Sent: Thursday, October 26, 2006 1:21 PM To: [EMAIL PROTECTED] Subject: Re: [R] help fo finding R package "utils" Respected sir I have a great problem for data entry in R. using sessionInfo() utils has appeared. but while entring data there is a message of no object found. Sir is there any option for spreadsheet for data entry. I shall be thankful to you. Amina On 10/25/06, Christos Hatzis <[EMAIL PROTECTED]> wrote: Amina, utils is a base package and should be already installed in a functional R system. You can try sessionInfo() to verify that it is loaded. -Christos -Original Message- From: [EMAIL PROTECTED] [mailto: [EMAIL PROTECTED] On Behalf Of amna khan Sent: Thursday, October 26, 2006 12:11 AM To: R-help@stat.math.ethz.ch <mailto:R-help@stat.math.ethz.ch> Subject: [R] help fo finding R package "utils" Sir I have a problem that from which country i can assess R package "utils" for data entry for loading it in R Please help me in this regard. -- AMINA SHAHZADI Department of Statistics GC University Lahore, Pakistan. Email: [EMAIL PROTECTED] [EMAIL PROTECTED] <mailto:[EMAIL PROTECTED]> [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch <mailto:R-help@stat.math.ethz.ch> mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html <http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code. -- AMINA SHAHZADI Department of Statistics GC University Lahore, Pakistan. Email: [EMAIL PROTECTED] [EMAIL PROTECTED] [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help fo finding R package "utils"
Amina, utils is a base package and should be already installed in a functional R system. You can try sessionInfo() to verify that it is loaded. -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of amna khan Sent: Thursday, October 26, 2006 12:11 AM To: R-help@stat.math.ethz.ch Subject: [R] help fo finding R package "utils" Sir I have a problem that from which country i can assess R package "utils" for data entry for loading it in R Please help me in this regard. -- AMINA SHAHZADI Department of Statistics GC University Lahore, Pakistan. Email: [EMAIL PROTECTED] [EMAIL PROTECTED] [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] avoiding a loop
Try this (essentially the trick is to shift the invector to get the y[i-i]
effect):
constructLt<-function(invector, a=1) {
invector[invector aa <- c(1,1,0.5,2,3,0.4,4,5)
> aa
[1] 1.0 1.0 0.5 2.0 3.0 0.4 4.0 5.0
> constructLt(aa)
[1] 1.0 1.0 0.5 2.0 3.0 1.2 4.0 5.0
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Leeds, Mark (IED)
Sent: Tuesday, October 24, 2006 2:36 PM
To: R-help@stat.math.ethz.ch
Subject: [R] avoiding a loop
I think I asked a similar question 3 years ago to the Splus list and I
think the answer was no or noone answered so noone should spend more than 5
minutes on this because it could definitely be a waste of time.
My question is whether the function below can be rewritten without a for
loop. apply is fine if it can be done that way but i doubt it. I call it a
lot and would prefer to not loop.
#---
--
constructLt<-function(invector) {
outvector<-invector
for ( i in 2:length(invector) ) {
if ( invector[i] < 1 ) {
outvector[i]<-invector[i]*outvector[i-1]
}
}
return(outvector)
}
#---
-
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Re: [R] Plotmath expression
Excellent! Thank you. I am reading now your article in R-News (2/3 p.32-34) to understand the logic behind substitute in the expression. -Christos -Original Message- From: Uwe Ligges [mailto:[EMAIL PROTECTED] Sent: Tuesday, October 24, 2006 12:27 PM To: [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Plotmath expression Christos Hatzis wrote: > Hello, > > I've been trying to plot a subscript in a text formula using plotmath > but I haven't been able to do so. > > In my example below I would like the text label to show X[min] = 10.1 > +/- 5.5 > > Here is the code: > > ll <- c(x=10.1, sde=5.5) > plot(1:10) > text(x=9, y=2, pos=2, expression(paste(X[min], "=", paste(ll, > collapse="+/-" plot(1:10) text(x=9, y=2, pos=2, substitute(X[min] == x %+-% sde, as.list(ll))) Uwe Ligges > This works fine up to the inner paste call that collapses the vector values. > I also tried assigning the value of the inner paste to a variable and > then use it inside the expression but that did not work either: > > ll.txt <- paste(ll, collapse="+/-") > text(x=9, y=2, pos=2, expression(paste(X[min], "=", ll.txt))) > > Any help is much appreciated. > > Thanks. > > Christos Hatzis, Ph.D. > Nuvera Biosciences, Inc. > 400 West Cummings Park > Suite 5350 > Woburn, MA 01801 > Tel: 781-938-3830 > www.nuverabio.com > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plotmath expression
Hello, I've been trying to plot a subscript in a text formula using plotmath but I haven't been able to do so. In my example below I would like the text label to show X[min] = 10.1 +/- 5.5 Here is the code: ll <- c(x=10.1, sde=5.5) plot(1:10) text(x=9, y=2, pos=2, expression(paste(X[min], "=", paste(ll, collapse="+/-" This works fine up to the inner paste call that collapses the vector values. I also tried assigning the value of the inner paste to a variable and then use it inside the expression but that did not work either: ll.txt <- paste(ll, collapse="+/-") text(x=9, y=2, pos=2, expression(paste(X[min], "=", ll.txt))) Any help is much appreciated. Thanks. Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with coef
Tom, coef returns a "named" vector, which is a vector with an extra attribute called "names". To remove the extra attribute you can: names(a) <- NULL# through the accessor function [EMAIL PROTECTED] <- NULL # directly accessing the attribute names or by creating a new vector as you did without setting its names attribute: as.vector(a) -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of tom soyer Sent: Saturday, October 21, 2006 1:00 AM To: Gabor Grothendieck Cc: r-help Subject: Re: [R] help with coef Gabor, Thanks for the code example, but it seems that BOD is not needed. I still don't understand what is going on with the data structure returned by coef(). The strangness is illustrated by the following example: > a=coef(lm(y~miles)) > is.vector(a) [1] TRUE > a[2] miles -7.2875 > a=as.vector(a) > is.vector(a) [1] TRUE > a[2] [1] -7.2875 As you can see, although coef() returns a vector already, only after as.vector(a) is used, did a[2] include the slope without the name of the slope. Why is that, and what happened to the name of the slope (names(a) returns NULL)? Tom On 10/20/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote: > > Using the builtin BOD data frame: > > as.vector(coef(lm(demand ~ Time, BOD)))[2] > > > On 10/21/06, tom soyer <[EMAIL PROTECTED]> wrote: > > Hi, > > > > I am trying to get R to return just the slope of a linear regression > line, > > but it seems that R has to return both the slope and the name of the > slope. > > For example, > > > > > a=coef(lm(y~miles)) > > > a > > (Intercept) miles > > 360.3778 -7.2875 > > > names(a) > > [1] "(Intercept)" "miles" > > > a[1] > > (Intercept) > > 360.3778 > > > a[2] > > miles > > -7.2875 > > > > I don't understand the data structure that's returned from coef(). > names(a) > > seems to suggest that coef() returns two columns of data, column one > > is > the > > Intercept, and column two miles. But R keeps telling me that the > > return value from coef() has only one dimension, i.e., a[,2] doesn't > > work, but > a[2] > > works. However, a[2] contains more than the slope, it also has the > > name > of > > the slope. Does anyone know how to access just the slope without its > name? > > > > Thanks, > > > > Tom > > > >[[alternative HTML version deleted]] > > > > __ > > R-help@stat.math.ethz.ch mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. > > > [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sort question in a dataset?
Another way to do this is: o <- order(a[,"y"],-a[,"x"], decreasing=TRUE) a[o,] -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Richard M. Heiberger Sent: Sunday, October 15, 2006 1:32 AM To: zhijie zhang Cc: R-help@stat.math.ethz.ch Subject: Re: [R] sort question in a dataset? Your desired answer just interchanges the sequence of the steps x <- c(2, 9, 18, 3, 2) y <- c(2,9,8,9,8) z <- c(21,5,5,19,7) a <- cbind(x, y, z) #dataset bb <- a[order(a[,"x"], decreasing=FALSE),] bbb <- bb[order(bb[,"y"], decreasing=TRUE),] bbb >From ?sort Sort (or order) a vector or factor (partially) into ascending (or descending) order. For ordering along more than one variable, e.g., for sorting data frames, see order. >From ?order order returns a permutation which rearranges its first argument into ascending or descending order, breaking ties by further arguments. tmp <- c(10,15,12,7) sort(tmp) order(tmp) tmp[order(tmp)] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] combinatorics
Hi Robin,
This approach first generates all combinations and then eliminates the
non-feasible ones. It should work fine for smallish vectors but might not
scale well for larger vectors. Hopefully it gives you what you need for
this problem.
xx <- c("A","A","B","B","C")
yy <- 1:length(xx)
zz <- expand.grid(yy,yy,yy,yy,yy)
ss <- zz[ apply(zz, 1, FUN=function(x) length(unique(x))) == length(xx), ]
ss <- as.matrix(ss)
pp <- apply(ss, 1, FUN=function(x,v) paste(v[as.vector(x)], collapse=""),
v=xx)
res <- unique(pp)
> res
[1] "CBBAA" "BCBAA" "BBCAA" "CBABA" "BCABA" "CABBA" "ACBBA" "BACBA" "ABCBA"
"BBACA" "BABCA"
[12] "ABBCA" "CBAAB" "BCAAB" "CABAB" "ACBAB" "BACAB" "ABCAB" "CAABB" "ACABB"
"AACBB" "BAACB"
[23] "ABACB" "AABCB" "BBAAC" "BABAC" "ABBAC" "BAABC" "ABABC" "AABBC"
> length(res)
[1] 30
-Christos
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Robin Hankin
Sent: Friday, October 13, 2006 10:19 AM
To: [EMAIL PROTECTED]
Subject: [R] combinatorics
Hi
How do I generate all ways of ordering sets of indistinguishable items?
suppose I have two A's, two B's and a C.
Then I want
AABBC
AABCB
AACBC
ABABC
. . .snip...
BBAAC
. . .snip...
CBBAA
[there are 5!/(2!*2!) = 30 arrangements. Note AABBC != BBAAC]
How do I do this?
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton European Way, Southampton SO14
3ZH, UK
tel 023-8059-7743
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Re: [R] The W statistic in wilcox.exact
Jue, On a second look, it appears that wilcox.test does report the offset-adjusted statistic U, as also mentioned in the help page. wilcox.test returns W=6 (instead of 12 as your example showed, unless "wilcox_test" is a different function). > wilcox.test( 1:5 ~ c(1,1,0,0,0) )$statistic # or wilcox.test( 1:5 ~ factor(c(1,1,0,0,0)) )$statistic W 6 So there does not appear to be a difference between the two methods. Did I miss something? -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Christos Hatzis Sent: Thursday, October 05, 2006 4:41 PM To: [EMAIL PROTECTED]; r-help@stat.math.ethz.ch Subject: Re: [R] The W statistic in wilcox.exact Probably because of the offset: U = W - n*(n+1)/2 In your example, W=12 (=3+4+5) as reported by wilcox.test. The offset is 6 (=3*4/2) and therefore U=6. I am not certain as I haven't installed the exactRankTests package, but it seems that wilcox.exact reports U instead of W. -Christos Hatzis -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED] Sent: Thursday, October 05, 2006 1:35 PM To: r-help@stat.math.ethz.ch Subject: [R] The W statistic in wilcox.exact Does anyone know why wilcox.exact gives W-statistic 6 instead of 12 as indicated below. 12 is the rank sum of group 0 of x, which is the linear statistic computed by wilcox_test. y<-c(1,2,3,4,5) x<-c(1,1,0,0,0) (a) wilcox.exact wilcox.exact(y~x) Exact Wilcoxon rank sum test data: y by x W = 6, p-value = 0.2 alternative hypothesis: true mu is not equal to 0 (b) wilcox_test tt<-wilcox_test(y~factor(x),distribution="exact") statistic(tt,"linear") 0 12 Jue Wang, Biostatistician Contracted Position for Preclinical & Research Biostatistics PrO Unlimited (908) 231-3022 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] The W statistic in wilcox.exact
Probably because of the offset: U = W - n*(n+1)/2 In your example, W=12 (=3+4+5) as reported by wilcox.test. The offset is 6 (=3*4/2) and therefore U=6. I am not certain as I haven't installed the exactRankTests package, but it seems that wilcox.exact reports U instead of W. -Christos Hatzis -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED] Sent: Thursday, October 05, 2006 1:35 PM To: r-help@stat.math.ethz.ch Subject: [R] The W statistic in wilcox.exact Does anyone know why wilcox.exact gives W-statistic 6 instead of 12 as indicated below. 12 is the rank sum of group 0 of x, which is the linear statistic computed by wilcox_test. y<-c(1,2,3,4,5) x<-c(1,1,0,0,0) (a) wilcox.exact wilcox.exact(y~x) Exact Wilcoxon rank sum test data: y by x W = 6, p-value = 0.2 alternative hypothesis: true mu is not equal to 0 (b) wilcox_test tt<-wilcox_test(y~factor(x),distribution="exact") statistic(tt,"linear") 0 12 Jue Wang, Biostatistician Contracted Position for Preclinical & Research Biostatistics PrO Unlimited (908) 231-3022 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] inserting columns in the middle of a dataframe
Sorry, I guess I did not explain at all how append could work in a one-liner: data.frame(df, v5)[append(1:4,5,2)] Your method is fine as well. The above might be more flexible if you need a more general solution, e.g. if you wanted to make it a function. -Christos -Original Message- From: Jon Minton [mailto:[EMAIL PROTECTED] Sent: Wednesday, September 13, 2006 6:43 PM To: [EMAIL PROTECTED]; r-help@stat.math.ethz.ch Cc: 'Jon Minton' Subject: RE: [R] inserting columns in the middle of a dataframe Thanks, but isn't that only for elements in vectors? I think I've found the following method to work: e.g. for df <- data.frame(v1,v2,v3,v4) use: df <- data.frame(df[1:2],v5,df[-c(1:2)]) I *believe* this is the one-line solution I was looking for. Can anyone see why this wouldn't work? Jon -----Original Message- From: Christos Hatzis [mailto:[EMAIL PROTECTED] Sent: 13 September 2006 23:22 To: 'Jon Minton'; r-help@stat.math.ethz.ch Subject: RE: [R] inserting columns in the middle of a dataframe See ?append -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Jon Minton Sent: Wednesday, September 13, 2006 5:14 PM To: r-help@stat.math.ethz.ch Cc: 'Jon Minton' Subject: Re: [R] inserting columns in the middle of a dataframe Dear R users: Is there a built-in and simple way to insert new columns after other columns in a dataframe? I.e. currently I have: V1 V2 V3 V4 [1,] [2,] Etc. But I want V1 V5 V2 V3 V4 [1,] [2,] Etc. Can this be done in one line? Jon Minton [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] inserting columns in the middle of a dataframe
See ?append -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Jon Minton Sent: Wednesday, September 13, 2006 5:14 PM To: r-help@stat.math.ethz.ch Cc: 'Jon Minton' Subject: Re: [R] inserting columns in the middle of a dataframe Dear R users: Is there a built-in and simple way to insert new columns after other columns in a dataframe? I.e. currently I have: V1 V2 V3 V4 [1,] [2,] Etc. But I want V1 V5 V2 V3 V4 [1,] [2,] Etc. Can this be done in one line? Jon Minton [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] rename cols
Try this: old.colnames <- colnames(my.439.vars.df) old.colnames[old.colnames=="fksm"] <- "new.name.a" old.colnames[old.colnames=="klmk"] <- "new.name.b" I don't think it would be too complicated to put this into a function. -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Anupam Tyagi Sent: Tuesday, September 12, 2006 12:34 AM To: r-help@stat.math.ethz.ch Subject: Re: [R] rename cols For a newcomer who wants to rename variable "fksm" and "klmk" in a dataframe of with 439 variables there is not easy and intuitive solution. That person has to spend a lot of time listing columns and counting columns or doing string searches or using brackets within brackets within brackets to get a simple thing done. Is there a simple function or solution to this in R without using an add-on package? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix multiplication using apply() or lappy() ?
See ?sweep
sweep(a, 2, a[1,],"/")
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
[EMAIL PROTECTED]
Sent: Wednesday, September 06, 2006 11:49 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Matrix multiplication using apply() or lappy() ?
I am trying to divide the columns of a matrix by the first row in the
matrix.
I have tried to get this using apply and I seem to be missing a concept
regarding the apply w/o calling a function but rather command args %*% /
etc. Would using apply be more efficient than this approach?
I have observed examples in the archives using this type of approach. Does
anybody have a snippet of a call to apply() that would accomplish this as
well?
Thanks!
seed=50
$a = array(rnorm(20),dim=c(4,5))
$b = matrix(a[1,],dim(a)[1],dim(a)[2],byrow=T)
$a
[,1] [,2] [,3][,4] [,5]
[1,] -1.3682810 -0.4314462 1.57572752 0.67928882 -0.3672346 [2,] 0.4328180
0.6556479 0.64289931 0.08983289 0.1852306 [3,] -0.8113932 0.3219253
0.08976065 -2.99309008 0.5818237 [4,] 1.4441013 -0.7838389 0.27655075
0.28488295 1.3997368
$a/b
[,1] [,2] [,3] [,4] [,5]
[1,] 1.000 1.000 1. 1.000 1.00 [2,] -0.3163225
-1.5196515 0.40800157 0.1322455 -0.504393 [3,] 0.5930018 -0.7461539
0.05696457 -4.4062113 -1.584338 [4,] -1.0554128 1.8167710 0.17550671
0.4193841 -3.811560
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Re: [R] Quick question about lm()
Say, my.lm <- lm(y ~ x, data=my.data) Then if you try: names(summary(my.lm)) you will see the components of the summary.lm object. The coefficients and t-statistics can be extracted by summary(my.lm)$coefficients and similarly for the r-squared and other statistics provided in the summary report. -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Tong Wang Sent: Tuesday, September 05, 2006 1:35 AM To: r-help@stat.math.ethz.ch Subject: [R] Quick question about lm() Hi, Feel awkward to ask , but really couldn't find a answer anywhere, How could I extract the R^2 and t-stat. from the result of lm()? Thanks a lot. best __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plots Without Displaying
Yes, you can do that for lattice-based plots. The functions in the lattice package produce objects of class "trellis" which can be stored in a list and processed or updated at a later time: library(lattice) attach(barley) plotList <- list(length=3) plotList[[1]] <- xyplot(yield ~ site, data=barley) plotList[[2]] <- xyplot(yield ~ variety, data=barley) plotList[[3]] <- xyplot(yield ~ year, data=barley) plotList plotList[[3]] <- update(plotList[[3]], yaxis="Yield (bushels/acre)") print(plotList[[3]]) Obviously, you can store any lattice-based plot in the list. HTH. -Christos Christos Hatzis, Ph.D. Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Lothar Botelho-Machado Sent: Wednesday, August 16, 2006 4:49 PM To: r-help@stat.math.ethz.ch Subject: Re: [R] Plots Without Displaying -BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Prof Brian Ripley wrote: > Yes, see > > ?jpeg > ?bitmap > > and as you didn't tell us your OS we don't know if these are available > to you. > > jpeg(file="test.jpg") > boxplot(sample(100)) > dev.off() > > may well work. > > 'An Introduction to R' explains about graphics devices, including these. > > > On Wed, 16 Aug 2006, Lothar Botelho-Machado wrote: > >> -BEGIN PGP SIGNED MESSAGE- >> Hash: SHA1 >> >> R Help Mailing List, >> >> >> I'd like to generate a plot that I could display and/or store it as e.g. >> jpeg. But unfortunately always a plotting window opens. Is it >> possible to prevent that? >> >> I tried the following: >> R> bp<-boxplot( sample(100), plot=FALSE) >> >> This works somehow, but it only stores data (as discribed in the >> help) in bp and it is not possible afaik to display bp later on or >> store them as a jpeg. >> >> The next: >> R> p<-plot(sample(100), sample(100), plot=FALSE) >> ..and also a variant using jpeg() didn't work at all. >> >> Is there a way to generally store the plots as object, without >> displaying them, or perhaps directly saving them to disc as jpeg? >> >> A "Yes" or "No" or any further help/links are appreciated!!! > > Thank you for the explanation and your patience in answering me this obviously very simple question!! Originally I tried to store plots directly in a list. So writing them directly to disc was just a good alternative. I knew that that jpeg() provides functionality for that, but didn't use it correctly. Hence, is it also possible to store a plot in a list, somehow? Kind regards, Lothar -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.3 (GNU/Linux) Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org iD8DBQFE44S8HRf7N9c+X7sRAgCpAKC3NhjCYwkteksOljUKWrO3166nCwCgsfLI EPGVIoqc2dla5t6s9mmZQqE= =h+Az -END PGP SIGNATURE- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] invisible() - does not return immediately as return() does
Hi,
The difference is in the _return_ value of the function.
E.g.
> foo <- function() { cat("before\n"); cat("after\n"); return("done")}
> foo()
before
after
[1] "done"
i.e. returns the return value "done".
However
> foo2 <- function() { cat("before\n"); cat("after\n"); invisible("done")}
> foo2()
before
after
does not show the return value (invisible), but it actually returns it
invisibly:
> x <- foo2()
before
after
> x
[1] "done"
HTH.
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Matthias Burger
Sent: Friday, August 11, 2006 4:45 PM
To: r-help@stat.math.ethz.ch
Subject: [R] invisible() - does not return immediately as return() does
Hi,
I stumbled across the following (unexpected for me) behavior after replacing
a return() statement in the middle of a function by invisible().
Example:
foo <- function() { cat("before\n"); return(); cat("after\n")}
>foo()
before
NULL
foo2 <- function() { cat("before\n"); invisible(TRUE); cat("after\n")}
>foo2()
before
after
I expected invisible to have the same behavior as return, namely immediately
return execution to the calling environment.
I rechecked ?invisible and ?return
and here I read in section 'See Also'
[...]
'invisible' for 'return(.)'ing _invisibly_.
Do I just misunderstand what this implies?
Put another way what is the intention behind invisible() continuing until
the last statement before returning? ?invisible does not hint at this.
Regards,
Matthias
>R.version.string
[1] "Version 2.3.1 (2006-06-01)"
same behavior in R 2.2.1 or
R.version.string
[1] "R version 2.4.0 Under development (unstable) (2006-07-29 r38715)"
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Re: [R] Pairwise n for large correlation tables?
Hi,
You can use complete.cases
It should run faster than the code you suggested.
See following example:
x <- matrix(runif(30),10,3)
# introduce missing values
x[sample(1:10,3),1] <- NA
x[sample(1:10,3),2] <- NA
x[sample(1:10,3),3] <- NA
cor(x,use="pairwise.complete.obs")
n <- ncol(x)
n.na <- matrix(0, n, n)
for (i in seq(1, n)) {
n.na[i,i] <- sum( complete.cases(x[, i]) )
for (j in seq(i+1, length=n-i)) {
ok <- sum( complete.cases(x[, i], x[, j]) )
n.na[i,j] <- n.na[j,i] <- ok
}
}
HTH
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Adam D. I. Kramer
Sent: Monday, August 07, 2006 10:04 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Pairwise n for large correlation tables?
Hello,
I'm using a very large data set (n > 100,000 for 7 columns), for which I'm
pretty happy dealing with pairwise-deleted correlations to populate my
correlation table. E.g.,
a <- cor(cbind(col1, col2, col3),use="pairwise.complete.obs")
...however, I am interested in the number of cases used to compute each cell
of the correlation table. I am unable to find such a function via google
searches, so I wrote one of my own. This turns out to be highly inefficient
(e.g., it takes much, MUCH longer than the correlations do). Any hints,
regarding other functions to use or ways to maket his speedier, would be
much appreciated!
pairwise.n <- function(df=stop("Must provide data frame!")) {
if (!is.data.frame(df)) {
df <- as.data.frame(df)
}
colNum <- ncol(df)
result <-
matrix(data=NA,nrow=colNum,ncol=ncolNum,dimnames=list(colnames(df),colnames(
df)))
for(i in 1:colNum) {
for (j in i:colNum) {
result[i,j] <- length(df[!is.na(df[i])&!is.na(df[j])])/colNum
}
}
result
}
--
Adam D. I. Kramer
University of Oregon
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Re: [R] inserting rows into a matrix
Hi Robin,
Ok. I see. Try this then:
f <- function(a){if( any(a==0) ) stop("Zeros in input vector") else
cbind(a,a+1,rev(a))}
a2 <- c(1,0,0,2,4,0,3)
f(a2) # error message
f2 <- function(a) {
indx.zero <- which(a==0)
indx.nz <- (1:length(a))[-indx.zero]
x <- f(a[indx.nz])
xx <- matrix(NA,length(a),ncol(x))
xx[ indx.zero, ] <- rep(0,ncol(x))
xx[ indx.nz, ] <- x
xx
}
> f2(a2)
[,1] [,2] [,3]
[1,]123
[2,]000
[3,]000
[4,]234
[5,]452
[6,]000
[7,]341
-Christos
-Original Message-
From: Robin Hankin [mailto:[EMAIL PROTECTED]
Sent: Thursday, July 27, 2006 10:16 AM
To: [EMAIL PROTECTED]
Cc: 'Robin Hankin'; 'RHelp'
Subject: Re: [R] inserting rows into a matrix
Hi Christos
thanks for this, but it won't work because in my application
f(A2) will fail if there are any zeroes in A2.
cheers
rksh
On 27 Jul 2006, at 15:10, Christos Hatzis wrote:
> This is not as elegant, but should work:
>
> a3 <- f(A2)
> a3[ which( apply(a3,1,prod) == 0 ), ] <- rep(0,ncol(a3))
> a3
>
> Essentially use the product to pick out the rows with at least one 0
> and replace these rows with 0s.
>
> HTH.
> -Christos
>
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Robin Hankin
> Sent: Thursday, July 27, 2006 9:54 AM
> To: RHelp
> Subject: [R] inserting rows into a matrix
>
> Hi
>
>
> I have a little vector function that takes a vector A of strictly
> positive integers and outputs a matrix M each of whose columns is the
> vector, modified in a complicated combinatorical way.
>
> Now I want to generalize the function so that A can include zeroes.
> Given A,
> I want to strip out the zeroes, pass it to my function, and pad M
> with rows at positions corresponding to the zeroes of A.
>
> Commented, minimal, self-contained, reproducible toy example follows.
>
>
> f <- function(a){cbind(a,a+1,rev(a))} #real function a ghastly
> nightmare
>
> A <- 1:5
> f(A)
> a
> [1,] 1 2 5
> [2,] 2 3 4
> [3,] 3 4 3
> [4,] 4 5 2
> [5,] 5 6 1
>
>
> # f() works as desired.
>
> # Now introduce A2, that includes zeroes. In my application, f(A2)
> would fail because of the zeroes.
>
> A2 <- c(1,0,0,2,4,0,3)
>
> I can strip the zeroes out and call f():
> f(A2[A2>0])
> a
> [1,] 1 2 3
> [2,] 2 3 4
> [3,] 4 5 2
> [4,] 3 4 1
>
> which is fine. How to put the zeroes back in in the appropriate rows
> and get the following:
>
>> cbind(c(1,0,0,2,4,0,3),c(2,0,0,3,5,0,4),c(3,0,0,4,2,0,1))
> [,1] [,2] [,3]
> [1,]123
> [2,]000
> [3,]000
> [4,]234
> [5,]452
> [6,]000
> [7,]341
>>
>
>
>
> anyone?
>
>
>
> --
> Robin Hankin
> Uncertainty Analyst
> National Oceanography Centre, Southampton European Way, Southampton
> SO14
> 3ZH, UK
> tel 023-8059-7743
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html and provide commented, minimal, self-contained,
> reproducible code.
>
>
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton European Way, Southampton SO14
3ZH, UK
tel 023-8059-7743
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] inserting rows into a matrix
This is not as elegant, but should work:
a3 <- f(A2)
a3[ which( apply(a3,1,prod) == 0 ), ] <- rep(0,ncol(a3))
a3
Essentially use the product to pick out the rows with at least one 0 and
replace these rows with 0s.
HTH.
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Robin Hankin
Sent: Thursday, July 27, 2006 9:54 AM
To: RHelp
Subject: [R] inserting rows into a matrix
Hi
I have a little vector function that takes a vector A of strictly positive
integers and outputs a matrix M each of whose columns is the vector,
modified in a complicated combinatorical way.
Now I want to generalize the function so that A can include zeroes.
Given A,
I want to strip out the zeroes, pass it to my function, and pad M
with rows at positions corresponding to the zeroes of A.
Commented, minimal, self-contained, reproducible toy example follows.
f <- function(a){cbind(a,a+1,rev(a))} #real function a ghastly nightmare
A <- 1:5
f(A)
a
[1,] 1 2 5
[2,] 2 3 4
[3,] 3 4 3
[4,] 4 5 2
[5,] 5 6 1
# f() works as desired.
# Now introduce A2, that includes zeroes. In my application, f(A2) would
fail because of the zeroes.
A2 <- c(1,0,0,2,4,0,3)
I can strip the zeroes out and call f():
f(A2[A2>0])
a
[1,] 1 2 3
[2,] 2 3 4
[3,] 4 5 2
[4,] 3 4 1
which is fine. How to put the zeroes back in in the appropriate rows and
get the following:
> cbind(c(1,0,0,2,4,0,3),c(2,0,0,3,5,0,4),c(3,0,0,4,2,0,1))
[,1] [,2] [,3]
[1,]123
[2,]000
[3,]000
[4,]234
[5,]452
[6,]000
[7,]341
>
anyone?
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton European Way, Southampton SO14
3ZH, UK
tel 023-8059-7743
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Re: [R] bilinear regression
It appears that you might have a latent (hidden) explanatory variable that causes the two-population appearance. If you have some ideas on what that other factor might be, you could try two separate linear regressions for each value of the latent factor and compare the slopes and intercepts. You can then do some formal tests on the slopes and intercepts to see if you can further simplify the model. Depending on what you find, you can formulate a linear regression model that incorporates such dependence on the slopes or intercepts to fit the "bilinear" trend. You might find helpful the discussion and example in Ch.10 of Venables & Ripley, 4th ed, that introduces the concepts behind random and mixed effects models. -Christos Christos Hatzis, Ph.D. Vice President, Technology Nuvera Biosciences, Inc. 400 West Cummings Park Suite 5350 Woburn, MA 01801 Tel: 781-938-3830 www.nuverabio.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Crabb, David Sent: Tuesday, July 18, 2006 1:36 PM To: r-help@stat.math.ethz.ch Subject: [R] bilinear regression I think this is an easy question, but I would be grateful for any advice on how to implement this in R. I simply have a response variable (y) that I am trying to predict with one explanatory variable (x) but the shape of the scatter plot is distinctly bilinear. It would be best described by two straight lines. Is there a way of fitting a linear model to give me a bilinear fit and (more importantly) automatically determine the 'cut off' point? I would also want some statistic to convince myself that the bilinear fit is better. Many thanks for your help. David Dr. David Crabb Department of Optometry and Visual Science, City University, Northampton Square, London EC1V OHB Tel: 44 207 040 0191 [EMAIL PROTECTED] http://www.city.ac.uk/optometry/html/david_crabb.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] String manipulation and formatting
Ok. You needed the format string.
xify(4.1) would be equivalent to the following:
paste(paste(rep("X",4),collapse=""),paste(rep("X",1),collapse=""),sep=".")
It should be straightforward to write the wrapper function for this.
HTH.
-Christos
_
From: Bashir Saghir (Aztek Global) [mailto:[EMAIL PROTECTED]
Sent: Monday, July 17, 2006 10:39 AM
To: '[EMAIL PROTECTED]'; [EMAIL PROTECTED]
Subject: RE: [R] String manipulation and formatting
Thanks.
I should have been clearer. The output is the string "XXX.XX" and "XXX".
So xify(4.1) should produce "XXX.X" as character string.
Any pointers? Or did I miss something with formatC?
-Original Message-
From: Christos Hatzis [mailto:[EMAIL PROTECTED]
Sent: Monday, July 17, 2006 16:21
To: Bashir Saghir (Aztek Global); [EMAIL PROTECTED]
Subject: RE: [R] String manipulation and formatting
See ?formatC
You might need to write a simple wrapper function to implement the interface
that you want.
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Bashir Saghir (Aztek
Global)
Sent: Monday, July 17, 2006 10:07 AM
To: '[EMAIL PROTECTED]'
Subject: [R] String manipulation and formatting
I'm trying to write a simple function that does the following:
[command] xify(5.2)
[output] XXX.XX
[command] xify(3)
[output] XXX
Any simple solutions (without using python/perl/unix script/...)?
Thanks,
Saghir
-
Legal Notice: This electronic mail and its attachments are i...{{dropped}}
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] String manipulation and formatting
See ?formatC
You might need to write a simple wrapper function to implement the interface
that you want.
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Bashir Saghir (Aztek
Global)
Sent: Monday, July 17, 2006 10:07 AM
To: '[EMAIL PROTECTED]'
Subject: [R] String manipulation and formatting
I'm trying to write a simple function that does the following:
[command] xify(5.2)
[output] XXX.XX
[command] xify(3)
[output] XXX
Any simple solutions (without using python/perl/unix script/...)?
Thanks,
Saghir
-
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Re: [R] Negative value on ternaryplot
Ternary plots are typically used to plot triplets of values (e.g. composition of 3 components) that add up to 100% or 1. In this context, I am not sure what a negative value means. Do the 3 numbers in your application still add up to 100%? Alternatively, would it be meaningful to translate the data along the dimension with negative values to shift them to the positive quadrant? -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Greg Snow Sent: Thursday, May 18, 2006 2:37 PM To: Poizot Emmanuel; r-help@stat.math.ethz.ch Subject: Re: [R] Negative value on ternaryplot There are also the functions: triangle.plot in package ade4, triplot in package TeachingDemos, tri in package cwhtool, and soil.texture in package plotrix. Perhaps one of these other functions will work better for you (all do the triangular plots, each with different bells and whistles). triplot from TeachingDemos has an option to add to an existing plot (the idea being that you already created one triplot and are now adding more points/lines to it). You could set up your own plot and axes then use triplot with add=TRUE to plot your points. If none of those work for you then you could look at the source code of any of them to see what modifications would help. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Poizot Emmanuel Sent: Thursday, May 18, 2006 7:05 AM To: r-help@stat.math.ethz.ch Subject: [R] Negative value on ternaryplot Dear all, I found a wonderful package (vcd) able to plot ternary diagrams, i.e. ternaryplot (thanks D. Meyer). The problem is that one of three variable has negative values. If I use the ternaryplot function but some points are outside the triangle, as value en negative. Is it possible to make the ternary diagram fit exactly the cloud points ? Regards Emmanuel Poizot Cnam/Intechmer B.P. 324 50103 Cherbourg Cedex Phone (Direct) : (00 33)(0)233887342 Fax : (00 33)(0)233887339 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] create a vector
paste("A", 1:300, sep="")
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of YIHSU CHEN
Sent: Thursday, May 18, 2006 12:56 AM
To: R-help@stat.math.ethz.ch
Subject: [R] create a vector
Dear R users:
I have an elementary question: how to creat a vector of [A1, A2, A3..
A300]? I know c(1:300) would give 1, 2, 3, , 300 but not sure how to
attch a A to each element.
Thank you
Yihsu Chen
The Johns Hopkins University
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Re: [R] everytime I download a new version of R, need I reinstall all packages?
The process that you outlined is described in the Windows R FAQ 2.8 "What's the best way to upgrade?". The same process should apply to other platforms as well, but I don't think it is in the general R FAQ. Perhaps it should. -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Berton Gunter Sent: Tuesday, May 16, 2006 8:16 PM To: 'Michael' Cc: R-help@stat.math.ethz.ch Subject: Re: [R] everytime I download a new version of R,need I reinstall all packages? > > I think you did not answer my question... I now upgraded my main R > program from 2.2.1 to 2.3.0 and I removed the 2.2.1 installation, but > all the Wait until after you use update.packages() to remove your previous installation. You can keep multiple versions of R simultaneously, so this is no problem. That is: 1) Install new R version 2) Run update.packages() on old library version 3) Copy updated old library to new library location (or point new library location to old) 4) Remove old R version (and its libraries if you copied them) There are probably better ways to do this, which this message may stimulate. -- Bert __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] remove Punctuation characters
Try
gsub('[[:punct:]]', '', str)
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Filipe Almeida
Sent: Tuesday, May 09, 2006 11:51 AM
To: r-help@stat.math.ethz.ch
Subject: [R] remove Punctuation characters
Hi,
I want to remove all punctuation characters in a string. I was trying it use
a regular expressions but it doesn't work.
Here is a sample os what i want:
str <- 'ABD - remove de punct, and dot characters.'
str <- gsub('[:punct:]','',str)
str
"'ABD remove de punct and dot characters"
is there any function that do this kind of thing?
Thanks to all.
Filipe Almeida
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Re: [R] predict.lm
I think you got it right. The mean of the (weighted) sum of a set of random variables is the (weighted) sum of the means and its variance is the (weighted) sum of the individual variances (using squared weights). Here you don't have to worry about weights. So what you proposed does exactly this. -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Bill Szkotnicki Sent: Tuesday, May 02, 2006 2:59 PM To: 'R-Help help' Subject: [R] predict.lm I have a model with a few correlated explanatory variables. i.e. > m1=lm(y~x1+x2+x3+x4,protdata) and I have used predict as follows: > x=data.frame(x=1:36) > yp=predict(m1,x,se.fit=T) > tprot=sum(yp$fit) # add up the predictions tprot tprot is the sum of the 36 predicted values and I would like the se of that prediction. I think > sqrt(sum(yp$se.fit^2)) is not correct. Would anyone know the correct approach? i.e. How to get the se of a function of predicted values (in this case sum) Thanks, Bill __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] random walk on graph
I did not look at your code in detail, but I see a potential problem in your
inner loop. I think you intended k to be in 2,3,...N+1. What you get from
your code is (2:N)+1 according to the operator precedence rules. You need
2:(N+1).
HTH,
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Eric Blabac
Sent: Thursday, April 27, 2006 11:28 AM
To: r-help@stat.math.ethz.ch
Subject: [R] random walk on graph
Hi all,
I'm having issues coding a random walk on a fully connected, undirected
graph G with probability transition matrix P = (p_ij). Here is what I have
so far ...
for(i in 1:n){
for(m in 1:M){
x <- as.vector(matrix(rep(0,N+1),nc=N+1))
x[1] <- i
for(k in 2:N+1){
y <- as.vector(matrix(rep(0,n),nc=n))
r <- runif(1)
c <- c(0,cumsum(P[x[k-1],]))
for(j in 1:n){
y[j] <- (r >= c[j] & r < c[j+1])
}
x[k] <- which(y == 1)
}
L[i,m] <- x[N+1]
}
}
its giving me the error:
Error in "[<-"(`*tmp*`, k, value = integer(0)) :
nothing to replace with
no matter what I do I cant fix it ... any suggestions ?
Thanks
Eric Blabac
PhD Student - ASU Dept of Mathematics/Statistics
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Re: [R] copying previously installed libraries to R 2.3.0
See Windows FAQ 2.8 - works well.
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Thomas Harte
Sent: Wednesday, April 26, 2006 2:54 PM
To: r-help@stat.math.ethz.ch
Subject: [R] copying previously installed libraries to R 2.3.0
hi all,
is there a new mechanism in R 2.3.0 for copying libraries from, say, R 2.2.1
to R 2.3.0? i ask because gabor grothendieck comments in his copydir.bat
(from gabor's batchfiles at:
http://cran.r-project.org/contrib/extra/batchfiles/batchfiles_0.2-5.zip ):
``:: I personally upgraded my 2.1.0 to 2.2.0 this way so it seems ok until
:: R replaces this with something better which is expected for 2.3.0.
'''
see also the posting below.
cheers,
thomas.
[R] copy contributed packages from R 2.2.0 to 2.2.1
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From: Ronnie Babigumira
Date: Fri, 23 Dec 2005 15:58:36 +0100
Hi Helli, this came up last week, Here are some of the replys posted
1.
In http://cran.r-project.org/contrib/extra/batchfiles/batchfiles_0.2-5.zip
are two Windows XP batch files:
movedir.bat
copydir.bat
which will move the packages (which is much faster and suitable if you don't
need the old version of R any more) or copy the packages (which takes
longer but preserves the old version).
2.
x <- installed.packages()[,1]
install.packages(x)
3.
This is one reason we normally recommend that you install into a separate
library. Then update.packages(checkBuilt =
TRUE) is all that is needed. However,
foo <- installed.packages()
as.vector(foo[is.na(foo[, "Priority"]), 1])
will give you a character vector which you can feed to install.packages(),
so it's not complex to do manually.
4.
If the previous installation is still alive, fire it up and
pS <- packageStatus()
pkgs <- pS$inst$Package[!pS$inst$Priority %in% c("base", "recommended")]
save(pkgs, file = "foo")
In the new installation,
load("foo")
install.packages(pkgs)
Helmut Kudrnovsky wrote:
> hi R-users,
>
> a few days ago R 2.2.1 came out. on my win xp i'installed R 2.2.0. along
the time i've installed a lot of contributed packages. my
internet-connection is not very fast.
>
> so my question: is it possible after installing R 2.2.1 to do copy/paste
the contributed packages from the C:\Programme\R221 to the
C:\Programme\R2.2.1- location in the files system?
>
> or have i to download and install the packages new?
>
>
> greetings from the snowy austria
> merry christmas
> helli
>
> system
> R.2.2.0
> win xp
>
> __
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
> http://www.R-project.org/posting-guide.html
>
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[R] variable labels in pairs
Hello,
I am using 'pairs' to produce a scatter plot matrix with a custom
upper.panel function to plot the Pearson's correlation coefficients for the
pairs of variables.
I would like to be able to use the actual variable names as subscripts in
rho in the printed text. I know these labels are accessible to diag.panel
but cannot find a good way to access them within panel.cor.
Any suggestions?
Thank you.
-Christos
### --- code snip ---
panel.cor <-
function(x, y, digits=3, subscripts, groups, cex.cor=2) {
usr <- par("usr"); on.exit(par(usr))
par(usr = c(0, 1, 0, 1))
r <- abs(cor(x, y))
txt <- format(c(r, 0.123456789), digits=digits)[1]
txt <- substitute(italic(rho) == txt)
text(0.5, 0.5, txt, cex = cex.cor, col="blue")
}
pairs(my.df, upper.panel=panel.cor)
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Re: [R] Store results of for loop
It is not very clear how you want to index your results vector.
If ss contains the indices of the results vector that you are trying to
change, this implies that you have a vector of length 9. In this case
results <- numeric(max(ss))
results[ss] <- ss + 1
will do the trick.
Or in case that ss contains the values that you want to augment by 1
results <- numeric(length(ss))
Results <- ss + 1
Am I missing something?
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Doran, Harold
Sent: Monday, April 24, 2006 4:32 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Store results of for loop
I have what I'm sure will turn out to be straightforward. I want to store
the results of a loop for some operations from a patterned vector.
For example, the following doesn't give what I would hope for
ss <- c(2,3,9)
results <- numeric(length(ss))
for (i in seq(along=ss)){
results[i] <- i + 1
}
The following does give what I expect, but creates a vector of length 9.
ss <- c(2,3,9)
results <- numeric(length(ss))
for (i in ss){
results[i] <- i + 1
}
What I am hoping for is that results should be a vector of length 3.
Harold
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Re: [R] how to do Splus compare() function in R
As suggested below,
sign(ii*1e9-f)
will give you the same result as the S-plus
compare(ii*1e-9,f)
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Bellinger Instruments
P/L
Sent: Thursday, April 20, 2006 11:02 PM
To: 'Prof Brian Ripley'
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] how to do Splus compare() function in R
The sign() function requires one argument, and returns the sign of the
value, where I require a number -1, 0, or 1 depending on if the value is <,
==, or > the first argument.
This is the Splus function when listed,
> compare
function(e1, e2)
.Internal(compare(e1, e2), "do_op", T, 18)
unfortunately I do not understand the .internal function and cannot find any
description of this type of programming on the CRAN site
This is how I am using it in the source file.
zz <- readline()# keyboard input eg: f15
if (charmatch(substring(zz,1,1),"f",
nomatch=-1) >0){
ii <- as.numeric(substring(zz,2,99))
iii <- compare(ii*1e9,f) # f is a vector of length(146) frequencies
# compare() searches thru f and
# creates vector iii with
# -1 if numeric is < f[i],
# 0 if numeric is == f[i], and
# 1 if numeric is > f[i]
# the end result is a vector(iii) length(146)
# with -1,0,1 if the numeric is in f
i <- match(-1,c(iii,-1))-1 # if vector(iii) contains 0 assign the
index
}
# error traps for if 0 is not in vector(iii)
if (i<1){cat("Min n = 1\n")
i <- 1}
if (i>length(f)){cat("Max n =",length(f),"\n")
i <- length(f)}
-Original Message-
From: Prof Brian Ripley [mailto:[EMAIL PROTECTED]
Sent: Thursday, 20 April 2006 4:02 PM
Subject: Re: [R] how to do Splus compare() function in R
You would need to tell us what it does (and what the inputs are).
I think it is likely that compare(x, y) in S-PLUS is the same as sign(x-y)
in R, at least with numeric vector inputs.
--
Bob Kelly [EMAIL PROTECTED]
Metrologist & Microwave Eng.http://www.bellinger.com.au
Bellinger Instruments Pty Ltd Tel: 612 9684 1442
4 Muriel AveFax: 612 9638 4435
Rydalmere NSW
Australia 2116
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Re: [R] The contrary of command %in%
A flexible way for doing this is to define logical vectors for the types of samples that you want to include or exclude. You can then use logical negation to select the complementary set: inSet1 <- HData$H < 1.3 inSet2 <- HData$H < 8 & Hdata$DBH > 20 HDataPart1 <- Hdata[!inSet1, ] HDataPart2 <- Hdata[!inSet2, ] -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ghislain Vieilledent Sent: Thursday, April 20, 2006 8:55 AM To: r-help@stat.math.ethz.ch Subject: [R] The contrary of command %in% Dear r-list, I've got a data base: > HData[1:10,] NumTree Site Species Date Age DBHH IdentTree 11 Queige Spruce 2002 184 49 33.5 Queige 1 22 Queige Fir 2002 NA 5 4.6 Queige 2 33 Queige Fir 2002 25 8 6.6 Queige 3 44 Queige Spruce 2002 198 47 32.5 Queige 4 55 Queige Fir 2002 200 59 35.3 Queige 5 66 Queige Spruce 2002 80 16 9.4 Queige 6 77 Queige Fir 2002 NA 5 4.2 Queige 7 88 Queige Fir 2002 200 44 32.5 Queige 8 99 Queige Fir 2002 NA 5 3.4 Queige 9 10 10 Queige Spruce 2002 167 48 32.8 Queige 10 ... I want to remove particular points determined by > HDataPartP<-HData[H<1.30,] and > HDataPartP2<-HData[H<8&DBH>20,] That's why I want to use subset in a close form to: HData2<-subset(HData,HData$H>1.30&HData$IdentTree "not"%in%HDataPartP2$IdentTree) How should I do that ? Is there any R-syntax saying "not element of that object" ? Thanks for your help. Ghislain. -- Ghislain Vieilledent 06 24 62 65 07 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Basic vector operations was: Function to approximatecomplex integral
You get the same result with
sum(outer(a,b))
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of jim holtman
Sent: Wednesday, April 19, 2006 3:32 PM
To: Doran, Harold
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Basic vector operations was: Function to approximatecomplex
integral
You can always write your own function that will take the values, multiply
and then sum them:
> a <- c(2,3)
> b <- c(4,5,6)
> total <- 0
> for (i in a) total <- sum(total, i * b) total
[1] 75
>
On 4/19/06, Doran, Harold <[EMAIL PROTECTED]> wrote:
>
> Dear List
>
> I apologize for the multiple postings. After being in the weeds on this
> problem for a while I think my original post may have been a little
> cryptic. I think I can be clearer. Essentially, I need the following
>
> a <- c(2,3)
> b <- c(4,5,6)
>
> (2*4) + (2*5) + (2*6) + (3*4) + (3*5) +(3*6)
>
> But I do not know of a built in function that would do this. Any
> suggestions?
>
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Doran, Harold
> Sent: Wednesday, April 19, 2006 11:51 AM
> To: r-help@stat.math.ethz.ch
> Subject: [R] Function to approximate complex integral
>
>
>
> I am writing a small function to approximate an integral that cannot be
> evaluated in closed form. I am partially successful at this point and am
> experiencing one small, albeit important problem. Here is part of my
> function below.
>
> This is a psychometric problem for dichotomously scored test items where
> x is a vector of 1s or 0s denoting whether the respondent answered the
> item correctly (1) or otherwise (0), b is a vector of item difficulties,
> and theta is an ability estimate for the individual.
>
> rasch <- function(b,theta){
> 1 / ( 1 + exp(b - theta))
> }
>
> The function rasch gives the probability of a correct response to item i
> conditional on theta, the individuals ability estimate
>
> myfun <- function(x, b, theta){
> sum(rasch(b, theta)^x * ( 1 - rasch(b,theta) )^(1-x) * dnorm(theta))
> }
>
> This is the likelihood function assuming the data are Bernoulli
> distributed multiplied by a population distribution.
>
> Now, when x,b, and theta are of equal length the function works fine as
> below x <- c(1,1,0)
> b <- c(-2,-1,0)
> t <- c(-2,-1.5,-1)
> > myfun(x,b,t)
> [1] 0.2527884
>
> However, I want theta to be a vector of discrete values that will be
> larger than both x and b, something like
>
> t <- seq(-5, 0, by = .01)
>
> However, this gives me the error
> > myfun(x,b,t)
>
> Warning messages:
> 1: longer object length
>is not a multiple of shorter object length in: b - theta
>
> So, for the problem above, I want item 1 to be evaluated at theta 1
> through theta q and then item 2 is evaluated at theta 1 and through
> theta q and so forth for each item.
>
> Can anyone recommend a way for me to modify my function above to
> accomplish this?
>
> Harold
>
>
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>
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390 (Cell)
+1 513 247 0281 (Home)
What the problem you are trying to solve?
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Re: [R] how funciton "expression" produces subscript
See ?plotmath for the answer to your question and many more regarding math expression drawing. It might be worth printing this help page for future reference. -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Jing Yang Sent: Tuesday, April 18, 2006 10:24 AM To: r-help Subject: [R] how funciton "expression" produces subscript Dear R-users, does anyone know how funciton "expression" produces subscript? Best,Jing __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
