[R] Multiv. NR
Do you know if R has any multiv. Newton-Raphson routine implemented? __ [EMAIL PROTECTED] 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] Frailty Model - parameter inferences
Hi, I've been trying to find how to extract the inference info about \theta ... Is there any easy way to do it? Thanks, Benilton _ Create your own personal Web page with the info you use most, at My MSN. http://click.atdmt.com/AVE/go/onm00200364ave/direct/01/ __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Failing to install on Linux FC4
Hi everyone, I downloaded the source code available at: http://cran.fhcrc.org/src/base/R-2/R-2.4.0.tar.gz to a linux machine (Linux 2.6.11-1.1369_FC4smp #1 SMP). I successfully configured and compiled it, which means I'm able to run it from the R-2.4.0/bin directory. I want to do a system wide installation (via 'sudo make install') and help2man fails, ie, when I execute the installation command mentioned previously I get: make[1]: Entering directory `/home/bcarvalh/R-2.4.0/m4' make[1]: Nothing to be done for `install'. make[1]: Leaving directory `/home/bcarvalh/R-2.4.0/m4' make[1]: Entering directory `/home/bcarvalh/R-2.4.0/tools' make[1]: Nothing to be done for `install'. make[1]: Leaving directory `/home/bcarvalh/R-2.4.0/tools' make[1]: Entering directory `/home/bcarvalh/R-2.4.0/doc' installing doc ... help2man: can't get `--version' info from ../bin/R make[1]: *** [R.1] Error 255 make[1]: Leaving directory `/home/bcarvalh/R-2.4.0/doc' make: *** [install] Error 1 Searching the archive, I did find a similar report (http:// tolstoy.newcastle.edu.au/R/devel/06/01/3699.html) but the proposed solution did not work for me. Any suggestion? Thank you very much, Benilton Carvalho PhD Candidate Department of Biostatistics 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.
Re: [R] Selective subsetting
A=matrix(1:9,3) A[lower.tri(A)] b On Nov 10, 2006, at 4:50 PM, Davendra Sohal wrote: Hi all, Here's an interesting (for me, at least!) problem I came across: I have a correlation matrix, let's say with 6 variables, A to F, as column headings and the same 6 as row headings. The matrix is filled with correlation coefficients. Therefore, the diagonal is all 1's, and each of the two triangles formed by the diagonal has the same 15 correlation coefficients. I need to extract these 15 coefficients from this. I don't want the 1's and I don't want redundant values. I tried converting the matrix to a list, using c(m) and then I'm stuck. Does something like b = a[x, x+1:n] exist in R? SAS will do it. x can be looped to go from 1 to n (n here is 6, of course) and it will select the required values, but I cannot make it work. Please help. Many thanks, -DS. [[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] R Help
data = read.delim(lahore.txt) is enough for what you want to do. b On Nov 11, 2006, at 2:11 PM, amna khan wrote: Respected Sir I request you to please fill the following read.table function and read.csvfor my understanding by assuming my data attached with this maiL, because I am fail to run these functions using manual guidlines. read.table(file, header = FALSE, sep = , quote = \', dec = ., row.names, col.names, as.is = !stringsAsFactors, na.strings = NA, colClasses = NA, nrows = -1, skip = 0, check.names = TRUE, fill = !blank.lines.skip, strip.white = FALSE, blank.lines.skip = TRUE, comment.char = #, allowEscapes = FALSE, flush = FALSE, stringsAsFactors = default.stringsAsFactors()) read.csv(file, header = TRUE, sep = ,, quote=\, dec=., fill = TRUE, comment.char=, ...) read.delim(file, header = TRUE, sep = \t, quote=\, dec=., fill = TRUE, comment.char=, ...) I shall be really thankful to you. REGARDS -- AMINA SHAHZADI Department of Statistics GC University Lahore, Pakistan. Email: [EMAIL PROTECTED] [EMAIL PROTECTED] [EMAIL PROTECTED] lahore.txt __ 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] segfault 'memory not mapped', dual core problem?
Mea culpa. Please accept my apologies... The problem was fixed on R-2.4.0 (final) and what I reported was on R-2.4.0 alpha. Good job of Martin, by the way... The problem was fixed and I didn't notice (given that I was always using my work-around). benilton __ 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] I think a simple question
does
getIdx - function(tmpin, x){
tmp2 = tmpin + x
out = sort(c(tmpin, tmp2))
out = out[out=max(tmpin)]
return(out)
}
do what you want?
b
On Nov 12, 2006, at 5:20 PM, Leeds, Mark ((IED)) wrote:
I have index ( of a vector ) values of say
tempin-c(1 31 61 91 121 all the way upto 1411)
What I want is a function that takes in a number say, x = 5, and gives
me an new vector
of
tempout-1 6 31 36 91 96 121 126 .. 1411 1416
This can't be so hard but I can't get it and I've honestly tried.
Obviously, tempin + 5 gives me the missing values but I don't know how
to interwine them in the order above. Thanks for any help
you can provide.
mark
This is not an offer (or solicitation of an offer) to buy/se...
{{dropped}}
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Re: [R] indexing question
diff(tmp[idx])
cheers,
b
On Nov 13, 2006, at 3:06 PM, Leeds, Mark ((IED)) wrote:
I have the following set of indices, call it idx, that correspond
to the
indices of a vector say temp.
[1] 31 36 41 61 66 71 91 96 101 121 126 131 151
156 161 181 186 191 211 216 221 241 246 251 271 276 281
301 306 311 331 336 341 361 366
[36] 371 391 396 401 421 426 431 451 456 461 481 486 491
511 516 521 541 546 551 571 576 581 601 606 611 631 636
641 661 666 671 691 696 701 721
[71] 726 731 751 756 761 781 786 791 811 816 821 841 846
851 871 876 881 901 906 911 931 936 941 961 966 971 991
996 1001 1021 1026 1031 1051 1056 1061
[106] 1081 1086 1091 1116 1121 1141 1146 1151 1171 1176 1181 1201
1206 1211 1231 1236 1241 1261 1266 1271 1291 1296 1301 1321 1326 1331
1351 1356 1361 1381 1386 1391 1411 1416
[141] 1421
I want to calculate temp[36] - temp[31] and temp[41] - temp[36]
Similarly, temp[66] - temp[61] and temp[71] - temp[66]
.
.
.
.
Similarly temp[1416]-temp[1411]
temp[1421] - temp[1416]
I'm doing this because the above subractions represent pairs of
returns
and the correlations between them wil be calculated eventually.
In other words, eventually I will have
X_36_31 ( i.e : temp[36] - temp[31] )
X_66-61
X_96-91
.
.
.
.
.
.
.
X_1411-1416
as one vector and
Y_41-36
Y_71-66
Y_101-96
.
.
.
.
.
Y_1416_1421
as another vector.
and will calculate the correlation between the two vectors in order to
get one number.
The point is I am really only using the indices 31, 61, 91 etc as
anchor's so a regular diff(temp[idx]) won't work because it will diff
all
the elements that are next to each other ? This is a weird problem.
I'm
still thinking about it. I'm hoping to figure it out before someone
sends me something but I won't mind so much if I get an external
solution first. I have no pride.
This is not an offer (or solicitation of an offer) to buy/se...
{{dropped}}
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[R] Problem with file size
Hi everyone, I have 2 environments (2 different R sessions) as described below: Session 1: Name of the environment: CrlmmInfo Objects in the environment: index1: logical index - length 238304 index2: logical index - length 238304 priors: list of 4 - (matrix 6x6, 2 vectors of length 6, vector of length 2) - all num params: list of 4: centers [238304 x 3 x 2]: num scales [238304 x 3 x 2]: num N [238304 x 3]: num f0 [scalar]: num If I save this environment to a file, I get a file of 23MB. Great. Session 2: Analogous to Session 1, but replace 238304 by 262264. If I save the environment on Session 2, I get a file of 8.4GB. I applied object.size on each of the objects in each environment, and this is what I got: For Session 1: index1: 16204864 index2: 16204864 priors: 3336 params: 74353584 For Session 2: index1: 1049096 index2: 1049096 priors: 3336 params: 81829104 Is this increase from 23MB to 8.4GB expected to happen? Benilton sessionInfo() on both sessions is identical: sessionInfo() R version 2.4.0 (2006-10-03) x86_64-unknown-linux-gnu locale: LC_CTYPE=en_US.iso885915;LC_NUMERIC=C;LC_TIME=en_US.iso885915;LC_COLLATE =en_US.iso885915;LC_MONETARY=en_US.iso885915;LC_MESSAGES=en_US.iso885915 ;LC_PAPER=en_US.iso885915;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASU REMENT=en_US.iso885915;LC_IDENTIFICATION=C attached base packages: [1] methods stats graphics grDevices utils datasets [7] base version _ platform x86_64-unknown-linux-gnu arch x86_64 os linux-gnu system x86_64, linux-gnu status major 2 minor 4.0 year 2006 month 10 day03 svn rev39566 language R version.string R version 2.4.0 (2006-10-03) __ 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 check a R object's property
check class() but if all you want is to test whether it's a data.frame or not: is.data.frame() b On Nov 14, 2006, at 3:07 PM, Weiwei Shi wrote: Hi, I am writing a generic function and need to check if an arg is a data frame or not. I could use is.null(dim(x)) to get what i want. But i want to know if there is a function which can tell me whether it is a list, a numeric vector, a data frame, a factor and so on. Can R do that? thanks. -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. Did you always know? No, I did not. But I believed... ---Matrix III __ 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] Creating a table
tb = table(df$loc, cut(df$year, seq(1970, 1985, by=5), right=F)) rs = rowSums(tb) tb = cbind(tb, rs) cs = colSums(tb) tb = rbind(tb, cs) cheers, b On Nov 14, 2006, at 3:20 PM, Michael Graber wrote: Dear R List, I am a new to R, so my question may be easy to answer for you: I have a dataframe, for example: df-data.frame(loc=c(A,B,A,A,A), year=as.numeric(c(1970,1970,1970,1976,1980))) and I want to create the following table without using loops: 1970-74 ; 1975-79 ; 1980-85; rowsum A 2 1 1 4 B 1 00 1 colsum 31 15 so that the frequencies of df$loc are shown in the table for different time intervals. Thanks in advance for any hint, Michael Graber __ 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] trouble loading example package
How do you build your packages? have you tried R CMD build --binary mypkg b On Nov 16, 2006, at 12:32 AM, Charles Annis, P.E. wrote: Greetings: I've installed Rtools, MikTeX, perl, minGW, and HTML Help Workshop, and have succeeded in making, checking (using R CMD check mypkg) then building the simple example package.skeleton(list=c(f,g,d,e), name=mypkg) R CMD build mypkg produces a tarball. I don't know how to get a zip file. But when I try to Install package(s) from local zip files, I get this error message: Error in gzfile(file, r) : unable to open connection In addition: Warning messages: 1: error -1 in extracting from zip file 2: cannot open compressed file 'mypkg/DESCRIPTION' But when I click on that file in the tarball it opens and shows me what I expected. I had hoped that I had weathered the hard part - building the package - but I still need some help: 1) How do I get a zipped file, rather than a tarball, 2) How do I install what I've built? Thanks in advance. Charles Annis, P.E. [EMAIL PROTECTED] phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.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] modifying colnames of tables in a loop
how about:
for (x in l) colnames(get(x)) - lower.case(colnames(get(x)))
b
On Nov 16, 2006, at 9:01 AM, Werner Wernersen wrote:
Hi,
I have a list with the names of tables, e.g.
l - c(t1,t2,t3)
and I want to change the colnames of each of the
tables in a for loop like this:
for (x in l) {
colnames(eval(x)) - lower.case(colnames(eval(x)))
}
This does not work but could someone give me some help
to get on the right track?
Thanks a million,
Werner
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Re: [R] Counting zeros in a matrix
countZerosBeforeOnes - function(v) sum(v[1:max(which(v == 1))] == 0) apply(A, 1, countZerosBeforeOnes) cheers, b On Nov 28, 2006, at 4:20 PM, Guenther, Cameron wrote: Hi All, If you could help me with this problem I would greatly appreciate it. Suppose I have a matrix A: 1 1 1 1 0 1 1 1 1 1 1 1 0 1 0 1 0 0 1 0 1 0 0 1 0 0 0 1 1 0 0 0 0 1 0 0 I would like, for each row, to sum the number of times a 0 appears in front of a 1. So what I would like is to have Sum 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 0 1 0 0 2 1 0 1 0 0 1 0 0 0 2 1 1 0 0 0 0 1 0 0 1 I tried writing a function to do this but am getting mixed up in the [i,j] coding. This is just an example the real matrix is much larger. Thanks in advance. Cameron Guenther, Ph.D. 100 8th Ave. SE St. Petersburg, Fl 33701 727-896-8626 ext. 4305 [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 https://stat.ethz.ch/mailman/listinfo/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] tests for NULL objects
Hi Everyone, After searching the subject and not being successful, I was wondering if any you could explain me the idea behind the following fact: all(NULL == 2) ## TRUE any(NULL == 2) ## FALSE Thanks a lot, Benilton -- Benilton Carvalho PhD Candidate Department of Biostatistics 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.
Re: [R] tests for NULL objects
Thank you Bert, Sorry if it sounded like a complain, this is not what I meant. The situation I had was: given a vector v containing integers or NA's, I wanted to know if after excluding the NA's all the remaining observations were 2's. Naively, I assumed that: all(na.omit(v) == 2) would do the work, but now I see that it fails if, for example, v = rep(NA, 10) But that's okay, just a matter of adding an extra test (already done). When I asked for clarification about the reasons for this, I assumed that: if all(v) is TRUE == any(v) is TRUE; for all (logical) v... When actually: if all(v) is TRUE == any(v) is TRUE; for all (logical) v of length = 1. Thank you, benilton On Nov 29, 2006, at 6:35 PM, Bert Gunter wrote: Merely convention. NULL == 2 == logical(0), that is, a logical vector of length 0. It makes sense (at least to me) that any(logical(0)) is FALSE, since no elements of the vector are TRUE. all(logical(0)) is TRUE since no elements of the vector are FALSE. I think these are reasonable and fairly standard conventions, but even if you disagree, they are certainly not worth making a fuss over and certainly cannot be changed without breaking a lot of code, I'm sure. Bert Gunter Nonclinical Statistics 7-7374 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Benilton Carvalho Sent: Wednesday, November 29, 2006 2:21 PM To: R-Mailingliste Subject: [R] tests for NULL objects Hi Everyone, After searching the subject and not being successful, I was wondering if any you could explain me the idea behind the following fact: all(NULL == 2) ## TRUE any(NULL == 2) ## FALSE Thanks a lot, Benilton -- Benilton Carvalho PhD Candidate Department of Biostatistics 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] tests for NULL objects
On Nov 29, 2006, at 7:54 PM, Martin Morgan wrote: http://en.wikipedia.org/wiki/Empty_set which says, in part: Operations on the empty set ... 'any' is like 'sum', 'all' is like 'prod'. Martin Thank you very much Martin. It's clear now. Benilton __ 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] require(simecol) error
The output of sessionInfo() will be helpful here. But I can't reproduce that on R 2.4.0 Patched r40106 + simecol (0.3-11). Maybe you should upgrade the package? b On Dec 6, 2006, at 5:06 PM, Milton Cezar Ribeiro wrote: Hi there, I´m trying to use simecol package but I got the error showed below. I´m runnig R version 2.4.0 (2006-10-03). Kind regards, miltinho Brazil --- require(simecol) Loading required package: simecol Error in loadNamespace(package, c(which.lib.loc, lib.loc), keep.source = keep.source) : in 'simecol' methods specified for export, but none defined: fixInit, fixParms, fixTimes, plot, print, solver, solver-, out, inputs, inputs-, main, main-, equations, equations-, sim, parms, parms-, init, init-, times, times- [1] FALSE - __ 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] expression()
mtext(expression(beta[max]), side=1, line=2)
is it what you want?
b
On Dec 12, 2006, at 10:59 AM, javier garcia-pintado wrote:
Hi,
I'm trying to use expression() to write a text to a graphic in the
margin.
Using:
mtext(expression(beta),side=1,line=2)
writes a perfect beta greek character, but I need to add a subindex
max, and I'm trying:
mtext(paste(expression(beta),max),side=1,line=2)
simply writes beta max in the plot.
Please, Could you tell me what I'm doing wrong?
By the way, is there a way to add Latex expressions to graphics?
Then I
could use the Latex expression: $\beta_{\mathrm{max}}$. This also
would
be very useful for me for more complex expressions in plots.
Best regards,
Javier
--
Javier García-Pintado
Institute of Earth Sciences Jaume Almera (CSIC)
Lluis Sole Sabaris s/n, 08028 Barcelona
Phone: +34 934095410
Fax: +34 934110012
e-mail:[EMAIL PROTECTED]
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Re: [R] install.packages
install.packages(plotrix)
b.
On Dec 13, 2006, at 1:42 PM, Aimin Yan wrote:
I try to type this in my R-winEdt.
but I got these. Do you know?
Aimin
install.packages('http://rh-mirror.linux.iastate.edu/CRAN/bin/
windows/contrib/2.4/plotrix_2.1-6.zip')
Warning in download.packages(pkgs, destdir = tmpd, available =
available, :
no package
'http://rh-mirror.linux.iastate.edu/CRAN/bin/windows/contrib/2.4/
plotrix_2.1-6.zip'
at the repositories
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Re: [R] min() return factor class values
Milton,
have you looked at the structure of your data.frame?
str(especies.aicc)
Are you sure especies.aicc is defined as numeric?
b
On Jan 9, 2007, at 10:51 AM, Milton Cezar Ribeiro wrote:
Dear Peter,
I tryed something like
head(especies.aicc)
especie aicc
1 Attila.rufus 17.15934
2 Attila.rufus 11.41371
3 Attila.rufus 11.41371
4 Attila.rufus 19.55998
5 Attila.rufus 17.23780
6 Attila.rufus 19.22545
especies.min-aggregate.data.frame(especies.aicc,list
(Especie=especies.aicc$especie),max)
But it works fine only for mean FUN and not for min and
max. Also also, when I use mean I got the following warnings:
especies.min-aggregate.data.frame(especies.aicc,list
(Especie=especies.aicc$especie),mean)
Warning messages:
1: argument is not numeric or logical: returning NA in: mean.default
(X[[1]], ...)
2: argument is not numeric or logical: returning NA in: mean.default
(X[[2]], ...)
In fact I need only min() and max().
Miltinho
-
Peter Dalgaard [EMAIL PROTECTED] escreveu:
Milton Cezar Ribeiro wrote:
Hi R-friends
I don´t know why the min() function below return the min value
as factor. When i force the aicc.min using a as.numeric()
function, it return a factor index (1,2,..) and not min value as I
want. By the way, I included a sessionInfo() at the end of this e-
mail.
min() is not doing anything out of the ordinary, but cbind'ing it with
the character vector sp coerces it to character and rbind'ing to a
data
frame turns character vectors into factors...
The whole thing looks like it could be a straightforward
application of
aggregate().
In fact I had the same problem (values as factor) on other part of
my script and I noticed that it occour when I use cbind(). It is
real?
Any idea?
Kind regards,
Miltinho
especies.aicc.min-data.frame()
for (sp in levels(especies.aicc$especie))
+ {
+ sele-subset(especies.aicc,especie==sp)
+ especies.aicc.min-rbind(especies.aicc.min,cbind(sp,aicc.min=min
(sele$aicc)))
+ }
especies.aicc.min
sp aicc.min
1 Attila.rufus 6.7387056413613
2 Automolus.leucophthalmus 125.791300522824
class(especies.aicc.min$aicc.min)
[1] factor
---
sessionInfo()
R version 2.4.0 (2006-10-03)
i386-pc-mingw32
locale:
LC_COLLATE=English_Jamaica.1252;LC_CTYPE=English_Jamaica.
1252;LC_MONETARY=English_Jamaica.
1252;LC_NUMERIC=C;LC_TIME=English_Jamaica.1252
attached base packages:
[1] methods stats graphics grDevices utils datasets
base
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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Re: [R] R graphics with Linux (libpng)
I forgot to mention that bitmap() will do what you want without an X11 connection. b On Jan 11, 2007, at 1:09 PM, Rebecca Tagett wrote: Hello, I'm trying to adapt some R code that works on Windows so that it will work on a Linux machine. The command : png(myFile.png, width=600, height=600) fails claiming that it is impossible to establish a connection with X11. (Error messages are in French, so I'm not pasting them here!) I have libpng installed: rpm -qa libpng* libpng-1.0.12-2 libpng-devel-1.0.12-2 So I don't understand why R thinks I'm trying to connect to X11. I haven't been able to find many examples of R graphics code specifically for Linux, but I have the impression that if libpng is installed, the graphics commands are identical. The libpng manual is not useful, because it does not mention use of libpng commands in an R environment. I'm using R 2.4.1, which I recently installed. Do I have to install a more recent libpng ? If so, do I have to reconfigure and remake R ? I've also read that no graphics devices are available under R CMD BATCH. Does that really mean that I can't create graphics from some R code that I launch in noninteractive mode? Thanks in advance ! Rebecca Tagett [[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] R graphics with Linux (libpng)
Hi Rebecca, png (and also jpeg, for example) require an X11 connection. So, assuming you're working from the command line and that your X11 server is up, you would need to do something like: linux$ export DISPLAY=:0.0 before loading R... and if your linux machine is remote (and you're connecting through SSH), you'd need linux$ ssh -X remote.machine.somewhere b On Jan 11, 2007, at 1:09 PM, Rebecca Tagett wrote: Hello, I'm trying to adapt some R code that works on Windows so that it will work on a Linux machine. The command : png(myFile.png, width=600, height=600) fails claiming that it is impossible to establish a connection with X11. (Error messages are in French, so I'm not pasting them here!) I have libpng installed: rpm -qa libpng* libpng-1.0.12-2 libpng-devel-1.0.12-2 So I don't understand why R thinks I'm trying to connect to X11. I haven't been able to find many examples of R graphics code specifically for Linux, but I have the impression that if libpng is installed, the graphics commands are identical. The libpng manual is not useful, because it does not mention use of libpng commands in an R environment. I'm using R 2.4.1, which I recently installed. Do I have to install a more recent libpng ? If so, do I have to reconfigure and remake R ? I've also read that no graphics devices are available under R CMD BATCH. Does that really mean that I can't create graphics from some R code that I launch in noninteractive mode? Thanks in advance ! Rebecca Tagett [[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 dataframe by field
say your data.frame is called df df[order(df$evidence),] or df[order(df$evidence, decreasing=T),] # if you want the other way around. b On Jan 17, 2007, at 11:21 AM, Milton Cezar Ribeiro wrote: Hi there, How can I sort (order?) a data.frame using a dataframe field (evidence) as classifyer? My data.frame looks like: record model evidence 1 areatotha 6638.32581 2 areatotha_ca000 8111.01860 3 areatotha_ca000_Pais 1721.41828 4 areatotha_ca020 827.33097 5 areatotha_ca020_Pais 2212.40899 6 areatotha_ca040 3569.17169 7 areatotha_ca040_Pais 2940.01636 8 areatotha_ca060 992.62852 9 areatotha_ca060_Pais 4237.95709 10 areatotha_ca080 62.74915 11 areatotha_ca080_Pais 1726.55082 12 areatotha_Pais 52.02524 13 areatotha_ca000 3391.92930 14 areatotha_ca000_Pais 39.52170 15 areatotha_ca020 268.55875 16 areatotha_ca020_Pais 20.43317 17 areatotha_ca040 1698.75892 18 areatotha_ca040_Pais 43.90613 19 areatotha_ca060 350.79857 20 areatotha_ca060_Pais 51.04471 Cheers, Miltinho, Brazil __ [[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 get the index of entry with max value in an array?
which.max() b On Jan 17, 2007, at 11:20 PM, Feng Qiu 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 __ 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] Suggestion on how to improve efficiency when using MASS:::hubers on high-dimensional arrays
Hi Everyone,
Given the scenario I have, I was wondering if anyone would be able to
give me a hind on how to get the results from hubers() in a more
efficient way.
I have an outcome on an array [N x S x D].
I also have a factor (levels 1,2,3) stored on a matrix N x S.
My objective is to get mu and sigma for each of the N rows
(outcome) stratified by the factor (levels 1, 2 and 3) for each of
the D levels, but using MASS:hubers().
Ideally the final result would be an array [N x D x 3 x 2].
The following toy example demonstrates what I want to do, and I'd
like to improve the performance when working on my case, where S=400
and N 20
Thank you very much for any suggestion.
benilton
## begin toy example
set.seed(1)
N - 100
S - 5
D - 2
outcome - array(rnorm(N*S*D), dim=c(N, S, D))
classes - matrix(sample(c(1:3, NA), N*S, rep=T), ncol=S)
results - array(NA, dim=c(N, D, 3, 2))
library(MASS)
myHubers - function(x)
if (length(x)1) as.numeric(hubers(x)) else c(NA, NA)
for (n in 1:N)
for (d in 1:D){
tmp - outcome[n,,d]
grp - classes[n,]
results[n, d,,] - t(sapply(split(tmp, factor(grp, levels=1:3)),
myHubers))
}
## end
--
Benilton Carvalho
PhD Candidate
Department of Biostatistics
Johns Hopkins University
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Re: [R] Bartlett test
The null hypothesis is that the variances do not differ across groups. The Bartlett test is sensitive to non-normality and that might lead you to consider something more robust (eg, Levene's test). b On Jan 19, 2007, at 9:04 AM, Matthieu Mourroux wrote: Bonjour, Je voudrais tester l'homoscédasdicité entre des groupes. J'aurais alors aimé savoir quelle était l'hypothèse nulle du test de bartlett. Merci pour votre aide. Matthieu. __ 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] Installing packages...
well, i don't use RGui, but trying: install.packages(neural, type=source) ## from the R command line i could not reproduce the error (which may be a problem with the copy on the mirror you're using)... b On Jan 23, 2007, at 3:09 PM, Nüzhet Dalfes wrote: Hi, I am a total newbie to R. I am using R (2.4.1) on Mac OS X 10.4.8 and trying to install some packages using GUI Packages Data/Package Installer interface... Every time I get: trying URL 'http://umfragen.sowi.uni-mainz.de/CRAN/bin/macosx/universal/ contrib/2.4/neural_1.4.1.tgz' Content type 'application/x-tar' length 18920 bytes opened URL == downloaded 18Kb Error in gzfile(file, r) : unable to open connection In addition: Warning message: cannot open compressed file 'neural/DESCRIPTION' What am I doing wrong? Any help will be much appreciated. Nüzhet Dalfes Istanbul Tech. Univ. __ 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] package servers
go to R / Preferences / Startup than you'll know what to do when you see: Default CRAN mirror. b On Jan 23, 2007, at 8:03 PM, Paul Gowder wrote: Hi everyone... Here's a good one. I'm using the R installation for mac osx. An overly helpful friend tried to download a couple of packages for me (via the menu bar), and in the process, set my default package server. To a server that is not responding. I can't seem to get R to give me the option to set a new server. Does anyone know how to tell R to stop looking for the default server? thanks, -Paul __ 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] filling the area
take a look at: ?polygon b On Jan 25, 2007, at 8:18 AM, Mauricio Cardeal wrote: Please, how to fill the area under the curve? x - c(1:10) y - c(rnorm(10)) plot(x,y) lines(x,y) Thanks, Mauricio Cardeal __ 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] strange behaviour with equality after simple subtraction
In addition to Mike's comment: x-c(0.1,0.9) 1-x[2] [1] 0.1 x[1]==1-x[2] [1] FALSE all.equal(x[1], 1-x[2]) [1] TRUE b On Jan 26, 2007, at 11:40 AM, Mike Prager wrote: Not at all strange, an expected property of floating-point arithmetic and one of the most frequently asked questions here. print(0.1, digits=17) [1] 0.1 print(1 - 0.9, digits=17) [1] 0.09998 A simple description of the issue is at http://docs.python.org/tut/node16.html In most cases, it suffices to test for approximate difference or relative difference. The former would look like this if (abs(x[1] - x[2]) eps)) ... with eps set to something you think is an insignificant difference, say 1.0e-10. -- 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] Inverse fuction of ecdf
?quantile b On Jan 28, 2007, at 5:41 PM, Geoffrey Zhu wrote: Hi Everyone, I want to generate some random numbers according to some empirical distribution. Therefore I am looking for the inverse of an empirical cumulative distribution function. I haven't found any in R. Can anyone give a pointer? Thanks, Geoffrey __ 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] lmer2 error under Mac OS X on PowerPC G5 but not on Dual-Core Intel Xeon
So, I decided to give it a try (and just now noticed that this is the example in lmer2) I just gave it a try on a PPC G4 and it worked as expected. I'm copying R-sig-mac (sorry for the crosspost) as the experts there might give you a better suggestion. fm1 - lmer2(Reaction ~ Days + (Days|Subject), sleepstudy) fm1 Linear mixed-effects model fit by REML Formula: Reaction ~ Days + (Days | Subject) Data: sleepstudy AIC BIC logLik MLdeviance REMLdeviance 1754 1770 -871.8 1752 1744 Random effects: Groups NameVariance Std.Dev. Corr Subject (Intercept) 612.128 24.7412 Days 35.049 5.9202 0.066 Residual 654.970 25.5924 Number of obs: 180, groups: Subject, 18 Fixed effects: Estimate Std. Error t value (Intercept) 251.405 6.825 36.84 Days 10.467 1.5456.77 Correlation of Fixed Effects: (Intr) Days -0.137 sessionInfo() R version 2.5.0 Under development (unstable) (2007-01-03 r40349) powerpc-apple-darwin8.8.0 locale: C attached base packages: [1] stats graphics grDevices utils datasets methods [7] base other attached packages: lme4 Matrix lattice 0.9975-11 0.9975-8 0.14-16 On Jan 29, 2007, at 7:40 AM, Michael Kubovy wrote: On Jan 28, 2007, at 9:39 PM, Benilton Carvalho wrote: This seems to be due to the fact that you didn't have enough memory when running lmer2. I might be wrong, but I think Calloc tries to get contiguous memory, so this might the problem. If you are positive that you have enough memory, a gc() might help. I have 2 GB memory on this machine. Should be enough, no? gc() used (Mb) gc trigger (Mb) max used (Mb) Ncells 1008175 27.01476915 39.5 1368491 36.6 Vcells 540055 4.21031040 7.9 1031026 7.9 (fm1 - lmer2(Reaction ~ Days + (Days|Subject), sleepstudy)) Error in as.double(start) : Calloc could not allocate (903190944 of 4) memory On Jan 28, 2007, at 8:35 PM, Michael Kubovy wrote: (fm1 - lmer2(Reaction ~ Days + (Days|Subject), sleepstudy)) Error in as.double(start) : Calloc could not allocate (888475968 of 4) memory * sessionInfo() R version 2.4.1 (2006-12-18) powerpc-apple-darwin8.8.0 locale: C attached base packages: [1] grid datasets stats graphics grDevices utils methods [8] base other attached packages: lme4 Matrix xtable latticeExtra lattice gridBase MASS 0.9975-11 0.9975-8 1.4-3 0.1-40.14-16 0.4-3 7.2-31 JGR iplots JavaGDrJava 1.4-15 1.0-5 0.3-5 0.4-13 * lmer runs the example w/o a problem I just tried to run it on on Intel-based MacPro, and lmer2 ran without a hitch. _ 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.
Re: [R] jump in sequence
is it sth like: as.integer(sapply(seq(4, 22, by=9), seq, length.out=3)) you're looking for? b On Jan 30, 2007, at 9:29 AM, Adrian DUSA wrote: Dear list, This should be a simple one, I just cannot see it. I need to generate a sequence of the form: 4 5 6 13 14 15 22 23 24 That is: starting with 4, make a 3 numbers sequence, jump 6, then another 3 and so on. I can create a whole vector with: myvec - rep(rep(c(F, T, F), rep(3, 3)), 3) Then see which are TRUE: which(myvec) [1] 4 5 6 13 14 15 22 23 24 I'd like to avoid creating the whole vector if possible; for very large ones it can be time consuming. There should be a way to only create the proper indexes... Thanks for any hint, 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 __ 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] Quick Question about R
suppressWarnings(a - as.numeric(c(1, 2, pi, a, 9, z))) b On Jan 31, 2007, at 2:35 PM, Konrad wrote: Hello, Is there a way to convert a character to a number with out getting a warning? I have a vector that has both numbers and letters in it and I need to convert it to only numbers. At the moment I'm using as.numeric but it is generating a warning when it converts a letter. Is there another function out there that will do what I need or is there a way to turn off the warnings as I don't want the warning to be displayed to the end user? Konrad Hammel Engineer Prilink LTD [EMAIL PROTECTED] 905.305.1096 [[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 efficient double sum of max (X_i, Y_i) (X Y vectors)
Well, a reproducible example would be nice =)
not tested:
x = rnorm(10)
y = rnorm(20)
mymax - function(t1, t2) apply(cbind(t1, t2), 1, max)
sum(outer(x, y, mymax))
is this sth like what you need?
b
On Feb 1, 2007, at 1:18 PM, Jeffrey Racine wrote:
Greetings.
For R gurus this may be a no brainer, but I could not find pointers to
efficient computation of this beast in past help files.
Background - I wish to implement a Cramer-von Mises type test
statistic
which involves double sums of max(X_i,Y_j) where X and Y are
vectors of
differing length.
I am currently using ifelse pointwise in a vector, but have a nagging
suspicion that there is a more efficient way to do this. Basically, I
require three sums:
sum1: \sum_i\sum_j max(X_i,X_j)
sum2: \sum_i\sum_j max(Y_i,Y_j)
sum3: \sum_i\sum_j max(X_i,Y_j)
Here is my current implementation - any pointers to more efficient
computation greatly appreciated.
nx - length(x)
ny - length(y)
sum1 - 0
sum3 - 0
for(i in 1:nx) {
sum1 - sum1 + sum(ifelse(x[i]x,x[i],x))
sum3 - sum3 + sum(ifelse(x[i]y,x[i],y))
}
sum2 - 0
sum4 - sum3 # symmetric and identical
for(i in 1:ny) {
sum2 - sum2 + sum(ifelse(y[i]y,y[i],y))
}
Thanks in advance for your help.
-- Jeff
--
Professor J. S. Racine Phone: (905) 525 9140 x 23825
Department of EconomicsFAX:(905) 521-8232
McMaster Universitye-mail: [EMAIL PROTECTED]
1280 Main St. W.,Hamilton, URL:
http://www.economics.mcmaster.ca/racine/
Ontario, Canada. L8S 4M4
`The generation of random numbers is too important to be left to
chance'
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Re: [R] Affymetrix data analysis
The bioconductor mailing list is probably a better place to ask this type of question. [EMAIL PROTECTED] But we also need to know what arrays are you working with, what the errors are, what your sessionInfo() is Let us know, ok? b On Feb 1, 2007, at 5:46 PM, Tristan Coram wrote: Hi, I am trying to read in my Affymetrix CEL files (48 files, total ~600 MB) but I keep getting memory errors. Can somebody please help me with this. Or is therea remote server I can send my data to for computation? Any help is much appreciated. Thanks Dr. Tristan Coram Postdoctoral Research Associate Research Plant Pathologist/Geneticist United States Department of Agriculture Agricultural Research Service Wheat Genetics, Quality Physiology Disease Research 209 Johnson Hall Washington State University Pullman, WA 99163 Office: +1 509 335-1596 Fax: +1 509 335-2553 Email: [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.
[R] predict on biglm class
Hi Everyone, I often use the 'safe prediction' feature available through glm(). Now, I'm at a situation where I must use biglm:::bigglm. ## begin example library(splines) library(biglm) ff - log(Volume)~ns(log(Girth), df=5) fit.glm - glm(ff, data=trees) fit.biglm - bigglm(ff, data=trees) predict(fit.glm, newdata=data.frame(Girth=2:5)) ## -1.3161465 -0.2975659 0.4251285 0.9856938 predict(fit.biglm, newdata=data.frame(Girth=2:5)) ## Error in predict(fit.biglm, newdata = data.frame(Girth = 2:5)) : ##no applicable method for predict ## end example So, it is my understanding that there is no 'predict' method for 'bigglm' class. That suggests me that I need to create my own prediction method, right? What would be an efficient way of making these predictions that use ns() on a very large dataset? My initial thought is that saving the Boundary.knots and knots, I could create the linear predictor by chunks (and therefore get the predictions). Is there a better way of doing this? Thank you very much. Benilton Carvalho 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.
Re: [R] Start and Restart R over SSH
Hi Nils, if the server you're using is *NIX, this is what you can do: example ssh [EMAIL PROTECTED] screen R do what you need in R close the terminal without quitting R ssh [EMAIL PROTECTED] screen -r continue working in R %% end example the problem is if you need X... it works until you quit the terminal, but screen -r doesn't reconnect the X11. b On Feb 19, 2007, at 9:40 AM, Douglas Bates wrote: On 2/19/07, Nils Höller [EMAIL PROTECTED] wrote: Hi, I have some big calculations in R to be done. Since I can use R on a server with ssh, i was wondering if I can reopen a R Shell after exiting ssh. I don't want to use the batch mode and nohup doesn't work. I want to use something like ssh [EMAIL PROTECTED] R ---do something in R and start calculation --- close ssh but let R remain on the server, doing the calculation ssh [EMAIL PROTECTED] open the existing R Shell / Process Has anyone done something similiar? Can you help me or suggest an other solution ? I don't think it is possible to reattach to a process started in one ssh session from another ssh session. However, you can put a session into the background with the -f flag to ssh. You haven't told us what operating system you are starting the ssh connection on and what system will run the R process. If I had an X server running on the local system and the remote system provided X clients like xterm I would do this by ssh -X -f [EMAIL PROTECTED] xterm This should spring up an autonomous xterm window on the local machine after which you can run R in it. I hope this helps. __ 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] linux gplots install unhappy
well, it's complaining because you don't have gtools installed. how about: install.packages(gplots, dep=T) ? b On Feb 20, 2007, at 1:17 PM, Randy Zelick wrote: Hello all, I use R on both windows and a mainframe linux installation (RedHat enterprise 3.0, which they tell me is soon to be upgraded to 4.0). On windows I installed the package gplots without trouble, and it works fine. When I attempted to do the same on the unix computer, the following error message was forthcoming: downloaded 216Kb * Installing *source* package 'gplots' ... ** R ** data ** inst ** preparing package for lazy loading Loading required package: gtools Warning in library(pkg, character.only = TRUE, logical = TRUE, lib.loc = lib.loc) : there is no package called 'gtools' Error: package 'gtools' could not be loaded Execution halted ERROR: lazy loading failed for package 'gplots' ** Removing '/n/fs/disk/resuser02/u/zelickr/R/library/gplots' The downloaded packages are in /tmp/RtmpikM2JW/downloaded_packages Warning messages: 1: installation of package 'gplots' had non-zero exit status in: install.packages(gplots, lib = ~/R/library) 2: cannot create HTML package index in: tools:::unix.packages.html(.Library) Can someone provide the bit of information I need to progress with this? Thanks very much, =Randy= R. Zelick email: [EMAIL PROTECTED] Department of Biology voice: 503-725-3086 Portland State University fax: 503-725-3888 mailing: P.O. Box 751 Portland, OR 97207 shipping: 1719 SW 10th Ave, Room 246 Portland, OR 97201 __ 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] returns from dnorm and dmvnorm
well, nobody said that the density must be smaller than 1, right? :-) it's just the value of the normal density function at the point you asked. you may try doing that by hand and, with the correct math, you'll get the same thing. b On Feb 26, 2007, at 3:03 PM, A Hailu wrote: Hi All, Why would calls to dnorm and dmvnorm return values that are above 1? For example, dnorm(0.3,mean=0, sd=0.1) [1] 3.989423 This is happening on two different installations of R that I have. Thank you. Hailu __ 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] Is there a quick way to count the number of times each element in a vector appears?
is this what you mean? tmp - combinations(3, 3, rep=TRUE) colSums(apply(tmp, 1, duplicated))+1 b On Mar 6, 2007, at 1:16 AM, Dylan Arena wrote: Hi there, I'm writing a function that calculates the probability of different outcomes of dice rolls (e.g., the sum of the highest three rolls of five six-sided dice). I'm using the combinations function from the gtools package, which is great: it gives me a matrix with all of the possible combinations (with repetitions allowed). Now I want to count the number of times each element appears in each arrangement so I can calculate the number of permutations of that arrangement. E.g., if I get output like: combinations(3,3, rep=TRUE) [,1] [,2] [,3] [1,]111 [2,]112 [3,]113 [4,]122 [5,]123 [6,]133 [7,]222 [8,]223 [9,]233 [10,]333 I'd like to be able to determine that the first row has 3 repetitions, yielding 3!/3! = 1 permutation, while the second row has 3 repetitions, yielding 3!/2! = 3 permutations, etc. (This gets harder when there are large numbers of dice with many faces.) I know there are simple things to do, like iterating over the rows with for loops, but I've heard that for loops are sub-optimal in R, and I'd like to see what an elegant solution would look like. E.g., I might like to use sapply() with whatever function I come up with; I thought of using something like duplicated() and just counting the number of TRUEs that are returned for each vector (since the elements are always returned in non-decreasing order), but I'm optimistic that there is a better (faster/cleaner) way. So here is my question in a nutshell: Does anyone have ideas for how I might efficiently process a matrix like that returned by a call to combinations(n, r, rep=TRUE) to determine the number of repetitions of each element in each row of the matrix? If so, I'd love to hear them! Thanks very much for your time, Dylan Arena (Statistics M.S. student) __ 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] Is there a quick way to count the number of times each element in a vector appears?
sorry, i forgot to mention that you will need an extra test |-) tmp - combinations(3, 3, rep=TRUE) out - colSums(apply(tmp, 1, duplicated))+1 out[out == 1] - 0 but now, re-reading your message, you say (..) want to count the number of times each element appears in each arrangement (...) apply(tmp, 1, function(v) table(factor(v, levels=1:3))) might be what you actually meant. sorry for the confusion, b On Mar 6, 2007, at 2:00 AM, Benilton Carvalho wrote: is this what you mean? tmp - combinations(3, 3, rep=TRUE) colSums(apply(tmp, 1, duplicated))+1 b On Mar 6, 2007, at 1:16 AM, Dylan Arena wrote: Hi there, I'm writing a function that calculates the probability of different outcomes of dice rolls (e.g., the sum of the highest three rolls of five six-sided dice). I'm using the combinations function from the gtools package, which is great: it gives me a matrix with all of the possible combinations (with repetitions allowed). Now I want to count the number of times each element appears in each arrangement so I can calculate the number of permutations of that arrangement. E.g., if I get output like: combinations(3,3, rep=TRUE) [,1] [,2] [,3] [1,]111 [2,]112 [3,]113 [4,]122 [5,]123 [6,]133 [7,]222 [8,]223 [9,]233 [10,]333 I'd like to be able to determine that the first row has 3 repetitions, yielding 3!/3! = 1 permutation, while the second row has 3 repetitions, yielding 3!/2! = 3 permutations, etc. (This gets harder when there are large numbers of dice with many faces.) I know there are simple things to do, like iterating over the rows with for loops, but I've heard that for loops are sub-optimal in R, and I'd like to see what an elegant solution would look like. E.g., I might like to use sapply() with whatever function I come up with; I thought of using something like duplicated() and just counting the number of TRUEs that are returned for each vector (since the elements are always returned in non-decreasing order), but I'm optimistic that there is a better (faster/cleaner) way. So here is my question in a nutshell: Does anyone have ideas for how I might efficiently process a matrix like that returned by a call to combinations(n, r, rep=TRUE) to determine the number of repetitions of each element in each row of the matrix? If so, I'd love to hear them! Thanks very much for your time, Dylan Arena (Statistics M.S. student) __ 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] substitute variable
say your data frame is called tmp tmp$PRODUCTS[tmp$PRODUCTS 70] - NA b On Mar 26, 2007, at 12:31 PM, Sergio Della Franca wrote: Dear R-Helpers, I want to substitute the contents of a variable under some contitions. I.e., I have this data set: YEAR PRODUCTS 1 80 2 90 3 50 4 60 5 30 I want to perform this condition: if products 70 then products=NA else products=products. I'd like to achive the seguent result: YEAR PRODUCTS 1 NA 2 NA 3 50 4 60 5 30 How can i develop this? Thank you in advance. Sergio Della Franca [[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] substitute variable
condition1 | condition2 (or) condition1 condition2 (and) tmp$PRODUCTS[tmp$PRODUCTS 70 | tmp$PRODUCTS 20] - NA b On Mar 26, 2007, at 12:47 PM, Sergio Della Franca wrote: Ok, this run correctly. Now i want to perform much more conditions, i.e.: tmp$PRODUCTS[tmp$PRODUCTS 70] - NA and tmp$PRODUCTS[tmp$PRODUCTS 20] - NA. How can i perform this double condition in the same code? 2007/3/26, Benilton Carvalho [EMAIL PROTECTED]: say your data frame is called tmp tmp$PRODUCTS[tmp$PRODUCTS 70] - NA b On Mar 26, 2007, at 12:31 PM, Sergio Della Franca wrote: Dear R-Helpers, I want to substitute the contents of a variable under some contitions. I.e., I have this data set: YEAR PRODUCTS 1 80 2 90 3 50 4 60 5 30 I want to perform this condition: if products 70 then products=NA else products=products. I'd like to achive the seguent result: YEAR PRODUCTS 1 NA 2 NA 3 50 4 60 5 30 How can i develop this? Thank you in advance. Sergio Della Franca [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to drop variables using a wildcard and logic...
if 'test' is your data frame... test[, grep([tT]$, names(test))] b On Mar 26, 2007, at 3:06 PM, [EMAIL PROTECTED] wrote: Dear R users I would like to make a new dataframe from an existing dataframe, retaining ONLY those variables that end in the letter t I have searched the help archives and consulted several reference books but cannot seem to find an example. Any ideas...? Thanks! Mark [[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] .duplicate question
have you tried replacing
.duplicate()
in your my.causality() by
vars:::.duplicate()
?
b
On Mar 28, 2007, at 3:11 PM, Leeds, Mark ((IED)) wrote:
I am using the vars package and it calls a function causality() which
then calls something
called .duplicate. I had to modify the causality function slightly for
my purposes and
I called it my.causality() but now the .duplicate function is no
longer
known to the my.causality function.
I'm fairly certain that this is due to my lack of expertise in R
but if
someone could tell me how
to make the my.causality function know about .duplicate, it would be
appreciated. Thanks.
Mark
This is not an offer (or solicitation of an offer) to buy/se...
{{dropped}}
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Re: [R] string into command
cmd = mylist = list(a = 5, b = 7) (eval(parse(text=cmd))) b On Mar 28, 2007, at 11:38 AM, Brian Dolan wrote: Hello, I would like to take the string mylist = list(a = 5, b = 7) and evaluate it as a list. I have attempted to use parse and several other functions with no success. Thanks for your time. -brian dolan ~~~ may all your sequences converge [[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] pattern search
how about: length(gregexpr(the, Text)[[1]]) ? b On Apr 4, 2007, at 10:49 AM, Schmitt, Corinna wrote: Dear R-experts, I'm looking for an easy possibility for pattern search. I have got a string and a special pattern. I would like to know if the pattern appears in the string, if yes where does it appear (position) and the number of appearance. Example: Text = c(If the sun shines, no clouds should be seen in the sky!) SearchPattern = c(the) NumberOfAppearance = 0 NumberOfLetters = 0 -- NumberOfAppearance should be 2 in the end How can I get the string length (=number of letters stored in Text)? Has anyone a good idea? Thanks, Corinna __ 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 do I print a string without the initial [1]?
x=a\n cat(x) a On Apr 18, 2007, at 5:31 PM, steve wrote: If I print a sting I get an initial [1]: xx=a xx [1] a How do I get it to print just a __ 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] like apply(x,1,sum), but using multiplication?
see ?prod b On May 7, 2007, at 2:25 PM, Jose Quesada wrote: Hi, I need to multiply all columns in a matrix so something like apply(x,2,sum), but using multiplication should do. I have tried apply(x,2,*) I know this must be trivial, but I get: Error in FUN(newX[, i], ...) : invalid unary operator The help for apply states that unary operators must be quoted. I tried single quotes too, with the same results. Thanks, -Jose -- Jose Quesada, PhD. http://www.andrew.cmu.edu/~jquesada __ 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] 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] Efficient computation of trimmed stats?
Hi everyone, I was wondering if there is anything already implemented for efficient (row-wise) computation of group-specific trimmed stats (mean and sd on the trimmed vector) on large matrices. For example: set.seed(1) nc = 300 nr = 25 x = matrix(rnorm(nc*nr), ncol=nc) g = matrix(sample(1:3, nr*nc, rep=T), ncol=nc) trimmedMeanByGroup - function(y, grp, trim=.05) tapply(y, factor(grp, levels=1:3), mean, trim=trim) sapply(1:10, function(i) trimmedMeanByGroup(x[i,], g[i,])) works fine... but: system.time(sapply(1:nr, function(i) trimmedMeanByGroup(x[i,], g [i,]))) user system elapsed 399.928 0.019 399.988 does not look interesting for me. Maybe some package has some implementation of the above? Thank you very much, -b -- Benilton Carvalho PhD Candidate Department of Biostatistics Bloomberg School of Public Health Johns Hopkins University [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.
Re: [R] conversion into capital letter
help(toupper) b On May 15, 2007, at 2:41 AM, [EMAIL PROTECTED] wrote: Dear all, I would need a function which convert small letter into capital letter (at least the first letter of a character variable). Does such a function exist in R ? Thanks by advance Jessica -- Benilton Carvalho PhD Candidate Department of Biostatistics Bloomberg School of Public Health Johns Hopkins University [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.
Re: [R] help with executing instruction every i-th run of loop
if (i %% 1000 == 0)
b
On May 17, 2007, at 10:56 AM, Mark W Kimpel wrote:
I am running a very long loop and would like to save intermediate
results in case of a system or program crash. Here is the skeleton of
what my code would be:
for (i in 1:zillion)
{
results[[i]]-do.something.function()
if (logical.test(i)) {save(results, results.tmp)}
}
logical.test would test to see if i/1000 has no remainder. What R
function would test that?
Is there an even better way to address my need?
Thanks,
Mark
--
---
Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry
Indiana University School of Medicine
15032 Hunter Court, Westfield, IN 46074
(317) 490-5129 Work, Mobile VoiceMail
(317) 663-0513 Home (no voice mail please)
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--
Benilton Carvalho
PhD Candidate
Department of Biostatistics
Bloomberg School of Public Health
Johns Hopkins University
[EMAIL PROTECTED]
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Re: [R] error message
install.packages(RMySQL, dep=T) should fix it for you. b ps: The error says RMySQL is the problem... it is not complaining about R itself (although it would not be a bad idea, given that the latest R is v 2.5.0, so it would be a better idea to start by upgrading your R) On May 22, 2007, at 11:59 AM, karen power wrote: Hi, I am trying to install the package exonmap and RMySQL however I keep getting the following error: Error in library(pkg, character.only = TRUE) : 'RMySQL' is not a valid package -- installed 2.0.0? I have R version 2.4.1 so I know its not a version issue. I deleted and reinstalled the folders again and the same thing happened. Has anyone any ideas? Thanks, Karen __ 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 maximum size of a matrix?
AFAIK, it only depends on how much free memory you have. b On Jun 1, 2007, at 5:05 PM, Guanrao Chen wrote: hi, Rers I tried to find out the max size (# of rows, # of columns) of a matrix that is allowed by R but failed. Can anybody let me know? Thanks! Guanrao __ 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] Windows source in Linux
iconv on your linux box should do the work. b On Jun 1, 2007, at 7:19 PM, Alberto Vieira Ferreira Monteiro wrote: I have a windows source file.r, with the default charset of windows. I can't use it in Linux as source(file.r), because Linux's default is Unicode. How can I read it? Alberto Monteiro __ 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] Why is the R mailing list so hard to figure out?
Well, I have my email client to organize everything by thread... which does the work for me... ethz: Eidgenössische Technische Hochschule Zürich ch: Confœderatio Helvetica best, b On Jun 4, 2007, at 7:25 PM, Robert Wilkins wrote: Why does the R mailing list need such an unusual and customized user interface? Last January, I figured out how to read Usenet mailing lists ( or Usenet groups ) and they all pretty much work the same, learn to use one, you've learned to use them all ( gnu.misc.discuss , comp.lang.lisp , and so on ). What's the best way to view and read discussions in this group for recent days? Can I view the postings for the current day via Google Groups? I hope I'm posting correctly. What does ethz and ch stand for? Is ch for Switzerland? Robert __ 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] name of the variable that will contain the result of a function
Hi everyone, say I have a function called 'foo', which takes the argument arg1. Is there any mechanism that I can use to learn about the variable where foo(arg1) is going to be stored? For example: x - foo(arg1) so, inside foo() I'd like to be able to get the string x. if, foo(arg1) was used insted, I'd like to get NA. thank you very much, b -- Benilton Carvalho PhD Candidate Department of Biostatistics Bloomberg School of Public Health Johns Hopkins University [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.
Re: [R] How to do clustering
sorry, I hit send before finishing my thoughts... and as for clustering microarray data, you might want to consider the bioconductor mailing list... [EMAIL PROTECTED] b On Jun 7, 2007, at 10:42 PM, ssls sddd wrote: Dear List, I have another question to bother you about how to do clustering. My data consists of 49 columns (49 variables) and 238804 rows. I would like to do hierarchical clustering (unsupervised clustering and PCA). So far I tried pvclust (www.is.titech.ac.jp/~shimo/prog/ *pvclust* /) but I always had the problem like for R like cannot allocate the memory. I am curious about what else packages can perform the clustering analysis while memory efficient. Meanwhile, is there any way that I can extract the features of each cluster. In other words, I would like to identify which are responsible for classifying these variables (samples). Thanks a lot! Sincerely, Alex [[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 do clustering
Hi Alex, just in case you're trying to get genotypes from the Affymetrix 500K set, you might want to check the oligo package available on BioConductor. best, b On Jun 7, 2007, at 10:42 PM, ssls sddd wrote: Dear List, I have another question to bother you about how to do clustering. My data consists of 49 columns (49 variables) and 238804 rows. I would like to do hierarchical clustering (unsupervised clustering and PCA). So far I tried pvclust (www.is.titech.ac.jp/~shimo/prog/ *pvclust* /) but I always had the problem like for R like cannot allocate the memory. I am curious about what else packages can perform the clustering analysis while memory efficient. Meanwhile, is there any way that I can extract the features of each cluster. In other words, I would like to identify which are responsible for classifying these variables (samples). Thanks a lot! Sincerely, Alex [[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] find position
which(a == .4)[1] b On Jun 10, 2007, at 4:45 AM, gallon li wrote: find the position of the first value who equals certain number in a vector: Say a=c(0,0,0,0,0.2, 0.2, 0.4,0.4,0.5) i wish to return the index value in a for which the value in the vector is equal to 0.4 for the first time. in this case, it is 7. [[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] p-value from GEE why factor 2*pnorm?
the recommendation was to use lower.tail=FALSE. b On Jun 11, 2007, at 11:21 AM, Carmen Meier wrote: I got an answer for the other question (thank you) But there is another question (I am afraid this is a basic question ...) In this tread there is a hint hwo to calculate the p-vlue of an GEE: _http://finzi.psych.upenn.edu/R/Rhelp02a/archive/74150.html_ Then, get the P values using a normal approximation for the distribution of z: / 2 * pnorm(abs(coef(summary(fm1))[,5]), lower.tail = FALSE) / (Intercept) TPTLD 0. 0.04190831 1. why is the result multiplicated with 2? There is a P-value between 1 and 2 with the results below and multiplicated with 2: 2*pnorm(c (1.8691945,0.5882351,2.4903091,1.9287802,2.3172983,2.2092593,2.2625959 ,1.6395695), lower.tail =TRUE) Regards Carmen __ 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] p-value from GEE why factor 2*pnorm?
Well, AFAIK, the definition of a p-value is the probability of observing something at least as extreme as the observed data. If you observed z, and Z follows a std-normal p-value = P( Z -abs(z) ) + P( Z abs(z) ) = 2*P ( Z abs(z) ) = 2*pnorm(z, lower.tail=FALSE) try z=0 (you should get 1) and z=1.96 (you should get 5%) b On Jun 11, 2007, at 11:34 AM, Carmen Meier wrote: Benilton Carvalho schrieb: the recommendation was to use lower.tail=FALSE. b O but then the results are significant and this does not match the observation. The results are matching the observations if the formula is pnorm(c (1.8691945,0.5882351,2.4903091,1.9287802,2.3172983,2.2092593,2.2625959 ,1.6395695), lower.tail =TRUE) so I have any unknown problem anywhere :-( REgards Carmen __ 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] Rearranging Capture History Data in R
date = c(1, 1, 1, 1, 2, 2, 3, 3, 4) tag = c(1, 2, 3, 4, 2, 4, 1, 2, 4) table(factor(tag, levels=1:4), factor(date, levels=1:4)) (not sure how you got Tag 1/Date 4 = 1) On Jun 11, 2007, at 3:11 PM, [EMAIL PROTECTED] wrote: What code can i use to convert a table like this: Tag#Date 1 1 2 1 3 1 4 1 2 2 4 2 1 3 2 3 4 4 Into one like this: Tag 1 2 3 4 #Date header 1 1 0 0 1 2 1 1 1 0 3 1 0 0 0 4 1 1 0 1 Thanks, Ben Cox Research Assistant (M.S.) Montana Cooperative Fishery Research Unit 301 Lewis Hall Montana State University Bozeman, MT 59717 (406)994-6643 __ 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] Build Windows pkgs from source - online
Hi, First of all, I apologize for sending out this message, as I'm sure the answer is on the archives, but I just can't find it. Not long ago, there was a discussion about building Windows packages from the source code and someone posted a link to a website to which we could submit the source and get the Windows version a while later. Could someone point me to that website or to where I can find such info? Thanks, -benilton __ 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] plot via xyplot not being saved
Hi everyone, it's been a while I've been trying to save a plot created via lattice:::xyplot if I have a file tst.R with the following code: y - rnorm(100) x - rnorm(100) z - sample(letters[1:4], 100, rep=T) library(lattice) bitmap(tst.png) xyplot(y~x|z) dev.off() and I source it, I get the tst.png file, which is a blank page. If I copy and paste instead, I get the correct plot. Any suggestion? Thank you very much, b sessionInfo() R version 2.5.0 (2007-04-23) x86_64-unknown-linux-gnu locale: LC_CTYPE=en_US.iso885915;LC_NUMERIC=C;LC_TIME=en_US.iso885915;LC_COLLATE =en_US.iso885915;LC_MONETARY=en_US.iso885915;LC_MESSAGES=en_US.iso885915 ;LC_PAPER=en_US.iso885915;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASU REMENT=en_US.iso885915;LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods [7] base other attached packages: lattice 0.15-4 __ 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] plot via xyplot not being saved
So, if those statements are inside a function, I have to make my
function to have an 'echo' argument/functionality? eg.:
## begin test.R
test - function(n){
y - rnorm(n)
x - rnorm(n)
z - sample(letters[1:4], n, rep=T)
library(lattice)
bitmap(tst.png)
xyplot(y~x|z)
dev.off()
}
test(100)
## end test.R
source(test.R, echo=T)
also fails in this case...
thanks a lot,
b
On Jun 15, 2007, at 8:53 PM, [EMAIL PROTECTED] wrote:
On 6/15/07, Benilton Carvalho [EMAIL PROTECTED] wrote:
Hi everyone,
it's been a while I've been trying to save a plot created via
lattice:::xyplot
if I have a file tst.R with the following code:
y - rnorm(100)
x - rnorm(100)
z - sample(letters[1:4], 100, rep=T)
library(lattice)
bitmap(tst.png)
xyplot(y~x|z)
dev.off()
and I source it, I get the tst.png file, which is a blank page.
If I copy and paste instead, I get the correct plot.
Any suggestion?
Use
source(..., echo = TRUE)
-Deepayan
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Re: [R] plot via xyplot not being saved
Thank you Deepayan,
I understand the behavior of not printing out the results inside the
functions.
What I didn't know was that for xyplot() saving the plot actually
meant save the result I see, which does not happen with plot(), in
which case my function test() works just fine if I replaced xyplot()
by plot().
Thank you very much,
b
On Jun 16, 2007, at 12:26 AM, [EMAIL PROTECTED] wrote:
On 6/15/07, Benilton Carvalho [EMAIL PROTECTED] wrote:
So, if those statements are inside a function, I have to make my
function to have an 'echo' argument/functionality? eg.:
## begin test.R
test - function(n){
y - rnorm(n)
x - rnorm(n)
z - sample(letters[1:4], n, rep=T)
library(lattice)
bitmap(tst.png)
xyplot(y~x|z)
dev.off()
}
test(100)
## end test.R
source(test.R, echo=T)
also fails in this case...
Yes. The following will produce some output (the values of x + y and x
- y) if you type it out at the R prompt:
x - rnorm(10)
y - rnorm(10)
x + y
x - y
If you put that in a file and source it, nothing will get printed,
unless you have echo=TRUE. If you define
test - function(){
x - rnorm(10)
y - rnorm(10)
x + y
x - y
}
calling test() at the R prompt will only print x - y and not x + y,
and so on.
This is all standard R behaviour. If you want something to be printed
irrespective of context, use print(), e.g.
print(x + y)
or
print(xyplot(y~x|z))
This is also mentioned in the R FAQ.
-Deepayan
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Re: [R] Creating directory under Windows R session
?dir.create b On Jun 20, 2007, at 8:07 AM, Milton Cezar Ribeiro wrote: Hi all, How can I create (and check the existence of) a directory in a R session under Windows(xp)? Kind regards, Miltinho __ 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 matrix from comparing two vectors
outer(test, fac, ) -b On Jun 26, 2007, at 2:13 PM, Van Campenhout Bjorn wrote: hi all, sorry for this basic question, I think I know I should use ? apply, but it is really confusing me... I want to create a matrix by comparing two vectors. Eg: test-seq(1:10) fac-c(3,6,9) and i want to end up with a 10*3 matrix with a boolean that tests if testfac, so something like: 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 I can't find the solution without using a loop... B [[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] stack multiple plots on one page
or just type: pairs.default b On Jun 28, 2007, at 2:36 PM, Henrique Dallazuanna wrote: https://svn.r-project.org/R/branches/R-2-5-branch/src/library/ graphics/R/pairs.R -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O On 28/06/07, Jiong Zhang, PhD [EMAIL PROTECTED] wrote: Hi All, I typed pairs to see its code but did not get what I want. How do I see its code? What I am trying to do, is to stack about 10 scatter plots on one page as the way pairs does. I have about 150 variables in my table. Instead of plotting 150X150 pairs using pairs, I only need to plot 10 pairs. Thanks. jiong The email message (and any attachments) is for the sole use of the intended recipient(s) and may contain confidential information. Any unauthorized review, use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply email and destroy all copies of the original message (and any attachments). Thank You. [[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] why this doesn't work for qqnorm
qqnorm(table[,1]) is what you want, isn't it? and other forms would include: par(ask=TRUE) results = apply(table, 2, qqnorm) par(ask=FALSE) b On Jun 28, 2007, at 9:50 PM, Jiong Zhang, PhD wrote: I want to qqnorm every column in a table. When I try the first column using qqnorm(table$column1), it worked. But when I use qqnorm(table[1]), it tells me Error in stripchart(x1, ...) : invalid plotting method. What happen? How can I make a function that qqnorms every column? thanks a lot. -jiong __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Comparison: glm() vs. bigglm()
Hi, Until now, I thought that the results of glm() and bigglm() would coincide. Probably a naive assumption? Anyways, I've been using bigglm() on some datasets I have available. One of the sets has 15M observations. I have 3 continuous predictors (A, B, C) and a binary outcome (Y). And tried the following: m1 - bigglm(Y~A+B+C, family=binomial(), data=dataset1, chunksize=10e6) m2 - bigglm(Y~A*B+C, family=binomial(), data=dataset1, chunksize=10e6) imp - m1$deviance-m2$deviance For my surprise imp was negative. I then tried the same models, using glm() instead... and as I expected, imp was positive. I also noticed differences on the coefficients estimated by glm() and bigglm() - small differences, though, and CIs for the coefficients (a given coefficient compared across methods) overlap. Are such incrongruences expected? What can I use to check for convergence with bigglm(), as this might be one plausible cause for a negative difference on the deviances? Thank you very much, -benilton sessionInfo() R version 2.5.0 (2007-04-23) x86_64-unknown-linux-gnu locale: LC_CTYPE=en_US.iso885915;LC_NUMERIC=C;LC_TIME=en_US.iso885915;LC_COLLATE =en_US.iso885915;LC_MONETARY=en_US.iso885915;LC_MESSAGES=en_US.iso885915 ;LC_PAPER=en_US.iso885915;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASU REMENT=en_US.iso885915;LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods [7] base other attached packages: biglm 0.4 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Comparison: glm() vs. bigglm()
Hi Peter, thank you very much for your feedback. As for your observations, I do realize that I'm using 1.5 chunks for this particular case (10e6 gives around 8 chunks on other sets). I just noticed that I didn't add the difference in the deviances that I observed: m1$deviance-m2$deviance [1] -93196.69 Thank you very much for the suggestion, I'll give it a try. Best, benilton On Jun 29, 2007, at 11:05 AM, Peter Dalgaard wrote: Benilton Carvalho wrote: Hi, Until now, I thought that the results of glm() and bigglm() would coincide. Probably a naive assumption? Anyways, I've been using bigglm() on some datasets I have available. One of the sets has 15M observations. I have 3 continuous predictors (A, B, C) and a binary outcome (Y). And tried the following: m1 - bigglm(Y~A+B+C, family=binomial(), data=dataset1, chunksize=10e6) m2 - bigglm(Y~A*B+C, family=binomial(), data=dataset1, chunksize=10e6) imp - m1$deviance-m2$deviance For my surprise imp was negative. I then tried the same models, using glm() instead... and as I expected, imp was positive. I also noticed differences on the coefficients estimated by glm() and bigglm() - small differences, though, and CIs for the coefficients (a given coefficient compared across methods) overlap. Are such incrongruences expected? What can I use to check for convergence with bigglm(), as this might be one plausible cause for a negative difference on the deviances? It doesn't sound right, but I cannot reproduce your problem on a similar sized problem (it pretty much killed my machine...). Some observations: A: You do realize that you are only using 1.5 chunks? (15M vs. 10e6 chunksize) B: Deviance changes are O(1) under the null hypothesis but the deviances themselves are O(N). In a smaller variant (N=1e5), I got m1$deviance [1] 138626.4 m2$deviance [1] 138626.4 m2$deviance - m1$deviance [1] -0.05865785 This does leave some scope for roundoff to creep in. You may want to play with a lower setting of tol=... -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ 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 More efficient method?
C1 - rep(-1, length(Cat))
C1[Cat == b]] - 1
b
On Jul 4, 2007, at 9:44 AM, Keith Alan Chamberlain wrote:
Dear Rhelpers,
Is there a faster way than below to set a vector based on values from
another vector? I'd like to call a pre-existing function for this,
but one
which can also handle an arbitrarily large number of categories.
Any ideas?
Cat=c('a','a','a','b','b','b','a','a','b')# Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
# Cycle through each column and set C1 to corresponding value of Cat.
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1 1 -1 -1 1
Cat
[1] a a a b b b a a b
Sincerely,
KeithC.
Psych Undergrad, CU Boulder (US)
RE McNair Scholar
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Re: [R] Source inside source
the smarter thing is to write a package, so you don't need source() at all. but your problem will be fixed if you use the full path instead: source(/home/user/dir1/dir2/file2.R) (and obviously you could pass the path as an argument to whatever function you're using...) b On Jul 16, 2007, at 6:11 PM, Alberto Monteiro wrote: Is there a way to know where is the source, so as to make a source call inside another source smarter? As an example: file1.R is in directory /files/dir1/ file2.R is in directory /files/dir1/dir2/ In file1.R, there is this line: source(dir2/file2.R) So, if I setwd to /files/dir1/, and then I call source(file1.R), it will run correctly. However, if I setwd to /files, then call source(dir1/file1.R), it will give an error when trying to source file2.R Alberto Monteiro __ 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] data.restore() in R 2.5.1 for Windows 95 and later
The R Data Import/Export also says that this function is in the foreign package. :-) b On Jul 16, 2007, at 11:38 PM, Brad Christoffersen wrote: To Whom It May Concern: I want to read in an S-PLUS data dump and I used data.restore(filepath/filename) (in R 2.5.1 for Windows 95 and later) and I get the message Error: could not find function data.restore I have also tried read.S() and read.dta() with the same result. I cannot find any of these functions in the R Help for package base, although data.restore() is mentioned in the R Data Import/Export manual. I have also tried apropos() and character(0) returns. Additionally, I tried getS3method(data,restore) but get Error in getS3method(data, restore) : no function 'data' could be found In addition, I tried dget(filepath/filename) and it was taking a very long time to import the desired data dump objects which previously took a shorter time using data.restore in SPlus. Any help is much appreciated! Thanks, Brad Christoffersen Graduate Student University of Arizona __ 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] Classification
maybe: x = c(.2, .1, .8, .3, .7, .6, .01, .2, .5, 1, 1) breaks = seq(0, 1, .2) LETTERS[1:(length(breaks)-1)][cut(x, breaks)] b On Jul 18, 2007, at 1:50 PM, Doran, Harold wrote: Michael Assume your data frame is called data and your variable is called V1. Converting this to a factor is: data$V1 - factor(data$V1) Creating the classes can be done using ifelse(). Something like data$class - ifelse(data$V1 .21, A, ifelse(data$V1 .41, B, C)) Harold -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ing. Michal Kneifl, Ph.D. Sent: Wednesday, July 18, 2007 1:37 PM To: r-help@stat.math.ethz.ch Subject: [R] Classification Hi, I am also a quite new user of R and would like to ask you for help: I have a data frame where all columns are numeric variables. My aim is to convert one columnt in factors. Example: MD 0.2 0.1 0.8 0.3 0.7 0.6 0.01 0.2 0.5 1 1 I want to make classes: 0-0.2 A 0.21-0.4 B 0.41-0.6 C . and so on So after classification I wil get: MD A A D B . . . and so on Please could you give an advice to a newbie? Thanks a lot in advance.. Michael __ 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] Fw: Do GLM by groups
Check the following example for by(): require(stats) attach(warpbreaks) by(warpbreaks, tension, function(x) lm(breaks ~ wool, data = x)) or just type: example(by) b On Jul 19, 2007, at 1:18 PM, Hongmei Jia wrote: Dear All, I'm trying to do 'glm' analysis by groups just like in SAS you use by variable. I don't know how to do it in R, anyone can help with this? i.e. groupline rep value 1 1 1 0.2 1 1 2 0.3 1 1 3 0.23 1 2 1 0.2 1 2 2 0.3 1 2 3 0.23 2 1 1 0.2 2 1 2 0.3 2 1 3 0.23 2 2 1 0.2 2 2 2 0.3 2 2 3 0.23 in SAS we say: model value=line rep; by group; How can I do this in R? Thanks, Hongmei Jia __ 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] Dataframe of factors transform speed?
it looks like that whatever method you used to genotype the 1002
samples on the STY array gave you a transposed matrix of genotype
calls. :-)
i'd use:
genoT = read.table(yourFile, stringsAsFactors = FALSE)
as a starting point... but I don't think that would be efficient (as
you'd need to fix one column at a time - lapply).
i'd preprocess yourFile before trying to load it:
cat yourFile | sed -e 's/AA/1/g' | sed -e 's/AB/2/g' | sed -e 's/BB/3/
g' outFile
and, now, in R:
genoT = read.table(outFile, header=TRUE)
b
On Jul 19, 2007, at 11:51 PM, Latchezar Dimitrov wrote:
Hello,
This is a speed question. I have a dataframe genoT:
dim(genoT)
[1] 1002 238304
str(genoT)
'data.frame': 1002 obs. of 238304 variables:
$ SNP_A.4261647: Factor w/ 3 levels 0,1,2: 3 3 3 3 3 3 3 3 3 3
...
$ SNP_A.4261610: Factor w/ 3 levels 0,1,2: 1 1 3 3 1 1 1 2 2 2
...
$ SNP_A.4261601: Factor w/ 3 levels 0,1,2: 1 1 1 1 1 1 1 1 1 1
...
$ SNP_A.4261704: Factor w/ 3 levels 0,1,2: 3 3 3 3 3 3 3 3 3 3
...
$ SNP_A.4261563: Factor w/ 3 levels 0,1,2: 3 1 2 1 2 3 2 3 3 1
...
$ SNP_A.4261554: Factor w/ 3 levels 0,1,2: 1 1 NA 1 NA 2 1 1
2 1
...
$ SNP_A.4261666: Factor w/ 3 levels 0,1,2: 1 1 2 1 1 1 1 1 1 2
...
$ SNP_A.4261634: Factor w/ 3 levels 0,1,2: 3 3 2 3 3 3 3 3 3 2
...
$ SNP_A.4261656: Factor w/ 3 levels 0,1,2: 1 1 2 1 1 1 1 1 1 2
...
$ SNP_A.4261637: Factor w/ 3 levels 0,1,2: 1 3 2 3 2 1 2 1 1 3
...
$ SNP_A.4261597: Factor w/ 3 levels AA,AB,BB: 2 2 3 3 3 2 1
2 2 3
...
$ SNP_A.4261659: Factor w/ 3 levels AA,AB,BB: 3 3 3 3 3 3 3
3 3 3
...
$ SNP_A.4261594: Factor w/ 3 levels AA,AB,BB: 2 2 2 1 1 1 2
2 2 2
...
$ SNP_A.4261698: Factor w/ 2 levels AA,AB: 1 1 1 1 1 1 1 1 1
1 ...
$ SNP_A.4261538: Factor w/ 3 levels AA,AB,BB: 2 3 2 2 3 2 2
1 1 2
...
$ SNP_A.4261621: Factor w/ 3 levels AA,AB,BB: 1 1 1 1 1 1 1
1 1 1
...
$ SNP_A.4261553: Factor w/ 3 levels AA,AB,BB: 1 1 2 1 1 1 1
1 1 1
...
$ SNP_A.4261528: Factor w/ 2 levels AA,AB: 1 1 1 1 1 1 1 1 1
1 ...
$ SNP_A.4261579: Factor w/ 3 levels AA,AB,BB: 1 1 1 1 1 2 1
1 1 2
...
$ SNP_A.4261513: Factor w/ 3 levels AA,AB,BB: 2 1 2 2 2 NA 1
NA 2
1 ...
$ SNP_A.4261532: Factor w/ 3 levels AA,AB,BB: 1 2 2 1 1 1 3
1 1 1
...
$ SNP_A.4261600: Factor w/ 2 levels AB,BB: 2 2 2 2 2 2 2 2 2
2 ...
$ SNP_A.4261706: Factor w/ 2 levels AA,BB: 1 1 1 1 1 1 1 1 1
1 ...
$ SNP_A.4261575: Factor w/ 3 levels AA,AB,BB: 1 1 1 1 1 1 1
2 2 1
...
Its columns are factors with different number of levels (from 1 to 3 -
that's what I got from read.table, i.e., it dropped missing levels). I
want to convert it to uniform factors with 3 levels. The 1st 10 rows
above show already converted columns and the rest are not yet
converted.
Here's my attempt wich is a complete failure as speed:
system.time(
+ for(j in 1:(10 )){ #-- this is to try 1st 10 cols and
measure the time, it otherwise is ncol(genoT) instead of 10
+gt-genoT[[j]] #-- this is to avoid 2D indices
+for(l in 1:length([EMAIL PROTECTED])){
+ levels(gt)[l] - switch([EMAIL PROTECTED],AA=0,AB=1,BB=2)
#-- convert levels to 0,1, or 2
+ genoT[[j]]-factor(gt,levels=0:2) #-- make a 3-level
factor
and put it back
+}
+ }
+ )
[1] 785.085 4.358 789.454 0.000 0.000
789s for 10 columns only!
To me it seems like replacing 10 x 3 levels and then making a
factor of
1002 element vector x 10 is a negligible amount of operations
needed.
So, what's wrong with me? Any idea how to accelerate significantly the
transformation or (to go to the very beginning) to make read.table
use a
fixed set of levels (AA,AB, and BB) and not to drop any
(missing)
level?
R-devel_2006-08-26, Sun Solaris 10 OS - x86 64-bit
The machine is with 32G RAM and AMD Opteron 285 (2.? GHz) so it's not
it.
Thank you very much for the help,
Latchezar Dimitrov,
Analyst/Programmer IV,
Wake Forest University School of Medicine,
Winston-Salem, North Carolina, USA
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Re: [R] Dataframe of factors transform speed?
set.seed(123)
genoT = lapply(1:24, function(i) factor(sample(c(AA, AB,
BB), 1000, prob=sample(c(1, 1000, 1000), 3), rep=T)))
names(genoT) = paste(snp, 1:24, sep=)
genoT = as.data.frame(genoT)
dim(genoT)
class(genoT)
system.time(out - lapply(genoT, function(x) match(x, c(AA, AB,
BB))-1))
##
##
user system elapsed
119.288 0.004 119.339
(for all 240K)
best,
b
ps: note that out is a list.
On Jul 20, 2007, at 2:01 AM, Latchezar Dimitrov wrote:
Hi,
-Original Message-
From: Benilton Carvalho [mailto:[EMAIL PROTECTED]
Sent: Friday, July 20, 2007 12:25 AM
To: Latchezar Dimitrov
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Dataframe of factors transform speed?
it looks like that whatever method you used to genotype the
1002 samples on the STY array gave you a transposed matrix of
genotype calls. :-)
It only looks like :-)
Otherwise it is correctly created dataframe of 1002 samples X (big
number) of columns (SNP genotypes). It worked perfectly until I
decided
to put together to cohorts independently processed in R already. I got
stuck with my lack of foreseeing. Otherwise I would have put 3 dummy
lines w/ AA,AB, and AB on each one to make sure all 3 genotypes are
present and that's it! Lesson for the future :-)
Maybe I am not using columns and rows appropriately here but the
dataframe is correct (I have not used FORTRAN since FORTRAN IV ;-)
- as
str says 1002 observ. of (big number) vars.
i'd use:
genoT = read.table(yourFile, stringsAsFactors = FALSE)
as a starting point... but I don't think that would be
efficient (as you'd need to fix one column at a time - lapply).
No it was not efficient at all. 'matter of fact nothing is more
efficient then loading already read data, alas :-(
i'd preprocess yourFile before trying to load it:
cat yourFile | sed -e 's/AA/1/g' | sed -e 's/AB/2/g' | sed -e
's/BB/3/ g' outFile
and, now, in R:
genoT = read.table(outFile, header=TRUE)
... Too late ;-) As it must be clear now I have two dataframes I
want to
put together with rbind(geno1,geno2). The issue again is
uniformization of factor variables w/ missing factors - they
ended up
like levels AA,BB on one of the and levels AB,BB on the other which
means as.numeric of AA is 1 on the 1st and as.numeric of AB is 1 on
the
second - complete mess. That's why I tried to make both uniform, i.e.
levels AA,AB, and BB for every SNP and then rbind works.
In any case my 1st questions remains: What's wrong with me? :-)
Thanks,
Latchezar
b
On Jul 19, 2007, at 11:51 PM, Latchezar Dimitrov wrote:
Hello,
This is a speed question. I have a dataframe genoT:
dim(genoT)
[1] 1002 238304
str(genoT)
'data.frame': 1002 obs. of 238304 variables:
$ SNP_A.4261647: Factor w/ 3 levels 0,1,2: 3 3 3 3 3
3 3 3 3 3
...
$ SNP_A.4261610: Factor w/ 3 levels 0,1,2: 1 1 3 3 1
1 1 2 2 2
...
$ SNP_A.4261601: Factor w/ 3 levels 0,1,2: 1 1 1 1 1
1 1 1 1 1
...
$ SNP_A.4261704: Factor w/ 3 levels 0,1,2: 3 3 3 3 3
3 3 3 3 3
...
$ SNP_A.4261563: Factor w/ 3 levels 0,1,2: 3 1 2 1 2
3 2 3 3 1
...
$ SNP_A.4261554: Factor w/ 3 levels 0,1,2: 1 1 NA 1 NA 2 1 1
2 1
...
$ SNP_A.4261666: Factor w/ 3 levels 0,1,2: 1 1 2 1 1
1 1 1 1 2
...
$ SNP_A.4261634: Factor w/ 3 levels 0,1,2: 3 3 2 3 3
3 3 3 3 2
...
$ SNP_A.4261656: Factor w/ 3 levels 0,1,2: 1 1 2 1 1
1 1 1 1 2
...
$ SNP_A.4261637: Factor w/ 3 levels 0,1,2: 1 3 2 3 2
1 2 1 1 3
...
$ SNP_A.4261597: Factor w/ 3 levels AA,AB,BB: 2 2 3 3 3 2 1
2 2 3
...
$ SNP_A.4261659: Factor w/ 3 levels AA,AB,BB: 3 3 3 3 3 3 3
3 3 3
...
$ SNP_A.4261594: Factor w/ 3 levels AA,AB,BB: 2 2 2 1 1 1 2
2 2 2
...
$ SNP_A.4261698: Factor w/ 2 levels AA,AB: 1 1 1 1 1 1 1 1 1
1 ...
$ SNP_A.4261538: Factor w/ 3 levels AA,AB,BB: 2 3 2 2 3 2 2
1 1 2
...
$ SNP_A.4261621: Factor w/ 3 levels AA,AB,BB: 1 1 1 1 1 1 1
1 1 1
...
$ SNP_A.4261553: Factor w/ 3 levels AA,AB,BB: 1 1 2 1 1 1 1
1 1 1
...
$ SNP_A.4261528: Factor w/ 2 levels AA,AB: 1 1 1 1 1 1 1 1 1
1 ...
$ SNP_A.4261579: Factor w/ 3 levels AA,AB,BB: 1 1 1 1 1 2 1
1 1 2
...
$ SNP_A.4261513: Factor w/ 3 levels AA,AB,BB: 2 1 2
2 2 NA 1 NA
2
1 ...
$ SNP_A.4261532: Factor w/ 3 levels AA,AB,BB: 1 2 2 1 1 1 3
1 1 1
...
$ SNP_A.4261600: Factor w/ 2 levels AB,BB: 2 2 2 2 2 2 2 2 2
2 ...
$ SNP_A.4261706: Factor w/ 2 levels AA,BB: 1 1 1 1 1 1 1 1 1
1 ...
$ SNP_A.4261575: Factor w/ 3 levels AA,AB,BB: 1 1 1 1 1 1 1
2 2 1
...
Its columns are factors with different number of levels
(from 1 to 3 -
that's what I got from read.table, i.e., it dropped missing
levels). I
want to convert it to uniform factors with 3 levels. The
1st 10 rows
above show already converted columns and the rest are not yet
converted.
Here's my attempt wich is a complete failure as speed:
system.time(
+ for(j in 1:(10 )){ #-- this is to try 1st 10 cols and
measure the time, it otherwise is ncol(genoT) instead of 10
Re: [R] automatically jpeg output
jpeg(...) ##if you have X11 bitmap(..., type=jpeg) ##otherwise b On Jul 20, 2007, at 11:34 AM, Ding, Rebecca wrote: Dear R users, I used R to draw many histograms and I would like to automatically save them into a jpeg file. I tried the following code since I know .ps file could be saved like this way: postscript(AYA_ELA.jpeg,horizontal=F,onefile=T) ..#some funtions inside here dev.off() There was a jpeg file, however, there is no pictures inside. Any suggestion? Thanks. Rebecca __ 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] Column-mean-values for targeted rows
set.seed(123)
N = 3
K = 400
theData = matrix(rnorm(N*K), ncol=K)
theData = as.data.frame(theData)
theData = cbind(indicator = sample(0:1, N, rep=T), theData)
system.time(results - colMeans(subset(theData, indicator == 1)))
user system elapsed
2.309 1.319 3.853
b
On Jul 20, 2007, at 6:17 PM, Diogo Alagador wrote:
Hi all,
I'm handling massive data.frames and matrices in R (3 x 400).
In the 1st column, say, I have 0s and 1s indicating rows that
matter; other columns have probability values.
One simple task I would like to do would be to get the column mean
values for signaled rows (the ones with 1)
As a very fresh programmer I have build a simple function in R
which should not be very efficient indeed! It works well for
current-dimension matrices, but it just not goes so well in huge ones.
meanprob-function(Robj){
NLINE-dim(Robj)[1];
NCOLUMN-dim(Robj)[2];
mprob-c(rep(0,(NCOLUMN-1)));
for (i in 2:NCOLUMN){
sumprob-0;
pa-0;
for (j in 1:NLINE){
if(Robj[j,1]!=0){
pa-pa+1;
sumprob-Robj[j,i]+sumprob;
}
}
mprob[i-1]-sumprob/pa;
}
return(mprob);
}
So I only see 3 ways to get through the problem:
- to reformulate the function to gain efficiency;
- to establish a C-routine (for example), where loops are more
speedy, and then interfacing with R;
- to find some function/ package that already do that.
Can anybody illuminate my way here,
Mush thanks,
Diogo Andre' Alagador
[[alternative HTML version deleted]]
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Re: [R] avoiding timconsuming for loop renaming identifiers
as.integer(factor(dta[[school_id]]))
b
On Jul 20, 2007, at 9:26 PM, [EMAIL PROTECTED] wrote:
Hi All
I was wondering if I can avoid a time-consuming for loop on my
60 obs dataset.
school_id y
8 9.87
8 8.89
8 7.89
8 8.88
20 6.78
20 9.99
20 8.79
31 10.1
31 11
There are, say, 143 different schools in this 60 obs dataset.
I need to thave sequential identifiers, 1,2,3,4,5,...,143.
I was using an awkward for look that took 30 minutes to run.
sid = 1
dta$sid[1] = 1
for (i in 2:nrow(dta)) {
if (dta$school_id[i] != dta$school_[i-1]) sid = sid+1
dta$sid[i] = sid
}
Any hints appreciated.
Thanks Toby
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Re: [R] Write columns from within a list to a matrix?
test - lapply(1:3, function(i) cbind(runif(15), rnorm(15,2)))
sapply(test, [, 16:30)
b
On Jul 22, 2007, at 3:39 PM, [EMAIL PROTECTED] wrote:
Hello,
I think I have a mental block when it comes to working with lists.
lapply and sapply appear to do some magical things, but I can't
seem to master their usage.
As an example, I would like to convert a column within a list to a
matrix, with the list element corresponding to the new matrix column.
#Here is a simplified example: .
test=vector(list, 3)
for (i in 1:3){ test[[i]]=cbind(runif(15), rnorm(15,2)) } #create
example list (I'm sure there is a better way to do this too).
#Now, I wan to get the second column back out, converting it from a
list to a matrix. This works, but gets confusing/inefficient when
I have multiple complex lists I am trying to manage.
savecol2=matrix(0,15,0)
for (i in 1:3){
savecol2=cbind(savecol2, test[[i]][,1])
}
#Something like??: (of course this doesn't work)
savecol2=sapply(test, [[, function(x) x[2,])
Thank you!
Jeff
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Re: [R] Write columns from within a list to a matrix?
oh! and if you want to be less ad-hoc:
sapply(test, function(x) x[,2])
b
On Jul 22, 2007, at 3:50 PM, Benilton Carvalho wrote:
test - lapply(1:3, function(i) cbind(runif(15), rnorm(15,2)))
sapply(test, [, 16:30)
b
On Jul 22, 2007, at 3:39 PM, [EMAIL PROTECTED] wrote:
Hello,
I think I have a mental block when it comes to working with
lists. lapply and sapply appear to do some magical things, but I
can't seem to master their usage.
As an example, I would like to convert a column within a list to a
matrix, with the list element corresponding to the new matrix column.
#Here is a simplified example: .
test=vector(list, 3)
for (i in 1:3){ test[[i]]=cbind(runif(15), rnorm(15,2)) } #create
example list (I'm sure there is a better way to do this too).
#Now, I wan to get the second column back out, converting it from
a list to a matrix. This works, but gets confusing/inefficient
when I have multiple complex lists I am trying to manage.
savecol2=matrix(0,15,0)
for (i in 1:3){
savecol2=cbind(savecol2, test[[i]][,1])
}
#Something like??: (of course this doesn't work)
savecol2=sapply(test, [[, function(x) x[2,])
Thank you!
Jeff
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] apply incompatible dimensions error
are you positive that your function is doing what you expect it to do?
it looks like you want something like:
sapply(1:10, function(i) cor(mat1[i,], mat2[i,]))
b
On Jul 24, 2007, at 11:05 AM, Bernzweig, Bruce ((Consultant)) wrote:
Hi,
I've created the following two matrices (mat1 and mat2) and a function
(f) to calculate the correlations between the two on a row by row
basis.
mat1 - matrix(sample(1:500,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
mat2 - matrix(sample(501:1000,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
f - function(x,y) cor(x,y)
When the matrices are squares (# rows = # columns) I have no problems.
However, when they are not (as in the example above with 5 columns and
10 rows), I get the following error:
apply(mat1, 1, f, y=mat2)
Error in cor(x, y, na.method, method == kendall) :
incompatible dimensions
Any help would be appreciated. Thanks!
- Bruce
**
Please be aware that, notwithstanding the fact that the pers...
{{dropped}}
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Re: [R] apply incompatible dimensions error
that's garbor's suggestion then.
sorry for the misunderstanding. :-)
b
On Jul 24, 2007, at 11:35 AM, Bernzweig, Bruce ((Consultant)) wrote:
Thanks Benilton,
I know what I want to do, just not sure how to do it using R. The
help
documentation is not very clear.
What I am trying to do is calculate correlations on a row against row
basis: mat1 row1 x mat2 row1, mat1 row1 x mat2 row2, ... mat1 row1 x
mat2 row-n, mat1 row-n, mat2 row-n
- Bruce
-Original Message-
From: Benilton Carvalho [mailto:[EMAIL PROTECTED]
Sent: Tuesday, July 24, 2007 11:31 AM
To: Bernzweig, Bruce (Consultant)
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] apply incompatible dimensions error
are you positive that your function is doing what you expect it to do?
it looks like you want something like:
sapply(1:10, function(i) cor(mat1[i,], mat2[i,]))
b
On Jul 24, 2007, at 11:05 AM, Bernzweig, Bruce ((Consultant)) wrote:
Hi,
I've created the following two matrices (mat1 and mat2) and a
function
(f) to calculate the correlations between the two on a row by row
basis.
mat1 - matrix(sample(1:500,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
mat2 - matrix(sample(501:1000,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
f - function(x,y) cor(x,y)
When the matrices are squares (# rows = # columns) I have no
problems.
However, when they are not (as in the example above with 5 columns
and
10 rows), I get the following error:
apply(mat1, 1, f, y=mat2)
Error in cor(x, y, na.method, method == kendall) :
incompatible dimensions
Any help would be appreciated. Thanks!
- Bruce
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Re: [R] generating symmetric matrices
after dat.cor use: Rmat[lower.tri(Rmat)] - dat.cor Rmat - t(Rmat) Rmat[lower.tri(Rmat)] - dat.cor b On Jul 27, 2007, at 11:28 PM, Gregory Gentlemen wrote: Greetings, I have a seemingly simple task which I have not been able to solve today. I want to construct a symmetric matrix of arbtriray size w/o using loops. The following I thought would do it: p - 6 Rmat - diag(p) dat.cor - rnorm(p*(p-1)/2) Rmat[outer(1:p, 1:p, )] - Rmat[outer(1:p, 1:p, )] - dat.cor However, the problem is that the matrix is filled by column and so the resulting matrix is not symmetric. I'd be grateful for any adive and/or solutions. Gregory - [[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] xtable with vector
a - matrix(1:6, nr=1) colnames(a) - paste(col, 1:6) xtable(a) On Jul 28, 2007, at 12:39 PM, Stefan Nachtnebel wrote: Hello, Is there a possibility to use xtable with a vector to generate a latex table? I always get an error, that no applicable method is available. For example: b-1:12 dim(b)-c(2,6) dimnames(b)[[2]]-paste(col,1:6) xtable(b) works fine and does not raise an error, but a-1:6 names(a)-paste(col,1:6) xtable(b) does not work. Regards, Stefan __ 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] Positioning text in top left corner of plot
maybe this is what you want?
plot(rnorm(10))
legend(topleft, A), bty=n)
?
b
On Aug 7, 2007, at 11:08 AM, Daniel Brewer wrote:
Simple question how can you position text in the top left hand
corner of
a plot? I am plotting multiple plots using par(mfrow=c(2,3)) and
all I
want to do is label these plots a), b), c) etc. I have been fiddling
around with both text and mtext but without much luck. text is
fine but
each plot has a different scale on the axis and so this makes it
problematic. What is the best way to do this?
Many thanks
Dan
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Re: [R] length of a string
As long as you keep in mind Prof. Ripley's comment, you're going to be fine with nchar(). http://tolstoy.newcastle.edu.au/R/e2/devel/07/05/3450.html Remember that what you want exactly is given by nchar(obj, type=chars), which is **NOT** the default on R 2.5.1 (only on R-2.6.0). In your particular situation, assuming R-2.5.1, nchar(obj) works, but i'm afraid it's only a coincidence. b On Sep 5, 2007, at 11:05 AM, (Ted Harding) wrote: On 05-Sep-07 13:50:57, João Fadista wrote: Dear all, I would like to know how can I compute the length of a string in a dataframe. Example: SEQUENCE ID TGCTCCCATCTCCACGGHR04FS00645 ACTGAACTCCCATCTCCAAT HR0595847847 I would like to know how to compute the length of each SEQUENCE. Best regards, João Fadista nchar(ACTGAACTCCCATCTCCAAT) [1] 20 seems to work. Find it, and related functions, with help.search(character) As it happens, help.search(string) will not help! Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 05-Sep-07 Time: 15:05:22 -- XFMail -- __ 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] SQL like function?
observation %in% ID b On Sep 7, 2007, at 1:40 AM, Takatsugu Kobayashi wrote: Hi RUsers, I am wonder if I can search observations whose IDs matches any of the values in another vector, such as in MySQL. While I am learing MySQL for future database management, I appreciate if anyone could give me a hint. Suppose I have one 5*1 vector containing observation IDs and frequencies, and one 3*1 vector containing observation IDs. observation-c(1,2,3,4,5) ID-c(1,3,4) Then, I would like to program a code that returns a results showing matched observations like result: TRUE FALSE TRUE TRUE FALSE I am reading S programming, but I cannot find a way to do this. Thank you very much. Taka __ 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.
