Re: [R] Creating bibtex file of all installed packages?
On 12/11/2009 08:41 AM, Achim Zeileis wrote: On Fri, 11 Dec 2009, Rainer M Krug wrote: Hi is there an easy and fast way, to generate a BibTeX file of all installed / loaded packages and R? I know about toBibtex(citation()) to extract the BibTeX for a single package, but how can I generate a file containg citations for all installed / loaded packages? I don't think that there is a way other than calling citation() for each of the installed.packages(). You could do something like this: ## try to get BibTeX for each of the installed packages b - lapply(installed.packages()[,1], function(x) try(toBibtex(citation(x ## omit failed citation calls b - b[-which(sapply(b, class) == try-error)] I would use logical indexing instead of which because if none actually fail, you end up indexing by integer(0) so b is empty. b - b[!(sapply(b, class) == try-error)] ## unify to list of Bibtex b - lapply(b, function(x) if(inherits(x, Bibtex)) list(x) else x) ## list of unique entries b - unique(do.call(c, b)) ## write everything to a single .bib file writeLines(do.call(c, lapply(b, as.character)), Rpackages.bib) hth, Z If you then want to do the reversed operation, read a bibtex file into a citationList object, you can use the unreleased(*) package bibtex. install.packages(bibtex, repos=http://R-Forge.R-project.org;) require( bibtex ) Loading required package: bibtex bib - read.bib( Rpackages.bib ) There were 50 or more warnings (use warnings() to see the first 50) (*) because I have been lazy The warnings are all about the lack of keys in the entries cooked by toBibtex. no big deal. Romain Cheers, Rainer -- NEW GERMAN FAX NUMBER!!! Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Natural Sciences Building Office Suite 2039 Stellenbosch University Main Campus, Merriman Avenue Stellenbosch South Africa Cell: +27 - (0)83 9479 042 Fax: +27 - (0)86 516 2782 Fax: +49 - (0)321 2125 2244 email: rai...@krugs.de Skype: RMkrug Google: r.m.k...@gmail.com [[alternative HTML version deleted]] -- Romain Francois Professional R Enthusiast +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr |- http://tr.im/Gq7i : ohloh |- http://tr.im/FtUu : new package : highlight `- http://tr.im/EAD5 : LondonR slides __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to draw three line on the same picture ?
On Thu, 10-Dec-2009 at 10:14PM -0800, z_axis wrote:
|
| The following is sampling data:
| No V1 V2 V3
| 1 0.23 0.12 0.89
| 2 0.11 0;56 0.12
| ...
|
| I just want to draw three lines on same picture according to value of V1, V2
| and V3.
?lines
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_ Average minds discuss events
(:_~*~_:) Small minds discuss people
(_)-(_) . Eleanor Roosevelt
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
__
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[R] copyING directories and files
Hi, I am using the windows version of R. I wanted to copy a directory (containing several files) to another directory. Is there any command in R that will let me do this (something like the 'cp' command in UNIX)? I have looked at 'file.copy', but as the name implies I think it only copies one file at a time. thanks! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating bibtex file of all installed packages?
On Fri, Dec 11, 2009 at 9:41 AM, Achim Zeileis achim.zeil...@wu-wien.ac.atwrote: On Fri, 11 Dec 2009, Rainer M Krug wrote: Hi is there an easy and fast way, to generate a BibTeX file of all installed / loaded packages and R? I know about toBibtex(citation()) to extract the BibTeX for a single package, but how can I generate a file containg citations for all installed / loaded packages? I don't think that there is a way other than calling citation() for each of the installed.packages(). You could do something like this: ## try to get BibTeX for each of the installed packages b - lapply(installed.packages()[,1], function(x) try(toBibtex(citation(x ## omit failed citation calls b - b[-which(sapply(b, class) == try-error)] ## unify to list of Bibtex b - lapply(b, function(x) if(inherits(x, Bibtex)) list(x) else x) ## list of unique entries b - unique(do.call(c, b)) ## write everything to a single .bib file writeLines(do.call(c, lapply(b, as.character)), Rpackages.bib) Thanks a lot Achim. That works. It would actually be nice, if the citation() function could take more then one package name, and probably have an option to save the citations into a BibTeX file. Thanks, Rainer hth, Z Cheers, Rainer -- NEW GERMAN FAX NUMBER!!! Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Natural Sciences Building Office Suite 2039 Stellenbosch University Main Campus, Merriman Avenue Stellenbosch South Africa Cell: +27 - (0)83 9479 042 Fax:+27 - (0)86 516 2782 Fax:+49 - (0)321 2125 2244 email: rai...@krugs.de Skype: RMkrug Google: r.m.k...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- NEW GERMAN FAX NUMBER!!! Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Natural Sciences Building Office Suite 2039 Stellenbosch University Main Campus, Merriman Avenue Stellenbosch South Africa Cell: +27 - (0)83 9479 042 Fax:+27 - (0)86 516 2782 Fax:+49 - (0)321 2125 2244 email: rai...@krugs.de Skype: RMkrug Google: r.m.k...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to figure out which the version of split is used?
On Thu, 10 Dec 2009 13:56:12 -0600 Peng Yu pengyu...@gmail.com wrote: type data.frame, split.data.frame will be called? Is it the case that if the argument is not of type data.frame, Date or POSIXct, split.default will be called? Yes. See ?UseMethod I tried it. But I'm not sure how to use it. Would you please give me a working example? Typing 'split' shows a nice example ... :-) -- Karl Ove Hufthammer __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Gamma correlation
Dear all, does anybody know if it is possible, and how, to compute the Gamma correlation in R? Thank you very much in advance! Manuel __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to draw three line on the same picture ?
thanks for your answer ! Would you mind giving me an example using my data ?
Sincerely!
Patrick Connolly-4 wrote:
On Thu, 10-Dec-2009 at 10:14PM -0800, z_axis wrote:
|
| The following is sampling data:
| No V1 V2 V3
| 1 0.23 0.12 0.89
| 2 0.11 0;56 0.12
| ...
|
| I just want to draw three lines on same picture according to value of
V1, V2
| and V3.
?lines
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_ Average minds discuss events
(:_~*~_:) Small minds discuss people
(_)-(_). Eleanor Roosevelt
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
__
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Re: [R] mutlidimensional in.convex.hull (was multidimensional point.in.polygon??)
2009/12/10 Charles C. Berry cbe...@tajo.ucsd.edu:
[snipped]
Many?
set.seed(1234)
ps - matrix(rnorm(4000),ncol=4)
phull - convhulln(ps)
xs - matrix(rnorm(1200),ncol=4)
phull2 - convhulln(rbind(ps,xs))
nrp - nrow(ps)
nrx - nrow(xs)
outside - unique(phull2[phull2nrp])-nrp
done - FALSE
while(!done){
+ phull3 - convhulln(rbind(ps,xs[-(outside),]))
+ also.outside - (1:nrx)[-outside][unique(phull3[phull3nrp])-nrp]
+ print(length(also.outside))
+ outside - c(outside,also.outside)
+ done - length(also.outside)==0
+ }
[1] 3
[1] 0
phull2 was evaluated once, phull3 twice.
Any point that is in the convex hull of rbind(ps,xs) is either in or outside
the convex hull of ps. Right? So, just recursively eliminate points that are
in the convex hull of the larger set.
If I'm not mistaken this method is efficient only because the two
point distributions are very similar (drawn from rnorm, so they look
like two concentric balls). If one of the convex hulls is very
distorted along one axis, say, I believe the method will involve many
more iterations and in the limit will require computing a convex hull
for each test point as Duncan suggested.
Such a pathological of test points example might be,
xs - matrix(0,ncol=4,nrow=100)
xs[,1] - seq(1,100)
Or did I completely miss something? (quite possible)
Regarding the inhull Matlab code, I came to the opposite conclusion:
it should be easily ported to R. 1) it is a very short piece of code
(even more so if one disregards the various checks and handling of
special cases), with no Matlab-specific objects (only integers,
booleans, matrices and vectors). 2) The core of the program relies on
the qhull library, and the same applies to R I think. 3) Matlab and R
use very similar indexing for matrices and similar linear algebra in
general.
That said, I'm a bit short on time to give it a go myself. I think the
open-source Octave could run this code too, so it might help in
checking the code step-by-step.
All the best,
baptiste
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Re: [R] How to get the runtime as well as the result?
Peng Yu wrote: On Thu, Dec 10, 2009 at 7:03 PM, Dirk Eddelbuettel e...@debian.org wrote: On 10 December 2009 at 18:12, Peng Yu wrote: | If I use system.time() to measure the runtime of an expression, I will | not get the result. Is there a way to measure the runtime and get the | result as well? Use an assignment inside system.time(): R system.time( m - max(rnorm(1e6)) ) user system elapsed 0.328 0.008 0.340 R m [1] 4.993 R I see. In this case, I'll have to use '-' rather than '='. Thanks to point it out on the list. This is one of the reasons not to use = as assignment operator at all: it is confusing. Best, Uwe Ligges __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Greek symbols on ylab= using barchart() {Lattice}
On Wed, 2009-12-09 at 18:46 +0100, baptiste auguie wrote: barchart(1:2, ylab=expression(Concentration (*mu*g/m^3*))) or barchart(1:2, ylab=expression(Concentration ~ (mu*g/m^3))) That way you don't have to worry about quoting parts of the expression. G 2009/12/9 Peng Cai pengcaimaill...@gmail.com: Hi Baptiste and Others, Thanks for your help. I'm writing: ylab=expression(Concentration(mu*g/m^3)) And its working fine, but is it possible to add a space between Concentration and (mu*g/m^3). Thanks again, Peng Cai On Wed, Dec 9, 2009 at 12:02 PM, baptiste auguie baptiste.aug...@googlemail.com wrote: Hi, try this, barchart(1:2, ylab=expression(mu*g/m^3)) ?plotmath baptiste 2009/12/9 Peng Cai pengcaimaill...@gmail.com: Hi All, I'm trying to write ug/m3 as y-label, with greek letter mu replacing u AND 3 going as a power. These commands works in general: plot.new() text(0.5, 0.5, expression(symbol(m))) But, I'm sure about how to do it using barchart() from Lattice. Can anyone help please? Thanks, Peng Cai [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Warning for data.table (with ref)?
This is the intended startup message for that package (and not a warning), nothing to fix. Uwe Ligges Peng Yu wrote: I have following the message dim(refdata) and dimnames(refdata) no longer allow parameter ref=TRUE, use dim(derefdata(refdata)), dimnames(derefdata(refdata)) instead when I loaded data.table. Is it from the package ref? Could it be fixed? Or there is something wrong with my installation? library(data.table) Loading required package: ref dim(refdata) and dimnames(refdata) no longer allow parameter ref=TRUE, use dim(derefdata(refdata)), dimnames(derefdata(refdata)) instead sessionInfo() R version 2.10.0 (2009-10-26) x86_64-unknown-linux-gnu locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8 [7] LC_PAPER=en_US.UTF-8 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] data.table_1.2 ref_0.97 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to draw three line on the same picture ?
try this:
x - read.table(textConnection(No V1 V2 V3
1 0.23 0.12 0.89
2 0.11 0.56 0.12), header=TRUE)
matplot(x[,1], x[,-1], type='l')
On Fri, Dec 11, 2009 at 3:39 AM, z_axis z_a...@163.com wrote:
thanks for your answer ! Would you mind giving me an example using my data
?
Sincerely!
Patrick Connolly-4 wrote:
On Thu, 10-Dec-2009 at 10:14PM -0800, z_axis wrote:
|
| The following is sampling data:
| No V1 V2 V3
| 1 0.23 0.12 0.89
| 2 0.11 0;56 0.12
| ...
|
| I just want to draw three lines on same picture according to value of
V1, V2
| and V3.
?lines
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_ Average minds discuss events
(:_~*~_:) Small minds discuss people
(_)-(_). Eleanor Roosevelt
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
View this message in context:
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__
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] copyING directories and files
I'd use a shell() command to call xcopy or robocopy, or cp given you have some unix tools installed. Uwe Ligges Paul Evans wrote: Hi, I am using the windows version of R. I wanted to copy a directory (containing several files) to another directory. Is there any command in R that will let me do this (something like the 'cp' command in UNIX)? I have looked at 'file.copy', but as the name implies I think it only copies one file at a time. thanks! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Recoding factor labels that are lists into first element of list
try this:
x - data.frame(a=c('cat', 'cat,dog', 'dog', 'dog,cat'))
x
a
1 cat
2 cat,dog
3 dog
4 dog,cat
levels(x$a)
[1] cat cat,dog dog dog,cat
# change the factors
x$a - factor(sapply(strsplit(as.character(x$a), ','), '[[', 1))
x
a
1 cat
2 cat
3 dog
4 dog
levels(x$a)
[1] cat dog
On Thu, Dec 10, 2009 at 10:53 PM, Jennifer Walsh walsh...@umich.edu wrote:
Hi all,
I've Googled far and wide but don't think I know the correct terms to
search for to find an answer.
I have a massive dataset where one of the factors is made up of both
individual items and lists of items (for example, cat and cat, dog,
bird). I would like to recode this factor somehow into only the first
element of the list (so every list starting with cat, plus the
observations that were already just cat would all be set equal to cat).
I would ideally like to do this in some simple way that does not require me
to write hundreds of different sets of code (since the lists probably start
with 300+ different items). Is this possible? Extremely complicated?
Also, I am sure this is much simpler, but I cannot seem to get rid of
levels of a factor that have no observations. I have tried setting the
levels of the factor to only the ones with observations that I am interested
in, but every time I summarize the variable there are still 100+ labels all
with 0 as their count. This hasn't happened to me before; is there an
explanation for it?
Thanks very much,
Jen
---
Jennifer Walsh
Graduate Student, Developmental Psychology
University of Michigan
2020 East Hall, 530 Church St.
Ann Arbor, MI 48109-1043
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
[[alternative HTML version deleted]]
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Re: [R] Help with beanplot fromatting
Easiest way is probably to make grouprep a factor and insert a level without observations such as in: mydata$grouprep - factor(mydata$grouprep, levels = paste(rep(LETTERS[1:5], each=2), paste(grp, 1:2, sep=))) beanplot(val~grouprep, data = mydata, ll = 0.04, main = example, ylab = Size, xlab = tank, side = both, border = NA, col = list(c(blue, white), c(red, yellow))) Uwe Ligges James Widman wrote: Sorry for the previous error. Dear Helpful R Users, I am graphing some data using the beanplot, but I am having trouble getting the output I desire. I have five tanks (A-E) and 2 groups for each tank grp1 or grp2, except tank C where there is only grp1. (I only changed the grouprep to C grp1 for the example) When I plot them, I would like A B C(only grp1 - half of the bean plot) then D and E (as full beans). I assume there is some way to do this, but searching the help archives, I haven't been able to find it. I would also like to thank all of you, for your answers to others questions, I have been gathering many useful tips. # example library(beanplot) mydata - data.frame(tank = rep(c(A,B,C,D,E), c(100,100,50,100,100)), group = rep(c(grp1, grp2)), val = rnorm(450)) mydata$grouprep - paste( mydata$tank, mydata$group) mydata$grouprep -with(mydata, ifelse((tank==C), C grp1, grouprep)) beanplot(val~grouprep, data = mydata, ll = 0.04, main = example, ylab = Size, xlab = tank, side = both, border = NA, col = list(c(blue, white), c(red, yellow))) Thanks, Jim James C. Widman Jr. Research Fishery Biologist NOAA Fisheries Milford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] copyING directories and files
What about this as a start :
dir.copy - function( source = getwd(), target ){
files - list.files( source, recursive = TRUE )
dirs - unique( gsub( /[^/]+$, , files[grepl(/, files)] ) )
if( !file.exists( target ) ){
dir.create( target )
}
for( d in dirs){
dir.create( file.path( target, d) , recursive = TRUE )
}
for( f in files ){
file.copy( file.path(source, f) , file.path( target, f ) )
}
invisible(NULL)
}
# what I am copying :
system( tree )
.
└── bar
├── blabla.txt
├── bla.txt
└── foobar
└── blabla.txt
2 directories, 3 files
dir.copy( getwd(), /tmp/target )
system( tree /tmp/target )
/tmp/target
└── bar
├── blabla.txt
├── bla.txt
└── foobar
└── blabla.txt
2 directories, 3 files
Romain
On 12/11/2009 11:18 AM, Uwe Ligges wrote:
I'd use a shell() command to call xcopy or robocopy, or cp given you
have some unix tools installed.
Uwe Ligges
Paul Evans wrote:
Hi,
I am using the windows version of R. I wanted to copy a directory
(containing several files) to another directory. Is there any command
in R that will let me do this (something like the 'cp' command in
UNIX)? I have looked at 'file.copy', but as the name implies I think
it only copies one file at a time.
thanks!
--
Romain Francois
Professional R Enthusiast
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr
|- http://tr.im/Gq7i : ohloh
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Re: [R] Reducing the dimension of a list object
Bogaso wrote:
Please consider following code :
set.seed(1)
res - vector(list)
for (i in 1:5) {
res1 - vector(list)
res1[[1]] - letters[1:5]
res1[[2]] - rnorm(5)
res1[[3]] - rnorm(5); res[[i]] - res1 }
res[[1]]
# Now I want to reduce the dimension of res through creating a data frame
like that
mat - data.frame(nrow=5, ncol=2)
for (i in 1:5) {
mat[i,1] - res[[i]][[1]][1]
mat[i,2] - res[[i]][[2]][1] }; mat
Here I am looking for more efficient code by avoiding the loop. Is there
any smart way to do that ?
Perhaps directly rewritten without thinking:
mat - data.frame(
letters = sapply(res, [[, c(1,1)),
numbers = sapply(res, [[, c(2,1)))
which should be roughly 2.5 times faster.
Uwe Ligges
Thanks,
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Re: [R] Recoding factor labels that are lists into first element of list
Or this which removing the comma and everything thereafter in each
level that has a comma:
levels(x$a) - sub(,.*, , levels(x$a))
On Fri, Dec 11, 2009 at 5:21 AM, jim holtman jholt...@gmail.com wrote:
try this:
x - data.frame(a=c('cat', 'cat,dog', 'dog', 'dog,cat'))
x
a
1 cat
2 cat,dog
3 dog
4 dog,cat
levels(x$a)
[1] cat cat,dog dog dog,cat
# change the factors
x$a - factor(sapply(strsplit(as.character(x$a), ','), '[[', 1))
x
a
1 cat
2 cat
3 dog
4 dog
levels(x$a)
[1] cat dog
On Thu, Dec 10, 2009 at 10:53 PM, Jennifer Walsh walsh...@umich.edu wrote:
Hi all,
I've Googled far and wide but don't think I know the correct terms to
search for to find an answer.
I have a massive dataset where one of the factors is made up of both
individual items and lists of items (for example, cat and cat, dog,
bird). I would like to recode this factor somehow into only the first
element of the list (so every list starting with cat, plus the
observations that were already just cat would all be set equal to cat).
I would ideally like to do this in some simple way that does not require me
to write hundreds of different sets of code (since the lists probably start
with 300+ different items). Is this possible? Extremely complicated?
Also, I am sure this is much simpler, but I cannot seem to get rid of
levels of a factor that have no observations. I have tried setting the
levels of the factor to only the ones with observations that I am interested
in, but every time I summarize the variable there are still 100+ labels all
with 0 as their count. This hasn't happened to me before; is there an
explanation for it?
Thanks very much,
Jen
---
Jennifer Walsh
Graduate Student, Developmental Psychology
University of Michigan
2020 East Hall, 530 Church St.
Ann Arbor, MI 48109-1043
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[R] Literature analysis
Dear all, i am new in R. I am writing a review paper about batteries. However, i am interested in analyzing all the papers by keywords, author, references and year. This could be done by refviz a software, which is only running on windows machines and which is not free. So my question to you is, is it somehow possible to write a script that can do all of this work? And if yes, with what i should start? Thanks a lot in advance, Schwan -- --- Hosseiny, MSc. S.S. (Seyed Schwan) University of Twente Science and Technology Meander, ME 322 P.O. Box 217 7500 AE Enschede The Netherlands Phone +31 534892869 Email: s.s.hosse...@utwente.n __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] copyING directories and files
Thanks Romain, Uwe!
From: Romain Francois romain.franc...@dbmail.com
To: Uwe Ligges lig...@statistik.tu-dortmund.de
Sent: Fri, December 11, 2009 5:35:57 AM
Subject: Re: [R] copyING directories and files
What about this as a start :
dir.copy - function( source = getwd(), target ){
files - list.files( source, recursive = TRUE )
dirs - unique( gsub( /[^/]+$, , files[grepl(/, files)] ) )
if( !file.exists( target ) ){
dir.create( target )
}
for( d in dirs){
dir.create( file.path( target, d) , recursive = TRUE )
}
for( f in files ){
file.copy( file.path(source, f) , file.path( target, f ) )
}
invisible(NULL)
}
# what I am copying :
system( tree )
.
âââ bar
âââ blabla.txt
âââ bla.txt
âââ foobar
âââ blabla.txt
2 directories, 3 files
dir.copy( getwd(), /tmp/target )
system( tree /tmp/target )
/tmp/target
âââ bar
âââ blabla.txt
âââ bla.txt
âââ foobar
âââ blabla.txt
2 directories, 3 files
Romain
On 12/11/2009 11:18 AM, Uwe Ligges wrote:
I'd use a shell() command to call xcopy or robocopy, or cp given you
have some unix tools installed.
Uwe Ligges
Paul Evans wrote:
Hi,
I am using the windows version of R. I wanted to copy a directory
(containing several files) to another directory. Is there any command
in R that will let me do this (something like the 'cp' command in
UNIX)? I have looked at 'file.copy', but as the name implies I think
it only copies one file at a time.
thanks!
-- Romain Francois
Professional R Enthusiast
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr
|- http://tr.im/Gq7i : ohloh
|- http://tr.im/FtUu : new package : highlight
`- http://tr.im/EAD5 : LondonR slides
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Literature analysis
Schwan wrote: Dear all, i am new in R. I am writing a review paper about batteries. However, i am interested in analyzing all the papers by keywords, author, references and year. This could be done by refviz a software, which is only running on windows machines and which is not free. So my question to you is, is it somehow possible to write a script that can do all of this work? Describing what you mean by all of this work would be useful as the above is rather vague as you don't describe what analysis refviz actually performs. Schwan wrote: And if yes, with what i should start? Start learning how to use R. There are good links from the R-project homepage under the Wiki, Other and Books section. I've found Braun Murdoch A First Course in Statistical Programming with R to be a good book to get me going. Neil -- View this message in context: http://n4.nabble.com/Literature-analysis-tp960960p960968.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Notification of false convergence with lmer()
I am running the lmer() command in a for loop and occasionally a particular iteration is producing the false convergence warning. I would like to be able to mark these iterations with a dummy variable, but I can't find any other notification besides the warning message, which, in a for loop, only is printed after the loop is finished (which does not allow me to see which iteration it happened on). Is there any way I can mark which iteration in the loop produces the false convergence? Thank you. -- View this message in context: http://n4.nabble.com/Notification-of-false-convergence-with-lmer-tp960969p960969.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mutlidimensional in.convex.hull (was multidimensionalpoint.in.polygon??)
baptiste auguie baptiste.aug...@googlemail.com wrote in message news:de4e29f50912110200g7e43551kef5e8053fbf6e...@mail.gmail.com... 2009/12/10 Charles C. Berry cbe...@tajo.ucsd.edu: [snipped] Many? snip Charles' elegant coding of the algorithm he summarises as: Any point that is in the convex hull of rbind(ps,xs) is either in or outside the convex hull of ps. Right? So, just recursively eliminate points that are in the convex hull of the larger set. Baptiste commented: If I'm not mistaken this method is efficient only because the two point distributions are very similar (drawn from rnorm, so they look like two concentric balls). If one of the convex hulls is very distorted along one axis, say, I believe the method will involve many more iterations and in the limit will require computing a convex hull for each test point as Duncan suggested. Such a pathological of test points example might be, xs - matrix(0,ncol=4,nrow=100) xs[,1] - seq(1,100) Or did I completely miss something? (quite possible) Until now I thought the same as Baptiste and (mea culpe) had rejected that algorithm without testing it. Now I've tried it and it works! Here's the result on my real test data; sorry it's long, but it shows some important features: a) ps is quite pathological?; some substantial correlations b) xs - expand.grid(lapply(ps, unique)); this is the reason I'm doing it in the first place. I want to expand.grid without 'extrapolating' beyound the (convex hull of) the original data c) xs has lost the correlation structure of ps d) 1170 'outside' points removed in 23 iterations e) xs[-(outside)] has regained (some of) the correlation structure of ps ### begin example source(.trPaths[5], echo=TRUE, max.deparse.length=150) describe(ps) ps 5 Variables 2637 Observations - t n missing uniqueMean .05 .10 .25 .50 .75 .90 .95 2637 0 35 136.2 0 0 30 120 194 312 360 lowest : 0 2 3 4 24, highest: 312 336 360 384 504 - pH n missing uniqueMean 2637 0 4 5.707 4.6 (727, 28%), 5.4 (729, 28%), 6.2 (624, 24%), 7 (557, 21%) - T n missing uniqueMean 2637 0 5 10.75 2 5 8 15 22 Frequency 447 510 593 537 550 % 17 19 22 20 21 - S n missing uniqueMean 2637 0 4 3.097 0 (631, 24%), 2 (648, 25%), 4 (638, 24%), 6 (720, 27%) - N n missing uniqueMean 2637 0 4 118.6 0 (701, 27%), 80 (636, 24%), 160 (628, 24%), 240 (672, 25%) - cor(ps) t pHT SN t 1.0 0.02255541 -0.425911455 0.05541686 0.004447023 pH 0.022555414 1. -0.029277466 0.05630345 -0.031032641 T -0.425911455 -0.02927747 1.0 0.05948337 0.003595186 S 0.055416859 0.05630345 0.059483366 1. 0.014074045 N 0.004447023 -0.03103264 0.003595186 0.01407404 1.0 xs - expand.grid(lapply(ps, unique)) describe(xs) xs 5 Variables 11200 Observations - t n missing uniqueMean .05 .10 .25 .50 .75 .90 .95 11200 0 35 141.7 2 4 48 123 194 336 384 lowest : 0 2 3 4 24, highest: 312 336 360 384 504 - pH n missing uniqueMean 11200 0 4 5.8 4.6 (2800, 25%), 5.4 (2800, 25%), 6.2 (2800, 25%), 7 (2800, 25%) - T n missing uniqueMean 11200 0 510.4 258 15 22 Frequency 2240 2240 2240 2240 2240 % 20 20 20 20 20 - S n missing uniqueMean 11200 0 4 3 0 (2800, 25%), 2 (2800, 25%), 4 (2800,
Re: [R] Literature analysis
Hi, from what I understand, you may be interested in text mining, so perhaps you want to look at the tm package. Then again, depending on what you are really trying to do, you may be better served with perl, awk and similar tools than with R... HTH, Stephan Schwan schrieb: Dear all, i am new in R. I am writing a review paper about batteries. However, i am interested in analyzing all the papers by keywords, author, references and year. This could be done by refviz a software, which is only running on windows machines and which is not free. So my question to you is, is it somehow possible to write a script that can do all of this work? And if yes, with what i should start? Thanks a lot in advance, Schwan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how can generate from trunceted gamma distribution in R ?
Hi, all How can generate a sample from truncated inverse gamma distribution in R? thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Literature analysis
Thanks for all the comments, and sorry about the unstructured question! I am trying to: 1: analyze keywords, names from Authors and year of publication from citations (with abstracts) i downloaded fron various sites(these downloads can be converted into .txt files as well) 2: to cluster these literature according to the analyzed keywords, authors or year of publication The software Refviz I was referring to earlier can be found here: http://refviz.com/ As I said, I have never worked with R before so I can not send any example. Hope this helps to understand my question better. Cheers On Fri, 2009-12-11 at 13:06 +0100, Stephan Kolassa wrote: Hi, from what I understand, you may be interested in text mining, so perhaps you want to look at the tm package. Then again, depending on what you are really trying to do, you may be better served with perl, awk and similar tools than with R... HTH, Stephan Schwan schrieb: Dear all, i am new in R. I am writing a review paper about batteries. However, i am interested in analyzing all the papers by keywords, author, references and year. This could be done by refviz a software, which is only running on windows machines and which is not free. So my question to you is, is it somehow possible to write a script that can do all of this work? And if yes, with what i should start? Thanks a lot in advance, Schwan -- --- Hosseiny, MSc. S.S. (Seyed Schwan) University of Twente Science and Technology Meander, ME 322 P.O. Box 217 7500 AE Enschede The Netherlands Phone +31 534892869 Email: s.s.hosse...@utwente.n __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Literature analysis
Schwan wrote: Thanks for all the comments, and sorry about the unstructured question! I am trying to: 1: analyze keywords, names from Authors and year of publication from citations (with abstracts) i downloaded fron various sites(these downloads can be converted into .txt files as well) Sorry to bang the drum but what do you mean by analyse, is it to simply count keywords and rank them, investigate which keywords occur together, look at trends in keywords by year or something else completely. Its a very general term (at least for me as I've no knowledge or experience of text mining)! Neil -- View this message in context: http://n4.nabble.com/Literature-analysis-tp960960p961019.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lattice - 2 graphs per panel with 2 differents colours
Hello,
I would like to get a lattice plot of 8 panels (unique(df$fact)=8) with 2
graphs (df$y1 and df$y2 as a function of df$x) and 1 red point at (500,
ymax) per panel.
The script below is quite ok but I'm not able to define two different colors
for the two graphs.
If you have an idea how to use the col function in order to attribute the
colors, it will be very kind of you to share it with a newbie.
Have a nice week-end,
Ptit Bleu.
x11(15,12)
xyplot(df$y1 + df$y1/df$coeff ~ df$x | df$fact,
panel = function(x, y) {
panel.grid(h=-1, v=-1, col=gray)
panel.xyplot(x, y, type=p, pch=20)
panel.points(500, ymax[panel.number()], col=red, pch=20,
cex=1.6)
},
xlab=X, ylab=Y)
--
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Re: [R] Literature analysis
Ok good question I havent explain! Well,lets keep it simple for the begining. By analyzing the keywords(which can also include the authors name) I mean, 1) investigate in which paper the keyword occur and how often 2) investigate if keywords occur together X axis can show the paper title (or authors name) y axis shows the keywords if a keyword occurs more often in a paper then the dot on the graph changes the color if keywords occur together then a symbol the same symbol appears. On Fri, 2009-12-11 at 04:33 -0800, nshephard wrote: Schwan wrote: Thanks for all the comments, and sorry about the unstructured question! I am trying to: 1: analyze keywords, names from Authors and year of publication from citations (with abstracts) i downloaded fron various sites(these downloads can be converted into .txt files as well) Sorry to bang the drum but what do you mean by analyse, is it to simply count keywords and rank them, investigate which keywords occur together, look at trends in keywords by year or something else completely. Its a very general term (at least for me as I've no knowledge or experience of text mining)! Neil -- --- Hosseiny, MSc. S.S. (Seyed Schwan) University of Twente Science and Technology Meander, ME 322 P.O. Box 217 7500 AE Enschede The Netherlands Phone +31 534892869 Email: s.s.hosse...@utwente.n __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] export anova table
Hi everybody, It might be a stupid question, but please excuse it! I would just like to export my ANOVA table from that kind of code: stats - lm(Asfc~TO_POS, ssfa) anova(stats) into a *.csv or *.txt file. Can anyone help please? Thanks in advance Ivan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] incorrect multiple outputs
The lack of a reproducible example is a bigger problem than any lack
of clarity of the explanation.
-Ista
On Fri, Dec 11, 2009 at 2:08 AM, bawa...@googlemail.com wrote:
Apologies,
I didn't explain this clearly. The Rscript is called by a perl script, which
creates input_file.txt by inserting 288 lines of (reformatted) data for
each data file in the directory. So the Rscript will (and is doing) run the
loop a number of times equal to the number of files the perl script read in.
The problem is that it should only create the data.frame and write to the
file after the last iteration, but it's (creating and)writing the complete
data.frame every iteration.
Sent using BlackBerry® from Orange
-Original Message-
From: jim holtman jholt...@gmail.com
Date: Thu, 10 Dec 2009 18:00:54
To: biscuitbawa...@googlemail.com
Cc: r-help@r-project.org
Subject: Re: [R] incorrect multiple outputs
If I rad you code right, file.rows is equal to 1 and your 'for' loop will
only iterate once. Is that what you were expecting?
No reproducible code provided, so that is my best guess.
file.rows- c(nrow(file)/288) # input_file.txt contains 288 reformatted
lines for each original data file
...
for (k in 1:file.rows){ # iterates code for each 288 line block of
input_file.txt
...
On Thu, Dec 10, 2009 at 11:39 AM, biscuit bawa...@googlemail.com wrote:
HI,
I'm having trouble with a piece of Rscript which keeps outputting
incorrectly. it's something like this: the code reads in from a file which
contains (reformated) input
file-read.table(file=input_file.txt,sep=\t)[,c(1,3:5)]
file.rows- c(nrow(file)/288) # input_file.txt contains 288 reformatted
lines for each original data file
...
for (k in 1:file.rows){ # iterates code for each 288 line block of
input_file.txt
...
cv[k] - 100*(sd(x.blank)/mean(x.blank))
t[k] -
(mean(x.note)-mean(x.blank))/sqrt(((sd(x.note)^2)/8)+((sd(x.blank)^2)/16))
t11[k] -
(sqrt(8)*(mean(x.note11)-mean(x.blank)))/sqrt(sd(x.note11)^2+sd(x.blank)^2)
}
all.data-data.frame(barcodes,t=format(as.numeric(t),digits=3),t11=format(as.numeric(t11),digits=3),cv=format(as.numeric(cv),digits=3))
write.table(all.data, file=
R_drug_plot.log,append=TRUE,sep=\t,row.names=FALSE)
this all works correctly except that I believed it would output to file
after completing the loop, instead it's writing to file every iteration. so
the output file looks like:
headers
a1
headers
a1
a2
headers
a1
a2
a3
...
I have checked the missing sections of code and can confirm there are no
missing/additional brackets. Has anyone any idea why this is happening and
what I can do about it?
--
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--
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Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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--
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org
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Re: [R] Literature analysis
It sounds pretty simple so far. Just put the citation info in a data frame, and plot it. I would use ggplot2 for plotting, but it could be done in base or lattice too. On Fri, Dec 11, 2009 at 8:04 AM, Schwan s.s.hosse...@utwente.nl wrote: Ok good question I havent explain! Well,lets keep it simple for the begining. By analyzing the keywords(which can also include the authors name) I mean, 1) investigate in which paper the keyword occur and how often 2) investigate if keywords occur together X axis can show the paper title (or authors name) y axis shows the keywords if a keyword occurs more often in a paper then the dot on the graph changes the color if keywords occur together then a symbol the same symbol appears. On Fri, 2009-12-11 at 04:33 -0800, nshephard wrote: Schwan wrote: Thanks for all the comments, and sorry about the unstructured question! I am trying to: 1: analyze keywords, names from Authors and year of publication from citations (with abstracts) i downloaded fron various sites(these downloads can be converted into .txt files as well) Sorry to bang the drum but what do you mean by analyse, is it to simply count keywords and rank them, investigate which keywords occur together, look at trends in keywords by year or something else completely. Its a very general term (at least for me as I've no knowledge or experience of text mining)! Neil -- --- Hosseiny, MSc. S.S. (Seyed Schwan) University of Twente Science and Technology Meander, ME 322 P.O. Box 217 7500 AE Enschede The Netherlands Phone +31 534892869 Email: s.s.hosse...@utwente.n __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] get the enclosing function name
Hi,
Is there a way to get the enclosing function name within a function?
For example, I would like to have a function getEnclosingFunctionName().
It works like below
f = function(){
print(getEnclosingFunctionName())
}
f() # will print f
Thanks
Jeff
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Re: [R] get the enclosing function name
Try placing this in your function:
Name - match.call()[[1]]
On Fri, Dec 11, 2009 at 8:50 AM, Hao Cen h...@andrew.cmu.edu wrote:
Hi,
Is there a way to get the enclosing function name within a function?
For example, I would like to have a function getEnclosingFunctionName().
It works like below
f = function(){
print(getEnclosingFunctionName())
}
f() # will print f
Thanks
Jeff
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Re: [R] export anova table
On Dec 11, 2009, at 8:17 AM, Ivan Calandra wrote: Hi everybody, It might be a stupid question, but please excuse it! I would just like to export my ANOVA table from that kind of code: stats - lm(Asfc~TO_POS, ssfa) anova(stats) into a *.csv or *.txt file. capture.output(anova(model), file=test.txt) Can anyone help please? Thanks in advance Ivan David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] copyING directories and files
copyDirectory() in the R.utils package. /Henrik On Fri, Dec 11, 2009 at 1:01 AM, Paul Evans p.evan...@yahoo.com wrote: Hi, I am using the windows version of R. I wanted to copy a directory (containing several files) to another directory. Is there any command in R that will let me do this (something like the 'cp' command in UNIX)? I have looked at 'file.copy', but as the name implies I think it only copies one file at a time. thanks! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] get the enclosing function name
Try this;
f - function()as.character(sys.call())
On Fri, Dec 11, 2009 at 11:50 AM, Hao Cen h...@andrew.cmu.edu wrote:
Hi,
Is there a way to get the enclosing function name within a function?
For example, I would like to have a function getEnclosingFunctionName().
It works like below
f = function(){
print(getEnclosingFunctionName())
}
f() # will print f
Thanks
Jeff
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and provide commented, minimal, self-contained, reproducible code.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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Re: [R] Subset sum problem.
Hans, you're my personal hero today !
The function seems to work fine for the tests I did already.
Thank you very much !
Geert
On Thursday 10 December 2009, Hans W Borchers wrote:
Geert Janssens janssens-geert at telenet.be writes:
On Wednesday 9 December 2009, Hans W Borchers wrote:
Geert Janssens janssens-geert at telenet.be writes:
[ ... ]
Has anybody tackled this issue before in R ? If so, I would be very
grateful if you could share your solution with me.
Is it really true that you only want to see a Yes or No answer to
this question whether a subset sums up to s --- without learning which
numbers this subset is composed of (the pure SUBSET SUM problem)?
Then the following procedure does that in a reasonable amount of time
(returning 'TRUE' or 'FALSE' instead of Y-or-N):
Unfortunately no. I do need the numbers in the subset. But thank you for
presenting this code.
Geert
Okay then, here we go. But don't tell later that your requirement was to
generate _all_ subsets that add up to a certain amount. I will generate
only one (with largest elements).
For simplicity I assume that the set is prepared s.t. it is decreasingly
ordered, has no elements larger than the amount given, and has a total sum
larger than this amount.
# Assume S decreasing, no elements t, total sum = t
solveSubsetSum - function(S, t) {
L - c(0)
inds - NULL
for (i in 1:length(S)) {
# L - unique(sort(c(L, L + S[i])))
L - c(L, L+S[i])
L - L[L = t]
if (max(L) == t) {
inds - c(i)
t - t - S[i]
while (t 0) {
K - c(0)
for (j in 1:(i-1)) {
K - c(K, K+S[j])
if (t %in% K) break
}
inds - c(inds, j)
t - t - S[j]
}
break
}
}
return(inds)
}
# former example
amount - 4748652
products -
c(30500,30500,30500,30500,42000,42000,42000,42000,
42000,42000,42000,42000,42000,42000,71040,90900,
76950,35100,71190,53730,456000,70740,70740,533600,
83800,59500,27465,28000,28000,28000,28000,28000,
26140,49600,77000,123289,27000,27000,27000,27000,
27000,27000,8,33000,33000,55000,77382,48048,
51186,4,35000,21716,63051,15025,15025,15025,
15025,80,111,59700,25908,829350,1198000,1031655)
# prepare set
prods - products[products = amount] # no elements amount
prods - sort(prods, decreasing=TRUE) # decreasing order
# now find one solution
system.time(is - solveSubsetSum(prods, amount))
# user system elapsed
# 0.320 0.032 0.359
prods[is]
# [1] 70740 70740 71190 76950 77382 8 83800
# [8] 90900 456000 533600 829350 111 1198000
sum(prods[is]) == amount
# [1] TRUE
Note that running times and memory needs will be much higher when more
summands are needed. To mention that too: I have not tested the code
extensively.
Regards
Hans Werner
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[R] Please help with a basic function
Hello,
I am learning how to use functions, but I'm running into a roadblock.
I would like my function to do two things: 1) convert an object to a
dataframe, 2) and then subset the dataframe. Both of these commands work
fine outside the function, but I would like to wrap them in a function so I
can apply the code iteratively to many such objects.
Here's what I wrote, but it doesn't work:
convert-function(d) {
d-data.frame(d); #convert object to dataframe
d-subset(d,select=c(time,coords.x1,coords.x2)) #select some columns
}
convert(data) #the problem is that data is the same as it was before
running the function
The objects being processed through my function are SpatialPointsDataFrames
but I'm quite sure that's not my problem, as I can process these outside of
the function (using the above code) ... it's when I try to wrap the code in
a function that it doesn't work.
Thanks, Mark
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[R] Fonts and axes using persp3d
Greetings, I am making 3D plots using persp3d, and would like to set z-axis limits and make axis labels (the automatic numbers at tick marks) bold. I have tried zlim, but this does not seem to force the plot to restrain itself within certain bounds (e.g., 0-1). The surface I am plotting (z values) does contain some values outside the range I am setting. Maybe this overrides the zlim? Is there a way to fix this without manually removing negative z values? Also, is there any way to make the numerical axis labels bold, or generally darker or larger? My code currently reads: xtemp - 6:22 ylight - seq(from=-7.5, to=-5, by=0.5) wDeltaT - 0 code- 1 grid.tld - expand.grid(temp=xtemp, logwm2=ylight, DeltaT=wDeltaT, code=code) YaoRasPred-predict(YaoRas.Distribution.T.L.DT.gamm$gam,newdata=grid.tld,se.fit=T) Rel.Dens - matrix(YaoRasPred$fit, nrow=17 , byrow=F) # use predict instead here library(rgl) persp3d(xtemp,ylight,Rel.Dens, zlim=c(min=0.0,max=0.032),xlab=,ylab=,zlab=,col=gray) Thanks for any advice, and thanks again to Duncan Murdoch for suggesting persp3d for my purposes! Sincerely, Paul Simonin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] get the enclosing function name
Hi,
.NotYetImplemented gives an example,
function ()
stop(gettextf('%s' is not implemented yet,
as.character(sys.call(sys.parent())[[1L]])),
call. = FALSE)
environment: namespace:base
HTH,
baptiste
2009/12/11 Hao Cen h...@andrew.cmu.edu:
Hi,
Is there a way to get the enclosing function name within a function?
For example, I would like to have a function getEnclosingFunctionName().
It works like below
f = function(){
print(getEnclosingFunctionName())
}
f() # will print f
Thanks
Jeff
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Re: [R] Literature analysis
Thanks, but how should I put the citation inside a data frame? data.frame(first txt file, second txt file...) plot (what should I insert here) type=p And how should I load the txt files anyway inside the frame? On Fri, 2009-12-11 at 08:37 -0500, Ista Zahn wrote: It sounds pretty simple so far. Just put the citation info in a data frame, and plot it. I would use ggplot2 for plotting, but it could be done in base or lattice too. On Fri, Dec 11, 2009 at 8:04 AM, Schwan s.s.hosse...@utwente.nl wrote: Ok good question I havent explain! Well,lets keep it simple for the begining. By analyzing the keywords(which can also include the authors name) I mean, 1) investigate in which paper the keyword occur and how often 2) investigate if keywords occur together X axis can show the paper title (or authors name) y axis shows the keywords if a keyword occurs more often in a paper then the dot on the graph changes the color if keywords occur together then a symbol the same symbol appears. On Fri, 2009-12-11 at 04:33 -0800, nshephard wrote: Schwan wrote: Thanks for all the comments, and sorry about the unstructured question! I am trying to: 1: analyze keywords, names from Authors and year of publication from citations (with abstracts) i downloaded fron various sites(these downloads can be converted into .txt files as well) Sorry to bang the drum but what do you mean by analyse, is it to simply count keywords and rank them, investigate which keywords occur together, look at trends in keywords by year or something else completely. Its a very general term (at least for me as I've no knowledge or experience of text mining)! Neil -- --- Hosseiny, MSc. S.S. (Seyed Schwan) University of Twente Science and Technology Meander, ME 322 P.O. Box 217 7500 AE Enschede The Netherlands Phone +31 534892869 Email: s.s.hosse...@utwente.n __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- --- Hosseiny, MSc. S.S. (Seyed Schwan) University of Twente Science and Technology Meander, ME 322 P.O. Box 217 7500 AE Enschede The Netherlands Phone +31 534892869 Email: s.s.hosse...@utwente.n __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Literature analysis
They are in Bibtex
For example:
@ARTICLE{adsdifvanadiumcationexchange,
author = {Jin-qing Chen and Bao-guo Wang and Ji-chu Yang},
title = {Adsorption and Diffusion of VOsup2+/sup and
VOsub2/sub sup+/sup
across Cation Membrane for All-Vanadium Redox Flow Battery},
journal = {Solvent Extraction and Ion Exchange},
year = {2009},
volume = {27},
pages = {312--327},
number = {2},
abstract = {A method based on a selectivity coefficient and the
Nernst-Planck
equation is proposed to determine diffusion coefficients of vanadium
ions across a cation exchange membrane in VOsup2+/sup/Hsup+/sup
and VOsub2/sub sup+/sup/Hsup+/sup systems. This simplified
method can be applied to high concentrations of vanadium ions. Three
cation exchange membranes were studied. The logarithmic value of
the selectivity coefficient was linearly dependent on the molar
fraction
of vanadium ions in solution. The diffusion coefficient of vanadium
ions decreased with decreasing water content. The membrane with the
lowest diffusion coefficient was selected as a battery separator
and showed the lowest capacity loss of the studied membranes.},
issn = {0736-6299},
publisher = {Taylor \ Francis},
url = {http://www.informaworld.com/10.1080/07366290802674614}
}
@ARTICLE{Chieng1992,
author = {Chieng, S.C. and Kazacos, M. and Skyllas-Kazacos, M.},
title = {Modification of Daramic, microporous separator, for redox
flow battery
applications},
journal = {Journal of Membrane Science},
year = {1992},
volume = {75},
pages = {81--91},
number = {1-2},
month = dec,
issn = {0376-7388},
keywords = {Daramic, microporous separator, redox flow cell and
battery},
owner = {schwan},
timestamp = {2009.11.30},
url =
{http://www.sciencedirect.com/science/article/B6TGK-43S71CR-7K/2/06f90d391c0eff0ff5df3f282ad5fe28}
}
On Fri, 2009-12-11 at 15:37 +0100, Gustaf Rydevik wrote:
On Fri, Dec 11, 2009 at 3:04 PM, Schwan s.s.hosse...@utwente.nl
wrote:
Thanks, but how should I put the citation inside a data frame?
data.frame(first txt file, second txt file...)
plot (what should I insert here) type=p
And how should I load the txt files anyway inside the frame?
Can you give an example of a couple of text files? Are they in a
standardised format (i.e. bibTEX or similar)?
/Gustaf
--
Gustaf Rydevik, M.Sci.
tel: +46(0)703 051 451
address:Essingetorget 40,112 66 Stockholm, SE
skype:gustaf_rydevik
--
---
Hosseiny, MSc. S.S. (Seyed Schwan)
University of Twente Science and Technology
Meander, ME 322
P.O. Box 217 7500 AE Enschede
The Netherlands
Phone +31 534892869
Email: s.s.hosse...@utwente.n
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Re: [R] plotting with varying dot sizes
I think I located the problem. the data frame associated with my spatial object (Insitu_sp) contains some NAs. And in this case, the function plot does not like NAs. It simply behaved strangely. The code below did the trick : idx = is.na(Insitu[,SPM]) ptsize = 0.35*(sqrt(Insitu[!idx,SPM])+1) plot(Insitu_sp[!idx,SPM],col=red,pch=19,cex=ptsize) Does the symbols function supports spatial (sp) objects? I could not manage to make it work with my sp data. Quoting Greg Snow greg.s...@imail.org: The symbols function may work better than plot for this situation. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Literature analysis
On Fri, Dec 11, 2009 at 3:04 PM, Schwan s.s.hosse...@utwente.nl wrote: Thanks, but how should I put the citation inside a data frame? data.frame(first txt file, second txt file...) plot (what should I insert here) type=p And how should I load the txt files anyway inside the frame? Can you give an example of a couple of text files? Are they in a standardised format (i.e. bibTEX or similar)? /Gustaf -- Gustaf Rydevik, M.Sci. tel: +46(0)703 051 451 address:Essingetorget 40,112 66 Stockholm, SE skype:gustaf_rydevik [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] foreach and deparse(substitute(x))
Hi,
I would like to get the actual object name passed as a parameter of a
function and am using deparse(substitute(x)) to do that. It doesn't work
when it is used along with the foreach package. Appreciate if any one can
give some suggestions on how to make it work with foreach.
FUN.aaa - function() {
}
ff = function( FUN){
foreach(i = 1:3) %dopar% {
print(sprintf(deparse(substitute(FUN)) = %s,
deparse(substitute(FUN
return(NULL)
}
}
ff(FUN.aaa)
# will print
ff(FUN.aaa)
[1] deparse(substitute(FUN)) = FUN
[1] deparse(substitute(FUN)) = FUN
[1] deparse(substitute(FUN)) = FUN
[[1]]
NULL
[[2]]
NULL
[[3]]
NULL
If there is no foreach, it works
FUN.aaa - function() {
}
ff = function( FUN){
print(sprintf(deparse(substitute(FUN)) = %s,
deparse(substitute(FUN
}
}
ff(FUN.aaa)
#It works and prints
ff(FUN.aaa)
ff(FUN.aaa)
[1] deparse(substitute(FUN)) = FUN.aaa
thanks
Jeff
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Re: [R] Literature analysis
On Fri, Dec 11, 2009 at 2:04 PM, Schwan s.s.hosse...@utwente.nl wrote: Thanks, but how should I put the citation inside a data frame? data.frame(first txt file, second txt file...) plot (what should I insert here) type=p And how should I load the txt files anyway inside the frame? Check out the resources I recommended in my first reply. These are aspects of using R that are fundamental and completely independent of what analysis you want to do. Neil -- The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey (1986), Sunset salvo. The American Statistician 40(1). Email - nsheph...@gmail.com Website - http://slack.ser.man.ac.uk/ Photos - http://www.flickr.com/photos/slackline/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] non-linear regression
Katharine,
The problem of estimation of parameters in R is that you have to know the
value of the initial estimates very accurately, otherwise it does not
converge.
The example below could be resolved in Excel, however in does not converge.
How to solve the problem?
I made the chart on a logarithmic scale to better visualize the differences.
Send the data file attached.
The commands are below:
tx.br - read.table('c:/tx.br.H.txt',header=F,dec=',')
tx.br -tx.br[,1]
id-1:100
qx.suav - function(id,A,B,C,D,E,F,G,H,K)
(A^((id+B)^C)+(D*exp(-E*(log(id)-log(F))^2))+(G*H^id)/(1+(K*G*H^id)))
HP - nls(tx.br~qx.suav(id,A,B,C,D,E,F,G,H,K),
data=data.frame(id=id,tx.br=tx.br),
trace=TRUE,nls.control(maxiter=5,warnOnly=TRUE,minFactor =
0.1),
algorithm='port',
start=list(A=0.000644,B=0.016761290,C=0.10927095582,D=0.00094877,
E=5.949082737,F=24.526811,G=0.46733960,H=1.0970550987,K=0.771722501657),
lower=list(A=0,B=0,C=0,D=0,E=0,F=0,G=0,H=0,K=0))
HP
matplot(cbind(log(fitted(HP)), log(tx.br)),type=l)
- Original Message -
From: Katharine Mullen k...@few.vu.nl
To: AneSR citb...@terra.com.br
Cc: r-help@r-project.org
Sent: Thursday, December 10, 2009 9:55 PM
Subject: Re: [R] non-linear regression
You did not provide the data or a way of generating it.
I would guess that Excel finds the same solution (the same residual sum-of
squares) as nls but that it uses a weak convergence criterion and/or does
not give you information regarding why it terminates.
Regarding the step size: you can set the minimum step size factor via the
minFactor argument of control.
qx.suav - function(id,A,B,C,D,E,F,G,H,K)
(A^((id+B)^C)+(D*exp(-E*(log(id)-log(F))^2))+(G*H^id)/(1+(K*G*H^id)))
## make noisy data from model
id - 1:1000
tx.br - qx.suav(id,A=0.0006347,B=0.0453814,C=0.1353538,D=0.1353538,
E=0.0002127,F=38.5448420,G=0.115,H=1.1109286,
K=0.382070638)
set.seed(1)
tx.br - tx.br + rnorm(length(tx.br),0,.0001)
HP - nls(tx.br~qx.suav(id,A,B,C,D,E,F,G,H,K),
data=data.frame(id=id,tx.br=tx.br),
trace=TRUE,nls.control(maxiter=5000,warnOnly=TRUE),
algorithm='port',
start=list(A=0.0006347,B=0.0453814,C=0.1353538,D=0.1353538,
E=0.0002127,F=38.5448420,G=0.115,H=1.1109286,
K=0.382070638),
lower=list(A=0,B=0,C=0,D=0,E=0,F=0,G=0,H=0,K=0))
matplot(cbind(fitted(HP), tx.br),type=l)
On Thu, 10 Dec 2009, AneSR wrote:
I have a non-linear regression with 8 parameters to solve however it
does not converge ... easily solves the excel ... including the initial
estimates used in the R were found in the excel ... Another question is
how
to establish the increments of R by the parameters in the search ..
qx.suav-function(id,A,B,C,D,E,F,G,H,K){(A^((id+B)^C)+(D*exp(-E*(log(id)-log(F))^2))+(G*H^id)/(1+(K*G*H^id)))}
HP-nls(tx.br~qx.suav(id,A,B,C,D,E,F,G,H,K),data=data.frame(id=id,tx.br=tx.br),
trace=TRUE,nls.control(maxiter=5000),algorithm='port',start=list(A=0.0006347,B=0.0453814,C=0.1353538,D=0.1353538,E=0.0002127,F=38.5448420,G=0.115,H=1.1109286,K=0.382070638),lower=list(A=0,B=0,C=0,D=0,E=0,F=0,G=0,H=0,K=0))
summary(HP)
How to solve this problem in R?
Ane
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0,007637
0,008968
0,010612
0,010907
0,011986
0,013174
0,014316
0,015557
0,017824
0,018740
0,020338
0,022034
0,022339
0,025929
0,030967
0,033520
0,039556
0,041168
0,044811
0,049932
0,047401
0,060318
0,063559
0,064084
0,077614
0,080425
0,094073
0,102383
0,116288
0,112727
0,114192
0,137330
0,159065
0,138378
0,135844
0,148244
0,165927
0,180803
0,217185
0,241570
0,249358
0,309202
0,269479
0,325142
0,461620
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Re: [R] How to draw three line on the same picture ?
It works ! thank you very much.
Moreover, How about reading data from file ? The file is formatted as:
No V1 V2 V3
1 0.23 0.12 0.89
2 0.11 0.56 0.12
..
jholtman wrote:
try this:
x - read.table(textConnection(No V1 V2 V3
1 0.23 0.12 0.89
2 0.11 0.56 0.12), header=TRUE)
matplot(x[,1], x[,-1], type='l')
On Fri, Dec 11, 2009 at 3:39 AM, z_axis z_a...@163.com wrote:
thanks for your answer ! Would you mind giving me an example using my
data
?
Sincerely!
Patrick Connolly-4 wrote:
On Thu, 10-Dec-2009 at 10:14PM -0800, z_axis wrote:
|
| The following is sampling data:
| No V1 V2 V3
| 1 0.23 0.12 0.89
| 2 0.11 0;56 0.12
| ...
|
| I just want to draw three lines on same picture according to value
of
V1, V2
| and V3.
?lines
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_ Average minds discuss events
(:_~*~_:) Small minds discuss people
(_)-(_). Eleanor Roosevelt
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
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Re: [R] Subset sum problem.
Hans, you're my personal hero today !
The function seems to work fine for the tests I did already.
Thank you very much !
Geert
On Thursday 10 December 2009, Hans W Borchers wrote:
Geert Janssens janssens-geert at telenet.be writes:
On Wednesday 9 December 2009, Hans W Borchers wrote:
Geert Janssens janssens-geert at telenet.be writes:
[ ... ]
Has anybody tackled this issue before in R ? If so, I would be very
grateful if you could share your solution with me.
Is it really true that you only want to see a Yes or No answer to
this question whether a subset sums up to s --- without learning which
numbers this subset is composed of (the pure SUBSET SUM problem)?
Then the following procedure does that in a reasonable amount of time
(returning 'TRUE' or 'FALSE' instead of Y-or-N):
Unfortunately no. I do need the numbers in the subset. But thank you for
presenting this code.
Geert
Okay then, here we go. But don't tell later that your requirement was to
generate _all_ subsets that add up to a certain amount. I will generate
only one (with largest elements).
For simplicity I assume that the set is prepared s.t. it is decreasingly
ordered, has no elements larger than the amount given, and has a total sum
larger than this amount.
# Assume S decreasing, no elements t, total sum = t
solveSubsetSum - function(S, t) {
L - c(0)
inds - NULL
for (i in 1:length(S)) {
# L - unique(sort(c(L, L + S[i])))
L - c(L, L+S[i])
L - L[L = t]
if (max(L) == t) {
inds - c(i)
t - t - S[i]
while (t 0) {
K - c(0)
for (j in 1:(i-1)) {
K - c(K, K+S[j])
if (t %in% K) break
}
inds - c(inds, j)
t - t - S[j]
}
break
}
}
return(inds)
}
# former example
amount - 4748652
products -
c(30500,30500,30500,30500,42000,42000,42000,42000,
42000,42000,42000,42000,42000,42000,71040,90900,
76950,35100,71190,53730,456000,70740,70740,533600,
83800,59500,27465,28000,28000,28000,28000,28000,
26140,49600,77000,123289,27000,27000,27000,27000,
27000,27000,8,33000,33000,55000,77382,48048,
51186,4,35000,21716,63051,15025,15025,15025,
15025,80,111,59700,25908,829350,1198000,1031655)
# prepare set
prods - products[products = amount] # no elements amount
prods - sort(prods, decreasing=TRUE) # decreasing order
# now find one solution
system.time(is - solveSubsetSum(prods, amount))
# user system elapsed
# 0.320 0.032 0.359
prods[is]
# [1] 70740 70740 71190 76950 77382 8 83800
# [8] 90900 456000 533600 829350 111 1198000
sum(prods[is]) == amount
# [1] TRUE
Note that running times and memory needs will be much higher when more
summands are needed. To mention that too: I have not tested the code
extensively.
Regards
Hans Werner
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[R] cluster size
hi r-help, i am doing kmeans clustering in stats. i tried for five clusters clustering using: kcl1 - kmeans(as1[,c(contlife,somlife,agglife,sexlife, rellife,hordlife,doutlife,symtlife,washlife, chcklife,rptlife,countlife,coltlife,ordlife)], 5, iter.max = 10, nstart = 1, algorithm = Hartigan-Wong) table(kcl1$cluster) every time i am getting five clusters of different sizes like first time with cluster sizes table(kcl1$cluster) 1 2 3 4 5 140 72 105 98 112 second time with cluster sizes table(kcl1$cluster) 1 2 3 4 5 91 149 106 76 105 and so on. I wish to know that whether there is any function to get same sizes of clusters everytime when we do kmeans clustering. Thanks in advance. regards, Ms.Karunambigai M PhD Scholar Dept. of Biostatistics NIMHANS Bangalore India The INTERNET now has a personality. YOURS! See your Yahoo! Homepage. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Literature analysis
I think I my original response was bad. And I also realize that I
don't really understand what you want to do.
Here is what I was thinking:
Format your bibliography as CSV (you can convert common bibliography
formats to .csv using Tellico or similar software). Then read the
references into a data frame:
Refs - read.csv(textConnection('Title,Entry Type,Author,Bibtex
Key,Book
Title,Editor,Organization,Publisher,Address,Edition,Pages,Year,ISBN#,Journal,DOI,Month,Number,How
Published,Chapter,Series,Volume,Cross-Reference,Keywords,URL,Abstract,Notes,ID,Date
Created,Date Modified
Unconscious authorship ascription: The effects of success and
effect-specific information priming on experienced
authorship,article,Aarts, H.,Aarts2007,,,119–126,2007,,Journal
of Experimental Social Psychology,,,43,,Lexical Decision;
unconscious; Perception; presentation of words; effect-specific
information; experienced authorship; authorship ascription; priming;
lexical decisions; subliminal perception; Self
Perception,http://localhost/refbase/Papers/Aarts/Unconscious%20authorship%20ascription:%20The%20effects%20of%20success.pdf,Abstract
goes here,exported from refbase
(http://ista.scp.rochester.edu/refbase/show.php?record=14), last
updated on Tue, 17 Nov 2009 10:23:27 -0500,0,2009-12-11,2009-12-11
Goal contagion: Perceiving is for pursuing,article,Aarts, H.;
Gollwitzer, P.; Hassin, R.,Aarts_etal2004,,,23–37,2004,,Journal
of Personality and Social Psychology,,,87,,Contagion; goal
contagion; automatic goal adion; Motivation; Perception; implied
behavioral goals; goal perception; Theories; Automatism; behavioral
information; goal directedness; Attention; Stimulus Parameters; Goals;
goal
characteristics,http://localhost/refbase/Papers/Aarts/Goal%20contagion:%20Perceiving%20is%20for%20pursuing.pdf,Abstract
goes here.,exported from refbase
(http://ista.scp.rochester.edu/refbase/show.php?record=15), last
updated on Tue, 17 Nov 2009 10:23:27 -0500,1,2009-12-11,2009-12-11'))
closeAllConnections()
(sorry, I know that's going to get all screwed up by word wrapping).
Next split out the keywords into separate columns, and then combine
them again into a value column:
library(ggplot2)
Refs - as.data.frame(cbind(Refs, colsplit(Refs$Keywords, split=; ,
names=Keyword)))
Refs - melt(Refs, measure.vars=31:45)
And then that's were I get stuck. I have authors and keywords for each
publication, but I'm not sure how you want this represented.
-Ista
On Fri, Dec 11, 2009 at 9:04 AM, Schwan s.s.hosse...@utwente.nl wrote:
Thanks, but how should I put the citation inside a data frame?
data.frame(first txt file, second txt file...)
plot (what should I insert here) type=p
And how should I load the txt files anyway inside the frame?
On Fri, 2009-12-11 at 08:37 -0500, Ista Zahn wrote:
It sounds pretty simple so far. Just put the citation info in a data
frame, and plot it. I would use ggplot2 for plotting, but it could be
done in base or lattice too.
On Fri, Dec 11, 2009 at 8:04 AM, Schwan s.s.hosse...@utwente.nl wrote:
Ok good question I havent explain!
Well,lets keep it simple for the begining. By analyzing the
keywords(which can also include the authors name) I mean,
1) investigate in which paper the keyword occur and how often
2) investigate if keywords occur together
X axis can show the paper title (or authors name)
y axis shows the keywords
if a keyword occurs more often in a paper then the dot on the graph
changes the color
if keywords occur together then a symbol the same symbol appears.
On Fri, 2009-12-11 at 04:33 -0800, nshephard wrote:
Schwan wrote:
Thanks for all the comments,
and sorry about the unstructured question! I am trying to:
1: analyze keywords, names from Authors and year of publication from
citations (with abstracts) i downloaded fron various sites(these
downloads can be converted into .txt files as well)
Sorry to bang the drum but what do you mean by analyse, is it to simply
count keywords and rank them, investigate which keywords occur together,
look at trends in keywords by year or something else completely.
Its a very general term (at least for me as I've no knowledge or
experience
of text mining)!
Neil
--
---
Hosseiny, MSc. S.S. (Seyed Schwan)
University of Twente Science and Technology
Meander, ME 322
P.O. Box 217 7500 AE Enschede
The Netherlands
Phone +31 534892869
Email: s.s.hosse...@utwente.n
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P.O.
[R] memory problem on Suse
Dear all, I am meeting some problems with memory allocation. I know it is an old issue, I'm sorry. I looked for a solution in the FAQs and manuals, mails, but without finding the working answer. I really hope you can help me. For instance, if I try to read micorarray data I get: mab=ReadAffy(cdfname=hgu133plus2cdf) Error: cannot allocate vector of size 858.0 Mb I get similar errors with smaller objects, smaller data sets or other procedures (Error: cannot allocate vector of size 123.0 Mb). I'm running R with Suse 11.1 Linux OS, on two Xeon processors (8 cores), 32 GB RAM. I suppose I have enough resources to manage these objects and data files Any suggestions or hints will be really appreciated! Many thanks in advance. Alessandro __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Array of legend text with math symbols from predefined variables
Hello, I am trying to include legend text with math symbols from a predefined character variable that is read in from a file. If there is only one line of text in the legend, the following, although cumbersome, works for me: LegendText = 'U' [infinity], '=10 m/s' # (read in from a file) LegendName = paste(bquote(paste(,LegendText, ))) plot( c(1,2,3), c(1,2,3) ) legend(‘topleft’,1, eval(parse(text=LegendName)) ) If I now have more than one line in a plot and hence want to include more than one line of legendtext, I run into problems because legend(‘topleft’,1, c( eval(parse(text=LegendName1)),eval(parse(text=LegendName2)) ) will not result in evaluation of the character string but just a paste of the string and legend(‘topleft’,1, eval(parse(text=c(LegendName1, LegendName2))) ) will only evaluate the first element of the array. Am I trying to do the impossible or is my approach totally wrong? Any help would be greatly appreciated! Koen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to draw three line on the same picture ?
%cat test.csv
No,HS,ZangF,ZenF
1,0.25,0.5,0.57
2,0.10,0.23,0.12
3,0.20,0.25,0.1
..
d - read.csv(000301.txt)
matplot(d[,1], d[,-1], type='l')
It works. However, i hope the legend is displayed.
Sincerely!
z_axis wrote:
It works ! thank you very much.
Moreover, How about reading data from file ? The file is formatted as:
No V1 V2 V3
1 0.23 0.12 0.89
2 0.11 0.56 0.12
..
jholtman wrote:
try this:
x - read.table(textConnection(No V1 V2 V3
1 0.23 0.12 0.89
2 0.11 0.56 0.12), header=TRUE)
matplot(x[,1], x[,-1], type='l')
On Fri, Dec 11, 2009 at 3:39 AM, z_axis z_a...@163.com wrote:
thanks for your answer ! Would you mind giving me an example using my
data
?
Sincerely!
Patrick Connolly-4 wrote:
On Thu, 10-Dec-2009 at 10:14PM -0800, z_axis wrote:
|
| The following is sampling data:
| No V1 V2 V3
| 1 0.23 0.12 0.89
| 2 0.11 0;56 0.12
| ...
|
| I just want to draw three lines on same picture according to value
of
V1, V2
| and V3.
?lines
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_ Average minds discuss events
(:_~*~_:) Small minds discuss people
(_)-(_). Eleanor Roosevelt
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
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--
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Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
[[alternative HTML version deleted]]
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[R] Problem with calibrate function
Hi, I´m trying to use the calibrate function from rms package (made by prof. Harrell) after fitting a model using cph. But it returns the following error message: calibrate(modelo1,B=200,bw=F,u=13) Using Cox survival estimates at 13 Days Convergence problems stopping addition Error in hare(S[, 1], S[, 2], fun(est.surv), maxdim = maxdim, ...) : no convergence this is serious! Has anybody experienced the same problem? Can anyone help me? many thanks, Rodrigo -- View this message in context: http://n4.nabble.com/Problem-with-calibrate-function-tp961100p961100.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] question about centroid-linkage (cluster analysis) (2)
Dear R community,
just in case some haven't noticed my previous email.
I realize hclust relies on a Fortran routine, but I hoped
some of you might exactly know how that Y.hc_c$height
is computed. And, thus, explain the anomaly I found.
Thank you.
J
Dr James Foadi PhD
Membrane Protein Laboratory (MPL)
Diamond Light Source Ltd
Diamond House
Harewell Science and Innovation Campus
Chilton, Didcot
Oxfordshire OX11 0DE
Email: james.fo...@diamond.ac.uk
Alt Email: j.fo...@imperial.ac.uk
-Original Message-
From: r-help-boun...@r-project.org on behalf of james.fo...@diamond.ac.uk
Sent: Thu 10/12/2009 13:26
To: r-help@r-project.org
Subject: [R] question about centroid-linkage (cluster analysis)
Dear R community,
I would be greatful if somebody could shed light on the following.
I have created a set of 6 points to check how centroid
agglomeration works in cluster analysis:
Y - data.frame(x=c(-1,1,1,-1,10,12),y=c(1,1,-1,-1,0,0))
It is quite intuitive to understand that the last clusters to be joined will be
{1,2,3,4} with {5,6}. Now, the centroid for the first cluster has coordinates
(0,0),
while the centroid for the second cluster has coordinates (11,0). Therefore, the
distance between these two cluster should be 11. But:
Y.dist - dist(Y)
Y.hc_c - hclust(Y.dist,method=centroid)
Y.hc_c$merge
[,1] [,2]
[1,] -1 -2
[2,] -31
[3,] -42
[4,] -5 -6
[5,]34
Y.hc_c$height
[1] 2.00 1.914214 1.517428 2.00 9.692575
So, from this it would appear that the distance between the last two clusters
is 9.692575!
How can it be?
J
Dr James Foadi PhD
Membrane Protein Laboratory (MPL)
Diamond Light Source Ltd
Diamond House
Harewell Science and Innovation Campus
Chilton, Didcot
Oxfordshire OX11 0DE
Email: james.fo...@diamond.ac.uk
Alt Email: j.fo...@imperial.ac.uk
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[R] shared axes in multipanel plot
Hello
I've created a function to make a plot with multiple pannels from columns
of data that are created in a previous function. In the example below the
number of columns is 8, giving 4 pannels, but in general it takes data
with any number of columns and figures out a nice layout.
The panels all have the same axes, and so I wonder what functions are
avialable to create axes only on the left and bottom of the whole plot
rather than each pannel.
I'd really like a generic way to do this for any number of plots, but was
even having trouble figuring out how to do it manually for this example;
How are pannels referred to, in a layout context?
That is, how do I say,
if(current.pannel==4) {do stuff}
Here's a simple version of the code.
baseline - (1:20)/20#example data
dat1 - matrix(baseline,20,8)
dat - dat1+matrix(rnorm(20*8)/30, 20,8)
nstrat - ncol(dat)
rows - ceiling(nstrat/4)
layout(matrix(1:(rows*2), rows, 2, T))
par(oma=c(4,4,3,1))
par(mar=c(1,1,0,1))
for(i in which(1:nstrat%%2!=0)){
plot(baseline, type=l, col=grey, lwd=2,
xlab=, ylab=, ylim=c(0,1), xaxt='n', yaxt='n')
axis(1, labels=F); axis(2, labels=F)
points(dat[,i], type=l, lty=2)
points(dat[,i+1], type=l, lty=2)
}
Thank you muchly
Jennifer Young
PS: I am a subscriber, but can't for the life of me figure out how to send
an email while logged in so that the moderators don't have to take the
time to read it over. I always get the please wait while we check it
over email. Likely I'm being dumb.
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Re: [R] incorrect multiple outputs
Ista,
Here is the full code:
file-read.table(file=input_file.txt,sep=\t)[,c(1,3:5)]
file.rows- c(nrow(file)/288)
full.array - array(0,dim=c(8,file.rows,12))
cellnames - rep(A,file.rows)
barcodes - rep(A,file.rows)
t - rep(A,file.rows)
t11 - rep(A,file.rows)
cv - rep(A,file.rows)
for (k in 1:file.rows){
plate- file[(((k-1)*288)+1),1]
barcodes[k] - levels(plate)[plate]
l - k*288
a - l-287
x-file[a:l,]
x.array-matrix(0,ncol=12,nrow=8)
n.array-matrix(0,ncol=12,nrow=8)
x.blank2 - matrix(0,ncol=3,nrow=24)
x.nodrug2 - matrix(0,ncol=6,nrow=24)
value - data.frame(conc=0,fluor=2)
# reformat (mean) input data into 96 well format
q-1
for (m in 1:8){ for (n in 1:12){ p- 3*q; x.array[m,n] -
mean(as.matrix(x[(p-2):p,2:4])); q - q+1; if (n == 1){z-3*m;
x.blank2[(z-2):z,]-as.matrix(x[(p-2):p,2:4])}; if
(n==11){z-3*m;x.nodrug2[(z-2):z,1:3]-as.matrix(x[(p-2):p,2:4])}; if
(n==12){z-3*m; x.nodrug2[(z-2):z,4:6]-as.matrix(x[(p-2):p,2:4])}}}
x.blank-c(x.blank2[,1],x.blank2[,3],x.blank2[,2])
x.nodrug
-c(x.nodrug2[,1],x.nodrug2[,2],x.nodrug2[,3],x.nodrug2[,4],x.nodrug2[,5],x.nodrug2[,6])
x.nodrug11 -c(x.nodrug2[,1],x.nodrug2[,2],x.nodrug2[,3])
# calculate output values
cv[k] - 100*(sd(x.blank)/mean(x.blank))
t[k] -
(mean(x.nodrug)-mean(x.blank))/sqrt(((sd(x.nodrug)^2)/8)+((sd(x.blank)^2)/16))
t11[k] -
(sqrt(8)*(mean(x.nodrug11)-mean(x.blank)))/sqrt(sd(x.nodrug11)^2+sd(x.blank)^2)
}
all.data
-data.frame(barcodes,t=format(as.numeric(t),digits=3),t11=format(as.numeric(t11),digits=3),cv=format(as.numeric(cv),digits=3))
#write(format(t,digits=6), file= R_drug_plot.log,append=TRUE,sep=\t)
write.table(all.data, file=
R_drug_plot.log,append=TRUE,sep=\t,row.names=FALSE)
When I first tried the mail list wouldn't let me send it as it contains the
word drug too many times! so I only included the problematic parts.
Rich
2009/12/11 Ista Zahn istaz...@gmail.com
The lack of a reproducible example is a bigger problem than any lack
of clarity of the explanation.
-Ista
On Fri, Dec 11, 2009 at 2:08 AM, bawa...@googlemail.com wrote:
Apologies,
I didn't explain this clearly. The Rscript is called by a perl script,
which creates input_file.txt by inserting 288 lines of (reformatted) data
for each data file in the directory. So the Rscript will (and is doing) run
the loop a number of times equal to the number of files the perl script read
in. The problem is that it should only create the data.frame and write to
the file after the last iteration, but it's (creating and)writing the
complete data.frame every iteration.
Sent using BlackBerry® from Orange
-Original Message-
From: jim holtman jholt...@gmail.com
Date: Thu, 10 Dec 2009 18:00:54
To: biscuitbawa...@googlemail.com
Cc: r-help@r-project.org
Subject: Re: [R] incorrect multiple outputs
If I rad you code right, file.rows is equal to 1 and your 'for' loop will
only iterate once. Is that what you were expecting?
No reproducible code provided, so that is my best guess.
file.rows- c(nrow(file)/288) # input_file.txt contains 288
reformatted
lines for each original data file
...
for (k in 1:file.rows){ # iterates code for each 288 line block of
input_file.txt
...
On Thu, Dec 10, 2009 at 11:39 AM, biscuit bawa...@googlemail.com
wrote:
HI,
I'm having trouble with a piece of Rscript which keeps outputting
incorrectly. it's something like this: the code reads in from a file
which
contains (reformated) input
file-read.table(file=input_file.txt,sep=\t)[,c(1,3:5)]
file.rows- c(nrow(file)/288) # input_file.txt contains 288
reformatted
lines for each original data file
...
for (k in 1:file.rows){ # iterates code for each 288 line block of
input_file.txt
...
cv[k] - 100*(sd(x.blank)/mean(x.blank))
t[k] -
(mean(x.note)-mean(x.blank))/sqrt(((sd(x.note)^2)/8)+((sd(x.blank)^2)/16))
t11[k] -
(sqrt(8)*(mean(x.note11)-mean(x.blank)))/sqrt(sd(x.note11)^2+sd(x.blank)^2)
}
all.data-data.frame(barcodes,t=format(as.numeric(t),digits=3),t11=format(as.numeric(t11),digits=3),cv=format(as.numeric(cv),digits=3))
write.table(all.data, file=
R_drug_plot.log,append=TRUE,sep=\t,row.names=FALSE)
this all works correctly except that I believed it would output to file
after completing the loop, instead it's writing to file every iteration.
so
the output file looks like:
headers
a1
headers
a1
a2
headers
a1
a2
a3
...
I have checked the missing sections of code and can confirm there are no
missing/additional brackets. Has anyone any idea why this is happening
and
what I can do about it?
--
View this message in context:
http://n4.nabble.com/incorrect-multiple-outputs-tp957192p957192.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide
Re: [R] Please help with a basic function
Hi Mark,
You're on the right track. You just need your function to return dataframe. Try
convert-function(d) {
d-data.frame(d); #convert object to dataframe
d-subset(d,select=c(time,coords.x1,coords.x2)) #select some columns
return(d)
}
-Ista
On Fri, Dec 11, 2009 at 9:19 AM, Mark Na mtb...@gmail.com wrote:
Hello,
I am learning how to use functions, but I'm running into a roadblock.
I would like my function to do two things: 1) convert an object to a
dataframe, 2) and then subset the dataframe. Both of these commands work
fine outside the function, but I would like to wrap them in a function so I
can apply the code iteratively to many such objects.
Here's what I wrote, but it doesn't work:
convert-function(d) {
d-data.frame(d); #convert object to dataframe
d-subset(d,select=c(time,coords.x1,coords.x2)) #select some columns
}
convert(data) #the problem is that data is the same as it was before
running the function
The objects being processed through my function are SpatialPointsDataFrames
but I'm quite sure that's not my problem, as I can process these outside of
the function (using the above code) ... it's when I try to wrap the code in
a function that it doesn't work.
Thanks, Mark
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] incorrect multiple outputs
Sorry, the nasty for (min... loop is probably easier to read like this:
for (m in 1:8){
for (n in 1:12){
p- 3*q; x.array[m,n] - mean(as.matrix(x[(p-2):p,2:4])); q - q+1;
if (n == 1){
z-3*m; x.blank2[(z-2):z,]-as.matrix(x[(p-2):p,2:4])
}; if (n==11){
z-3*m; x.nodrug2[(z-2):z,1:3]-as.matrix(x[(p-2):p,2:4])
}; if (n==12){
z-3*m; x.nodrug2[(z-2):z,4:6]-as.matrix(x[(p-2):p,2:4])
}
}
}
Richard
2009/12/11 Richard Thompson bawa...@googlemail.com
Ista,
Here is the full code:
file-read.table(file=input_file.txt,sep=\t)[,c(1,3:5)]
file.rows- c(nrow(file)/288)
full.array - array(0,dim=c(8,file.rows,12))
cellnames - rep(A,file.rows)
barcodes - rep(A,file.rows)
t - rep(A,file.rows)
t11 - rep(A,file.rows)
cv - rep(A,file.rows)
for (k in 1:file.rows){
plate- file[(((k-1)*288)+1),1]
barcodes[k] - levels(plate)[plate]
l - k*288
a - l-287
x-file[a:l,]
x.array-matrix(0,ncol=12,nrow=8)
n.array-matrix(0,ncol=12,nrow=8)
x.blank2 - matrix(0,ncol=3,nrow=24)
x.nodrug2 - matrix(0,ncol=6,nrow=24)
value - data.frame(conc=0,fluor=2)
# reformat (mean) input data into 96 well format
q-1
for (m in 1:8){ for (n in 1:12){ p- 3*q; x.array[m,n] -
mean(as.matrix(x[(p-2):p,2:4])); q - q+1; if (n == 1){z-3*m;
x.blank2[(z-2):z,]-as.matrix(x[(p-2):p,2:4])}; if
(n==11){z-3*m;x.nodrug2[(z-2):z,1:3]-as.matrix(x[(p-2):p,2:4])}; if
(n==12){z-3*m; x.nodrug2[(z-2):z,4:6]-as.matrix(x[(p-2):p,2:4])}}}
x.blank-c(x.blank2[,1],x.blank2[,3],x.blank2[,2])
x.nodrug
-c(x.nodrug2[,1],x.nodrug2[,2],x.nodrug2[,3],x.nodrug2[,4],x.nodrug2[,5],x.nodrug2[,6])
x.nodrug11 -c(x.nodrug2[,1],x.nodrug2[,2],x.nodrug2[,3])
# calculate output values
cv[k] - 100*(sd(x.blank)/mean(x.blank))
t[k] -
(mean(x.nodrug)-mean(x.blank))/sqrt(((sd(x.nodrug)^2)/8)+((sd(x.blank)^2)/16))
t11[k] -
(sqrt(8)*(mean(x.nodrug11)-mean(x.blank)))/sqrt(sd(x.nodrug11)^2+sd(x.blank)^2)
}
all.data
-data.frame(barcodes,t=format(as.numeric(t),digits=3),t11=format(as.numeric(t11),digits=3),cv=format(as.numeric(cv),digits=3))
#write(format(t,digits=6), file= R_drug_plot.log,append=TRUE,sep=\t)
write.table(all.data, file=
R_drug_plot.log,append=TRUE,sep=\t,row.names=FALSE)
When I first tried the mail list wouldn't let me send it as it contains the
word drug too many times! so I only included the problematic parts.
Rich
2009/12/11 Ista Zahn istaz...@gmail.com
The lack of a reproducible example is a bigger problem than any lack
of clarity of the explanation.
-Ista
On Fri, Dec 11, 2009 at 2:08 AM, bawa...@googlemail.com wrote:
Apologies,
I didn't explain this clearly. The Rscript is called by a perl script,
which creates input_file.txt by inserting 288 lines of (reformatted) data
for each data file in the directory. So the Rscript will (and is doing) run
the loop a number of times equal to the number of files the perl script read
in. The problem is that it should only create the data.frame and write to
the file after the last iteration, but it's (creating and)writing the
complete data.frame every iteration.
Sent using BlackBerry® from Orange
-Original Message-
From: jim holtman jholt...@gmail.com
Date: Thu, 10 Dec 2009 18:00:54
To: biscuitbawa...@googlemail.com
Cc: r-help@r-project.org
Subject: Re: [R] incorrect multiple outputs
If I rad you code right, file.rows is equal to 1 and your 'for' loop
will
only iterate once. Is that what you were expecting?
No reproducible code provided, so that is my best guess.
file.rows- c(nrow(file)/288) # input_file.txt contains 288
reformatted
lines for each original data file
...
for (k in 1:file.rows){ # iterates code for each 288 line block of
input_file.txt
...
On Thu, Dec 10, 2009 at 11:39 AM, biscuit bawa...@googlemail.com
wrote:
HI,
I'm having trouble with a piece of Rscript which keeps outputting
incorrectly. it's something like this: the code reads in from a file
which
contains (reformated) input
file-read.table(file=input_file.txt,sep=\t)[,c(1,3:5)]
file.rows- c(nrow(file)/288) # input_file.txt contains 288
reformatted
lines for each original data file
...
for (k in 1:file.rows){ # iterates code for each 288 line block of
input_file.txt
...
cv[k] - 100*(sd(x.blank)/mean(x.blank))
t[k] -
(mean(x.note)-mean(x.blank))/sqrt(((sd(x.note)^2)/8)+((sd(x.blank)^2)/16))
t11[k] -
(sqrt(8)*(mean(x.note11)-mean(x.blank)))/sqrt(sd(x.note11)^2+sd(x.blank)^2)
}
all.data-data.frame(barcodes,t=format(as.numeric(t),digits=3),t11=format(as.numeric(t11),digits=3),cv=format(as.numeric(cv),digits=3))
write.table(all.data, file=
R_drug_plot.log,append=TRUE,sep=\t,row.names=FALSE)
this all works correctly except that I believed it would output to file
after completing the loop, instead it's writing to file every
iteration. so
the output file looks like:
headers
a1
headers
a1
a2
headers
a1
[R] ggplot: Problem with legend background
Dear R-users, I am preparing graphs for an upcoming article using the different functions of the ggplot2 package and I've been having problems with the legend background. It doesn't seem to scale when the text size is increased. Here's the mandatory reproducible example: library(ggplot2) repFrame - data.frame(A= 1:10, B= rnorm(1:10), groupNum = rep(c(First group, Second group),each=5)) testPlot - ggplot(repFrame, aes(x=A, y = B, group = groupNum)) + opts(legend.position=c(0.85,0.3), legend.background = theme_rect(fill=white), legend.text=theme_text(size=16), legend.title=theme_text(size=20)) testPlot + geom_point(aes(colour= groupNum)) As you can see, the text doesn't fit in the white rectangle. I suspect there is a theme setting I could modify to fix this, but I can't seem to find which one. I sincerely thank you for your time and assistance. Luc -- View this message in context: http://n4.nabble.com/ggplot-Problem-with-legend-background-tp961142p961142.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] non-linear regression
The problem of estimation of parameters in R is that you have to know the
value of the initial estimates very accurately, otherwise it does not
converge.
This was discussed on R-help in the last 2 weeks; see the thread on
'Starting estimates for nls Exponential Fit'.
The example below could be resolved in Excel, however in does not converge.
How to solve the problem?
Can you consider a model with a different functional form or do you need
to fit this exact function?
If you want to minimize log(data) log(model) RSS, then:
tx.br - read.table('tx.br.H.txt',header=F,dec=',')
tx.br - tx.br[,1]
id - 1:100
qx.suav - function(id,A,B,C,D,E,F,G,H,K)
(A^((id+B)^C)+(D*exp(-E*(log(id)-log(F))^2))+(G*H^id)/(1+(K*G*H^id)))
newD - log(tx.br)
HP - nls(newD~log(qx.suav(id,A,B,C,D,E,F,G,H,K)),
data=data.frame(id=id,newD=newD),
trace=TRUE,
nls.control(maxiter=5,warnOnly=TRUE),
algorithm='port',
start=list(A=0.000644,B=0.016761290,C=0.10927095582,D=0.00094877,
E=5.949082737,F=24.526811,G=0.46733960,H=1.0970550987,
K=0.771722501657),
lower=list(A=0,B=0,C=0,D=0,E=0,F=0,G=0,H=0,K=0))
HP
par(mfrow=c(2,1))
matplot(cbind(fitted(HP), newD),type=l, main=model and fit)
matplot(cbind(exp(fitted(HP)), exp(newD)), type=l,
main=transformed back to original space)
I made the chart on a logarithmic scale to better visualize the differences.
Send the data file attached.
The commands are below:
tx.br - read.table('c:/tx.br.H.txt',header=F,dec=',')
tx.br -tx.br[,1]
id-1:100
qx.suav - function(id,A,B,C,D,E,F,G,H,K)
(A^((id+B)^C)+(D*exp(-E*(log(id)-log(F))^2))+(G*H^id)/(1+(K*G*H^id)))
HP - nls(tx.br~qx.suav(id,A,B,C,D,E,F,G,H,K),
data=data.frame(id=id,tx.br=tx.br),
trace=TRUE,nls.control(maxiter=5,warnOnly=TRUE,minFactor =
0.1),
algorithm='port',
start=list(A=0.000644,B=0.016761290,C=0.10927095582,D=0.00094877,
E=5.949082737,F=24.526811,G=0.46733960,H=1.0970550987,K=0.771722501657),
lower=list(A=0,B=0,C=0,D=0,E=0,F=0,G=0,H=0,K=0))
HP
matplot(cbind(log(fitted(HP)), log(tx.br)),type=l)
- Original Message -
From: Katharine Mullen k...@few.vu.nl
To: AneSR citb...@terra.com.br
Cc: r-help@r-project.org
Sent: Thursday, December 10, 2009 9:55 PM
Subject: Re: [R] non-linear regression
You did not provide the data or a way of generating it.
I would guess that Excel finds the same solution (the same residual sum-of
squares) as nls but that it uses a weak convergence criterion and/or does
not give you information regarding why it terminates.
Regarding the step size: you can set the minimum step size factor via the
minFactor argument of control.
qx.suav - function(id,A,B,C,D,E,F,G,H,K)
(A^((id+B)^C)+(D*exp(-E*(log(id)-log(F))^2))+(G*H^id)/(1+(K*G*H^id)))
## make noisy data from model
id - 1:1000
tx.br - qx.suav(id,A=0.0006347,B=0.0453814,C=0.1353538,D=0.1353538,
E=0.0002127,F=38.5448420,G=0.115,H=1.1109286,
K=0.382070638)
set.seed(1)
tx.br - tx.br + rnorm(length(tx.br),0,.0001)
HP - nls(tx.br~qx.suav(id,A,B,C,D,E,F,G,H,K),
data=data.frame(id=id,tx.br=tx.br),
trace=TRUE,nls.control(maxiter=5000,warnOnly=TRUE),
algorithm='port',
start=list(A=0.0006347,B=0.0453814,C=0.1353538,D=0.1353538,
E=0.0002127,F=38.5448420,G=0.115,H=1.1109286,
K=0.382070638),
lower=list(A=0,B=0,C=0,D=0,E=0,F=0,G=0,H=0,K=0))
matplot(cbind(fitted(HP), tx.br),type=l)
On Thu, 10 Dec 2009, AneSR wrote:
I have a non-linear regression with 8 parameters to solve however it
does not converge ... easily solves the excel ... including the initial
estimates used in the R were found in the excel ... Another question is
how
to establish the increments of R by the parameters in the search ..
qx.suav-function(id,A,B,C,D,E,F,G,H,K){(A^((id+B)^C)+(D*exp(-E*(log(id)-log(F))^2))+(G*H^id)/(1+(K*G*H^id)))}
HP-nls(tx.br~qx.suav(id,A,B,C,D,E,F,G,H,K),data=data.frame(id=id,tx.br=tx.br),
trace=TRUE,nls.control(maxiter=5000),algorithm='port',start=list(A=0.0006347,B=0.0453814,C=0.1353538,D=0.1353538,E=0.0002127,F=38.5448420,G=0.115,H=1.1109286,K=0.382070638),lower=list(A=0,B=0,C=0,D=0,E=0,F=0,G=0,H=0,K=0))
summary(HP)
How to solve this problem in R?
Ane
--
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Literature analysis
OK Thanks again for the help.
I have tried
example - read.csv(/PATH/test.cvs)
If i call for the results R is plotting all the data separated by title,
Keyword, author...)
However, I dont know how to tell R that it just should look for e.g.
author,keywords and year and how to plot these for example on x axis the
author and y axis the keywords and on z axis the year?
On Fri, 2009-12-11 at 10:06 -0500, Ista Zahn wrote:
I think I my original response was bad. And I also realize that I
don't really understand what you want to do.
Here is what I was thinking:
Format your bibliography as CSV (you can convert common bibliography
formats to .csv using Tellico or similar software). Then read the
references into a data frame:
Refs - read.csv(textConnection('Title,Entry Type,Author,Bibtex
Key,Book
Title,Editor,Organization,Publisher,Address,Edition,Pages,Year,ISBN#,Journal,DOI,Month,Number,How
Published,Chapter,Series,Volume,Cross-Reference,Keywords,URL,Abstract,Notes,ID,Date
Created,Date Modified
Unconscious authorship ascription: The effects of success and
effect-specific information priming on experienced
authorship,article,Aarts, H.,Aarts2007,,,119–126,2007,,Journal
of Experimental Social Psychology,,,43,,Lexical Decision;
unconscious; Perception; presentation of words; effect-specific
information; experienced authorship; authorship ascription; priming;
lexical decisions; subliminal perception; Self
Perception,http://localhost/refbase/Papers/Aarts/Unconscious%20authorship%20ascription:%20The%20effects%20of%20success.pdf,Abstract
goes here,exported from refbase
(http://ista.scp.rochester.edu/refbase/show.php?record=14), last
updated on Tue, 17 Nov 2009 10:23:27 -0500,0,2009-12-11,2009-12-11
Goal contagion: Perceiving is for pursuing,article,Aarts, H.;
Gollwitzer, P.; Hassin, R.,Aarts_etal2004,,,23–37,2004,,Journal
of Personality and Social Psychology,,,87,,Contagion; goal
contagion; automatic goal adion; Motivation; Perception; implied
behavioral goals; goal perception; Theories; Automatism; behavioral
information; goal directedness; Attention; Stimulus Parameters; Goals;
goal
characteristics,http://localhost/refbase/Papers/Aarts/Goal%20contagion:%20Perceiving%20is%20for%20pursuing.pdf,Abstract
goes here.,exported from refbase
(http://ista.scp.rochester.edu/refbase/show.php?record=15), last
updated on Tue, 17 Nov 2009 10:23:27 -0500,1,2009-12-11,2009-12-11'))
closeAllConnections()
(sorry, I know that's going to get all screwed up by word wrapping).
Next split out the keywords into separate columns, and then combine
them again into a value column:
library(ggplot2)
Refs - as.data.frame(cbind(Refs, colsplit(Refs$Keywords, split=; ,
names=Keyword)))
Refs - melt(Refs, measure.vars=31:45)
And then that's were I get stuck. I have authors and keywords for each
publication, but I'm not sure how you want this represented.
-Ista
On Fri, Dec 11, 2009 at 9:04 AM, Schwan s.s.hosse...@utwente.nl wrote:
Thanks, but how should I put the citation inside a data frame?
data.frame(first txt file, second txt file...)
plot (what should I insert here) type=p
And how should I load the txt files anyway inside the frame?
On Fri, 2009-12-11 at 08:37 -0500, Ista Zahn wrote:
It sounds pretty simple so far. Just put the citation info in a data
frame, and plot it. I would use ggplot2 for plotting, but it could be
done in base or lattice too.
On Fri, Dec 11, 2009 at 8:04 AM, Schwan s.s.hosse...@utwente.nl wrote:
Ok good question I havent explain!
Well,lets keep it simple for the begining. By analyzing the
keywords(which can also include the authors name) I mean,
1) investigate in which paper the keyword occur and how often
2) investigate if keywords occur together
X axis can show the paper title (or authors name)
y axis shows the keywords
if a keyword occurs more often in a paper then the dot on the graph
changes the color
if keywords occur together then a symbol the same symbol appears.
On Fri, 2009-12-11 at 04:33 -0800, nshephard wrote:
Schwan wrote:
Thanks for all the comments,
and sorry about the unstructured question! I am trying to:
1: analyze keywords, names from Authors and year of publication from
citations (with abstracts) i downloaded fron various sites(these
downloads can be converted into .txt files as well)
Sorry to bang the drum but what do you mean by analyse, is it to
simply
count keywords and rank them, investigate which keywords occur together,
look at trends in keywords by year or something else completely.
Its a very general term (at least for me as I've no knowledge or
experience
of text mining)!
Neil
--
---
Hosseiny, MSc. S.S. (Seyed Schwan)
University of Twente Science and Technology
Re: [R] cluster size
Dear Ms Karunambigai, the kmeans algorithm depends on random initialisation. There are two basic strategies that can be applied in order to make your results reproducible: 1) Fix the random number generator by means of set.seed (see ?set.seed) before you run kmeans. The problem with this is that your solution can only be reproduced using the same random seed; it technically still is random. 2) Specify fixed initial centers, using the centers argument in kmeans. (Sensible initial centers may be obtained by running hclust using Ward's method, obtain the desired number of clusters using cutree and compute the centers of the resulting clusters; sorry that I don't have the time right now to explain how to do that precisely; the help pages and hopefully some understanding of what is going on may help you further.) An alternative strategy that will not absolutely guarantee reproducibility but make your results more stable is to use kmeansruns in library fpc, which is a wrapper that runs kmeans several times and gives you the optimal solution. That should reproduce its outcome with higher probability (though not precisely 1). I don't know whether the default value runs=100 is sufficient to give a stable solution for your data, but increasing the runs parameter may help. Cheers, Christian On Fri, 11 Dec 2009, karuna m wrote: hi r-help, i am doing kmeans clustering in stats. i tried for five clusters clustering using: kcl1 - kmeans(as1[,c(contlife,somlife,agglife,sexlife, rellife,hordlife,doutlife,symtlife,washlife, chcklife,rptlife,countlife,coltlife,ordlife)], 5, iter.max = 10, nstart = 1, algorithm = Hartigan-Wong) table(kcl1$cluster) every time i am getting five clusters of different sizes like first time with cluster sizes table(kcl1$cluster) 1 2 3 4 5 140 72 105 98 112 second time with cluster sizes table(kcl1$cluster) 1 2 3 4 5 91 149 106 76 105 and so on. I wish to know that whether there is any function to get same sizes of clusters everytime when we do kmeans clustering. Thanks in advance. regards, Ms.Karunambigai M PhD Scholar Dept. of Biostatistics NIMHANS Bangalore India The INTERNET now has a personality. YOURS! See your Yahoo! Homepage. [[alternative HTML version deleted]] *** --- *** Christian Hennig University College London, Department of Statistical Science Gower St., London WC1E 6BT, phone +44 207 679 1698 chr...@stats.ucl.ac.uk, www.homepages.ucl.ac.uk/~ucakche__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to creat a matrix
Dear R family I am attempting to create a matrix. e.g., 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 How could I write a R program? Later I want to extend it to a N by N case. Thanks in advance best Moohwan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Please help with a basic function
Hi Mark,
This question would probably be better suited for the r-sig-geo mailing
list. In addition, please read the posting guide and provide a piece of
code that reproduces the problem.
library(sp)
convert-function(d) {
d-data.frame(d); #convert object to dataframe
d-subset(d,select=c(zinc,x,y)) #select some columns
d # - add this, or alternatively 'return(d)'
}
data(meuse)
coordinates(meuse) = ~x+y
convert(meuse)
But maybe better, subsetting a SPDF can be done using:
meuse[zinc] # Remains an SPDF
# Returns a data.frame
data.frame(coordinates(meuse), zinc = meuse$zinc)
And some unrequested advice :). To process multiple files, take a look
at lapply, both for reading and processing.
all_data = lapply(list_of_files, function(file) {
bla = read.table(file)
coordinates(bla) = ~coor.x1 + coor.x2
return(bla)
}
# all data is now a list wit the SPDF's
processed_data = lapply(all_data, function(dat) {
return(data.frame(coordinates(dat), zinc = dat$zinc))
}
ofcourse you can include the latter lapply stuff inside the first
'loading' lapply.
all_data = lapply(list_of_files, function(file) {
bla = read.table(file)
bla = subset(bla, select = select=c(time,coords.x1,coords.x2))
coordinates(bla) = ~coor.x1 + coor.x2
return(bla)
}
hope this helps and good luck,
Paul
Mark Na wrote:
Hello,
I am learning how to use functions, but I'm running into a roadblock.
I would like my function to do two things: 1) convert an object to a
dataframe, 2) and then subset the dataframe. Both of these commands work
fine outside the function, but I would like to wrap them in a function so I
can apply the code iteratively to many such objects.
Here's what I wrote, but it doesn't work:
convert-function(d) {
d-data.frame(d); #convert object to dataframe
d-subset(d,select=c(time,coords.x1,coords.x2)) #select some columns
}
convert(data) #the problem is that data is the same as it was before
running the function
The objects being processed through my function are SpatialPointsDataFrames
but I'm quite sure that's not my problem, as I can process these outside of
the function (using the above code) ... it's when I try to wrap the code in
a function that it doesn't work.
Thanks, Mark
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Phone: +3130 253 5773 Wed-Fri
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Re: [R] memory problem on Suse
On Dec 11, 2009, at 6:24 AM, Ambrosi Alessandro wrote: Dear all, I am meeting some problems with memory allocation. I know it is an old issue, I'm sorry. I looked for a solution in the FAQs and manuals, mails, but without finding the working answer. I really hope you can help me. For instance, if I try to read micorarray data I get: mab=ReadAffy(cdfname=hgu133plus2cdf) Error: cannot allocate vector of size 858.0 Mb I get similar errors with smaller objects, smaller data sets or other procedures (Error: cannot allocate vector of size 123.0 Mb). I'm running R with Suse 11.1 Linux OS, on two Xeon processors (8 cores), 32 GB RAM. I suppose I have enough resources to manage these objects and data files Any suggestions or hints will be really appreciated! Many thanks in advance. Alessandro Well, you are running into a situation where there is not a contiguous chunk of RAM available in the sizes referenced, for allocation to the vector. Presuming that you are running a 64 bit version of SUSE (what does 'uname -a' show in a system console), you should also check to be sure that you are also running a 64 bit version of R. What does: .Machine$sizeof.pointer show? If it returns 4, then you are running a 32 bit version of R, which cannot take advantage of your 64 bit platform. You should install a 64 bit version of R. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to creat a matrix
Try this: N - 5 diag(1, N)[c(N, 1:(N - 1)),] On Fri, Dec 11, 2009 at 1:47 PM, Moohwan Kim kmhl...@gmail.com wrote: Dear R family I am attempting to create a matrix. e.g., 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 How could I write a R program? Later I want to extend it to a N by N case. Thanks in advance best Moohwan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] (no subject)
Hi,
i am new to the R-project but until now i have found solutions for every
problem in toturials, R Wikis and this mailing list, but now i have some
problems which I can't solve with this knowledge.
I have some data like this:
# sample data
head1 = a;b;c;d;e;f;g;h;i;k;l;m;n;o
data1 = 1;1;1;1;1;1;1;1;1;1;1;1;1;1
data2 = 2;2;2;2;2;2;2;2;2;2;2;2;2;2
data3 = 3;3;3;3;3;3;3;3;3;3;3;3;3;3
datastring = paste(, head1,data1,data2,data3,,sep=\n)
# import operation
res = read.table(textConnection(datastring), header=TRUE, sep = c(;))
closeAllConnections()
# I use these two lines in a for-loop like this:
#for( j in 1:length(data)) {
# res[j] = read.table(textConnection(datastring[j]),
header=TRUE, sep = c(;))
# closeAllConnections()
#}
I get these strings from a file which contains about 50 to 1000 of them, so I
can read them all into a list. I am not sure if there is a better way to do
this, but it works for me. Maybe you have some suggestions for a better
solution.
Now after this short introduction to the r-program I use, I have two problems
with this approach.
1) warnings
i get warnings like unused connection 3 (datastring) closed after some other
operations from time to time. But all connections should already be closed, and
I doesn't create new ones.
2) ram usage and program shutdowns
length(data) is usually between 50 to 1000. So it takes some space in ram
(approx 100-200 mb) which is no problem but I use some analysis code which
results in about 500-700 mb ram usage, also not a real problem.
The results are matrixes of (50x14 to 1000x14) so they are small enough to work
with them afterwards: create plots, or make some more analysis.
So i wrote a function which do the analysis one file after another and keep
only the results in a list. But after some about 2-4 files my R process uses
about 1500MB and then the troubles begin. The R console terminates or prints
the error that no more space can be allocated. So i have to do each file
separate and save each result in a file and restart R after 2 processed files.
And do that 3-5 times so that all files are processed, which is a bit anoying.
I did some research on this problem and i find out that
-) after I import the data in the same variable the ram usage goes up each time
about 100-200mb instead of reusing or purging the old data, which should be
overwritten since they are no longer available after i import a new file.
-) the same occures with the analysis functions which uses much more space and
also doesn't release the old no longer used variables. But ls() doesn't shows
them at all.
-) also after I cleared all variables with rm(list=ls(all=TRUE)) the used ram
space is still the same.
So is there a possibility to get the ram space back? So i can do all the
analysis in one session and don't have to mess around with additional files?
Thanks for your help
Tom
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[R] some problems with ram usage and warnings
Hi,
i am new to the R-project but until now i have found solutions for every
problem in toturials, R Wikis and this mailing list, but now i have some
problems which I can't solve with this knowledge.
I have some data like this:
# sample data
head1 = a;b;c;d;e;f;g;h;i;k;l;m;n;o
data1 = 1;1;1;1;1;1;1;1;1;1;1;1;1;1
data2 = 2;2;2;2;2;2;2;2;2;2;2;2;2;2
data3 = 3;3;3;3;3;3;3;3;3;3;3;3;3;3
datastring = paste(, head1,data1,data2,data3,,sep=\n)
# import operation
res = read.table(textConnection(datastring), header=TRUE, sep = c(;))
closeAllConnections()
# I use these two lines in a for-loop like this:
#for( j in 1:length(data)) {
# res[j] = read.table(textConnection(datastring[j]),
header=TRUE, sep = c(;))
# closeAllConnections()
#}
I get these strings from a file which contains about 50 to 1000 of them, so I
can read them all into a list. I am not sure if there is a better way to do
this, but it works for me. Maybe you have some suggestions for a better
solution.
Now after this short introduction to the r-program I use, I have two problems
with this approach.
1) warnings
i get warnings like unused connection 3 (datastring) closed after some other
operations from time to time. But all connections should already be closed, and
I doesn't create new ones.
2) ram usage and program shutdowns
length(data) is usually between 50 to 1000. So it takes some space in ram
(approx 100-200 mb) which is no problem but I use some analysis code which
results in about 500-700 mb ram usage, also not a real problem.
The results are matrixes of (50x14 to 1000x14) so they are small enough to work
with them afterwards: create plots, or make some more analysis.
So i wrote a function which do the analysis one file after another and keep
only the results in a list. But after some about 2-4 files my R process uses
about 1500MB and then the troubles begin. The R console terminates or prints
the error that no more space can be allocated. So i have to do each file
separate and save each result in a file and restart R after 2 processed files.
And do that 3-5 times so that all files are processed, which is a bit anoying.
I did some research on this problem and i find out that
-) after I import the data in the same variable the ram usage goes up each time
about 100-200mb instead of reusing or purging the old data, which should be
overwritten since they are no longer available after i import a new file.
-) the same occures with the analysis functions which uses much more space and
also doesn't release the old no longer used variables. But ls() doesn't shows
them at all.
-) also after I cleared all variables with rm(list=ls(all=TRUE)) the used ram
space is still the same.
So is there a possibility to get the ram space back? So i can do all the
analysis in one session and don't have to mess around with additional files?
Thanks for your help
Tom
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Re: [R] how to creat a matrix
Hi try R library(magic) R ashift(diag(5),1) HTH rksh enrique Dallazuanna wrote: Try this: N - 5 diag(1, N)[c(N, 1:(N - 1)),] On Fri, Dec 11, 2009 at 1:47 PM, Moohwan Kim kmhl...@gmail.com wrote: Dear R family I am attempting to create a matrix. e.g., 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 How could I write a R program? Later I want to extend it to a N by N case. Thanks in advance best Moohwan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Robin K. S. Hankin Uncertainty Analyst University of Cambridge 19 Silver Street Cambridge CB3 9EP 01223-764877 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 creat a matrix
Hi Him. Did you read the Help/Manual in PDF/A Instroduction to R.PDF? MyVect-scan() 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 MyVect MyMat-matrix(MyVect, ncol=5, byrow=T) MyMat bests milton On Fri, Dec 11, 2009 at 10:47 AM, Moohwan Kim kmhl...@gmail.com wrote: Dear R family I am attempting to create a matrix. e.g., 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 How could I write a R program? Later I want to extend it to a N by N case. Thanks in advance best Moohwan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] obtain intermediate estimate using optim
Doing a hessian estimate at each Nelder-Mead iteration is rather like going from den Haag to Delft as a pedestrian walking and swimming via San Francisco. The structure of the algorithm means the Hessian estimate is done in addition to the NM work. While my NM code was used for optim(), I didn't do the interfacing. The reporting choices are reasonably good, but don't necessarily suit your current needs. I'd recommend going to r-forge and installing my updated BFGS code. See http://r-forge.r-project.org/R/?group_id=395 for a list of the codes -- Rvmmin is the one you want) which is all in R so you can put in output where you choose. It also has bounds constraints, which are quite useful to avoid roaming into unsuitable areas of the parameter space. While Rvmmin works best with analytic gradients, it does OK most of the time with numeric approximations. It keeps an approximate inverse hessian, but I would not assume that bears too much resemblance to the real hessian. Package ucminf uses essentially the same algorithm (unconstrained only), and the detailed tactics seem to be well-thought out. However, I don't know how well reporting can be controlled (it is R interfaced to Fortran). A derivative free method that may be worth a try is bobyqa in the minqa package at the same site as above. This is Mike Powell's code. The output can be set quite detailed by pushing the reporting control (iprint) higher. Again R - Fortran interface (thanks to Kate Mullen). Ravi Varadhan has several NM versions in R also, but I don't think they are yet on r-forge. If you try any of these, you can help us improve them by reporting success/failure off list. We believe that they are in pretty good shape, but there are always interfacing and tuning issues. Cheers, JN Message: 1 Date: Thu, 10 Dec 2009 12:40:17 +0100 From: Lisanne Sanders lisan_sand...@hotmail.com Subject: [R] obtain intermediate estimate using optim To: r-help@r-project.org Message-ID: col110-w27ffa963674aa82fe4bc2395...@phx.gbl Content-Type: text/plain Hi, Currently I am trying to solve a minimization problem using optim as method Nelder-Mead. However, Neldel-Mead needs many iterations until it finally converges. I have set $control.trace and $control.report such that I can see the value of the function at each iteration. I do see that I set the convergence criteria to strict in the sense that the function value does not change much. However, before loosening my convergence criteria, I was wondering how to progamm that I can see the estimates of the true parameters and of the hessian such that I can see whether they do not change much either. Than I can adjust my convergence criteria such that he ends at that point. I do know how to adjust the convergence parameters but I do not know how to obtain intermediate estimates of the parameters. I was wondering whether someone can help me with this. Kind regards, Lisanne Sanders __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with calibrate function
Rodrigo wrote: Hi, I´m trying to use the calibrate function from rms package (made by prof. Harrell) after fitting a model using cph. But it returns the following error message: calibrate(modelo1,B=200,bw=F,u=13) Using Cox survival estimates at 13 Days Convergence problems stopping addition Error in hare(S[, 1], S[, 2], fun(est.surv), maxdim = maxdim, ...) : no convergence this is serious! Has anybody experienced the same problem? Can anyone help me? many thanks, Rodrigo Rodrigo, This is a new feature in rms and there are probably a few warnings I need to put in the help file. How many events are in the dataset used to develop your model? Why did you specify bw=FALSE (which is the default)? Did you use any variable selection when building the model? Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How could I find the inverse of a matrix?
Dear R family I have a following question. Suppose I have a matrix as follows, for instance: tau= 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 I want to have the inverse of the above matrix and then add some exponent to it. That is, I want to calculate tau to the (-m). For example, m=893. Thanks in advance Best regards Moohwan Kim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] memory problem on Suse
Ask on the bioconductpr mailing list, where you will be diirected to several solutions for analyzing what I guess are 100's is cel files http://bioconductor.org -- Martin Morgan On Dec 11, 2009, at 8:02 AM, Marc Schwartz marc_schwa...@me.com wrote: On Dec 11, 2009, at 6:24 AM, Ambrosi Alessandro wrote: Dear all, I am meeting some problems with memory allocation. I know it is an old issue, I'm sorry. I looked for a solution in the FAQs and manuals, mails, but without finding the working answer. I really hope you can help me. For instance, if I try to read micorarray data I get: mab=ReadAffy(cdfname=hgu133plus2cdf) Error: cannot allocate vector of size 858.0 Mb I get similar errors with smaller objects, smaller data sets or other procedures (Error: cannot allocate vector of size 123.0 Mb). I'm running R with Suse 11.1 Linux OS, on two Xeon processors (8 cores), 32 GB RAM. I suppose I have enough resources to manage these objects and data files Any suggestions or hints will be really appreciated! Many thanks in advance. Alessandro Well, you are running into a situation where there is not a contiguous chunk of RAM available in the sizes referenced, for allocation to the vector. Presuming that you are running a 64 bit version of SUSE (what does 'uname -a' show in a system console), you should also check to be sure that you are also running a 64 bit version of R. What does: .Machine$sizeof.pointer show? If it returns 4, then you are running a 32 bit version of R, which cannot take advantage of your 64 bit platform. You should install a 64 bit version of R. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 could I find the inverse of a matrix?
Hi Moohwan, On Dec 11, 2009, at 11:26 AM, Moohwan Kim wrote: Dear R family I have a following question. Suppose I have a matrix as follows, for instance: tau= 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 I want to have the inverse of the above matrix and then add some exponent to it. That is, I want to calculate tau to the (-m). For example, m=893. If you *really* want the inverse, use the `solve` function without a second parameter: R tau - c(0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0) R tau - matrix(tau, ncol=5, byrow=TRUE) R tau %*% solve(tau) [,1] [,2] [,3] [,4] [,5] [1,]10000 [2,]01000 [3,]00100 [4,]00010 [5,]00001 -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Why a list of NULL's are reduced to NULL?
The following examples are confusing to me. It is OK, to assigned NULL
to one element in a list. The result is still a list. However, a list
of NULL's are reduced to NULL. I don't understand how this conversion
occurs. Could somebody let me know what is going on?
X=matrix(1:8, nr=4)
apply(X,1, function(x) {if(x[[1]]==3){NULL}else{x[[1]]}})
[[1]]
[1] 1
[[2]]
[1] 2
[[3]]
NULL
[[4]]
[1] 4
apply(X,1, function(x) {NULL})
NULL
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and provide commented, minimal, self-contained, reproducible code.
[R] Sources for open sourced homework questions for R?
Hi, I am teaching a one month class in applied statistics and want to bring my students up to speed in R without devoting much/any lecture time to R instruction. I think that the best way to do this is to provide them with a lot of R questions for homework. These questions would be numerous (there is a lot of material to cover), go from very simple to somewhat complex, and focus on all the commands and options that will be useful in applied work. Here are some of my initial questions: Q: Load the data from the cars data frame into the local workspace. A: data(cars) Q: Find information about the cars data frame. A: help(cars) Q: Calculate the dimensions of the data frame. A: dim(cars) Q: What are the names of the variables? A: names(cars) Needless to say, the questions will become more complex, including the writing of simple functions. I also want to provide answers to all the questions that, in theory, could be used in an automated fashion to check the students work. My current plan is to load these questions (somehow) into the quiz module in Moodle (http://moodle.org/). Ideally, I would like this system to be usable by very large classes and even in the context of distance learning. Student goes to a web page, logs in and is presented with a page of questions (or a single question). She figures out the answer in her R session and pastes in the command (or result) into the answer slot on the webpage and pushes a button (or does it for ten questions first). The server then determines which questions she got right and which she got wrong. It might then provide clues to the ones that she has wrong. Once she is done, the professor gets a list of her results (how many right, how many wrong, how many required more than one try and so on). For now, I am not building that system. (Has anyone already done so?) Instead, I am just creating the collection of R questions/answers that might go into such a system. I am aiming for around 1,000 questions. So: Does anyone know of open sourced collections of R questions like this which I might use? Thanks, Dave Kane Adjunct Instructor, Williams College __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 a list of NULL's are reduced to NULL?
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Peng Yu
Sent: Friday, December 11, 2009 8:44 AM
To: r-h...@stat.math.ethz.ch
Subject: [R] Why a list of NULL's are reduced to NULL?
The following examples are confusing to me. It is OK, to assigned NULL
to one element in a list. The result is still a list. However, a list
of NULL's are reduced to NULL. I don't understand how this conversion
occurs. Could somebody let me know what is going on?
The simplification algorithm for reformatting
the output of apply and sapply is handy in the
common case when you know that FUN will return
the same sort of thing each time it it called.
The algorithm is not very useful when FUN may return
objects of various classes or lengths. sapply has
a simplify=FALSE argument to avoid the simplification
(so it acts like lapply) but apply doesn't.
I suggest you either change your function to always
return one class and length of object or use lapply()
or sapply(simplify=FALSE,...) when you must use a function
with variable output type. E.g., instead of
apply(X, 1, function(row){f(row)})
use
lapply(seq_len(nrow(X)), function(rowIndex){f(X[rowIndex,])})
or
lapply(split(X, row(X)), function(row){f(row)})
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
X=matrix(1:8, nr=4)
apply(X,1, function(x) {if(x[[1]]==3){NULL}else{x[[1]]}})
[[1]]
[1] 1
[[2]]
[1] 2
[[3]]
NULL
[[4]]
[1] 4
apply(X,1, function(x) {NULL})
NULL
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Looking for categorization method/module in R
All, I'm relatively new to using R, having used it thus far for some simple statistics and plotting. However, I'm not new to programming by any measure. I've been looking at the various modules available for clustering, factor analysis, etc. and find that I need advice on which modules I should be focusing on and their application. I have a data set comprised of columns of both quantitative and qualitative / non-numeric attributes. I would like to perform two operations on this data: identify correlations between attributes, and cluster the records by attribute. All of the clustering algorithms that I've looked at so far are based on numerical distance functions, and it's not clear to me how I'd apply them to qualitative attributes. It's not appropriate to simple convert discrete qualitative attributes (e.g., native language) to numerical values or independent columns with binary values. Is there a module that provides such an algorithm or that can be adapted to this purpose? I can wrap my head around the problem of looking for cross-correlation between the attributes, but would appreciate any insight in how to do it most efficiently and present the results. Thank you. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] SVM regression
Dear R users,
I am trying to apply SVM regression for a set of microarray data. I am using
the function svm() under the package {e1071}. Can anyone tell me what
the *residuals
*value represents? I have some observed values *y_obs* for the parameter
that I want to estimate and I would expect that *svm$residuals = y_obs -
svm$fitted.
*However, this does not happen...Does anyone have any idea on that?
Thanks a lot!
Eleni C.* *
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] The correct way to set an element in a list to NULL? (FAQ is not clear)
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-set-components-of-a-list-to-NULL_003f The explanation on this FAQ entry is not clear. It says '... similarly for named components...'. What I understood was x[i]-list(NULL) is the same as x$a_name-list(NULL). But, they are not. As the example below shows, x$a_name-list(NULL) is the same as x[[i]]-list(NULL). x=list(a=1:3,b=NULL,c=2:5,d=NULL) x $a [1] 1 2 3 $b NULL $c [1] 2 3 4 5 $d NULL x[[3]]=list(NULL) x $a [1] 1 2 3 $b NULL $c $c[[1]] NULL $d NULL x$c=list(NULL) x $a [1] 1 2 3 $b NULL $c $c[[1]] NULL $d NULL x[[3]]=7:8 x $a [1] 1 2 3 $b NULL $c [1] 7 8 $d NULL x$c=7:8 x $a [1] 1 2 3 $b NULL $c [1] 7 8 $d NULL What seems confusing to me is: even 'x[i]-list(NULL)' and 'x[[i]]-list(NULL)' are different, why x[i]-NULL and x[[i]]-NULL are the same? Shouldn't the meaning of 'x[[i]]-NULL' be defined as the set the i'th element NULL, rather than deleting the i'th element? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 figure out which the version of split is used?
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Karl Ove Hufthammer Sent: Thursday, December 10, 2009 1:46 AM To: r-h...@stat.math.ethz.ch Subject: Re: [R] How to figure out which the version of split is used? On Wed, 9 Dec 2009 19:20:47 -0600 Peng Yu pengyu...@gmail.com wrote: Is there a way to figure out which of these variants is actually dispatched to when I call split? I know that if the argument is of the type data.frame, split.data.frame will be called? Is it the case that if the argument is not of type data.frame, Date or POSIXct, split.default will be called? Yes. See ?UseMethod You can also use trace() to see what actually happens in test cases. E.g., invisible(lapply(methods(split), function(method)trace(method, bquote(cat(Entering, .(method), x=, class(x), f=, class(f), \n) Tracing function split.data.frame in package base Tracing function split.Date in package base Tracing function split.default in package base Tracing function split.POSIXct in package base split(data.frame(x=1:3,y=1:3), f=c(10,10,20)) Tracing split.data.frame(data.frame(x = 1:3, y = 1:3), f = c(10, 10, on entry Entering split.data.frame x= data.frame f= numeric Tracing split.default(seq_len(nrow(x)), f, drop = drop, ...) on entry Entering split.default x= integer f= numeric $`10` x y 1 1 1 2 2 2 $`20` x y 3 3 3 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -- Karl Ove Hufthammer __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] SVM regression
Hi Eleni,
On Dec 11, 2009, at 12:04 PM, Eleni Christodoulou wrote:
Dear R users,
I am trying to apply SVM regression for a set of microarray data. I am using
the function svm() under the package {e1071}. Can anyone tell me what
the *residuals
*value represents? I have some observed values *y_obs* for the parameter
that I want to estimate and I would expect that *svm$residuals = y_obs -
svm$fitted.
*However, this does not happen...Does anyone have any idea on that?
This actually is what's happening. The $residuals that are reported in the
model are against your *scaled* y-vector.
So, with your data:
R m - svm(x,y)
R all(scale(y) - predict(m,x) == m$residuals)
[1] TRUE
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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 a list of NULL's are reduced to NULL?
On Fri, Dec 11, 2009 at 11:01 AM, William Dunlap wdun...@tibco.com wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Peng Yu
Sent: Friday, December 11, 2009 8:44 AM
To: r-h...@stat.math.ethz.ch
Subject: [R] Why a list of NULL's are reduced to NULL?
The following examples are confusing to me. It is OK, to assigned NULL
to one element in a list. The result is still a list. However, a list
of NULL's are reduced to NULL. I don't understand how this conversion
occurs. Could somebody let me know what is going on?
The simplification algorithm for reformatting
the output of apply and sapply is handy in the
common case when you know that FUN will return
the same sort of thing each time it it called.
The algorithm is not very useful when FUN may return
objects of various classes or lengths. sapply has
a simplify=FALSE argument to avoid the simplification
(so it acts like lapply) but apply doesn't.
I suggest you either change your function to always
return one class and length of object or use lapply()
or sapply(simplify=FALSE,...) when you must use a function
with variable output type. E.g., instead of
apply(X, 1, function(row){f(row)})
use
lapply(seq_len(nrow(X)), function(rowIndex){f(X[rowIndex,])})
or
lapply(split(X, row(X)), function(row){f(row)})
Change my function to always returning one class may not always be
possible as I may call a third party R package that is not made by me
and does this kind of wired things of trying to 'simplify'. And I may
not know all the cases where the third party R package 'simplify' the
results, which does not always return the same type. In this case, I
can not be sure the return type is always the same. How do you deal
with this problem?
X=matrix(1:8, nr=4)
apply(X,1, function(x) {if(x[[1]]==3){NULL}else{x[[1]]}})
[[1]]
[1] 1
[[2]]
[1] 2
[[3]]
NULL
[[4]]
[1] 4
apply(X,1, function(x) {NULL})
NULL
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] shared axes in multipanel plot
Try this using xyplot.zoo in the zoo package. We define the baseline
and a panel function. The panel function just performs the default
action to display the graphs and adds the baseline. The screens
variable is 1,1,2,2,3,3,4,4. We create a zoo object from dat and use
screens to name the columns according to their group. Finally we call
xyplot.zoo passing it screens so that the successive columns go in the
indicated panels and also passing the other items. See ?xyplot.zoo in
zoo and ?xyplot in lattice.
library(zoo)
library(lattice)
baseline - 1:nrow(dat)/nrow(dat)
pnl - function(x, ...) {
panel.plot.default(x, ...)
panel.lines(x, baseline, lwd = 2, col = grey(0.5))
}
nc - ncol(dat)
screens - rep(1:(nc/2), each = 2)
z - zoo(dat)
colnames(z) - paste(Group, screens)
xyplot(z, screens = screens , layout = c(2, 2), col = black, lty =
2, scales = list(y = list(relation = same)), panel = pnl)
On Fri, Dec 11, 2009 at 10:02 AM, Jennifer Young
jennifer.yo...@math.mcmaster.ca wrote:
Hello
I've created a function to make a plot with multiple pannels from columns
of data that are created in a previous function. In the example below the
number of columns is 8, giving 4 pannels, but in general it takes data
with any number of columns and figures out a nice layout.
The panels all have the same axes, and so I wonder what functions are
avialable to create axes only on the left and bottom of the whole plot
rather than each pannel.
I'd really like a generic way to do this for any number of plots, but was
even having trouble figuring out how to do it manually for this example;
How are pannels referred to, in a layout context?
That is, how do I say,
if(current.pannel==4) {do stuff}
Here's a simple version of the code.
baseline - (1:20)/20 #example data
dat1 - matrix(baseline,20,8)
dat - dat1+matrix(rnorm(20*8)/30, 20,8)
nstrat - ncol(dat)
rows - ceiling(nstrat/4)
layout(matrix(1:(rows*2), rows, 2, T))
par(oma=c(4,4,3,1))
par(mar=c(1,1,0,1))
for(i in which(1:nstrat%%2!=0)){
plot(baseline, type=l, col=grey, lwd=2,
xlab=, ylab=, ylim=c(0,1), xaxt='n', yaxt='n')
axis(1, labels=F); axis(2, labels=F)
points(dat[,i], type=l, lty=2)
points(dat[,i+1], type=l, lty=2)
}
Thank you muchly
Jennifer Young
PS: I am a subscriber, but can't for the life of me figure out how to send
an email while logged in so that the moderators don't have to take the
time to read it over. I always get the please wait while we check it
over email. Likely I'm being dumb.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] The correct way to set an element in a list to NULL? (FAQ is not clear)
Hi, On Dec 11, 2009, at 12:07 PM, Peng Yu wrote: http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-set-components-of-a-list-to-NULL_003f The explanation on this FAQ entry is not clear. It says '... similarly for named components...'. What I understood was x[i]-list(NULL) is the same as x$a_name-list(NULL). But, they are not. You're right, x[i] is not the same as x$a_name: R x - list(a=1:3, b='hello') R identical(x[1], x$a) [1] FALSE R identical(x[1], x['a']) [1] TRUE I think that's what it means by similarly for named components. You see, x$a_name is really x[[1]], which does the destructive-null-assignment-thing: R identical(x[[1]], x$a) [1] TRUE -steve As the example below shows, x$a_name-list(NULL) is the same as x[[i]]-list(NULL). x=list(a=1:3,b=NULL,c=2:5,d=NULL) x $a [1] 1 2 3 $b NULL $c [1] 2 3 4 5 $d NULL x[[3]]=list(NULL) x $a [1] 1 2 3 $b NULL $c $c[[1]] NULL $d NULL x$c=list(NULL) x $a [1] 1 2 3 $b NULL $c $c[[1]] NULL $d NULL x[[3]]=7:8 x $a [1] 1 2 3 $b NULL $c [1] 7 8 $d NULL x$c=7:8 x $a [1] 1 2 3 $b NULL $c [1] 7 8 $d NULL What seems confusing to me is: even 'x[i]-list(NULL)' and 'x[[i]]-list(NULL)' are different, why x[i]-NULL and x[[i]]-NULL are the same? Shouldn't the meaning of 'x[[i]]-NULL' be defined as the set the i'th element NULL, rather than deleting the i'th element? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] The correct way to set an element in a list to NULL? (FAQ is not clear)
On Fri, Dec 11, 2009 at 11:27 AM, Steve Lianoglou mailinglist.honey...@gmail.com wrote: Hi, On Dec 11, 2009, at 12:07 PM, Peng Yu wrote: http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-set-components-of-a-list-to-NULL_003f The explanation on this FAQ entry is not clear. It says '... similarly for named components...'. What I understood was x[i]-list(NULL) is the same as x$a_name-list(NULL). But, they are not. You're right, x[i] is not the same as x$a_name: R x - list(a=1:3, b='hello') R identical(x[1], x$a) [1] FALSE R identical(x[1], x['a']) [1] TRUE I think that's what it means by similarly for named components. You see, x$a_name is really x[[1]], which does the destructive-null-assignment-thing: R identical(x[[1]], x$a) [1] TRUE -steve As the example below shows, x$a_name-list(NULL) is the same as x[[i]]-list(NULL). x=list(a=1:3,b=NULL,c=2:5,d=NULL) x $a [1] 1 2 3 $b NULL $c [1] 2 3 4 5 $d NULL x[[3]]=list(NULL) x $a [1] 1 2 3 $b NULL $c $c[[1]] NULL $d NULL x$c=list(NULL) x $a [1] 1 2 3 $b NULL $c $c[[1]] NULL $d NULL x[[3]]=7:8 x $a [1] 1 2 3 $b NULL $c [1] 7 8 $d NULL x$c=7:8 x $a [1] 1 2 3 $b NULL $c [1] 7 8 $d NULL What seems confusing to me is: even 'x[i]-list(NULL)' and 'x[[i]]-list(NULL)' are different, why x[i]-NULL and x[[i]]-NULL are the same? Shouldn't the meaning of 'x[[i]]-NULL' be defined as the set the i'th element NULL, rather than deleting the i'th element? Do you have any comments on the above question? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 a list of NULL's are reduced to NULL?
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Peng Yu
Sent: Friday, December 11, 2009 9:18 AM
To: r-h...@stat.math.ethz.ch
Subject: Re: [R] Why a list of NULL's are reduced to NULL?
On Fri, Dec 11, 2009 at 11:01 AM, William Dunlap
wdun...@tibco.com wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Peng Yu
Sent: Friday, December 11, 2009 8:44 AM
To: r-h...@stat.math.ethz.ch
Subject: [R] Why a list of NULL's are reduced to NULL?
The following examples are confusing to me. It is OK, to
assigned NULL
to one element in a list. The result is still a list.
However, a list
of NULL's are reduced to NULL. I don't understand how this
conversion
occurs. Could somebody let me know what is going on?
The simplification algorithm for reformatting
the output of apply and sapply is handy in the
common case when you know that FUN will return
the same sort of thing each time it it called.
The algorithm is not very useful when FUN may return
objects of various classes or lengths. sapply has
a simplify=FALSE argument to avoid the simplification
(so it acts like lapply) but apply doesn't.
I suggest you either change your function to always
return one class and length of object or use lapply()
or sapply(simplify=FALSE,...) when you must use a function
with variable output type. E.g., instead of
apply(X, 1, function(row){f(row)})
use
lapply(seq_len(nrow(X)), function(rowIndex){f(X[rowIndex,])})
or
lapply(split(X, row(X)), function(row){f(row)})
Change my function to always returning one class may not always be
possible as I may call a third party R package that is not made by me
and does this kind of wired things of trying to 'simplify'. And I may
not know all the cases where the third party R package 'simplify' the
results, which does not always return the same type. In this case, I
can not be sure the return type is always the same. How do you deal
with this problem?
I don't understant your constraints. You say you cannot control
what FUN returns and you cannot control whether apply or lapply
is called. A reproducible set of examples would help.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
X=matrix(1:8, nr=4)
apply(X,1, function(x) {if(x[[1]]==3){NULL}else{x[[1]]}})
[[1]]
[1] 1
[[2]]
[1] 2
[[3]]
NULL
[[4]]
[1] 4
apply(X,1, function(x) {NULL})
NULL
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] The correct way to set an element in a list to NULL? (FAQ is not clear)
On Dec 11, 2009, at 12:36 PM, Peng Yu wrote: [snip] What seems confusing to me is: even 'x[i]-list(NULL)' and 'x[[i]]-list(NULL)' are different, why x[i]-NULL and x[[i]]-NULL are the same? Shouldn't the meaning of 'x[[i]]-NULL' be defined as the set the i'th element NULL, rather than deleting the i'th element? Do you have any comments on the above question? Sure. I think it has something to do with how memory is managed and allocated in R. You might try to read up on it a bit ... In all seriousness tho: No, I don't really have any comment on that question. The semantics of what x[i]-list(NULL) vs x[[i]]-list(NULL) seems quite reasonable to me ... I'm not sure what that has to do with anything. I also can't comment on why x[[i]] - NULL deletes the element (instead of setting it to NULL, like you want it to) .. it's just the way it is. That having been said. Once you get it (and I guess it's on the FAQ for a reason), then you can figure out how to deal with it. -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 a list of NULL's are reduced to NULL?
On Fri, Dec 11, 2009 at 11:43 AM, William Dunlap wdun...@tibco.com wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Peng Yu
Sent: Friday, December 11, 2009 9:18 AM
To: r-h...@stat.math.ethz.ch
Subject: Re: [R] Why a list of NULL's are reduced to NULL?
On Fri, Dec 11, 2009 at 11:01 AM, William Dunlap
wdun...@tibco.com wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Peng Yu
Sent: Friday, December 11, 2009 8:44 AM
To: r-h...@stat.math.ethz.ch
Subject: [R] Why a list of NULL's are reduced to NULL?
The following examples are confusing to me. It is OK, to
assigned NULL
to one element in a list. The result is still a list.
However, a list
of NULL's are reduced to NULL. I don't understand how this
conversion
occurs. Could somebody let me know what is going on?
The simplification algorithm for reformatting
the output of apply and sapply is handy in the
common case when you know that FUN will return
the same sort of thing each time it it called.
The algorithm is not very useful when FUN may return
objects of various classes or lengths. sapply has
a simplify=FALSE argument to avoid the simplification
(so it acts like lapply) but apply doesn't.
I suggest you either change your function to always
return one class and length of object or use lapply()
or sapply(simplify=FALSE,...) when you must use a function
with variable output type. E.g., instead of
apply(X, 1, function(row){f(row)})
use
lapply(seq_len(nrow(X)), function(rowIndex){f(X[rowIndex,])})
or
lapply(split(X, row(X)), function(row){f(row)})
Change my function to always returning one class may not always be
possible as I may call a third party R package that is not made by me
and does this kind of wired things of trying to 'simplify'. And I may
not know all the cases where the third party R package 'simplify' the
results, which does not always return the same type. In this case, I
can not be sure the return type is always the same. How do you deal
with this problem?
I don't understant your constraints. You say you cannot control
what FUN returns and you cannot control whether apply or lapply
is called. A reproducible set of examples would help.
A very common situation is that the users don't know all the possible
return types of 'some_third_party_function()'. If the users don't know
all the return types, he/she can not make sure the return type of
function(x) {...} be always the same. How do you deal with this case?
apply(X, 1, function(x) {
do something...
some_third_party_function(x)
}
)
X=matrix(1:8, nr=4)
apply(X,1, function(x) {if(x[[1]]==3){NULL}else{x[[1]]}})
[[1]]
[1] 1
[[2]]
[1] 2
[[3]]
NULL
[[4]]
[1] 4
apply(X,1, function(x) {NULL})
NULL
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Re: [R] Sources for open sourced homework questions for R?
Hi Dave, I have a few drills available from http://had.co.nz/stat405 - see the right hand column, about half way down. They seem similar in spirit to what you're thinking of. You might want to look at the Little Schemer for a similar approach with a different programming language. However, I'm not sure how pedagogically useful this approach is. If you break things down too finely, you don't teach the problem solving skills necessary to attack a new problem. Students will try and solve the problems as rapidly, using as little of their brain as possible. I also feel like these small problem fail to invoke any intellectually curiosity - why the heck should I care that mtcars has 32 observations and 11 rows? I'd suggest starting with a big problem that's of interest to the students - how do we detect spam? What determines the price of a used car on ebay? Do soap operas influence baby name trends? Are my facebook friends representative of the university as a whole? Then talk about how you might attack the problem in general, before getting to the concrete tools you'd use in R. Hadley On Fri, Dec 11, 2009 at 10:48 AM, David Kane d...@kanecap.com wrote: Hi, I am teaching a one month class in applied statistics and want to bring my students up to speed in R without devoting much/any lecture time to R instruction. I think that the best way to do this is to provide them with a lot of R questions for homework. These questions would be numerous (there is a lot of material to cover), go from very simple to somewhat complex, and focus on all the commands and options that will be useful in applied work. Here are some of my initial questions: Q: Load the data from the cars data frame into the local workspace. A: data(cars) Q: Find information about the cars data frame. A: help(cars) Q: Calculate the dimensions of the data frame. A: dim(cars) Q: What are the names of the variables? A: names(cars) Needless to say, the questions will become more complex, including the writing of simple functions. I also want to provide answers to all the questions that, in theory, could be used in an automated fashion to check the students work. My current plan is to load these questions (somehow) into the quiz module in Moodle (http://moodle.org/). Ideally, I would like this system to be usable by very large classes and even in the context of distance learning. Student goes to a web page, logs in and is presented with a page of questions (or a single question). She figures out the answer in her R session and pastes in the command (or result) into the answer slot on the webpage and pushes a button (or does it for ten questions first). The server then determines which questions she got right and which she got wrong. It might then provide clues to the ones that she has wrong. Once she is done, the professor gets a list of her results (how many right, how many wrong, how many required more than one try and so on). For now, I am not building that system. (Has anyone already done so?) Instead, I am just creating the collection of R questions/answers that might go into such a system. I am aiming for around 1,000 questions. So: Does anyone know of open sourced collections of R questions like this which I might use? Thanks, Dave Kane Adjunct Instructor, Williams College __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 a list of NULL's are reduced to NULL?
A very common situation is that the users don't know all the possible
return types of 'some_third_party_function()'. If the users don't know
all the return types, he/she can not make sure the return type of
function(x) {...} be always the same. How do you deal with this case?
It's not that common. It's pretty bad practice to return different
types from a function depending on the input parameters. In many
languages this isn't even possible.
The solution is to write a function that takes the output from the
first function, inspects it, and coerces all possibilities to the same
type.
Hadley
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Re: [R] confint for glm (general linear model)
I suspect that you don't know about 'profile' confidence intervals. If that's true then I can recommend the discussion in MASS (the book) in section 8.4. In a nutshell, I don't think that you want to do a profile confint calculation manually (unless typing instructions that use the function profile.glm() counts as 'manual'). -Peter Ehlers casperyc wrote: I think the help page are exactly the same... I just want to verify the confidence interval manually. That's all I want. Thanks. casper brestat wrote: This functions are different. I advice you study them: ?confint # profile likelihood ?confint.default # t-distribution Walmes Zeviani - Brazil casperyc wrote: Hi, I have a glm gives summary as follows, Estimate Std. Errorz valuePr(|z|) (Intercept) -2.03693352 1.449574526 -1.405194 0.159963578 A0.01093048 0.006446256 1.695633 0.089955471 N0.41060119 0.224860819 1.826024 0.067846690 S -0.20651005 0.067698863 -3.050421 0.002285206 then I use confint(k.glm) to obtain a confidnece interval for the estimates. confint(k.glm,level=0.97) Waiting for profiling to be done... 1.5 % 98.5 % (Intercept) -5.471345995 0.94716503 A -0.002340863 0.02631582 N -0.037028592 0.95590178 S -0.365570347 -0.06573675 while reading the help for 'confint', i found something like confint.glm for general linear model. I load the MASS package by clicking on the Menu( or otherwise how should I load the package?) then I still cant use the confint.glm command, what have I dont wrong? How do I calculate this confidence interval for glm estimate manually?? for A, I use 0.01093048 + c(-1,1) * 0.006446256 * qt(0.985,df=77) which is a different interval i got from the confint(k.glm,level=0.97) above. To be short, what's the right command to find the confidence interval for glm estimats? How do I verify it manully? Thanks. casper -- Peter Ehlers University of Calgary __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot: Problem with legend background
Hi Luc, You want: legend.title=theme_text(size=20, hjust = 0) So the legend title is left aligned, not centred. Hadley On Fri, Dec 11, 2009 at 9:26 AM, MUHC_Research villa...@dms.umontreal.ca wrote: Dear R-users, I am preparing graphs for an upcoming article using the different functions of the ggplot2 package and I've been having problems with the legend background. It doesn't seem to scale when the text size is increased. Here's the mandatory reproducible example: library(ggplot2) repFrame - data.frame(A= 1:10, B= rnorm(1:10), groupNum = rep(c(First group, Second group),each=5)) testPlot - ggplot(repFrame, aes(x=A, y = B, group = groupNum)) + opts(legend.position=c(0.85,0.3), legend.background = theme_rect(fill=white), legend.text=theme_text(size=16), legend.title=theme_text(size=20)) testPlot + geom_point(aes(colour= groupNum)) As you can see, the text doesn't fit in the white rectangle. I suspect there is a theme setting I could modify to fix this, but I can't seem to find which one. I sincerely thank you for your time and assistance. Luc -- View this message in context: http://n4.nabble.com/ggplot-Problem-with-legend-background-tp961142p961142.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] The correct way to set an element in a list to NULL? (FAQ is not clear)
On Fri, Dec 11, 2009 at 11:51 AM, Steve Lianoglou
mailinglist.honey...@gmail.com wrote:
On Dec 11, 2009, at 12:36 PM, Peng Yu wrote:
[snip]
What seems confusing to me is:
even 'x[i]-list(NULL)' and 'x[[i]]-list(NULL)' are different, why
x[i]-NULL and x[[i]]-NULL are the same?
Shouldn't the meaning of 'x[[i]]-NULL' be defined as the set the i'th
element NULL, rather than deleting the i'th element?
Do you have any comments on the above question?
Sure.
I think it has something to do with how memory is managed and allocated in R.
You might try to read up on it a bit ...
Which question do you refer by the first 'it'?
I have been asking a good reference on memory management in R. So far,
no one have given me any useful information. Do you have a good
reference?
In all seriousness tho:
No, I don't really have any comment on that question.
The semantics of what x[i]-list(NULL) vs x[[i]]-list(NULL) seems quite
reasonable to me ... I'm not sure what that has to do with anything.
I also can't comment on why x[[i]] - NULL deletes the element (instead of
setting it to NULL, like you want it to) .. it's just the way it is.
The design choice of x[[i]] - NULL deleting the element instead of
setting it to NULL might increase the complexity of the code. Suppose
that I set the i'th element of x by calling some_function(), which
never return NULL, the following code is perfectly fine.
x[[i]] - some_function()
However, when some_function() does return NULL, the i'th element will
be deleted.
In this case I have to do the following. I will have to use the
following code, when I don't know if some_function() can return NULL,
for the sake of safety. As you can see one line of code has been
expanded to 6 lines.
result=some_function()
if(NULL==result) {
x[i] - list(NULL)
} else {
x[[i]] - some_function()
}
For this reason, x[[i]]-NULL should be defined as setting it to NULL.
That having been said. Once you get it (and I guess it's on the FAQ for a
reason), then you can figure out how to deal with it.
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
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Re: [R] Why a list of NULL's are reduced to NULL?
On Fri, Dec 11, 2009 at 12:05 PM, hadley wickham h.wick...@gmail.com wrote:
A very common situation is that the users don't know all the possible
return types of 'some_third_party_function()'. If the users don't know
all the return types, he/she can not make sure the return type of
function(x) {...} be always the same. How do you deal with this case?
It's not that common. It's pretty bad practice to return different
types from a function depending on the input parameters. In many
languages this isn't even possible.
I know this is a bad practice. But R doesn't have a way to forbid such
thing happen. To program defensively, I have to test even uncommon
case, unless it is impossible. When you use a third party software in
your code, do you just ignore the possibility that a function could
return different types?
The solution is to write a function that takes the output from the
first function, inspects it, and coerces all possibilities to the same
type.
How do you figure out all the possibilities?
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Re: [R] Please help with a basic function
Many thanks for the replies to my call for help this morning. I didn't know
about return() and that helped quite a bit.
Best, Mark
On Fri, Dec 11, 2009 at 10:00 AM, Paul Hiemstra p.hiems...@geo.uu.nlwrote:
Hi Mark,
This question would probably be better suited for the r-sig-geo mailing
list. In addition, please read the posting guide and provide a piece of code
that reproduces the problem.
library(sp)
convert-function(d) {
d-data.frame(d); #convert object to dataframe
d-subset(d,select=c(zinc,x,y)) #select some columns
d # - add this, or alternatively 'return(d)'
}
data(meuse)
coordinates(meuse) = ~x+y
convert(meuse)
But maybe better, subsetting a SPDF can be done using:
meuse[zinc] # Remains an SPDF
# Returns a data.frame
data.frame(coordinates(meuse), zinc = meuse$zinc)
And some unrequested advice :). To process multiple files, take a look at
lapply, both for reading and processing.
all_data = lapply(list_of_files, function(file) {
bla = read.table(file)
coordinates(bla) = ~coor.x1 + coor.x2
return(bla)
}
# all data is now a list wit the SPDF's
processed_data = lapply(all_data, function(dat) {
return(data.frame(coordinates(dat), zinc = dat$zinc))
}
ofcourse you can include the latter lapply stuff inside the first 'loading'
lapply.
all_data = lapply(list_of_files, function(file) {
bla = read.table(file)
bla = subset(bla, select = select=c(time,coords.x1,coords.x2))
coordinates(bla) = ~coor.x1 + coor.x2
return(bla)
}
hope this helps and good luck,
Paul
Mark Na wrote:
Hello,
I am learning how to use functions, but I'm running into a roadblock.
I would like my function to do two things: 1) convert an object to a
dataframe, 2) and then subset the dataframe. Both of these commands work
fine outside the function, but I would like to wrap them in a function so
I
can apply the code iteratively to many such objects.
Here's what I wrote, but it doesn't work:
convert-function(d) {
d-data.frame(d); #convert object to dataframe
d-subset(d,select=c(time,coords.x1,coords.x2)) #select some columns
}
convert(data) #the problem is that data is the same as it was before
running the function
The objects being processed through my function are
SpatialPointsDataFrames
but I'm quite sure that's not my problem, as I can process these outside
of
the function (using the above code) ... it's when I try to wrap the code
in
a function that it doesn't work.
Thanks, Mark
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--
Drs. Paul Hiemstra
Department of Physical Geography
Faculty of Geosciences
University of Utrecht
Heidelberglaan 2
P.O. Box 80.115
3508 TC Utrecht
Phone: +3130 274 3113 Mon-Tue
Phone: +3130 253 5773 Wed-Fri
http://intamap.geo.uu.nl/~paul http://intamap.geo.uu.nl/%7Epaul
--
Mark Na
University of Saskatchewan
Saskatoon, Canada
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