Re: [R] how to get output from a nested loop
How about like this:
t1 - data.frame(row.names=c('c1','c2','c3','c4'), mk1=c(1,1,0,0),
mk2=c(0,0,0,1), mk3=c(1,1,1,1), mk4=c(0,0,0,0), mk5=c(0,0,0,1), S=c(4,5,3,2))
t1
mk1 mk2 mk3 mk4 mk5 S
c1 1 0 1 0 0 4
c2 1 0 1 0 0 5
c3 0 0 1 0 0 3
c4 0 1 1 0 1 2
apply(combn(1:5, 2), 2, function(x) t1[,c(x[1], 6, x[2])])
[[1]]
mk1 S mk2
c1 1 4 0
c2 1 5 0
c3 0 3 0
c4 0 2 1
... ...
On Wed, Jun 10, 2009 at 1:10 PM, Scott Hermannsherm...@bses.org.au wrote:
Dear all,
I imagine that this is a trival question, but it has perplexed for most of
the day. Any help would be greatly appreciated.
Below is an example of what I'm trying to do.
Essentially I want to produce all unique 1 x 1 combinations of certain
columns from a dataframe, and join these to other columns from the same
dataframe. I'm having problems with the nested loop as I can only output
data from the last cycle of the loop. I realise that the problem is with
the st1[[i]] but I'm not sure how to define it differently.
##I want to make a list file of all 1x1 combinations of mk columns, and add
clone and S to these combinations.
clone-c(c1,c2,c3,c4)
mk1-c(1,1,0,0)
mk2-c(0,0,0,1)
mk3-c(1,1,1,1)
mk4-c(0,0,0,0)
mk5-c(0,0,0,1)
S-c(4,5,3,2)
t1-as.data.frame(cbind(clone,mk1,mk2,mk3,mk4, mk5,S))
row.names(t1)-t1$clone
t1-t1[,-1]
t1
###A nested loop. I'm trying to get all combinations of columns 1:5 and join
each of them with column 7.
st1 - list()
for(i in 1:4) {
for(j in (i+1):5){
st1[[i]] - cbind(t1[,c(i,6)],t1[,j])
}
}
st1
Thanks for your help,
Scott
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Re: [R] R CMD check does not find a mistake
CG == Christophe Genolini cgeno...@u-paris10.fr on Tue, 09 Jun 2009 16:17:15 +0200 writes: CG Hi the list, I build a package. They was a mistake in CG it, but R CMD check did not find it. Is that normal ? CG Here is what Kurt gets (which is right, I did this CG mistake): CG --- 8 CG * checking for code/documentation mismatches ... WARNING CG S4 class codoc mismatches from documentation object 'LongData-class': CG Slots for class 'LongData' CG Code: id other time traj varName CG Docs: id time traj varName CG --- 8 Others have already hinted at the solution: The problem only shows in R-devel (2.10.x). Note that I had sent an explicit message to this list *exactly* in order to advise all package authors : Subject: [Rd] R-devel:codocClasses() now finds more -- R CMD check warnings Date: Wed, 3 Jun 2009 11:45:45 +0200 i.e. only 6 days before you saw the problem yes, I am sometimes too optimistic, assuming that people actually read what I write Regards anyway! Martin -- Martin Maechler, ETH Zurich __ 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] Asking for help while using Simplex to solve LP Problem.
Hello. I am a new user and want to use R to solve Linear Programming problem. I
met a problem while using function Simplex and I would kindly ask whether
someone could help to figure it out.
In my problem, there are 47 parameters and all limited to (0,1). M file has 147
lines and P file has 100 line. Each line in M indicated following inequality:
V1=V2*X1+V3*X2+V4*X3+...+V48*X47, where the top 47 lines 1=Xi (i=1 to 47) .
Each line in P indicated inequality: V1=V2*X1+V3*X2+V4*X3+...+V48*X47
The object function is (1,0,0...0) as I want to know the range of first
parameter.
So I wrote statements as following:
enj-c(1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
P-as.matrix(read.table(P,header=FALSE,sep=\t));
M-as.matrix(read.table(M,header=FALSE,sep=\t));
simplex(a=enj,A1=M[,2:48],b1=t(M[,1]),A2=P[,2:48],b2=t(P[,1]),maxi=TRUE)
Then I got the error message:
Error in while (!all(obfun -eps) (it = n.iter)) { :
missing value where TRUE/FALSE needed
Is anybody knows how to fix the error? Thank you.
Cheers.
Shubin Ren
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Re: [R] graphically representing frequency of words in a speech?
Hi, As Gregor Gorjanc mentioned, it's very inconvenient to let R decide the fontsize and placement of words in a plot. There have already been very mature applications of tag cloud; one of them I'm relatively familiar is the WordPress plugin wp-cumulus, which makes use of a Flash object to generate tag cloud, and it has fantastic 3D rotation effect of the cloud. I've spent a couple of hours porting it into R; see the source code and effect here: http://yihui.name/en/2009/06/creating-tag-cloud-using-r-and-flash-javascript-swfobject/ HTH. Regards, Yihui -- Yihui Xie xieyi...@gmail.com Phone: +86-(0)10-82509086 Fax: +86-(0)10-82509086 Mobile: +86-15810805877 Homepage: http://www.yihui.name School of Statistics, Room 1037, Mingde Main Building, Renmin University of China, Beijing, 100872, China On Mon, Jun 8, 2009 at 2:41 AM, Brown, Tony Nicholastony.n.br...@vanderbilt.edu wrote: Dear all, I recently saw a graph on television that displayed selected words/phrases in a speech scaled in size according to their frequency. So words/phrases that were often used appeared large and words that were rarely used appeared small. The closest thing I can find on the web to approximate what I saw can be found here: http://stateoftheunion.onetwothree.net/ The example at that website is more complicated but captures the general idea. Would someone point me in the right direction in terms of replicating such a graph. Thanks in advance, Tony - Tony N. Brown, Ph.D. Editor-Elect, American Sociological Review Associate Professor of Sociology and Human and Organizational Development (secondary) Program Faculty, Effective Health Communication and African American Diaspora Studies Faculty Head of Hank Ingram House, The Commons Vanderbilt University (615) 322-7518 (615) 322-7505 fax [[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] Unable to load package:lme4 [ Ubuntu 9.04 ]
GAR == Girish A R garam...@gmail.com
on Tue, 9 Jun 2009 22:00:01 -0700 (PDT) writes:
GAR Hi folks,
GAR When I try to load package 'lme4' on my Linux box (64-bit Ubuntu
GAR 9.04), I get the following error:
GAR -
library(lme4)
GAR Error in dyn.load(file, DLLpath = DLLpath, ...) :
GAR function 'cholmod_start' not provided by package 'Matrix'
GAR Error: package/namespace load failed for 'lme4'
GAR
GAR I removed and re-installed both 'Matrix' and 'lme4', but that doesn't
GAR seem to solve the problem. If this is something for the [R-sig-ME]
GAR mailing list, please let me know.
How did you re-install Matrix?
Your version Matrix_0.999375-17 (indicated below)
is a bit old, and indeed probably too old for the version of
lme4 you've tried.
Can you call
packageDescription(lme4)
without an error {probably not}, or at least
library(help = lme4)
the output of these would show you the version of Matrix you
need at least.
Ideally you'd upgrade your version of R (to 2.9.0) as well,
since the latest couple of versions of Matrix all have
R = 2.9.0 in their dependency requirements.
Regards,
Martin Maechler
GAR Thanks,
GAR -Girish
GAR
sessionInfo()
GAR R version 2.8.1 (2008-12-22)
GAR x86_64-pc-linux-gnu
GAR locale:
GAR
LC_CTYPE=en_IN;LC_NUMERIC=C;LC_TIME=en_IN;LC_COLLATE=en_IN;LC_MONETARY=C;LC_MESSAGES=en_IN;LC_PAPER=en_IN;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_IN;LC_IDENTIFICATION=C
GAR attached base packages:
GAR [1] splines stats graphics grDevices utils datasets
GAR methods
GAR [8] base
GAR other attached packages:
GAR [1] Matrix_0.999375-17 lattice_0.17-25Design_2.2-0
GAR survival_2.35-4
GAR [5] Hmisc_3.6-0nlme_3.1-92car_1.2-14
GAR MASS_7.2-47
GAR loaded via a namespace (and not attached):
GAR [1] cluster_1.12.0 grid_2.8.1 tools_2.8.1
GAR __
GAR R-help@r-project.org mailing list
GAR https://stat.ethz.ch/mailman/listinfo/r-help
GAR PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
GAR 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
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Re: [R] graphically representing frequency of words in a speech?
There is a similar discussion in statalist (http://n2.nabble.com/st%3A-Tag-clouds-in-Stata--tt2992551.html#none), I think they make a reasonable argument that tag cloud is not a good statistical graphic. 2009/6/10 Yihui Xie xieyi...@gmail.com: Hi, As Gregor Gorjanc mentioned, it's very inconvenient to let R decide the fontsize and placement of words in a plot. There have already been very mature applications of tag cloud; one of them I'm relatively familiar is the WordPress plugin wp-cumulus, which makes use of a Flash object to generate tag cloud, and it has fantastic 3D rotation effect of the cloud. I've spent a couple of hours porting it into R; see the source code and effect here: http://yihui.name/en/2009/06/creating-tag-cloud-using-r-and-flash-javascript-swfobject/ HTH. Regards, Yihui -- Yihui Xie xieyi...@gmail.com Phone: +86-(0)10-82509086 Fax: +86-(0)10-82509086 Mobile: +86-15810805877 Homepage: http://www.yihui.name School of Statistics, Room 1037, Mingde Main Building, Renmin University of China, Beijing, 100872, China On Mon, Jun 8, 2009 at 2:41 AM, Brown, Tony Nicholastony.n.br...@vanderbilt.edu wrote: Dear all, I recently saw a graph on television that displayed selected words/phrases in a speech scaled in size according to their frequency. So words/phrases that were often used appeared large and words that were rarely used appeared small. The closest thing I can find on the web to approximate what I saw can be found here: http://stateoftheunion.onetwothree.net/ The example at that website is more complicated but captures the general idea. Would someone point me in the right direction in terms of replicating such a graph. Thanks in advance, Tony - Tony N. Brown, Ph.D. Editor-Elect, American Sociological Review Associate Professor of Sociology and Human and Organizational Development (secondary) Program Faculty, Effective Health Communication and African American Diaspora Studies Faculty Head of Hank Ingram House, The Commons Vanderbilt University (615) 322-7518 (615) 322-7505 fax [[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. -- HUANG Ronggui, Wincent PhD Candidate Dept of Public and Social Administration City University of Hong Kong Home page: http://asrr.r-forge.r-project.org/rghuang.html __ 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] Unable to load package:lme4 [ Ubuntu 9.04 ]
Hi Martin,
I upgraded to R 2.9.0, but still have the same problem. The error
message, output of packageDescription(lme4), and sessionInfo() are
displayed below.
Thanks,
-Girish
--
library(lme4)
Loading required package: Matrix
Loading required package: lattice
Attaching package: 'Matrix'
The following object(s) are masked from package:stats :
xtabs
The following object(s) are masked from package:base :
rcond
Error in dyn.load(file, DLLpath = DLLpath, ...) :
function 'cholmod_start' not provided by package 'Matrix'
Error: package/namespace load failed for 'lme4'
packageDescription(lme4)
Package: lme4
Version: 0.999375-28
Date: 2008-12-13
Title: Linear mixed-effects models using S4 classes
Author: Douglas Bates ba...@stat.wisc.edu, Martin Maechler
maech...@r-project.org and Bin Dai d...@stat.wisc.edu
Maintainer: Douglas Bates ba...@stat.wisc.edu
Description: Fit linear and generalized linear mixed-effects models.
Depends: methods, R(= 2.7.0), Matrix(= 0.999375-11), lattice
LinkingTo: Matrix, stats
Imports: graphics, stats
Suggests: mlmRev, MEMSS
LazyLoad: yes
LazyData: yes
License: GPL (=2)
URL: http://lme4.r-forge.r-project.org/
Packaged: Sat Dec 13 20:07:38 2008; maechler
Built: R 2.8.0; x86_64-pc-linux-gnu; 2008-12-15 13:58:54; unix
-- File: /usr/lib/R/site-library/lme4/Meta/package.rds
sessionInfo()
R version 2.9.0 (2009-04-17)
x86_64-pc-linux-gnu
locale:
LC_CTYPE=en_IN;LC_NUMERIC=C;LC_TIME=en_IN;LC_COLLATE=en_IN;LC_MONETARY=C;LC_MESSAGES=en_IN;LC_PAPER=en_IN;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_IN;LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods
base
other attached packages:
[1] Matrix_0.999375-27 lattice_0.17-25
loaded via a namespace (and not attached):
[1] grid_2.9.0
==
On Jun 10, 1:43 pm, Martin Maechler maech...@stat.math.ethz.ch
wrote:
How did you re-install Matrix?
Your version Matrix_0.999375-17 (indicated below)
is a bit old, and indeed probably too old for the version of
lme4 you've tried.
Can you call
packageDescription(lme4)
without an error {probably not}, or at least
library(help = lme4)
the output of these would show you the version of Matrix you
need at least.
Ideally you'd upgrade your version of R (to 2.9.0) as well,
since the latest couple of versions of Matrix all have
R = 2.9.0 in their dependency requirements.
Regards,
Martin Maechler
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Unable to load package:lme4 [ Ubuntu 9.04 ]
GAR == Girish A R garam...@gmail.com
on Wed, 10 Jun 2009 02:06:57 -0700 (PDT) writes:
GAR Hi Martin,
GAR I upgraded to R 2.9.0, but still have the same problem. The error
GAR message, output of packageDescription(lme4), and sessionInfo() are
GAR displayed below.
GAR Thanks,
GAR -Girish
GAR --
library(lme4)
GAR Loading required package: Matrix
GAR Loading required package: lattice
GAR Attaching package: 'Matrix'
GAR The following object(s) are masked from package:stats :
GAR xtabs
GAR The following object(s) are masked from package:base :
GAR rcond
GAR Error in dyn.load(file, DLLpath = DLLpath, ...) :
GAR function 'cholmod_start' not provided by package 'Matrix'
GAR Error: package/namespace load failed for 'lme4'
GAR
packageDescription(lme4)
GAR Package: lme4
GAR Version: 0.999375-28
GAR Date: 2008-12-13
GAR Title: Linear mixed-effects models using S4 classes
GAR Author: Douglas Bates ba...@stat.wisc.edu, Martin Maechler
GAR maech...@r-project.org and Bin Dai d...@stat.wisc.edu
GAR Maintainer: Douglas Bates ba...@stat.wisc.edu
GAR Description: Fit linear and generalized linear mixed-effects models.
GAR Depends: methods, R(= 2.7.0), Matrix(= 0.999375-11), lattice
GAR LinkingTo: Matrix, stats
GAR Imports: graphics, stats
GAR Suggests: mlmRev, MEMSS
GAR LazyLoad: yes
GAR LazyData: yes
GAR License: GPL (=2)
GAR URL: http://lme4.r-forge.r-project.org/
GAR Packaged: Sat Dec 13 20:07:38 2008; maechler
GAR Built: R 2.8.0; x86_64-pc-linux-gnu; 2008-12-15 13:58:54; unix
^^
i.e. installed a while ago, and *not* reinstalled recently
Try to re-install both Matrix and then lme4 using your current R version
2.9.0.
Apart from that, yes, you are indeed using current versions of
Matrix and lme4...
Martin
GAR -- File: /usr/lib/R/site-library/lme4/Meta/package.rds
GAR
sessionInfo()
GAR R version 2.9.0 (2009-04-17)
GAR x86_64-pc-linux-gnu
GAR locale:
GAR
LC_CTYPE=en_IN;LC_NUMERIC=C;LC_TIME=en_IN;LC_COLLATE=en_IN;LC_MONETARY=C;LC_MESSAGES=en_IN;LC_PAPER=en_IN;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_IN;LC_IDENTIFICATION=C
GAR attached base packages:
GAR [1] stats graphics grDevices utils datasets methods
GAR base
GAR other attached packages:
GAR [1] Matrix_0.999375-27 lattice_0.17-25
GAR loaded via a namespace (and not attached):
GAR [1] grid_2.9.0
GAR
==
GAR On Jun 10, 1:43 pm, Martin Maechler maech...@stat.math.ethz.ch
GAR wrote:
How did you re-install Matrix?
Your version Matrix_0.999375-17 (indicated below)
is a bit old, and indeed probably too old for the version of
lme4 you've tried.
Can you call
packageDescription(lme4)
without an error {probably not}, or at least
library(help = lme4)
the output of these would show you the version of Matrix you
need at least.
Ideally you'd upgrade your version of R (to 2.9.0) as well,
since the latest couple of versions of Matrix all have
R = 2.9.0 in their dependency requirements.
Regards,
Martin Maechler
__
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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] filled.contour as image
Hi, Is there a way to make the filled.contour() function produce a plot similar to the image() function? i.e. not have smooth contours but rather distinct squares corresponding to the different cells of the matrix. The reason I ask is because since image doesn't have an option to display colour bars it is a lot less useful. Cheers Muri Soares Department of Maths and Applied Maths University of Cape Town South Africa [[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] Unable to load package:lme4 [ Ubuntu 9.04 ]
Hi Martin, Thanks for your prompt response! As suggested, I re-installed 'lme4' and am now able to load it without any problem. best, -Girish --- packageDescription(lme4) Package: lme4 Version: 0.999375-31 Date: 2009-05-20 Title: Linear mixed-effects models using S4 classes Author: Douglas Bates ba...@stat.wisc.edu and Martin Maechler maech...@r-project.org Maintainer: Douglas Bates ba...@stat.wisc.edu Description: Fit linear and generalized linear mixed-effects models. Depends: methods, R(= 2.9.0), Matrix(= 0.999375-24), lattice LinkingTo: Matrix, stats Imports: graphics, stats Suggests: mlmRev, MEMSS LazyLoad: yes LazyData: yes License: GPL (= 2) URL: http://lme4.r-forge.r-project.org/ Packaged: 2009-05-20 17:41:16 UTC; bates Repository: CRAN Date/Publication: 2009-05-24 08:36:20 Built: R 2.9.0; x86_64-pc-linux-gnu; 2009-06-10 09:26:57 UTC; unix ^ -- File: /usr/lib/R/site-library/lme4/Meta/package.rds On Jun 10, 2:15 pm, Martin Maechler maech...@stat.math.ethz.ch wrote: [.part of message truncated.] packageDescription(lme4) GAR Package: lme4 GAR Version: 0.999375-28 GAR Date: 2008-12-13 GAR Title: Linear mixed-effects models using S4 classes GAR Author: Douglas Bates ba...@stat.wisc.edu, Martin Maechler GAR maech...@r-project.org and Bin Dai d...@stat.wisc.edu GAR Maintainer: Douglas Bates ba...@stat.wisc.edu GAR Description: Fit linear and generalized linear mixed-effects models. GAR Depends: methods, R(= 2.7.0), Matrix(= 0.999375-11), lattice GAR LinkingTo: Matrix, stats GAR Imports: graphics, stats GAR Suggests: mlmRev, MEMSS GAR LazyLoad: yes GAR LazyData: yes GAR License: GPL (=2) GAR URL:http://lme4.r-forge.r-project.org/ GAR Packaged: Sat Dec 13 20:07:38 2008; maechler GAR Built: R 2.8.0; x86_64-pc-linux-gnu; 2008-12-15 13:58:54; unix ^^ i.e. installed a while ago, and *not* reinstalled recently Try to re-install both Matrix and then lme4 using your current R version 2.9.0. Apart from that, yes, you are indeed using current versions of Matrix and lme4... Martin __ 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] Using ADF.Test
ehxpieterse wrote: Hi, I am quite new to R and would appreciate some guidance, if possible. I have imported a csv file: spread - read.csv(Spread.csv) I get the following error when I try to run adf.test: adf.test(spread,alternative = c(stationary, explosive),0) Error in embed(y, k) : 'x' is not a vector or matrix Why is this? What does class(spread) give? You should try as.vector(spread) and see what happens -- View this message in context: http://www.nabble.com/Using-ADF.Test-tp23940824p23959465.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] Problem : solving a equation with R , fail with uniroot function
Thanks for your help !
I got another problem with my function :
test - function(x,bb0=-3,bb1=5,c0=2,r0=0) {
((exp(c0-r0)*(bb0+x)*(bb1-x))/((bb0+x+1)*(bb1-x-1))-1)}
With this :
curve(test(x),from=-10,to=10)
we can see that there are 2 roots : one in [3,4] and one in [5,10]
uniroot(test,c(5, 10))$root
[1] 5.243265
uniroot(test,c(3, 4))$root
[1] 3.069772
Even this works :
uniroot(test,c(3, 10))$root
[1] 5.243263
but it's unfortunate that :
uniroot(test3,c(-10, 10))$root
*Erreur dans uniroot(test3, c(-10, 10)) :
les valeurs de f() aux points extrêmes ne sont pas de signe opposé*
doesn't work, because I have to find the root for c0 in [-10,10] but I can't
change the range at each loop:
I want to do something like this, do you have any ideas ?
for (i in -10:10) {
+ z - uniroot(test,c(-10, 10), c0=i)$root
+ print(z)}
Yann
On Wed, Jun 10, 2009 at 2:33 AM, Ben Bolker bol...@ufl.edu wrote:
Peter Alspach wrote:
Tena koe Yann
It may not be the only problem, but you are missing an operator between
(bb0+x+1) and (bb1-x-1).
And fixing that typo and viewing the function shows that it
is a bit problematic on this interval anyway ... (diverges
at x = 1, closer look shows it's pretty nasty -- I haven't stopped
to think about its behavior very carefully, but it will clearly
take some effort. If you're lucky there is a typo and the
real function is better behaved!)
test - function(x,bb0=-3,bb1=5,c0=2,r0=0) {
((exp(c0-r0)*(bb0+x)*(bb1-x))/((bb0+x+1)*(bb1-x-1))-1)}
uniroot(test,c(-100,100))$root
curve(test(x),from=-100,to=100)
curve(test(x),from=1.1,to=10)
abline(h=0,col=2)
--
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Re: [R] poly regression
They have different coefficients because their model matrices are different but they both lead to the same predictions: fitted(lm(y~1+x+I(x^2))) 1 2 3 4 5 6 7 8 9 10 1 4 9 16 25 36 49 64 81 100 fitted(lm(y~poly(x,2))) 1 2 3 4 5 6 7 8 9 10 1 4 9 16 25 36 49 64 81 100 On Wed, Jun 10, 2009 at 1:41 AM, Ning Mapnin...@gmail.com wrote: hi, I want to do a polynomial regression of y on x of degree 2, as following x-1:10 y-x^2 lm(y~poly(x,2)) Call: lm(formula = y ~ poly(x, 2)) Coefficients: (Intercept) poly(x, 2)1 poly(x, 2)2 38.50 99.91 22.98 Which is not what i had expected. If I wrote the expression in an explicit form, y~1+x+I(x^2), I could get the expected result: lm(y~1+x+I(x^2)) Call: lm(formula = y ~ 1 + x + I(x^2)) Coefficients: (Intercept) x I(x^2) 0 0 1 What is the diff between them? __ 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] R on EC2 and R CMD javareconf
Java on Fedora 8 is a little funny so my guess is that is where the problem is (or that you are missing the Java development (*-devel) packages). I am (right now) building an Amazon image with Fedora 11 on x86_64 with R (and the Engineering and Scientific group) installed: let me know if you want me to share it once it is done even though you'd need a 40c/hour instance to run it. Allan. Saptarshi Guha wrote: Hello, I am using EC2 to launch several instances. On each instance, I perform the following commands yum -y install R (installs R-2.6) R CMD javareconf sleep 10 wget rpackage.tgz R CMD INSTALL rpackage.tgz Now, rpackage.tgz needs to be built with jni libraries. When the instance is fully up, i log in and check if rpackage was successfully installed and find out it wasn't. Running R CMD INSTALL rpackage.tgz again, i get the response that I should run R CMD javareconf, which I do and then successfully install rpackage (using R CMD...) Q: Why doesn't it work the first time? I am using a small instance and Fedora 8 AMI Regards Saptarshi __ 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] Using ADF.Test
On Wed, 10 Jun 2009, matifou wrote: ehxpieterse wrote: Hi, I am quite new to R and would appreciate some guidance, if possible. I have imported a csv file: spread - read.csv(Spread.csv) I get the following error when I try to run adf.test: adf.test(spread,alternative = c(stationary, explosive),0) Error in embed(y, k) : 'x' is not a vector or matrix Why is this? What does class(spread) give? data.frame, almost certainly, which is not a vector or a matrix. The original poster should access the relevant column, e.g. by spread[,1] or spread$foo etc. or (even better) turn it into a proper ts object! You should try as.vector(spread) and see what happens If spread is a 1-column data.frame, then that won't change anything. That's because is.vector(as.vector(...)) is not necessarily TRUE. Consider x - cars[, 1, drop=FALSE] is.vector(x) is.vector(as.vector(x)) identical(x, as.vector(x)) Z -- View this message in context: http://www.nabble.com/Using-ADF.Test-tp23940824p23959465.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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help with package simsalabim
I have attached a text file representing the centralized amplitude of a signal, sampled at 30Hz, whose length N = 6922 My goal is to remove the trend. I am using package simsalabim. I ran command decompSSA with L = length(Amps)/5 The reason is that I have SSA/MTM toolkit running in Mac/OS. SSA/MTM documentation, relative to SSA, recommends that N/10 = L = N/5. Documentation of simsalabim package recommens that L=N/2 I ran the command getSignal and felt uncomfortable at picking values for parameters C0 and r0. I would appreciate some suggestion/guidelines because these values seem to influence the trend extraction. I chose C0=0.0005, r0=0.0005 which are both below the Nyquist frequency 1/Amps.dc$L = 0.0007223346 As a consequence, getSignal finds two Eigenvectors capturing the trend. From my experience with toolkit SSA/MTM I know this output can be misleading. I am trying to figure out whether trend has really been found by double-checking the print-out data: Amps.signal$trend k smoothness explainedVariance 1 10.134495414 26.254096 11 12 0.006693601 1.627475 The above print-out marks EOF_1 and EOF_12 as capturing all the trend in the signal. I canNOT understand the difference between smoothness and explainedVariance I tried to compare such values with the leading frequency associated with EOF_1 and EOF_12. Unluckily, I am unable to draw any comclusion: Amps.dc$freq[1] #EOF_1 LEADING FREQUENCY [1] 0.0007225434 Amps.dc$freq[12] #EOF_12 LEADING FREQUENCY [1] 0.002890173 getSignal on-line documentation states about Trend smoothness: The proportion of variance explained by the frequencies =q omega0. But nowhere I could find what q is equal to. I guess the above sentence indicates a fraction of the nyquist frequency Omega_0 getSignal on-line documentation states about Trend explainedVariance: Percentage of the variance explained by the Eigenvalues with rank k Does the above variance refer to the signal variance I would greatly appreciate your help at using this package properly. Thank you in advance. Maura e tutti i telefonini TIM! Vai su e tutti i telefonini TIM! Vai su e tutti i telefonini TIM! Vai su __ 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] isolating Hour and minute form date and time
Hi everybody. I have a dataframe that contains a factor with the date information in the format like in the example below: 2009/05/12 11:22:31 AM I have been able to convert it to POSIXt using strptime Now I want to print only the date as one vector and the time in another vector but they must be real date and time classes so that i can use them in calculations to reconstruct my situation: date=as.factor(2009/05/12 11:22:31 AM) posix.date=strptime(date, %Y/%m/%d %I:%M:%S %p) I tried: time=format(posix.date, %A %H:%M) # For weekday, hour and minute but the class in character. Now it prints like i want but it cannot be used in calculations. I would appreciate any suggestions Thanks in advance Christiaan [[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] isolating Hour and minute form date and time
Try this and see R News 4/1 for more. now - Sys.time(); now [1] 2009-06-10 07:26:23 EDT library(chron) ch - as.chron(format(now)) dd - dates(ch); dd; dd+1 day 06/10/09 day 06/11/09 tt - times(ch) - times(dd); tt; tt+1/24 day 07:26:23 day 08:26:23 On Wed, Jun 10, 2009 at 7:13 AM, christiaan pauwcjp...@gmail.com wrote: Hi everybody. I have a dataframe that contains a factor with the date information in the format like in the example below: 2009/05/12 11:22:31 AM I have been able to convert it to POSIXt using strptime Now I want to print only the date as one vector and the time in another vector but they must be real date and time classes so that i can use them in calculations to reconstruct my situation: date=as.factor(2009/05/12 11:22:31 AM) posix.date=strptime(date, %Y/%m/%d %I:%M:%S %p) I tried: time=format(posix.date, %A %H:%M) # For weekday, hour and minute but the class in character. Now it prints like i want but it cannot be used in calculations. I would appreciate any suggestions Thanks in advance Christiaan [[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] Using ADF.Test
Achim Zeileis wrote: [...] is.vector(as.vector(...)) is not necessarily TRUE. Consider x - cars[, 1, drop=FALSE] is.vector(x) is.vector(as.vector(x)) identical(x, as.vector(x)) interesting. i wonder why as.vector does not give, at the very least, a warning when the result of its application is not a vector. the purpose of as.vector, one might guess, is to obtain a vector from whatever input -- or have the statement fail if such a conversion is impossible. the 'help' page for as.vector does not really explain what as.vector returns. the section 'value' talks about vector, exclusively: Value: For 'vector', a vector of the given length and mode. Logical vector elements are initialized to 'FALSE', numeric vector elements to '0', character vector elements to '', raw vector elements to 'nul' bytes and list elements to 'NULL'. All attributes are removed from the answer if it is of an atomic mode. fine, but what about as.vector? the section 'description' says: 'as.vector', a generic, attempts to coerce its argument into a vector of mode 'mode' (the default is to coerce to whichever mode is most convenient). fine, but what if the attempt is unsuccessful? the following fails: as.vector(as.vector) # Error in as.vector(x, mode) : # cannot coerce type 'closure' to vector of type 'any' but as.vector(data.frame()) 'works', even though the result is not a vector of type 'any' either. perhaps the 'help' page might be made more helpful? vQ __ 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] isolating Hour and minute form date and time
There is no class for times only in R. Your best bet is probably a difftime, e.g. time - difftime(posix.date, as.Date(posix.date), units=secs) Allan. christiaan pauw wrote: Hi everybody. I have a dataframe that contains a factor with the date information in the format like in the example below: 2009/05/12 11:22:31 AM I have been able to convert it to POSIXt using strptime Now I want to print only the date as one vector and the time in another vector but they must be real date and time classes so that i can use them in calculations to reconstruct my situation: date=as.factor(2009/05/12 11:22:31 AM) posix.date=strptime(date, %Y/%m/%d %I:%M:%S %p) I tried: time=format(posix.date, %A %H:%M) # For weekday, hour and minute but the class in character. Now it prints like i want but it cannot be used in calculations. I would appreciate any suggestions Thanks in advance Christiaan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ggplot, qplot: alpha channel for colors corresponding to factor
Hi, I have a qplot like the one in the minimal example below, except I also have faceting like this: qplot(jitter(Goodall),jitter(Better.adapt),colour=Second.adapt,facets=~Pol,data=d1) and with the real data I get quite a lot of overplotting, so I would like to add an alpha channel. In addition, I would like to be able to control which colors are used for each value of Second.adapt (which takes 7 different values). With my original data, qplot by default uses a range from blue to red, which is quite suitable, but I probably need to change this to shades of grey for publication. I tried qplot(jitter(Goodall),jitter(Better.adapt),colour=I(alpha(Second.adapt,1/5)),facets=~Pol,data=d1) but that does not give me what I want (There is no legend and I don't think anything corresponds to the seven values of Second.adapt). Minimal example: Goodall - c(rep(1:3,5)) Better.adapt - c(rep(1,7),rep(2,8)) Second.adapt - c(rep(1:5,3)) d1 - data.frame(Goodall=c(Goodall),Better.adapt=c(Better.adapt),Second.adapt=c(Second.adapt)) library(ggplot2) qplot(jitter(Goodall),jitter(Better.adapt),colour=Second.adapt,data=d1) # follwoing doesn't work: qplot(jitter(Goodall),jitter(Better.adapt),colour=I(alpha(Second.adapt,1/5)),data=d1) Thanks for any pointers, Marianne -- Marianne Promberger PhD http://www.psych.upenn.edu/~mpromber PGP/GnuPG public key ID 80AD9916 __ 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] filled.contour as image
Muri Soares wrote: Hi, Is there a way to make the filled.contour() function produce a plot similar to the image() function? i.e. not have smooth contours but rather distinct squares corresponding to the different cells of the matrix. The reason I ask is because since image doesn't have an option to display colour bars it is a lot less useful. Hi Muri, Have a look at color2D.matplot in the plotrix package. Jim __ 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] plot two variograms on a same graph
Hi, I would know how to plot two variograms on a same graph. I can plot one by one but I would draw both on the same one. Is it possible? Do i need any special package? Thanks! Cordialement Damien Landais __ 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, qplot: alpha channel for colors corresponding to factor
Dear Marianne, If find that a bit easier with ggplot() instead of qplot() d1 - data.frame(Goodall=c(rep(1:3,5)), Better.adapt = c(rep(1,7),rep(2,8)),Second.adapt=c(rep(1:5,3))) library(ggplot2) ggplot(d1, aes(x= Goodall, y = Better.adapt, colour=Second.adapt)) + geom_jitter(alpha = 0.2) + scale_colour_gradient(low = white, high = black) Another option would be to convert Second.adapt to a factor and use scale_colour_manual d1$Second - factor(d1$Second.adapt) ggplot(d1, aes(x= Goodall, y = Better.adapt, colour=Second)) + geom_jitter(alpha = 0.9) + scale_colour_manual(values = c(white, grey, black, green, red)) HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner 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 -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens Marianne Promberger Verzonden: woensdag 10 juni 2009 13:37 Aan: R-help Onderwerp: [R] ggplot, qplot: alpha channel for colors corresponding to factor Hi, I have a qplot like the one in the minimal example below, except I also have faceting like this: qplot(jitter(Goodall),jitter(Better.adapt),colour=Second.adapt,facets=~P ol,data=d1) and with the real data I get quite a lot of overplotting, so I would like to add an alpha channel. In addition, I would like to be able to control which colors are used for each value of Second.adapt (which takes 7 different values). With my original data, qplot by default uses a range from blue to red, which is quite suitable, but I probably need to change this to shades of grey for publication. I tried qplot(jitter(Goodall),jitter(Better.adapt),colour=I(alpha(Second.adapt,1 /5)),facets=~Pol,data=d1) but that does not give me what I want (There is no legend and I don't think anything corresponds to the seven values of Second.adapt). Minimal example: Goodall - c(rep(1:3,5)) Better.adapt - c(rep(1,7),rep(2,8)) Second.adapt - c(rep(1:5,3)) d1 - data.frame(Goodall=c(Goodall),Better.adapt=c(Better.adapt),Second.adapt= c(Second.adapt)) library(ggplot2) qplot(jitter(Goodall),jitter(Better.adapt),colour=Second.adapt,data=d1) # follwoing doesn't work: qplot(jitter(Goodall),jitter(Better.adapt),colour=I(alpha(Second.adapt,1 /5)),data=d1) Thanks for any pointers, Marianne -- Marianne Promberger PhD http://www.psych.upenn.edu/~mpromber PGP/GnuPG public key ID 80AD9916 __ 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. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ 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] R CMD check does not find a mistake
MM == Martin Maechler maech...@stat.math.ethz.ch on Wed, 10 Jun 2009 08:50:46 +0200 writes: CG == Christophe Genolini cgeno...@u-paris10.fr on Tue, 09 Jun 2009 16:17:15 +0200 writes: CG Hi the list, I build a package. They was a mistake in CG it, but R CMD check did not find it. Is that normal ? CG Here is what Kurt gets (which is right, I did this CG mistake): CG --- 8 CG * checking for code/documentation mismatches ... WARNING CG S4 class codoc mismatches from documentation object 'LongData-class': CG Slots for class 'LongData' CG Code: id other time traj varName CG Docs: id time traj varName CG --- 8 MM Others have already hinted at the solution: MM The problem only shows in R-devel (2.10.x). MM Note that I had sent an explicit message to this list this list (above) was not quite correct, excuse me. I meant the R-devel list, and I indeed would suggest that active R package authors *do* subscribe to R-devel (or read it very regularly through one of its mirrors/ list-interfaces). Martin MM *exactly* in order to advise all package authors : MM Subject: [Rd] R-devel:codocClasses() now finds more -- R CMD check warnings MM Date: Wed, 3 Jun 2009 11:45:45 +0200 MM i.e. only 6 days before you saw the problem MM MM yes, I am sometimes too optimistic, assuming that people MM actually read what I write MM Regards anyway! MM Martin MM -- MM Martin Maechler, ETH Zurich MM __ MM R-help@r-project.org mailing list MM https://stat.ethz.ch/mailman/listinfo/r-help MM PLEASE do read the posting guide http://www.R-project.org/posting-guide.html MM 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] plot two variograms on a same graph
Dear Damien, I tend do use ggplot2 for more advanced plotting. You only have to create a dataframe with all the data you need. Here are some examples. library(gstat) library(ggplot2) data(meuse) coordinates(meuse) = ~x+y g1 - gstat(id = Raw, formula = log(zinc)~1, data = meuse) g2 - gstat(id = Linear trend, formula = log(zinc)~x+y, data = meuse) vg1 - variogram(g1) vg2 - variogram(g2) ggplot(rbind(vg1, vg2), aes(x = dist, y = gamma, colour = id)) + geom_point() vgm1 - fit.variogram(vg1, vgm(1, Sph, 700, 1)) vgm2 - fit.variogram(vg2, vgm(1, Sph, 700, 1)) vgLine - rbind( cbind(variogramLine(vgm1, maxdist = max(vg1$dist)), id = Raw), cbind(variogramLine(vgm2, maxdist = max(vg2$dist)), id = Linear trend) ) ggplot(rbind(vg1, vg2), aes(x = dist, y = gamma, colour = id)) + geom_line(data = vgLine) + geom_point() HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner 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 -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens damien landais Verzonden: woensdag 10 juni 2009 13:55 Aan: R-help@r-project.org Onderwerp: [R] plot two variograms on a same graph Hi, I would know how to plot two variograms on a same graph. I can plot one by one but I would draw both on the same one. Is it possible? Do i need any special package? Thanks! Cordialement Damien Landais __ 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. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ 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] filled.contour as image
The image function in package Matrix plots a color legend by default. Searching on what seemed like the obvious strategy image legend in R Site Search would have identified this and a huge number of other alternatives, of which Lemon's color2D.matplot is the tenth. -- David Winsemius On Jun 10, 2009, at 5:28 AM, Muri Soares wrote: Hi, Is there a way to make the filled.contour() function produce a plot similar to the image() function? i.e. not have smooth contours but rather distinct squares corresponding to the different cells of the matrix. The reason I ask is because since image doesn't have an option to display colour bars it is a lot less useful. Cheers Muri Soares Department of Maths and Applied Maths University of Cape Town South Africa [[alternative HTML version deleted]] 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] ggplot, qplot: alpha channel for colors corresponding to factor
Many thanks for the quick reply! d1 - data.frame(Goodall=c(rep(1:3,5)), Better.adapt = c(rep(1,7),rep(2,8)),Second.adapt=c(rep(1:5,3))) library(ggplot2) ggplot(d1, aes(x= Goodall, y = Better.adapt, colour=Second.adapt)) + geom_jitter(alpha = 0.2) + scale_colour_gradient(low = white, high = black) Great, this is exactly what I was looking for. In case someone else needs this in the future, for my original data, I was able to add the faceting using: ggplot(d1, aes(x= Goodall, y = Better.adapt, colour=Second.adapt)) + geom_jitter(alpha = 0.6) + scale_colour_gradient(low = white, high = black) + facet_wrap(~ Pol, ncol = 4) (Pol has length 16, so it is nice to have it wrapped to four columns using facet_wrap instead of facet_grid) Thanks again, Marianne Another option would be to convert Second.adapt to a factor and use scale_colour_manual d1$Second - factor(d1$Second.adapt) ggplot(d1, aes(x= Goodall, y = Better.adapt, colour=Second)) + geom_jitter(alpha = 0.9) + scale_colour_manual(values = c(white, grey, black, green, red)) HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner 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 -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens Marianne Promberger Verzonden: woensdag 10 juni 2009 13:37 Aan: R-help Onderwerp: [R] ggplot, qplot: alpha channel for colors corresponding to factor Hi, I have a qplot like the one in the minimal example below, except I also have faceting like this: qplot(jitter(Goodall),jitter(Better.adapt),colour=Second.adapt,facets=~P ol,data=d1) and with the real data I get quite a lot of overplotting, so I would like to add an alpha channel. In addition, I would like to be able to control which colors are used for each value of Second.adapt (which takes 7 different values). With my original data, qplot by default uses a range from blue to red, which is quite suitable, but I probably need to change this to shades of grey for publication. I tried qplot(jitter(Goodall),jitter(Better.adapt),colour=I(alpha(Second.adapt,1 /5)),facets=~Pol,data=d1) but that does not give me what I want (There is no legend and I don't think anything corresponds to the seven values of Second.adapt). Minimal example: Goodall - c(rep(1:3,5)) Better.adapt - c(rep(1,7),rep(2,8)) Second.adapt - c(rep(1:5,3)) d1 - data.frame(Goodall=c(Goodall),Better.adapt=c(Better.adapt),Second.adapt= c(Second.adapt)) library(ggplot2) qplot(jitter(Goodall),jitter(Better.adapt),colour=Second.adapt,data=d1) # follwoing doesn't work: qplot(jitter(Goodall),jitter(Better.adapt),colour=I(alpha(Second.adapt,1 /5)),data=d1) Thanks for any pointers, Marianne -- Marianne Promberger PhD http://www.psych.upenn.edu/~mpromber PGP/GnuPG public key ID 80AD9916 __ 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. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. -- Marianne Promberger PhD http://www.psych.upenn.edu/~mpromber PGP/GnuPG public key ID 80AD9916 __ 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] arima modeling for multiple time series
Dear R People: Is there a package for arima modeling of multiple time series, please? I think that Dr. Paul Gilbert may have one, but I'm drawing a blank on the package name. Thanks in advance, Sincerely, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodg...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot two variograms on a same graph
Hi,
It is possible, but without you specifying in which pacakge you are
going to fit the variograms it is hard for us to provide an example. And
what do you mean by variogram, is that the sample variogram or the
fitted variogram model, or both? Try and keep to the posting guide next
time to make it possible for us to answer your question. When you use
gstat for geostatistics you can use somehting along the line of
(assuming that only the variogram model is different):
library(automap) # I use automap to fit the variogram, available on CRAN
library(lattice)
data(meuse)
coordinates(meuse) = ~x+y
av1 = autofitVariogram(log(zinc)~dist, meuse, model = Sph)
av2 = autofitVariogram(log(zinc)~dist, meuse, model = Exp)
# Make the plot
xyplot(gamma ~ dist, av1$exp_var,
panel = function(...) {
panel.xyplot(...)
# First model
vm1 = av1$var_model
ret = variogramLine(vm1, maxdist = 1600)
llines(ret$dist, ret$gamma)
# Second model
vm2 = av2$var_model
ret = variogramLine(vm2, maxdist = 1600)
llines(ret$dist, ret$gamma, col = red)
}
)
cheers,
Paul
ps There is a mailing list for geographical problems, r-sig-geo. These
kinds of problems are likely to get more response there.
damien landais wrote:
Hi,
I would know how to plot two variograms on a same graph. I can plot one by one
but I would draw both on the same one.
Is it possible? Do i need any special package?
Thanks!
Cordialement
Damien Landais
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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
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Re: [R] poly regression
To get the result that you were expecting, use the following (which uses the raw polynomial a + bx + cx^2 rather than the orthogonal polynomial of degree 2): lm(y~poly(x,2, raw=TRUE)) Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: rvarad...@jhmi.edu Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Varadhan.h tml -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Gabor Grothendieck Sent: Wednesday, June 10, 2009 6:37 AM To: Ning Ma Cc: r-help@r-project.org Subject: Re: [R] poly regression They have different coefficients because their model matrices are different but they both lead to the same predictions: fitted(lm(y~1+x+I(x^2))) 1 2 3 4 5 6 7 8 9 10 1 4 9 16 25 36 49 64 81 100 fitted(lm(y~poly(x,2))) 1 2 3 4 5 6 7 8 9 10 1 4 9 16 25 36 49 64 81 100 On Wed, Jun 10, 2009 at 1:41 AM, Ning Mapnin...@gmail.com wrote: hi, I want to do a polynomial regression of y on x of degree 2, as following x-1:10 y-x^2 lm(y~poly(x,2)) Call: lm(formula = y ~ poly(x, 2)) Coefficients: (Intercept) poly(x, 2)1 poly(x, 2)2 38.50 99.91 22.98 Which is not what i had expected. If I wrote the expression in an explicit form, y~1+x+I(x^2), I could get the expected result: lm(y~1+x+I(x^2)) Call: lm(formula = y ~ 1 + x + I(x^2)) Coefficients: (Intercept) x I(x^2) 0 0 1 What is the diff between them? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-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] arima modeling for multiple time series
An R Site Search on the obvious strategy arima Gilbert produces two
hits of which the first is:
dse-package {dse1}
On Jun 10, 2009, at 9:13 AM, Erin Hodgess wrote:
Dear R People:
Is there a package for arima modeling of multiple time series, please?
I think that Dr. Paul Gilbert may have one, but I'm drawing a blank on
the package name.
Thanks in advance,
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
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David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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[R] Predict GARCH
hello, i was trying to predict values for a garch, so i did: predict(fitgarch,n.ahead = 20) but this doesn't work. Someone can tell me how to get the 20 values ahead of a garch model. thanks in advance _ O Windows Live ajuda-o a manter-se em contacto com todos os seus amigos, num só local. http://www.microsoft.com/portugal/windows/windowslive/products/social-network-connector.aspx [[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] (sem assunto)
hello, i was trying to predict values for a garch, so i did: predict(fitgarch,n.ahead = 20) but this doesn't work. Someone can tell me how to get the 20 values ahead of a garch model. thanks in advance _ Obtenha 30 Emoticons grátis para o seu Windows Live Messenger http://www.livemessenger-emoticons.com/funfamily/pt-pt/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem : solving a equation with R , fail with uniroot function
yann lancien wrote:
Thanks for your help !
I got another problem with my function :
test - function(x,bb0=-3,bb1=5,c0=2,r0=0) {
((exp(c0-r0)*(bb0+x)*(bb1-x))/((bb0+x+1)*(bb1-x-1))-1)}
With this :
curve(test(x),from=-10,to=10)
we can see that there are 2 roots : one in [3,4] and one in [5,10]
uniroot(test,c(5, 10))$root
[1] 5.243265
uniroot(test,c(3, 4))$root
[1] 3.069772
Even this works :
uniroot(test,c(3, 10))$root
[1] 5.243263
but it's unfortunate that :
uniroot(test3,c(-10, 10))$root
*Erreur dans uniroot(test3, c(-10, 10)) :
les valeurs de f() aux points extrêmes ne sont pas de signe opposé*
doesn't work, because I have to find the root for c0 in [-10,10] but I
can't
change the range at each loop:
I want to do something like this, do you have any ideas ?
for (i in -10:10) {
+ z - uniroot(test,c(-10, 10), c0=i)$root
+ print(z)}
Yann
This is the point at which somebody's (I forget whose) R-help signature
comes to mind: what is the problem you are trying to solve?
You could develop an elaborate brute-force solution that combined
try() [to catch the function when it doesn't work] and attempts to
find roots in (for example) [-10,-9],[-9,-8],[-8,-7], ... [9,10] -- but
how are you going to decide which root you want when there is
more than one? I'm afraid this
is no longer an R problem ... you might have to go back and re-think
your approach to the problem ...
Ben Bolker
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Re: [R] Comparing R and SAS
Satish, There are nearly as many opinions as people on this question. So accept these as my views only. 1. I have written S packages, SAS procedures and SAS macros over my career. S was specifically designed for extensibility and it shows. The ratio of time to get a new statistical idea up and running is about 1:5:20 for S:macro:proc, using the therneau days metric. New ideas are the lifeblood for an academic; SAS makes no sense in that environment and so will always lag far behind. For simple things like a repetive recoding, a 2-10 line R function say, SAS macro will be competive with R in terms of programming effort. 2. The SAS programming model could be described as mini batch; the block of statements including and following a data statment is executed as a unit. For data manipulation this is turns out to be a very powerful programming paradym and many people, myself included, prefer it for data input and manipulation. At the time I was first using SAS (mid 70s) no other package or system was even close to SAS in this arena, and by the late 80s many disciplines with complex, large, or messy data had wedded themselves to SAS, e.g., drug company clinical trials. Most are still wedded. Terry Therneau __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating a specific skewed distribution
With skewed unimodal distributions, the mode can't equal the mean, so assuming you want the mean to be around 30, I find that a weibull function can get close to what you want: mean(rweibull(1e5,1.5,33)) [1] 29.77781 pweibull(60,1.5,33) [1] 0.9138475 I'm sure you can play with the parameters to try to get even closer to what you want. On Wed, Jun 10, 2009 at 2:48 AM, David Arnolddwarnol...@suddenlink.net wrote: All, Can someone help me create a skewed distribution, mean = 30, with probability of selecting a random number from the distribution greater than or equal 60 equal to 10%? I need the probability density function to equal zero at zero, and have a maximum height at or near 30. Is this possible? And if it is possible, how can I adjust the distribution so that the probability of selecting a random number greater than or equal to 60 is p. Thanks. No idea how to start. David [[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. -- Mike Lawrence Graduate Student Department of Psychology Dalhousie University Looking to arrange a meeting? Check my public calendar: http://tr.im/mikes_public_calendar ~ Certainty is folly... I think. ~ __ 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] Merge data frame and keep unmatched
Hi,
With two data sets, one complete and another one partial, I would like to
merge them and keep the unmatched lines. The problem is that merge() dosen't
keep the unmatched lines. Is there another function that I could use to
merge the data frames.
Example:
completedf - expand.grid(alpha=letters[1:3],beta=1:3)
partdf - data.frame(
alpha= c('a','a','c'),
beta = c(1,3,2),
val = c(2,6,4))
mergedf - merge(x=completedf, y=partdf, by=c('alpha','beta'))
# it only kept the common rows
nrow(mergedf)
Thanks,
Etienne
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Re: [R] Merge data frame and keep unmatched
On Jun 10, 2009, at 8:56 AM, Etienne B. Racine wrote:
Hi,
With two data sets, one complete and another one partial, I would
like to
merge them and keep the unmatched lines. The problem is that merge()
dosen't
keep the unmatched lines. Is there another function that I could use
to
merge the data frames.
Example:
completedf - expand.grid(alpha=letters[1:3],beta=1:3)
partdf - data.frame(
alpha= c('a','a','c'),
beta = c(1,3,2),
val = c(2,6,4))
mergedf - merge(x=completedf, y=partdf, by=c('alpha','beta'))
# it only kept the common rows
nrow(mergedf)
Thanks,
Etienne
Is this what you want?
merge(x=completedf, y=partdf, by=c('alpha','beta'), all = TRUE)
alpha beta val
1 a1 2
2 a2 NA
3 a3 6
4 b1 NA
5 b2 NA
6 b3 NA
7 c1 NA
8 c2 4
9 c3 NA
Note the 'all', 'all.x' and 'all.y' arguments...
HTH,
Marc Schwartz
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Re: [R] Merge data frame and keep unmatched
Try:
merge(completedf, partdf, all.x = TRUE)
or
library(sqldf) # see http://sqldf.googlecode.com
sqldf(select * from completedf left join partdf using(beta, alpha))
On Wed, Jun 10, 2009 at 9:56 AM, Etienne B. Racineetienn...@gmail.com wrote:
Hi,
With two data sets, one complete and another one partial, I would like to
merge them and keep the unmatched lines. The problem is that merge() dosen't
keep the unmatched lines. Is there another function that I could use to
merge the data frames.
Example:
completedf - expand.grid(alpha=letters[1:3],beta=1:3)
partdf - data.frame(
alpha= c('a','a','c'),
beta = c(1,3,2),
val = c(2,6,4))
mergedf - merge(x=completedf, y=partdf, by=c('alpha','beta'))
# it only kept the common rows
nrow(mergedf)
Thanks,
Etienne
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Re: [R] Creating a specific skewed distribution
You could also make some algebra. e.g. : ?rweibull gives the formula of the mean and of the cumulative distribution function in the Details section. So using your known parameters (i.e. mean=30 and p(10)=.10, i.e. cumulative function(10) =.90), I think it is sufficient to determine the exact values you want shape and scale to be. And you can do that with any probability function (some easier the calculate) as most (if not all) have these equations. Of course, it might be more pleasing to play with the parameters. Etienne Mike Lawrence wrote: With skewed unimodal distributions, the mode can't equal the mean, so assuming you want the mean to be around 30, I find that a weibull function can get close to what you want: mean(rweibull(1e5,1.5,33)) [1] 29.77781 pweibull(60,1.5,33) [1] 0.9138475 I'm sure you can play with the parameters to try to get even closer to what you want. On Wed, Jun 10, 2009 at 2:48 AM, David Arnolddwarnol...@suddenlink.net wrote: All, Can someone help me create a skewed distribution, mean = 30, with probability of selecting a random number from the distribution greater than or equal 60 equal to 10%? I need the probability density function to equal zero at zero, and have a maximum height at or near 30. Is this possible? And if it is possible, how can I adjust the distribution so that the probability of selecting a random number greater than or equal to 60 is p. Thanks. No idea how to start. David [[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. -- Mike Lawrence Graduate Student Department of Psychology Dalhousie University Looking to arrange a meeting? Check my public calendar: http://tr.im/mikes_public_calendar ~ Certainty is folly... I think. ~ __ 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. -- View this message in context: http://www.nabble.com/Creating-a-specific-skewed-distribution-tp23956114p23963140.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] MCMC validity question
Hello, I have quite a tough problem, which might be able to be solved by MCMC. I am fairly new to MCMC (in the learning process) - so apologize if the answer is totally obvious, and any hints, links etc are greatly appreciated. I'll illustrate the problem in a version cut-down to the essentials - the real problem is ways more complex. Suppose I have a Markovian series of poisson processes, which generate (or kill) some objects. At t0, say I have initially x = 3 objects. In the next time step, t1, the number of total objects is Poisson-distributed, subject to a function of x at t0, covariates, and parameters. So x_t1 ~ Pois, with E(x_t1) = f(x_t0, covariates, parameters). Let's choose a very simple function for f, say just f = x * par1 * Covariate1. Now let this process be repeated for say 6 times, always with the number of objects obtained in a previous step as input (x) for the next step. The problem is, at all time steps the total number of objects remains unobservable, because they are only detected with a certain, low probability (itself subject to covariates and parameters). So if you observe say 2 objects, the only thing you know is the figure must be = 2. Presume however that the detection prob is equal and independent for all objects at a time step, and so the observed number of objects is also Poisson-distributed. The likelihood-function is then built upon that figure. The main problem is the input to function f: In the first step, I know what x is (or even don't know that, might again just be from a distribution). From that step on, I have only a Poisson-distribution, and lack the concrete realization; all I know is a pure minimum value. In general f is not a simply thing, and is quite impracticable to input a distribution itself; moreover, because of the inflation of variance, the output could not be treated as lambda of a Poisson-distribution any more. So the Poisson-distribution is lost, although physical knowledge tells you it really is (and would be, were it x was known precisely). The question is therefore, how can I work around the fact that x is always known to be only from a distribution, not knowing the precise realization? Ignoring above issue, and keeping in mind that I have shown only a simplified version of the model, MCMC methods seem a reasonable choice. I had an idea, which looks so strange, that I have strong doubts if it's valid to so: Say I have a present parameter set. Is it then possible to estimate the Poisson-lambda for t1 using f, then draw a random number from that distribution, and now treat that drawn figure as (fixed) input for calculation for t2, and so on, repeating it until t6 ? At the end, propose new parameters based on MCMC sampling, and repeat all over again. Will a long sequence of MCMC iterations homogenize these fake (simulated) realizations of the poisson-distributions, and so the chain will converge to the posterior distribution of the parameters? many many thanks for any inputs and thoughts, cheers, Thomas __ 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 get the unique pairs of a set of pairs dataframe ?
Hi friends, Please can anyone help me with an easier solution of doing the below mentioned work. Suppose i have a dataset like this:--- i1 i2 i3 i4 i5 1 7 13 1 2 2 8 14 2 2 3 9 15 3 3 4 10 16 4 4 5 11 17 5 5 6 12 18 6 7 *i1,i2,i3,i4,i5 are my items.I am able to find all possible pairs i.e Say this dataframe is item_pairs **i1,i2 **i1,i3 **i1,i4 i1,i5 **i2,i1 **i2,i3 i2,i4 i2,i5 **i3,i1 **i3,i2 **i3,i4 **i3,i5 **i4,i1 **i4,i2 **i4,i3 i4,i5 i5,i1 i5,i2 i5,i3 i5,i4 Pairs like (i1,i1) or (i2,i2) are not required.Now pair (i1,i2) is same as pair (i2,i1) .How can i chop off the second pair which is identicle to the first one,only that sequence of item numbers are different ,otherwise my dataset is same. I thought of something...like running loops for all rows and columns of this item_pairs dataframe and check if first item in a pair matches with any second item of previous pair and similarly **second item in a pair matches with any first item of previous pair* and keep entering them in a dataframe and get the unique pairs. But is there any other easier way of chopping off the identicle pairs? *-- * Thanks in advance Moumita [[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] Using yum to install R-2.9
Hello, On Fedora 8, yum install R , installs R-2.6. I dont have much experience with yum, how do change the repos(?) to install R-2.9? Thank you Saptarshi __ 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 get the unique pairs of a set of pairs dataframe ?
Try this: do.call(rbind, apply(combn(names(x), 2), 2, function(n)expand.grid(x[,n[1]], x[,n[2]]))) On Wed, Jun 10, 2009 at 11:47 AM, Moumita Das das.moumita.onl...@gmail.comwrote: Hi friends, Please can anyone help me with an easier solution of doing the below mentioned work. Suppose i have a dataset like this:--- i1 i2 i3 i4 i5 1 7 13 1 2 2 8 14 2 2 3 9 15 3 3 4 10 16 4 4 5 11 17 5 5 6 12 18 6 7 *i1,i2,i3,i4,i5 are my items.I am able to find all possible pairs i.e Say this dataframe is item_pairs **i1,i2 **i1,i3 **i1,i4 i1,i5 **i2,i1 **i2,i3 i2,i4 i2,i5 **i3,i1 **i3,i2 **i3,i4 **i3,i5 **i4,i1 **i4,i2 **i4,i3 i4,i5 i5,i1 i5,i2 i5,i3 i5,i4 Pairs like (i1,i1) or (i2,i2) are not required.Now pair (i1,i2) is same as pair (i2,i1) .How can i chop off the second pair which is identicle to the first one,only that sequence of item numbers are different ,otherwise my dataset is same. I thought of something...like running loops for all rows and columns of this item_pairs dataframe and check if first item in a pair matches with any second item of previous pair and similarly **second item in a pair matches with any first item of previous pair* and keep entering them in a dataframe and get the unique pairs. But is there any other easier way of chopping off the identicle pairs? *-- * Thanks in advance Moumita [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to get the unique pairs of a set of pairs dataframe ?
?combn x [1] i1 i2 i3 i4 i5 combn(x,2) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] i1 i1 i1 i1 i2 i2 i2 i3 i3 i4 [2,] i2 i3 i4 i5 i3 i4 i5 i4 i5 i5 On Wed, Jun 10, 2009 at 10:47 AM, Moumita Das das.moumita.onl...@gmail.comwrote: Hi friends, Please can anyone help me with an easier solution of doing the below mentioned work. Suppose i have a dataset like this:--- i1 i2 i3 i4 i5 1 7 13 1 2 2 8 14 2 2 3 9 15 3 3 4 10 16 4 4 5 11 17 5 5 6 12 18 6 7 *i1,i2,i3,i4,i5 are my items.I am able to find all possible pairs i.e Say this dataframe is item_pairs **i1,i2 **i1,i3 **i1,i4 i1,i5 **i2,i1 **i2,i3 i2,i4 i2,i5 **i3,i1 **i3,i2 **i3,i4 **i3,i5 **i4,i1 **i4,i2 **i4,i3 i4,i5 i5,i1 i5,i2 i5,i3 i5,i4 Pairs like (i1,i1) or (i2,i2) are not required.Now pair (i1,i2) is same as pair (i2,i1) .How can i chop off the second pair which is identicle to the first one,only that sequence of item numbers are different ,otherwise my dataset is same. I thought of something...like running loops for all rows and columns of this item_pairs dataframe and check if first item in a pair matches with any second item of previous pair and similarly **second item in a pair matches with any first item of previous pair* and keep entering them in a dataframe and get the unique pairs. But is there any other easier way of chopping off the identicle pairs? *-- * Thanks in advance Moumita [[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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[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] checking and building ROracle on OS X
Hi, I'm running R 2.9.0 (from the DMG) on OS X 10.5 and am trying to get ROracle 0.5-9 to check and build. I have installed the full Oracle installation (10.2.0.4.0). The problem I'm facing is that I need to compile ROracle.so as 32 bit, but my environment needs to point to the 64 bit Oracle libs for the compilatin process to work (since the Pro*C compiler is 64 bit and needs the 64 bit libs). So my DYLD_LIBRARY_PATH poins to the 64 bit libs, but the package should be built with the 32 bit libs Looking at the configure.in file, I see that I can specify configure arguments to indicate a 32 bit build - but I can't see how I can specify these args via R CMD CHECK or R CMD BUILD. Any pointers would be appreciated. -- Rajarshi Guha [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Best way to plot a Matrix of all possible pair combinations
Hallo R Users, I have some distance matrix data like M[1:10,] [,1] [,2] [,3] [,4] [,5] [,6] [1,] 0.875 0.500 0.500 0.375 0.625 0. [2,] 0.8928571 1.000 0.000 0.8928571 0.1071429 0. [3,] 0.8928571 1.000 0.000 0.8928571 0.1071429 0. [4,] 0.8928571 1.000 0.000 0.8928571 0.1071429 0. [5,] 1.000 0.1304348 0.8695652 0.1304348 0.8695652 0. [6,] 1.000 0.1304348 0.8695652 0.1304348 0.8695652 0. [7,] 1.000 0.1304348 0.8695652 0.1304348 0.8695652 0. [8,] 0.9497743 0.7605977 0.2394023 0.7103720 0.2896280 0. [9,] 0.0754717 0.7169811 0.000 0.3584906 0.9245283 0.28301887 [10,] 0.3870968 0.9354839 0.000 0.4516129 0.6129032 0.06451613 where all the columns (1 to 6) are representing all possible pair-wise combination from four datasets, like cmbn [,1] [,2] [,3] [,4] [,5] [,6] [1,]111223 [2,]234344 and all the rows represent different species so; M[1,1] [1] 0.875## is the distance value for species 1 for combination data pair 1 2. Can any body please suggest me what will be the best way to represent these results in plot? Thanks a lot in advance, With best regard, Suparna --- [[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] GeoXp package
Dear all, I would like to use GeoXp package under Ubuntu Jaunty. I did install the required package GeoXp, for which I had to install before rgdal package. So when done, I want to load the library (library(GepXp), I got the error when GeoXP want to load rgdal package: library(sp) library(rgdal) Error in fun(...) : GDAL Error 1: libgrass_I.so: Ne peut ouvrir le fichier d'objet partagé: Aucun fichier ou dossier de ce type (sorry for the french :) ) I check on my system and libgrass_I.so is located at /usr/lib/grass/lib/libgrass_I.so. So how to specify the path of that file to allows rgdal loading ? Regards -- Emmanuel Poizot Cnam/Intechmer B.P. 324 50103 Cherbourg Cedex Phone (Direct) : (00 33)(0)233887342 Fax : (00 33)(0)233887339 __ 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] Delta in time series
Hi all, I have time data in the following format: 2009-03-09 12:00:00 2009-03-09 13:30:00 Now I want to compute the delta between the two time points, which results in this example in 1,5 h. I tried this via ts and the related functions but so far without sucess. Can anybody help me? Thank you very much in advance. -- View this message in context: http://www.nabble.com/Delta-in-time-series-tp23957835p23957835.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] R: Best way to plot a Matrix of all possible pair combinations
Hallo R Users, Please help I have some distance matrix data like M[1:10,] [,1] [,2] [,3] [,4] [,5] [,6] [1,] 0.875 0.500 0.500 0.375 0.625 0. [2,] 0.8928571 1.000 0.000 0.8928571 0.1071429 0. [3,] 0.8928571 1.000 0.000 0.8928571 0.1071429 0. [4,] 0.8928571 1.000 0.000 0.8928571 0.1071429 0. [5,] 1.000 0.1304348 0.8695652 0.1304348 0.8695652 0. [6,] 1.000 0.1304348 0.8695652 0.1304348 0.8695652 0. [7,] 1.000 0.1304348 0.8695652 0.1304348 0.8695652 0. [8,] 0.9497743 0.7605977 0.2394023 0.7103720 0.2896280 0. [9,] 0.0754717 0.7169811 0.000 0.3584906 0.9245283 0.28301887 [10,] 0.3870968 0.9354839 0.000 0.4516129 0.6129032 0.06451613 where all the columns (1 to 6) are representing all possible pair-wise combination from four datasets, like cmbn [,1] [,2] [,3] [,4] [,5] [,6] [1,]111223 [2,]234344 and all the rows represent different species so; M[1,1] [1] 0.875## is the distance value for species 1 for combination data pair 1 2. Can any body please suggest me what will be the best way to represent these results in plot? Thanks a lot in advance, With best regard, Suparna [[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] Delta in time series
koj wrote: Hi all, I have time data in the following format: 2009-03-09 12:00:00 2009-03-09 13:30:00 Now I want to compute the delta between the two time points, which results in this example in 1,5 h. I tried this via ts and the related functions but so far without sucess. Can anybody help me? Thank you very much in advance. Probably this is of interest for another users: I`ve found a solution, which computes the delta: (test-strptime(c(jens1$Z_Zeit[1], jens1$Z_Zeit[2]),%Y-%m-%d %H:%M:%S, tz=EST5EDT)) class(test) difftime(test[2],test[1],units=hours,tz=EST5EDT) -- View this message in context: http://www.nabble.com/Delta-in-time-series-tp23957835p23959207.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] graphically representing frequency of words in a speech?
Yihui, This is quite impressive, thanks for helping me think about how to make tag clouds in R. Tony -Original Message- From: Yihui Xie [mailto:xieyi...@gmail.com] Sent: Wednesday, June 10, 2009 3:15 AM To: Brown, Tony Nicholas Cc: r-help@r-project.org Subject: Re: [R] graphically representing frequency of words in a speech? Hi, As Gregor Gorjanc mentioned, it's very inconvenient to let R decide the fontsize and placement of words in a plot. There have already been very mature applications of tag cloud; one of them I'm relatively familiar is the WordPress plugin wp-cumulus, which makes use of a Flash object to generate tag cloud, and it has fantastic 3D rotation effect of the cloud. I've spent a couple of hours porting it into R; see the source code and effect here: http://yihui.name/en/2009/06/creating-tag-cloud-using-r-and-flash-javascript-swfobject/ HTH. Regards, Yihui -- Yihui Xie xieyi...@gmail.com Phone: +86-(0)10-82509086 Fax: +86-(0)10-82509086 Mobile: +86-15810805877 Homepage: http://www.yihui.name School of Statistics, Room 1037, Mingde Main Building, Renmin University of China, Beijing, 100872, China On Mon, Jun 8, 2009 at 2:41 AM, Brown, Tony Nicholastony.n.br...@vanderbilt.edu wrote: Dear all, I recently saw a graph on television that displayed selected words/phrases in a speech scaled in size according to their frequency. So words/phrases that were often used appeared large and words that were rarely used appeared small. The closest thing I can find on the web to approximate what I saw can be found here: http://stateoftheunion.onetwothree.net/ The example at that website is more complicated but captures the general idea. Would someone point me in the right direction in terms of replicating such a graph. Thanks in advance, Tony - Tony N. Brown, Ph.D. Editor-Elect, American Sociological Review Associate Professor of Sociology and Human and Organizational Development (secondary) Program Faculty, Effective Health Communication and African American Diaspora Studies Faculty Head of Hank Ingram House, The Commons Vanderbilt University (615) 322-7518 (615) 322-7505 fax [[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] Delta in time series
Try this: library(zoo) Lines - 2009-03-09 12:00:00,1 + 2009-03-09 13:30:00,2 z - read.zoo(textConnection(Lines), sep = ,, tz = ) 1/frequency(as.zooreg(z)) [1] 5400 noting that 5400 seconds is 1.5 hours. On Wed, Jun 10, 2009 at 5:04 AM, kojjens.k...@gmx.li wrote: Hi all, I have time data in the following format: 2009-03-09 12:00:00 2009-03-09 13:30:00 Now I want to compute the delta between the two time points, which results in this example in 1,5 h. I tried this via ts and the related functions but so far without sucess. Can anybody help me? Thank you very much in advance. -- View this message in context: http://www.nabble.com/Delta-in-time-series-tp23957835p23957835.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-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] SSOAP failing
Duncan Temple Lang a écrit : Olivier Cailloux wrote: Duncan Temple Lang a écrit : Olivier Cailloux wrote: Dear list, Sorry about that, my e-mail has been sent too soon by mistake (also from an incorrect exp. address). I was planning to add further details. So I'm running Debian Lenny. sessionInfo() gives: R version 2.7.1 (2008-06-23) i486-pc-linux-gnu locale: C attached base packages: [1] stats graphics grDevices utils datasets methods base Package: SSOAP Version: 0.5-0 Date: 2009/05/04 Package: XML Version: 2.3-0 Date: 2009/03/04 Package: RCurl Version: 0.97-3 Package: XMLSchema Version: 0.1-0 I must be stupid but I can't find XMLSchema on the omegahat website. (I just installed the 0.1-0 version a few days ago when trying to use SSOAP.) Google gives nothing useful for XMLSchema site:omegahat.org. http://www.omegahat.org/XMLSchema/XMLSchema_0.1-1.tar.gz or from a link in http://www.omegahat.org/XMLSchema/ or via the repository using install.packages(XMLSchema, repos = http://www.omegahat.org/R;) Thanks a lot: with the new version, calling my SOAP services work like a charm! It would be nice to add a link to XMLSchema from the main page, to make it easier to find for users... And for Google! (So that it will index the page.) Calling with the normal SOAP call works without trouble. However I can't use the genSOAPClientInterface way... See example hereunder. --- smg8Echo - SOAPServer(smg8.ulb.ac.be, axis2/services/EchoService, 8080) .SOAP(smg8Echo, doEchoOp, action=, xmlns=http://smg8.ulb.ac.be/echoSvc/;, toEchoInt=32) [1] 32 defEcho - processWSDL(http://smg8.ulb.ac.be:8080/axis2/services/EchoService?wsdl;) genSOAPClientInterface(def = defEcho) [1] TheString [1] TheInt Error in d...@operations[[1]] : subscript out of bounds length(defe...@operations) [1] 0 --- I can use the normal interface to my SOAP service anyway, so it is not *that* important if I can't make the genSOAPClientInterface call work. I just wanted to let you know, and also that way might reveal simpler for my user... Please tell me if I can try something else, if you have some time to look at it... Olivier I'm beginner in both web services and R programming (my knowledge is more in desktop CORBA Java programming). I am simply trying to make Java and R communicate (call Java code from R remotely), for use by someone else who works in R. I'd like to provide him with a fully functional example. I must recognize that having to step through SSOAP code to find where the problem lies might be too difficult considering my current knowledge (knowing I don't intend to become an expert R programmer). Please don't take it as rude or implying that I'm simply expecting someone to solve the problem for me: I am ready to investigate but I'd simply like to have an idea of how difficult it will be. If I have to patch part of SSOAP or if you think that the functions I need are not implemented yet, then I'd better not insist and try to communicate from R to Java using an other way. If on the other hand you think it should be working and is simply a matter of using the right version of the right package, or using an other web services platform on the Java side, or configuring something differently, etc., then I'll happily dig into it. Either of Simon Urbanek's rJava or Rserve packages are worth investigating. More generally, if someone has a suggestion on what to use to call Java over TCP/IP from R, I'd be interested. I am currently considering OSS as well as SSOAP. I'd prefer learning an approach that I can reuse for other languages, so possibly a standard like SOAP or CORBA would be best. Thanks for any pointer. Olivier D. I tried different action values, because I don't know what I should put there. But I always get the same error. Also, FYI, the following command produces the following output: smg8Def - processWSDL(http://smg8.ulb.ac.be:8080/axis2/services/Version?wsdl;, verbose=TRUE) processing (sub) schema http://axisversion.sample 1 ) Exception 2 ) Exception 3 ) getVersionResponse Warning message: In processWSDL(http://smg8.ulb.ac.be:8080/axis2/services/Version?wsdl;, : Ignoring additional serviceport ... elements and then I also tried a simpler service (WSDL manually created myself, but I am not sure it is correct although it validates): smg8EchoDef - processWSDL(http://smg8.ulb.ac.be:8080/axis2/services/EchoService?wsdl;, verbose=TRUE) processing (sub) schema http://smg8.ulb.ac.be/echoSvc/ 1 ) TheString genSOAPClientInterface(def = smg8EchoDef) Error in d...@operations[[1]] : subscript out of bounds Indeed: length(smg8echo...@operations) [1] 0 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide
Re: [R] Using yum to install R-2.9
On Jun 10, 2009, at 9:53 AM, Saptarshi Guha wrote: Hello, On Fedora 8, yum install R , installs R-2.6. I dont have much experience with yum, how do change the repos(?) to install R-2.9? Thank you Saptarshi Upgrade to a still supported version of Fedora. Fedora 8 went EOL back in January, which means that you have been without *any* updates (bug fixes, kernel updates, security patches, etc.) since then. With the release of Fedora 11 this week, Fedora 9 will go EOL in about a month, at which point, only Fedora 10 and 11 will be actively supported. If you are going to ride the Fedora train, you need to be willing to update your Fedora distribution relatively frequently. The current Fedora release and support schedule will generally allow you to skip one release, but not two. So if you are going to upgrade, you will want to move to at least Fedora 10. You could try to download and install the appropriate 32 or 64 bit F9 RPMs that Martyn prepares here: http://cran.r-project.org/bin/linux/redhat/fedora9/ But it may have F9 dependencies that can put you on a slippery slope of satisfying certain dependencies. I have not looked at the spec file. BTW, there is a Fedora specific R e-mail list: https://stat.ethz.ch/mailman/listinfo/r-sig-fedora which you might want to subscribe to and then post future Fedora/RH specific queries there. 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] Creating a specific skewed distribution
Etienne et al, This is exactly what I need. So I gave it an algebraic try and set: Mean=30=b*gamma(1+1/a), solve for b and substitute into F(60)=0.10. After a little algebra, this left me with [ 2*gamma(1+1/a) ]^a = -ln(0.10) Now, I don't think this has a closed form solution, at least not one I have the skill to find. So, how would I solve this using R? David. On Jun 10, 2009, at 7:09 AM, Etienne B. Racine wrote: You could also make some algebra. e.g. : ?rweibull gives the formula of the mean and of the cumulative distribution function in the Details section. So using your known parameters (i.e. mean=30 and p(10)=.10, i.e. cumulative function(10) =.90), I think it is sufficient to determine the exact values you want shape and scale to be. And you can do that with any probability function (some easier the calculate) as most (if not all) have these equations. Of course, it might be more pleasing to play with the parameters. Etienne Mike Lawrence wrote: With skewed unimodal distributions, the mode can't equal the mean, so assuming you want the mean to be around 30, I find that a weibull function can get close to what you want: mean(rweibull(1e5,1.5,33)) [1] 29.77781 pweibull(60,1.5,33) [1] 0.9138475 I'm sure you can play with the parameters to try to get even closer to what you want. On Wed, Jun 10, 2009 at 2:48 AM, David Arnolddwarnol...@suddenlink.net wrote: All, Can someone help me create a skewed distribution, mean = 30, with probability of selecting a random number from the distribution greater than or equal 60 equal to 10%? I need the probability density function to equal zero at zero, and have a maximum height at or near 30. Is this possible? And if it is possible, how can I adjust the distribution so that the probability of selecting a random number greater than or equal to 60 is p. Thanks. No idea how to start. David [[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. -- Mike Lawrence Graduate Student Department of Psychology Dalhousie University Looking to arrange a meeting? Check my public calendar: http://tr.im/mikes_public_calendar ~ Certainty is folly... I think. ~ __ 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. -- View this message in context: http://www.nabble.com/Creating-a-specific-skewed-distribution-tp23956114p23963140.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-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 a specific skewed distribution
Here is one way to solve the equation: require(BB) f - function(x) (2*gamma(1+1/x))^x + log (0.10) ans - dfsane(par=1, fn=f) ans Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: rvarad...@jhmi.edu Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Varadhan.h tml -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of David Arnold Sent: Wednesday, June 10, 2009 11:25 AM To: Etienne B. Racine Cc: r-help@r-project.org Subject: Re: [R] Creating a specific skewed distribution Etienne et al, This is exactly what I need. So I gave it an algebraic try and set: Mean=30=b*gamma(1+1/a), solve for b and substitute into F(60)=0.10. After a little algebra, this left me with [ 2*gamma(1+1/a) ]^a = -ln(0.10) Now, I don't think this has a closed form solution, at least not one I have the skill to find. So, how would I solve this using R? David. On Jun 10, 2009, at 7:09 AM, Etienne B. Racine wrote: You could also make some algebra. e.g. : ?rweibull gives the formula of the mean and of the cumulative distribution function in the Details section. So using your known parameters (i.e. mean=30 and p(10)=.10, i.e. cumulative function(10) =.90), I think it is sufficient to determine the exact values you want shape and scale to be. And you can do that with any probability function (some easier the calculate) as most (if not all) have these equations. Of course, it might be more pleasing to play with the parameters. Etienne Mike Lawrence wrote: With skewed unimodal distributions, the mode can't equal the mean, so assuming you want the mean to be around 30, I find that a weibull function can get close to what you want: mean(rweibull(1e5,1.5,33)) [1] 29.77781 pweibull(60,1.5,33) [1] 0.9138475 I'm sure you can play with the parameters to try to get even closer to what you want. On Wed, Jun 10, 2009 at 2:48 AM, David Arnolddwarnol...@suddenlink.net wrote: All, Can someone help me create a skewed distribution, mean = 30, with probability of selecting a random number from the distribution greater than or equal 60 equal to 10%? I need the probability density function to equal zero at zero, and have a maximum height at or near 30. Is this possible? And if it is possible, how can I adjust the distribution so that the probability of selecting a random number greater than or equal to 60 is p. Thanks. No idea how to start. David [[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. -- Mike Lawrence Graduate Student Department of Psychology Dalhousie University Looking to arrange a meeting? Check my public calendar: http://tr.im/mikes_public_calendar ~ Certainty is folly... I think. ~ __ 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. -- View this message in context: http://www.nabble.com/Creating-a-specific-skewed-distribution-tp239561 14p23963140.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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Two-sided log axis
Hi, I have two curves that are best displayed using a logarithmic y-axis. I want to plot the difference of these curves, which means I would need a log scaling on both the negative and the positive side (i.e. the y-axis should be -100 -10 -1 0 1 10 100). Zero values should not be discarded. Is there an easy way to do that? Thanks, John __ 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] Two-sided log axis
I would not have called such an axis logarithmic since the logarithm function is bounded below at 0, but I think that what you seek will be found among the examples that accompany Sarkar's book, chapter 8, figures 8.3 - 8.5: http://lmdvr.r-forge.r-project.org/figures/figures.html On Jun 10, 2009, at 12:05 PM, John Wiedenhoeft wrote: Hi, I have two curves that are best displayed using a logarithmic y- axis. I want to plot the difference of these curves, which means I would need a log scaling on both the negative and the positive side (i.e. the y-axis should be -100 -10 -1 0 1 10 100). Zero values should not be discarded. Is there an easy way to do that? Thanks, John 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] Two-sided log axis
On Jun 10, 2009, at 12:23 PM, David Winsemius wrote: I would not have called such an axis logarithmic since the logarithm function is bounded below at 0, (I meant to say that arguments to log are bounded at zero.) but I think that what you seek will be found among the examples that accompany Sarkar's book, chapter 8, figures 8.3 - 8.5: http://lmdvr.r-forge.r-project.org/figures/figures.html On Jun 10, 2009, at 12:05 PM, John Wiedenhoeft wrote: Hi, I have two curves that are best displayed using a logarithmic y- axis. I want to plot the difference of these curves, which means I would need a log scaling on both the negative and the positive side (i.e. the y-axis should be -100 -10 -1 0 1 10 100). Zero values should not be discarded. Is there an easy way to do that? Thanks, John 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.
[R] Analisys in Multidimensional contingency tables
Dear R-list,
Hi everyone, Im trying to make an analysis of multidimensional contingency
tables using R. I' working with the Agresti example where you have the
data from 3 categories. The thing is how can I do the analisys using the
G2 statistics. Somebody can send me an Idea?
I attach the program where you can find the data.
Best Regards,
prob1- data.frame(victim=c(rep('white',4),rep('black',4)),
+ perp=c(rep('white',2),rep('black',2),rep('white',2),rep('black',2)),
+ death=rep(c('yes','no'),4), count=c(19,132,11,52,0,9,6,97))
prob1
victim perp death count
1 white white yes19
2 white whiteno 132
3 white black yes11
4 white blackno52
5 black white yes 0
6 black whiteno 9
7 black black yes 6
8 black blackno97
ftable(xtabs(count ~ victim + perp + death, data = prob1))
death no yes
victim perp
black black97 6
white 9 0
white black52 11
white 132 19
O__ José Bustos Melo.
c/ /'_ --- Master in Applied Statistics
(*) \(*) -- B.S. in Marine Biology
-Science Faculty
-Catholic University of Concepcion
-Alonso de Ribera 2850 Concepción, Casilla 297
-Cell phone: +56 9 9 5939144
__
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] Predict GARCH
Suppose the GARCH(1,1) equation is : Sigma[t]^2 = w + a* Sigma[t-1]^2 + b*r[t-1]^2 One step ahead forecast : Sigma[t+1]^2 = w + a* Sigma[t]^2 + b*r[t]^2 All informations are available here Two step ahead forecast : Sigma[t+2]^2 = w + a* Sigma[t+1]^2 + b*r[t+1]^2 Here r[t+1] is not known at time t therefore is a r.v. Replacing this with it's expected value as r[t+1]^2 = E[r[t+1]^2] = sigma[t+1]^2, assuming E[r[t+1]] = 0 Therefore Sigma[t+2]^2 = w + a* Sigma[t+1]^2 + b*r[t+1]^2 = w + a* Sigma[t+1]^2 + b* Sigma[t+1]^2 = w + (a+b)* Sigma[t+1]^2 Carry on same procedure for next period forecast. Hope this helps. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Daniel Mail Sent: 10 June 2009 18:55 To: r-help@r-project.org Subject: [R] Predict GARCH hello, i was trying to predict values for a garch, so i did: predict(fitgarch,n.ahead = 20) but this doesn't work. Someone can tell me how to get the 20 values ahead of a garch model. thanks in advance _ O Windows Live ajuda-o a manter-se em contacto com todos os seus amigos, num ss local. http://www.microsoft.com/portugal/windows/windowslive/products/social-networ k-connector.aspx [[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] is it possible to combine multiple barplots?
Thank you for the helpful reply! I am relatively new to R (the software and the community) and was not aware of the example galler. Now I am. The example you found looks very close to what I am trying to do and I should be able to modify it. Best - P On Tue, Jun 9, 2009 at 4:45 PM, Titus von der Malsburgmalsb...@gmail.com wrote: On Tue, Jun 09, 2009 at 04:39:29PM +0200, Titus von der Malsburg wrote: is there a way to plot both of them in one plot, so that the bars for value 1 (dataA: 3, dataB: 5) would appear side by side, followed by the bars for value 2 etc.? Oh, I see you want something different. I should've read your message more closely. I found this example in the gallery: http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=54 Maybe it's close enough to what you want to do. Titus -- Sharing Nicely at www.bokaap.net __ 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] Analisys in Multidimensional contingency tables
Hi José, Hi everyone, Im trying to make an analysis of multidimensional contingency tables using R. I' working with the Agresti example where you have the data from 3 categories. The thing is how can I do the analisys using the G2 statistics. Somebody can send me an Idea? Please find below a link to an R companion for Agresti by Laura Thompson https://home.comcast.net/~lthompson221/Splusdiscrete2.pdf There are chances you find the example you would like to reproduce in there. HTH, Tobias __ 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] is it possible to combine multiple barplots?
philipp schmidt-3 wrote: i am working with two sets of likert scale type (4 distinct values) data: dataA - rep(1:4, c(3,2,2,4)) dataB - rep(1:4, c(5,4,3,2)) i can now (bar)plot both of these separately and compare the distributions. plot(table(dataA), type='h') plot(table(dataB), type='h') is there a way to plot both of them in one plot, so that the bars for value 1 (dataA: 3, dataB: 5) would appear side by side, followed by the bars for value 2 etc.? barplot(rbind(table(dataA),table(dataB)),beside=TRUE) or see plotrix::multhist -- View this message in context: http://www.nabble.com/is-it-possible-to-combine-multiple-barplots--tp23943970p23966848.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] Creating a specific skewed distribution
There are actually two roots to your equation: 0.329 and 1.385. f - function(x) (2*gamma(1+1/x))^x + log (0.10) x - seq(0.1, 5, length=500) plot(x, fn(x), type=l) abline(h = 0, lty=2, col=2) Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: rvarad...@jhmi.edu Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Varadhan.h tml -Original Message- From: Ravi Varadhan [mailto:rvarad...@jhmi.edu] Sent: Wednesday, June 10, 2009 11:53 AM To: 'David Arnold'; 'Etienne B. Racine' Cc: 'r-help@r-project.org' Subject: RE: [R] Creating a specific skewed distribution Here is one way to solve the equation: require(BB) f - function(x) (2*gamma(1+1/x))^x + log (0.10) ans - dfsane(par=1, fn=f) ans Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: rvarad...@jhmi.edu Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Varadhan.h tml -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of David Arnold Sent: Wednesday, June 10, 2009 11:25 AM To: Etienne B. Racine Cc: r-help@r-project.org Subject: Re: [R] Creating a specific skewed distribution Etienne et al, This is exactly what I need. So I gave it an algebraic try and set: Mean=30=b*gamma(1+1/a), solve for b and substitute into F(60)=0.10. After a little algebra, this left me with [ 2*gamma(1+1/a) ]^a = -ln(0.10) Now, I don't think this has a closed form solution, at least not one I have the skill to find. So, how would I solve this using R? David. On Jun 10, 2009, at 7:09 AM, Etienne B. Racine wrote: You could also make some algebra. e.g. : ?rweibull gives the formula of the mean and of the cumulative distribution function in the Details section. So using your known parameters (i.e. mean=30 and p(10)=.10, i.e. cumulative function(10) =.90), I think it is sufficient to determine the exact values you want shape and scale to be. And you can do that with any probability function (some easier the calculate) as most (if not all) have these equations. Of course, it might be more pleasing to play with the parameters. Etienne Mike Lawrence wrote: With skewed unimodal distributions, the mode can't equal the mean, so assuming you want the mean to be around 30, I find that a weibull function can get close to what you want: mean(rweibull(1e5,1.5,33)) [1] 29.77781 pweibull(60,1.5,33) [1] 0.9138475 I'm sure you can play with the parameters to try to get even closer to what you want. On Wed, Jun 10, 2009 at 2:48 AM, David Arnolddwarnol...@suddenlink.net wrote: All, Can someone help me create a skewed distribution, mean = 30, with probability of selecting a random number from the distribution greater than or equal 60 equal to 10%? I need the probability density function to equal zero at zero, and have a maximum height at or near 30. Is this possible? And if it is possible, how can I adjust the distribution so that the probability of selecting a random number greater than or equal to 60 is p. Thanks. No idea how to start. David [[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. -- Mike Lawrence Graduate Student Department of Psychology Dalhousie University Looking to arrange a meeting? Check my public calendar: http://tr.im/mikes_public_calendar ~ Certainty is folly... I think. ~ __ 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. -- View this message in context: http://www.nabble.com/Creating-a-specific-skewed-distribution-tp239561 14p23963140.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
[R] Extracting Sequence Data from a Vector
Thanks in advance. I have a vector of numbers which contain sections that are sequences which increase by a value of 1 followed by a gap in the data and then another sequence occurs, etc: x-c(1:3, 6: 7, 10:13) From the vector I need to extract 2 items of information A) the first number in the sequence (e.g., 1, 6, 10) and B) how many observations were in each sequence section (e.g., 3, 2, 4). v1 v2 1 3 6 2 104 It seems simple, but my R skills are still in their infancy so I very much appreciate the help. Eric -- Eric Vander Wal Ph.D. Candidate University of Saskatchewan, Department of Biology, 112 Science Place, Saskatoon, SK., S7N 5E2 Pluralitas non est ponenda sine neccesitate __ 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] Extracting Sequence Data from a Vector
try this: x [1] 1 2 3 6 7 10 11 12 13 # find breaks breaks - c(FALSE, diff(x) != 1) # now create matrix with groupings (just for visual) z - data.frame(x, cumsum(breaks)) # create list with first element of each seq and the length t(sapply(split(z, z[,2]), function(b) c(b[1,1], nrow(b [,1] [,2] 013 162 2 104 On Wed, Jun 10, 2009 at 1:28 PM, Eric Vander Wal eric.vander...@usask.cawrote: Thanks in advance. I have a vector of numbers which contain sections that are sequences which increase by a value of 1 followed by a gap in the data and then another sequence occurs, etc: x-c(1:3, 6: 7, 10:13) From the vector I need to extract 2 items of information A) the first number in the sequence (e.g., 1, 6, 10) and B) how many observations were in each sequence section (e.g., 3, 2, 4). v1 v2 1 3 6 2 104 It seems simple, but my R skills are still in their infancy so I very much appreciate the help. Eric -- Eric Vander Wal Ph.D. Candidate University of Saskatchewan, Department of Biology, 112 Science Place, Saskatoon, SK., S7N 5E2 Pluralitas non est ponenda sine neccesitate __ 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[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] rgl vertices
On 6/10/2009 1:44 PM, kapo coulibaly wrote: I'm trying to make sense of the following example in the package rgl: vertices - c( -1.0, -1.0, 0, 1.0, 1.0, -1.0, 0, 1.0, 1.0, 1.0, 0, 1.0, -1.0, 1.0, 0, 1.0 ) indices - c( 1, 2, 3, 4 ) open3d() wire3d( qmesh3d(vertices,indices) ) One would think that each vertex would have 3 coordinates (x,y,z) what does the fourth one in the definition of the variable vertices stand for. By default qmesh3d uses 4-coordinate homogeneous coordinates, because that's the coordinate system used by OpenGL. See the ?rgl::matrices help topic for a description of how they work. Duncan Murdoch __ 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] Extracting Sequence Data from a Vector
On Jun 10, 2009, at 1:28 PM, Eric Vander Wal wrote: Thanks in advance. I have a vector of numbers which contain sections that are sequences which increase by a value of 1 followed by a gap in the data and then another sequence occurs, etc: x-c(1:3, 6: 7, 10:13) From the vector I need to extract 2 items of information A) the first number in the sequence (e.g., 1, 6, 10) and B) how many observations were in each sequence section (e.g., 3, 2, 4). v1 v2 1 3 6 2 104 It seems simple, but my R skills are still in their infancy so I very much appreciate the help. Eric Yet another solution: x[which(c(-Inf, diff(x)) != 1)] [1] 1 6 10 diff(c(which(c(-Inf, diff(x)) != 1), length(x)+1) ) [1] 3 2 4 -- 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] Extracting Sequence Data from a Vector
jim holtman wrote: try this: Oh well, i spent the time writing this so i might as well post my (almost identical) solution, x-c(1:3, 6: 7, 10:13) breaks = c(TRUE, diff(x) != 1) data.frame(start = x[breaks], length = tabulate(cumsum(breaks))) Hoping this works, baptiste x [1] 1 2 3 6 7 10 11 12 13 # find breaks breaks - c(FALSE, diff(x) != 1) # now create matrix with groupings (just for visual) z - data.frame(x, cumsum(breaks)) # create list with first element of each seq and the length t(sapply(split(z, z[,2]), function(b) c(b[1,1], nrow(b [,1] [,2] 013 162 2 104 On Wed, Jun 10, 2009 at 1:28 PM, Eric Vander Wal eric.vander...@usask.cawrote: Thanks in advance. I have a vector of numbers which contain sections that are sequences which increase by a value of 1 followed by a gap in the data and then another sequence occurs, etc: x-c(1:3, 6: 7, 10:13) From the vector I need to extract 2 items of information A) the first number in the sequence (e.g., 1, 6, 10) and B) how many observations were in each sequence section (e.g., 3, 2, 4). v1 v2 1 3 6 2 104 It seems simple, but my R skills are still in their infancy so I very much appreciate the help. Eric -- Eric Vander Wal Ph.D. Candidate University of Saskatchewan, Department of Biology, 112 Science Place, Saskatoon, SK., S7N 5E2 Pluralitas non est ponenda sine neccesitate __ 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. __ 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] rgl vertices
I'm trying to make sense of the following example in the package rgl: vertices - c( -1.0, -1.0, 0, 1.0, 1.0, -1.0, 0, 1.0, 1.0, 1.0, 0, 1.0, -1.0, 1.0, 0, 1.0 ) indices - c( 1, 2, 3, 4 ) open3d() wire3d( qmesh3d(vertices,indices) ) One would think that each vertex would have 3 coordinates (x,y,z) what does the fourth one in the definition of the variable vertices stand for. 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] tracking progress of loading large dataset into R for updating progress bar
Hello UseRs, I am creating a GUI application in R using the RGtk2 library. I have incorporated a progress bar which I want to refresh periodically in keeping with the progress/completion of the read.csv function. I would like to show the progress bar in animation (pulsing of the bar) for as long as the read.csv function completes its task. I am not sure how I can track the progress of the read.csv function. As of now I just initialise the progress bar before the read.csv function and show the completed bar after the read.csv function has been executed. Is there a way to access the data.frame object and check for its completeness? Thank you for your feedback. Cheers Harsh Singhal Bangalore, India __ 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] by and by: using two indices in by() or tapply()
Hi everyone I want to apply a function by two indices. I have a number of surveyors submitting questionnaires. I want to check the time of the first submission for the day for each surveyor and also see a NA is no submission was done on a particular day. This generates a sample of the data: starttime=c(11:07:32,14:07:28,11:32:21,13:27:49,11:45:05, 12:30:06,10:27:07,10:18:07,15:29:36,16:29:23,13:46:45,10:45:26 ,09:21:14,10:29:51,12:32:56,11:06:02,12:41:36,11:03:47, 10:58:12,10:05:54) submitdate=c(2009-05-21,2009-06-02, 2009-05-12 ,2009-05-21, 2009-05-21, 2009-05-07, 2009-05-19 ,2009-05-13 ,2009-06-05, 2009-05-13, 2009-06-05, 2009-05-28, 2009-05-15, 2009-05-28, 2009-06-05, 2009-05-28, 2009-05-12, 2009-05-28, 2009-05-07, 2009-05-20) surveyor=rep(LETTERS[1:4],5) data=data.frame(surveyor, submitdate,starttime) I can generate a list of the earliest submission per day: tapply(starttime,submitdate,min) or of the earliest submission per surveyor: tapply(starttime,surveyor,min) or of the number of submissions per surveyor day: table(submitdate,surveyor) But what I want is the time of the earliest submission per surveyor per day (and NA's where applicable) Can anyone offer some advice Thanks Christiaan [[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] by and by: using two indices in by() or tapply()
Try this: tapply(starttime,list(submitdate, surveyor),min) On Wed, Jun 10, 2009 at 4:20 PM, christiaan pauw cjp...@gmail.com wrote: Hi everyone I want to apply a function by two indices. I have a number of surveyors submitting questionnaires. I want to check the time of the first submission for the day for each surveyor and also see a NA is no submission was done on a particular day. This generates a sample of the data: starttime=c(11:07:32,14:07:28,11:32:21,13:27:49,11:45:05, 12:30:06,10:27:07,10:18:07,15:29:36,16:29:23,13:46:45,10:45:26 ,09:21:14,10:29:51,12:32:56,11:06:02,12:41:36,11:03:47, 10:58:12,10:05:54) submitdate=c(2009-05-21,2009-06-02, 2009-05-12 ,2009-05-21, 2009-05-21, 2009-05-07, 2009-05-19 ,2009-05-13 ,2009-06-05, 2009-05-13, 2009-06-05, 2009-05-28, 2009-05-15, 2009-05-28, 2009-06-05, 2009-05-28, 2009-05-12, 2009-05-28, 2009-05-07, 2009-05-20) surveyor=rep(LETTERS[1:4],5) data=data.frame(surveyor, submitdate,starttime) I can generate a list of the earliest submission per day: tapply(starttime,submitdate,min) or of the earliest submission per surveyor: tapply(starttime,surveyor,min) or of the number of submissions per surveyor day: table(submitdate,surveyor) But what I want is the time of the earliest submission per surveyor per day (and NA's where applicable) Can anyone offer some advice Thanks Christiaan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Delta in time series
Or x=c(2009-03-09 12:00:00,2009-03-09 13:30:00) y=data.frame(x) attach(y) x=as.POSIXlt(x) x[1]-x[2] On Wed, Jun 10, 2009 at 5:04 AM, kojjens.k...@gmx.li wrote: I have time data in the following format: 2009-03-09 12:00:00 2009-03-09 13:30:00 [[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] GeoXp package
On Wed, Jun 10, 2009 at 08:21:06AM +0200, Poizot Emmanuel wrote: Error in fun(...) : GDAL Error 1: libgrass_I.so: Ne peut ouvrir le fichier d'objet partagé: Aucun fichier ou dossier de ce type (sorry for the french :) ) It would have been far more useful had you translated the error message to english than to have apologized. Try doing export LD_LIBRARY_PATH=/path/to/dir/where/so/is/located before starting R from the shell. __ 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] gpc.poly datatype
I have a list of polygons generated by the contourLines() command (each object of the list is a list in itself with two objects: a vector of x values, and a vector of y values for each vertex). I wish to convert that list into a gpc.poly object of multiple contours. How do I do this? gpclib apparently has no method of coercing lists into the gpc.poly object type. As well, when I have a gpc.poly object, I can view the points using the get.pts() command, but cannot find a way of only extracting the x values of a polygon (or the y values, for that matter). How do I do this? Any help would be greatly appreciated -- View this message in context: http://www.nabble.com/gpc.poly-datatype-tp23967664p23967664.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] Predict GARCH
Hello, On 6/10/09, Daniel Mail d20...@live.com.pt wrote: i was trying to predict values for a garch, so i did: predict(fitgarch,n.ahead = 20) but this doesn't work. Someone can tell me how to get the 20 values ahead of a garch model. You didn't specify what function you used to obtain the GARCH model. If you use rgarch [1], it provides functions for forecasting step-ahead values. Liviu [1] http://rgarch.r-forge.r-project.org/index.html __ 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] nonparametric test for row of a matrix
Hi, Are there any nonparametric test for ordinal categorical data for row of a matrix? Thank you very much, Stefo [[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] Difficulties with the simpest table
I have the data arranged in table (in Excel, or Notepad): x y 0 100 2 100 4 80 6 40 8 0 I need to transfer these data in R file and then going to operate with x and y as with variables. I should note, that I have 2.8.1 version. Thank you for advance. With regard, Dmitry. __ 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 extract from a matrix based on indices in a vector?
Sorry, there may be some lingo for describing the extraction problem I have, but I don't know it. I have a matrix of 2 rows and N columns, and a vector of N 1s and 2s. Matrix M: 1 2 3 4 ... N 1A C D G ... 2B D F H ... Vector v: 1 2 2 1 2 1 1 ... N I'd like to apply v to M to get Vector w: A D F G ... I.e. each element of v is interpreted as a row-index used to extract that row's value from the corresponding column in M into the corresponding column of w. Also eventually nrow(M) 2, in which case the value of the elements of v would range over 1:nrow(M). Seems it should be simple, but maybe not? Thanks! Doug PS : I originally posted this on R devel, then realized my mistake. Move thread didn't work when I tried it, though. -- View this message in context: http://www.nabble.com/How-to-extract-from-a-matrix-based-on-indices-in-a-vector--tp23967316p23967316.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] problem with PCA loading plot
Hi, I am a beginner with R. I would like to get a loading plot of PC 3 vs PC 1. For PC 1 vs PC 2 I use library(pls) loadingplot(pca.result, comps = 1:2, scatter = TRUE, labels=names) if I try loadingplot(pca.result, comps = 1:3, scatter = TRUE, labels=names) I get the loading plots of PC 1 vs PC 2, PC 1 vs PC 3 and PC 2 vs PC 3. What do I have to do to get just a single loading plot of PC 3 vs PC 1. Thanks for your help in advance ! -- View this message in context: http://www.nabble.com/problem-with-PCA-loading-plot-tp23967169p23967169.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] Vista + R = *!!?@
Hello People of R, Is there any way that I can get R to function properly using Vista. I get very strange output using lmer, as in no p-values. Is there ANY way I can fix this. Thank you for your time, John Townsend-Mehler PhD Candidate Department of Zoology Michigan State 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.
Re: [R] Vista + R = *!!?@
On Jun 10, 2009, at 1:57 PM, John Townsend-Mehler wrote: Hello People of R, Is there any way that I can get R to function properly using Vista. I get very strange output using lmer, as in no p-values. Is there ANY way I can fix this. That's not a Vista issue, it is a statistical theory issue. See the following R FAQ: http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-p_002dvalues-not-displayed-when-using-lmer_0028_0029_003f 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] Difficulties with the simpest table
Hello Dmitry, On 6/10/09, Dmitry Gospodaryov gospodar...@rambler.ru wrote: I have the data arranged in table (in Excel, or Notepad): x y 0 100 2 100 4 80 6 40 8 0 I need to transfer these data in R file and then going to operate with x and y as with variables. I should note, that I have 2.8.1 version. Thank you for advance. To beginners I often suggest to try Rcmdr, which is a graphical interface to R that smooths the transition to R. In your specific case I would export the data from Excel to a .csv file, and then import it using the Data Import data from text file menu in Rcmdr. Afterwards, you can manipulate the data either using the command line or Rcmdr's statistical menus. Liviu __ 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] Preventing .Call from printing a blank line to the R console on Windows
Hello everyone, I am using the CVODES integrator in the Rsundials package. Every time I call the integrator, a blank line is printed in the console. Using debug, I was able to isolate the problem to this line in the function cvodes(...): solutions = .Call(cvodes, PACKAGE = Rsundials, as.double(y), ... Since I am running the integrator thousands of times, I wind up with thousands of blank lines. This makes it hard to print out periodic updates from a for loop; they are lost among all the blank lines! Is it somehow possible to stop this .Call command from printing a blank line? Thank you for your help. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Difficulties with the simpest table
On Jun 10, 2009, at 11:27 AM, Dmitry Gospodaryov wrote: I have the data arranged in table (in Excel, or Notepad): x y 0 100 2 100 4 80 6 40 8 0 I need to transfer these data in R file and then going to operate with x and y as with variables. I should note, that I have 2.8.1 version. Thank you for advance. Dmitry, There is a complete R manual that describes how to get data into and out of R, called R Data Import/Export. It is available within your R distribution (if on Windows, via the R menus) or on the R web site at: http://cran.r-project.org/manuals.html There is also a page on the R Wiki at: http://wiki.r-project.org/rwiki/doku.php?id=tips:data-io:ms_windows which specifically covers R and Windows/Excel data exchange. A quick example, if the above is in Notepad, but with NO BLANK LINE between the x and y labels and the data. You can highlight the text, press CTRL-C to copy it to the clipboard and then use the following in R (see ?read.table): DF - read.table(clipboard, header = TRUE) That will give you a 'data frame' in R that contains the two columns, which can then be addressed as DF$x and DF$y. Further information is available in the R manual An Introduction to 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] Performance Data Analysis Class using R
GUERRILLA DATA ANALYSIS TECHNIQUES (GDAT) http://www.perfdynamics.com/Classes/Outlines/gdata.html with an emphasis on R and PDQ-R modeling tools http://www.perfdynamics.com/Tools/PDQ-R.html August 10-14 Instructors: Prof. David Lilja, Mr. Jim Holtman and Dr. Neil Gunther == LOCATION PLACE: Larkspur Landing, Pleasanton, in sunny California http://www.larkspurlanding.com/hotels/pleasanton/index.html RATES: See the schedule page for the special hotel room rates. http://www.perfdynamics.com/Classes/schedule.html *IMPORTANT * You must make your own hotel reservations online at the special group page set up the Larkspur Landing hotel. See our Schedule page http://www.perfdynamics.com/Classes/schedule.html for the links. Since this is a new hotel, if you run into difficulties, please call them at 925-463-1212 and identify yourself as a Performance Dynamics enrollee. == FOOD: Breakfast, lunch, and coffee breaks are included in the price. [[elided Yahoo spam]] ONLINE REGISTRATION === Register http://www.perfdynamics.com/regofax.html Contact http://www.perfdynamics.com/contact.html FON: 530-873-0575 FAX: 530-873-6697 EMAIL: clas...@perfdynamics.com We accept AMEX, VISA and MASTERCARD. Corporate discounts are available for enrollments of 3 or more. Registration processing may take up to 24 hours. All emails, faxes and phone calls will be acknowledged. Thanks for your patience. ___ PERFORMANCE DYNAMICS COMPANY(sm) http://www.perfdynamics.com/ ___ Consulting Services Educational Services 4061 East Castro Valley Blvd. P.O. Box 1238, Magalia Suite 110, Castro Valley California 95954, USA California 94552, USA Bookings and registrations FON: +1-510-537-5758 FON: +1-530-873-0575 FAX: Dial FON (Handshakes automagically) FAX: +1-530-873-6697 NET: i...@perfdynamics.comNET: clas...@perfdynamics.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] Comparing R and SAs
Satish, For a comparison of SAS and S, see the document An Introduction to S and the Hmisc and Design Libraries by Carlos Alzola and Frank E. Harrell. Frank Harrell is an expert in both SAS and R. You can download this document from http://www.r-project.org/, then click on manuals, and then contributed documentation. You can also look at the document written by Bob Muenchen (at http://RforSASandSPSSusers.com http://rforsasandspssusers.com/ (also a book published by Springer Verlag) for a comparison of SAS and R (and SPSS). I have been using both SAS and R. While my primary expertise is mainly in SAS, I have been using R more and more relative to SAS as my familiarity with it grows. From my point of view, cutting edge methodologies will always be implemented first in R (as you pointed out as well). SAS will follow several years later with some of these methodologies. Also, SAS has different products and users may not have all SAS products. Many firms have SAS/STAT but not other SAS products like SAS/ETS (economic time series), SAS/Enterpriser Miner or SAS/GRAPH. So in these situations R may be your only option. Even if you have these other SAS products you can do things more rapidly in R, if you take the time to learn it well, than you can with SAS. I have SAS/Enterprise Miner but still prefer R for neural networks, splines, decision trees, etc., as I can program R to produce several neural networks, etc. using for loops. SAS/Enterprise Miner cannot be programmed. R graphs are definitely superior to SAS graphics, and can be programmed very easily. I also use R for EDA (exploratory data analysis) prior to building predictive models/data mining. One area where SAS still excels is in processing huge files (over 30 GB in size - online data from vendors like double click with literally billions of records). But for statistical analysis you generally don't need to work with such large volumes of data. A much smaller random sample should suffice. If you have R running on Unix or Linux 64-bit operating systems (or Windows Vista?) and huge amounts of RAM handling large datasets in R is less of an issue. Also, if your data resides on mainframes, SAS is probably your only choice if you cannot download the mainframe data to your PC. I use R on a 32-bit Windows operating system with 3 GB of RAM, and I have not had any problem doing statistical analysis/data mining with R on around 25,000 or so records with anywhere from 25 to 50 variables. Hope this helps. Jude Satish wrote: Hi: For those of you who are adept at both SAS and R, I have the following questions: a) What are some reasons / tasks for which you would use R over SAS and vice versa? b) What are some things for which R is a must have that SAS cannot fulfill the requirements? I am on the ramp up on both of them. The general feeling that I am getting by following this group is that R updates to the product are at a much faster pace and therefore, this would be better for someone who wants the bleeding edge (correct me if I am wrong). But I am also interested in what is inherently better in R that SAS cannot offer perhaps because of the design. Thanks. Satish ___ Jude Ryan Director, Client Analytical Services Strategy Business Development UBS Financial Services Inc. 1200 Harbor Boulevard, 4th Floor Weehawken, NJ 07086-6791 Tel. 201-352-1935 Fax 201-272-2914 Email: jude.r...@ubs.com Please do not transmit orders or instructions regarding a UBS account electronically, including but not limited to e-mail, fax, text or instant messaging. The information provided in this e-mail or any attachments is not an official transaction confirmation or account statement. For your protection, do not include account numbers, Social Security numbers, credit card numbers, passwords or other non-public information in your e-mail. Because the information contained in this message may be privileged, confidential, proprietary or otherwise protected from disclosure, please notify us immediately by replying to this message and deleting it from your computer if you have received this communication in error. Thank you. UBS Financial Services Inc. UBS International Inc. UBS Financial Services Incorporated of Puerto Rico UBS AG UBS reserves the right to retain all messages. Messages are protected and accessed only in legally justified cases.__ 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] gpc.poly datatype
On 10/06/2009 1:53 PM, Derek Lacoursiere wrote: I have a list of polygons generated by the contourLines() command (each object of the list is a list in itself with two objects: a vector of x values, and a vector of y values for each vertex). I wish to convert that list into a gpc.poly object of multiple contours. How do I do this? gpclib apparently has no method of coercing lists into the gpc.poly object type. As well, when I have a gpc.poly object, I can view the points using the get.pts() command, but cannot find a way of only extracting the x values of a polygon (or the y values, for that matter). How do I do this? Any help would be greatly appreciated After running example(contourLines), this comes close: lines - contourLines(x,y,volcano) pts - lapply(lines, function(line) list(x=line$x, y=line$y, hole=F)) gp - new(gpc.poly, pts=pts) It is not quite right because some of the contours are incomplete, and this joins them up, but it should give you the idea. I don't know how gpc.poly handles partial polygons. Duncan Murdoch __ 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] Preventing .Call from printing a blank line to the R console on Windows
On 10/06/2009 5:16 PM, Joel wrote: Hello everyone, I am using the CVODES integrator in the Rsundials package. Every time I call the integrator, a blank line is printed in the console. Using debug, I was able to isolate the problem to this line in the function cvodes(...): solutions = .Call(cvodes, PACKAGE = Rsundials, as.double(y), ... Since I am running the integrator thousands of times, I wind up with thousands of blank lines. This makes it hard to print out periodic updates from a for loop; they are lost among all the blank lines! Is it somehow possible to stop this .Call command from printing a blank line? The best way is to fix the package so it doesn't do that, but a workaround using capture.output() or sink() is probably possible. Duncan Murdoch __ 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 extract from a matrix based on indices in a vector?
On Jun 10, 2009, at 12:36 PM, Logickle wrote: Sorry, there may be some lingo for describing the extraction problem I have, but I don't know it. I have a matrix of 2 rows and N columns, and a vector of N 1s and 2s. Matrix M: 1 2 3 4 ... N 1A C D G ... 2B D F H ... Vector v: 1 2 2 1 2 1 1 ... N I'd like to apply v to M to get Vector w: A D F G ... I.e. each element of v is interpreted as a row-index used to extract that row's value from the corresponding column in M into the corresponding column of w. Also eventually nrow(M) 2, in which case the value of the elements of v would range over 1:nrow(M). Seems it should be simple, but maybe not? Thanks! Doug Doug, Try this: # M is a character matrix M 1 2 3 4 1 A C D G 2 B D F H # v is a vector, equal in length to the number of columns in M v [1] 1 2 2 1 # Get the diagonal of the matrix that results from the combinations of indices diag(M[v, 1:ncol(M)]) [1] A D F G I created a larger 2 row matrix to test further: set.seed(1) M - matrix(sample(LETTERS, 40, replace = TRUE), 2, 20) M [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [1,] G O F Y Q F R U S J Y Q [2,] J X X R B E J M Z U F D [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20] [1,] G A W M M V U S [2,] K J I P E R C K and 'v': set.seed(2) v - sample(2, 20, replace = TRUE) v [1] 1 2 2 1 2 2 1 2 1 2 2 1 2 1 1 2 2 1 1 1 Then: diag(M[v, 1:ncol(M)]) [1] G X X Y B E R M S U F Q K A W P E [18] V U S Looks like it might work. 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 extract from a matrix based on indices in a vector?
On Jun 10, 2009, at 6:26 PM, Marc Schwartz wrote: On Jun 10, 2009, at 12:36 PM, Logickle wrote: Sorry, there may be some lingo for describing the extraction problem I have, but I don't know it. I have a matrix of 2 rows and N columns, and a vector of N 1s and 2s. Matrix M: 1 2 3 4 ... N 1A C D G ... 2B D F H ... Vector v: 1 2 2 1 2 1 1 ... N I'd like to apply v to M to get Vector w: A D F G ... I.e. each element of v is interpreted as a row-index used to extract that row's value from the corresponding column in M into the corresponding column of w. Also eventually nrow(M) 2, in which case the value of the elements of v would range over 1:nrow(M). Seems it should be simple, but maybe not? Thanks! Doug Doug, Try this: # M is a character matrix M 1 2 3 4 1 A C D G 2 B D F H # v is a vector, equal in length to the number of columns in M v [1] 1 2 2 1 # Get the diagonal of the matrix that results from the combinations of indices diag(M[v, 1:ncol(M)]) [1] A D F G I created a larger 2 row matrix to test further: set.seed(1) M - matrix(sample(LETTERS, 40, replace = TRUE), 2, 20) M [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [1,] G O F Y Q F R U S J Y Q [2,] J X X R B E J M Z U F D [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20] [1,] G A W M M V U S [2,] K J I P E R C K and 'v': set.seed(2) v - sample(2, 20, replace = TRUE) v [1] 1 2 2 1 2 2 1 2 1 2 2 1 2 1 1 2 2 1 1 1 Then: diag(M[v, 1:ncol(M)]) [1] G X X Y B E R M S U F Q K A W P E [18] V U S Looks like it might work. Actually, the solution can be simplified further: # Using the second 'M': diag(M[v, ]) [1] G X X Y B E R M S U F Q K A W P E [18] V U S Marc __ 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] smooth.spline() fucntion
Hi, all, I found that the smooth.spline() function produces different results between R and S-Plus. I was trying to play different parameters of the function without any success. The script of the function contains Fortran code, so it seems impossible to port the code from S-Plus to R (or I may be wrong). Can anyone suggest anything so that I can produce the same result in R as in S-plus? Thanks John __ 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 extract from a matrix based on indices in a vector?
Subscripting by a 2-column matrix
M[cbind(v, seq_len(ncol(M)))]
uses much less space (hence time) than making
the ncol(M) by ncol(M) intermediate matrix just
to extract its diagonal. E.g.
test - function(n, seed) {
if (!missing(seed))
set.seed(seed)
M - matrix(sample(LETTERS, 2*n, replace = TRUE), 2)
v - sample(2, n, replace=T)
t1-system.time(r1-M[cbind(v,seq_len(ncol(M)))])
t2-system.time(r2-diag(M[v, 1:ncol(M)]))
list(identical=identical(r1,r2), time(matrix subscript)=t1,
time(diag(big matrix))=t2)
}
test(100)
$identical
[1] TRUE
$`time(matrix subscript)`
user system elapsed
0.000 0.000 0.001
$`time(diag(big matrix))`
user system elapsed
0.001 0.000 0.001
test(1000)
$identical
[1] TRUE
$`time(matrix subscript)`
user system elapsed
0.000 0.000 0.001
$`time(diag(big matrix))`
user system elapsed
0.082 0.021 0.103
test(5000)
$identical
[1] TRUE
$`time(matrix subscript)`
user system elapsed
0.001 0.000 0.001
$`time(diag(big matrix))`
user system elapsed
3.379 0.552 3.932
Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Marc Schwartz
Sent: Wednesday, June 10, 2009 4:42 PM
To: Logickle
Cc: rhelp help
Subject: Re: [R] How to extract from a matrix based on
indices in a vector?
On Jun 10, 2009, at 6:26 PM, Marc Schwartz wrote:
On Jun 10, 2009, at 12:36 PM, Logickle wrote:
Sorry, there may be some lingo for describing the extraction
problem I have,
but I don't know it.
I have a matrix of 2 rows and N columns, and a vector of N
1s and 2s.
Matrix M:
1 2 3 4 ... N
1A C D G ...
2B D F H ...
Vector v:
1 2 2 1 2 1 1 ... N
I'd like to apply v to M to get
Vector w:
A D F G ...
I.e. each element of v is interpreted as a row-index used to
extract that
row's value from the corresponding column in M into the
corresponding column
of w.
Also eventually nrow(M) 2, in which case the value of the
elements of v
would range over 1:nrow(M).
Seems it should be simple, but maybe not?
Thanks!
Doug
Doug,
Try this:
# M is a character matrix
M
1 2 3 4
1 A C D G
2 B D F H
# v is a vector, equal in length to the number of columns in M
v
[1] 1 2 2 1
# Get the diagonal of the matrix that results from the
combinations
of indices
diag(M[v, 1:ncol(M)])
[1] A D F G
I created a larger 2 row matrix to test further:
set.seed(1)
M - matrix(sample(LETTERS, 40, replace = TRUE), 2, 20)
M
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] G O F Y Q F R U S J Y Q
[2,] J X X R B E J M Z U F D
[,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20]
[1,] G A W M M V U S
[2,] K J I P E R C K
and 'v':
set.seed(2)
v - sample(2, 20, replace = TRUE)
v
[1] 1 2 2 1 2 2 1 2 1 2 2 1 2 1 1 2 2 1 1 1
Then:
diag(M[v, 1:ncol(M)])
[1] G X X Y B E R M S U F Q K A
W P
E
[18] V U S
Looks like it might work.
Actually, the solution can be simplified further:
# Using the second 'M':
diag(M[v, ])
[1] G X X Y B E R M S U F Q K A
W P E
[18] V U S
Marc
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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 extract from a matrix based on indices in a vector?
On Jun 10, 2009, at 7:05 PM, William Dunlap wrote:
Subscripting by a 2-column matrix
M[cbind(v, seq_len(ncol(M)))]
uses much less space (hence time) than making
the ncol(M) by ncol(M) intermediate matrix just
to extract its diagonal. E.g.
test - function(n, seed) {
if (!missing(seed))
set.seed(seed)
M - matrix(sample(LETTERS, 2*n, replace = TRUE), 2)
v - sample(2, n, replace=T)
t1-system.time(r1-M[cbind(v,seq_len(ncol(M)))])
t2-system.time(r2-diag(M[v, 1:ncol(M)]))
list(identical=identical(r1,r2), time(matrix subscript)=t1,
time(diag(big matrix))=t2)
}
test(100)
$identical
[1] TRUE
$`time(matrix subscript)`
user system elapsed
0.000 0.000 0.001
$`time(diag(big matrix))`
user system elapsed
0.001 0.000 0.001
test(1000)
$identical
[1] TRUE
$`time(matrix subscript)`
user system elapsed
0.000 0.000 0.001
$`time(diag(big matrix))`
user system elapsed
0.082 0.021 0.103
test(5000)
$identical
[1] TRUE
$`time(matrix subscript)`
user system elapsed
0.001 0.000 0.001
$`time(diag(big matrix))`
user system elapsed
3.379 0.552 3.932
Nicely done comparison Bill.
Thanks,
Marc
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to extract from a matrix based on indices in a vector?
On Jun 10, 2009, at 1:36 PM, Logickle wrote: Sorry, there may be some lingo for describing the extraction problem I have, but I don't know it. I have a matrix of 2 rows and N columns, and a vector of N 1s and 2s. Matrix M: 1 2 3 4 ... N 1A C D G ... 2B D F H ... Vector v: 1 2 2 1 2 1 1 ... N I'd like to apply v to M to get Vector w: A D F G ... Using length 10 M and v w - apply(cbind(v, 1:length(v) ), 1, function(x) M[x[1], x[2] ]) [1] A D F G I K N P R S M - matrix(LETTERS[1:20], nrow=2) M [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] A C E G I K M O Q S [2,] B D F H J L N P R T v - c(1,2,2,1, sample(1:2, 6, replace=TRUE) ) v [1] 1 2 2 1 1 1 2 2 2 1 The number of rows in M should not be a problem. I.e. each element of v is interpreted as a row-index used to extract that row's value from the corresponding column in M into the corresponding column of w. Also eventually nrow(M) 2, in which case the value of the elements of v would range over 1:nrow(M). Seems it should be simple, but maybe not? Seems reasonably simple. 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.
[R] Help on drawing stars and radars in R
Hi I don't know if you can help. I am a 2nd year Bsc Cosmetic Science student and in R I need some help in drawing stars. The problem that I have is I want to recreate a radar diagram similar to the one in excel. I have put in these commands in a script window: stars(shampoo1[, 1:6], locations = c(0,0), radius = TRUE, key.loc=c(0,0), main = Ranked Results for the Sensory Evaluation of the Shampoo , frame.plot = TRUE, key.labels = c(Foam Texture, Skin Feel, Rinisability, Wet Combability, Wet Feel, Odour), lty = 2, ) This gives me a black and white radar which is fine. Now according to the reference manual to colour the stars i need to add: col.stars = rainbow (30) Which colours in each star segment. The problem I have with this is that the segments are filled with colour and the segments overlap each other and each segment is not visible. So I desperately need the command that only colours the outline of the segments or stars instead of filling it. Can u help? Thank You Caroline __ 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.
