Re: [R] getting started on Bayesian analysis
On Wed, 2004-09-15 at 03:27, HALL, MARK E wrote:
I've found
Bayesian Methods: A Social and Behavioral Sciences Approach
by Jeff Gill
useful as an introduction. The examples are written in R and S with generalized
scripts for doing
a variety of problems. (Though I never got change-point analysis to successfully in
R.)
Change point analysis? I haven't seen the book, but I read lecture
handouts of one Bayesian course over here in Finland (Antti Penttinen,
Jyväskylä), and translated his example to R during one (rare) warm
summer day in a garden. So do you mean this (binary case):
source(/mnt/flash/cb.update.R)
cb.update
function (y, A=1, B=1, C=1, D=1, N=1200, burnin=200)
{
n - length(y)
lambda - numeric(N)
mu - numeric(N)
k - numeric(N)
lambda[1] - A/(A+B)
mu[1] - C/(C+D)
k[1] - n/2
sn - sum(y)
for (i in 2:N) {
kold - k[i-1]
sk - sum(y[1:kold])
lambda[i] - rbeta(1, A+sk, B + kold - sk)
mu[i] - rbeta(1, C + sn - sk, D + n - sn + sk - kold )
knew - sample(n-1, 1)
sknew - sum(y[1:knew])
r - (sknew - sk) *
(log(lambda[i])-log(mu[i]))-(knew-kold)*(lambda[i]-mu[i])
if(min(0,r) log(runif(1))) k[i] - knew
else k[i] - k[i-1]
}
out - cbind(lambda, mu, k)
out[(burnin+1):N, ]
}
y - c(rbinom(60, 1, 0.8), rbinom(40, 1, 0.3))
uh - cb.update(y, N=5200)
colMeans(uh)
lambda mu k
0.8189303 0.4169367 59.077
mean(y[1:60])
[1] 0.783
mean(y[41:100])
[1] 0.45
plot(density(uh[,1]))
plot(density(uh[,2]))
plot(table(uh[,3]), type=h)
This was off-topic. So something about business: isn't the (Win)BUGS
author working with a R port?
cheers, jari oksanen
--
Jari Oksanen -- Dept Biology, Univ Oulu, 90014 Oulu, Finland
email [EMAIL PROTECTED], homepage http://cc.oulu.fi/~jarioksa/
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Re: [R] getting started on Bayesian analysis
Jari Oksanen wrote: On Wed, 2004-09-15 at 03:27, HALL, MARK E wrote: I've found Bayesian Methods: A Social and Behavioral Sciences Approach by Jeff Gill useful as an introduction. The examples are written in R and S with generalized scripts for doing a variety of problems. (Though I never got change-point analysis to successfully in R.) Change point analysis? I haven't seen the book, but I read lecture handouts of one Bayesian course over here in Finland (Antti Penttinen, Jyväskylä), and translated his example to R during one (rare) warm summer day in a garden. So do you mean this (binary case): snip This was off-topic. So something about business: isn't the (Win)BUGS author working with a R port? Yes, he is, and he's got it working in Windows. But if anyone wants to discuss how R does memory management with Andrew, he'll be all ears. In the mean time, there is the R2WinBUGS package on CRAN for you to enjoy. Bob -- Bob O'Hara Dept. of Mathematics and Statistics P.O. Box 68 (Gustaf Hällströmin katu 2b) FIN-00014 University of Helsinki Finland Telephone: +358-9-191 51479 Mobile: +358 50 599 0540 Fax: +358-9-191 51400 WWW: http://www.RNI.Helsinki.FI/~boh/ Journal of Negative Results - EEB: http://www.jnr-eeb.org __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Cancor
Dear Gabor, thank you for your answer related to the normalization of cancor coefficients. If you want to interpret the coefficients in terms of variables' contributions to canonical variables, loadings, redundancy measure, etc. , you have to normalize the results so that the canonical variables have identity variance matrix. Multiplying cancor coefficients xcoef and ycoef by as.numeric(sqrt(nrow(x)-1)) does the job. (I sometimes miss such information in the R help.) Best regards, Irena Komprej __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Bessel function
Dear all Currently, I'm implementing the generalized hyperbolic distribution into Splus. Unfortunately the Bessel function is not implemented in Splus. In R the Bessel function does exist but it is an internal function and I'm not able to look at the code. Is there any possibility to see the code of the Bessel function in R or does anybody has an implementation of the Bessel function in Splus? Thanks a lot for your help. Beat [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Bessel function
Hi Beat, take a look at this link, http://www.statsci.org/s/besseli0.html for the modified Bessel function. I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/16/396887 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat/ http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Wednesday, September 15, 2004 10:24 AM Subject: [R] Bessel function Dear all Currently, I'm implementing the generalized hyperbolic distribution into Splus. Unfortunately the Bessel function is not implemented in Splus. In R the Bessel function does exist but it is an internal function and I'm not able to look at the code. Is there any possibility to see the code of the Bessel function in R or does anybody has an implementation of the Bessel function in Splus? Thanks a lot for your help. Beat [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Cancor
On (she claimed) Thu, 1 Jan 1998, Irena Komprej wrote: Dear Gabor, This is R-help, not `Gabor', although you keep sending mail addressed to `Gabor' to this list. thank you for your answer related to the normalization of cancor coefficients. If you want to interpret the coefficients in terms of variables' contributions to canonical variables, loadings, redundancy measure, etc. , you have to normalize the results so that the canonical variables have identity variance matrix. Multiplying cancor coefficients xcoef and ycoef by as.numeric(sqrt(nrow(x)-1)) does the job. (I sometimes miss such information in the R help.) I think you miss it in the references given on the R help pages. Please do read them. (Any good book will tell you that the scaling of canonical variates is arbitrary.) Please also ask your local IT advisors to set your computer to a sensible time: the world is 6.7 years ahead of you. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Bessel function
On Wed, 15 Sep 2004 [EMAIL PROTECTED] wrote: Currently, I'm implementing the generalized hyperbolic distribution into Splus. Unfortunately the Bessel function is not implemented in Splus. In R the Bessel function does exist but it is an internal function and I'm not able to look at the code. Is there any possibility to see the code of the Bessel function in R or Of course. It *is* in the sources. What stops you looking at the sources? (Where they are is in the FAQ, for example.) BTW, there are four Bessel functions in R, and others do exist. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] testing goodness of fit of linear model
Dear R-users,
I've been reading a bunch of things on linear models but cannot quite find a
clear answer. How can one determine whether a linear model is significant or
not?
For background info, I am modelling the response of topographic slope to the
distance of a catchment's outlet. Some guys have shown that if there is a
significant fit to a linear model, one can deduce the dynamic state of the
basin, that is, whether erosion is as strong as rock uplift, erosion is
smaller than rock uplift, or erosion is greater than rock uplift. I am thus
to test 4 situations:
Situation 1: a linear model is inappropriate for describing the data, the
scatter is too large, and thus a linear model is unfit to explain the data.
Situation 2: the linear model of the kind y = b0 + b1 * x is fit to
describe the data, ie data points lie close to a straight line.
Situation 2a: the relationship between slope and distance is significantly
positive
Situation 2b: the relationship between slope and distance is significantly
null (ie data is clustered around a line with b1 non-significantly different
from 0)
Situation 2c: the relationship between slope and distance is significantly
negative
I am confused as to what test I should use for discriminating these
situations.
The glm offers an indication about the significance of regression
parameters. So in the case where b1 is significantly different from 0 (p
value =0.05 for a test where H0: b1=0; H1: b1 != 0), it is straightforward.
But I don't know how to discriminate between situation 1 and situation 2 (ie
whether a linear model is significant).
Any suggestion are welcome
Cheers,
Thomas
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Re: [R] Bessel function
Hi you might find package(hyperbolicDist) interesting. best wishes rksh Dear all Currently, I'm implementing the generalized hyperbolic distribution into Splus. Unfortunately the Bessel function is not implemented in Splus. In R the Bessel function does exist but it is an internal function and I'm not able to look at the code. Is there any possibility to see the code of the Bessel function in R or does anybody has an implementation of the Bessel function in Splus? Thanks a lot for your help. Beat [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Robin Hankin Uncertainty Analyst Southampton Oceanography Centre SO14 3ZH tel +44(0)23-8059-7743 [EMAIL PROTECTED] (edit in obvious way; spam precaution) __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Bessel function
Hi you might find package(hyperbolicDist) interesting. best wishes robin Dear all Currently, I'm implementing the generalized hyperbolic distribution into Splus. Unfortunately the Bessel function is not implemented in Splus. In R the Bessel function does exist but it is an internal function and I'm not able to look at the code. Is there any possibility to see the code of the Bessel function in R or does anybody has an implementation of the Bessel function in Splus? Thanks a lot for your help. Beat [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Robin Hankin Uncertainty Analyst Southampton Oceanography Centre SO14 3ZH tel +44(0)23-8059-7743 [EMAIL PROTECTED] (edit in obvious way; spam precaution) __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] loading error of the Rcmdr library on Debian Sid
Hi, it seems you have to install the rgl package and this demand the gl4 java lib what is necessary, too. http://www.jausoft.com/products/gl4java/gl4java_install.html christian Am Mittwoch, 15. September 2004 12:01 schrieb Thomas Schönhoff: Hello, I just tried to get Rcmdr package working, resulting in: -- library(Rcmdr) Loading required package: tcltk Loading required package: lattice Loading required package: foreign Loading required package: abind Loading required package: lmtest Loading required package: multcomp Loading required package: relimp Loading required package: effects Loading required package: rgl Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library /usr/lib/R/site-library/rgl/libs/rgl.so: libnvidia-tls.so.1: cannot handle TLS data Error in .C(symbol.C(rgl_quit), success = FALSE, PACKAGE = rgl) : C function name not in DLL for package rgl Loading required package: mgcv This is mgcv 1.1-1 Loading required package: car Error: Missing packages: rgl Error: .onLoad failed in loadNamespace Error in library(Rcmdr) : package/namespace load failed Missing rgl-package ? --- thomas dpkg -l |grep r- r-cran-abind 1.1.0-1 r-cran-car 1.0.13-1 r-cran-foreign 0.7-1 r-cran-lattice 0.9.16-1 r-cran-mgcv1.1.1.1-1 r-cran-rgl 0.64.13-1 r-cran-relimp 0.8.4-1 r-cran-rcmdr 0.9.11-1 r-cran-lmtest 0.9.6-2 r-cran-effects 1.0.5-1 r-cran-multcom 0.4.7-1 r-cran-mvtnorm 0.6.8-1 amongst other R related (basic and specific) packages. Am I still missing some required packages to run RCommander smoothly? My system: Debian Sid (unstable) GNU R 1.9.1 Xfree 4.3 Anyone else noticed this on Debian Sid whilst trying to run the R Commander !? So far I didn't find a related bug report for Linux. Maybe this is a Debian related problem, I really have no clue at the moment. Regards Thomas __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] loading error of the Rcmdr library on Debian Sid
Hello Christian, Christian Schulz schrieb: Hi, it seems you have to install the rgl package and this demand the gl4 java lib what is necessary, too. http://www.jausoft.com/products/gl4java/gl4java_install.html Seems like that this dependency is missing in the r-cran-rgl package!? So I am going to write a bug report to the debian maintainer! Thanks Thomas __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Density Estimation
Hi there, Sorry if this is a rather loing post. I have a simple list of single feature data points from which I would like to generate a probability that an unseen point comes from the same distribution. To do this I am trying to estimate the probability density of the list of points and use this to generate a probability for the new unseen points. I have managed to use the R density function to generate the density estimate but have not been able to do anything with this - i.e. generate a rpobability that a new point comes from the same distribution. Is there a function to do this, or am I way off the mark using the density function at all? Thanks in advance, Brian. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Read.fwf
Dear List I have a fixed width file with variables of varying width. The help is pretty transparent for this feature, but I can't seem to figure out how I can make effective use of the package with my data. In my dataset, the first 80 columns are of width 1 followed by other variables with width larger than 1. I think the correct way to do this, by brute force, would be read.fwf(dataset, 1,1,1,...,1,5,...). I'm sure I do not need to actually enter 80 1s, correct? How might I do this more effectively? Harold [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Density Estimation
Dear Brian, I can suggest you to use density() function to get an estimate of the pdf you're finding (I believe it's unknown). Then you can plot the point you got by density() using plot(). In this way you have a graphic representation of you unknown pdf. According its shape and helping by the graphic you could try to understand what kind of pdf it would be (normal, gamma, weibul, etc.) After you can estimate parameters of pdf using your data with LS or ML methods. Then you can calculate the goodness of fit for each model of pdf and use the best one. I hope I get you a little help. Cordially Vito Ricci [EMAIL PROTECTED] wrote: Hi there, Sorry if this is a rather loing post. I have a simple list of single feature data points from which I would like to generate a probability that an unseen point comes from the same distribution. To do this I am trying to estimate the probability density of the list of points and use this to generate a probability for the new unseen points. I have managed to use the R density function to generate the density estimate but have not been able to do anything with this - i.e. generate a rpobability that a new point comes from the same distribution. Is there a function to do this, or am I way off the mark using the density function at all? Thanks in advance, Brian. = Diventare costruttori di soluzioni Visitate il portale http://www.modugno.it/ e in particolare la sezione su Palese http://www.modugno.it/archivio/cat_palese.shtml ___ http://it.seriea.fantasysports.yahoo.com/ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Read.fwf
You can use rep() in specifying the widths argument to avoid typing all those ones. For example: read.fwf(file=c:\\mydata.dat, widths = c(rep(1,80), 5, 4, 6)) hope this helps, Chuck Doran, Harold wrote: I have a fixed width file with variables of varying width. The help is pretty transparent for this feature, but I can't seem to figure out how I can make effective use of the package with my data. In my dataset, the first 80 columns are of width 1 followed by other variables with width larger than 1. I think the correct way to do this, by brute force, would be read.fwf(dataset, 1,1,1,...,1,5,...). I'm sure I do not need to actually enter 80 1s, correct? How might I do this more effectively? -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 452-1424 (M, W, F) fax: (917) 438-0894 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Read.fwf
Doran, Harold wrote: Dear List I have a fixed width file with variables of varying width. The help is pretty transparent for this feature, but I can't seem to figure out how I can make effective use of the package with my data. In my dataset, the first 80 columns are of width 1 followed by other variables with width larger than 1. I think the correct way to do this, by brute force, would be read.fwf(dataset, 1,1,1,...,1,5,...). I'm sure I do not need to actually enter 80 1s, correct? How might I do this more effectively? ?read.fwf: widths integer vector, giving the widths of the fixed-width fields (of one line). Hence: read.fwf(dataset, c(rep(1, 80), 5, ...)) Uwe Ligges Harold [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Bessel function
Dear Beat, You might also want to look at the book by Zang and Jin - Computation of Special Functions, John Wiley. They have Fortran sources for all the special functions covered in there. Ravi. Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology, Johns Hopkins University, 2024 E. Monument Street, Suite 2-700 Baltimore, MD 21205. (410) 502 - 7806. From: [EMAIL PROTECTED] on behalf of [EMAIL PROTECTED] Sent: Wed 9/15/2004 4:24 AM To: [EMAIL PROTECTED] Subject: [R] Bessel function Dear all Currently, I'm implementing the generalized hyperbolic distribution into Splus. Unfortunately the Bessel function is not implemented in Splus. In R the Bessel function does exist but it is an internal function and I'm not able to look at the code. Is there any possibility to see the code of the Bessel function in R or does anybody has an implementation of the Bessel function in Splus? Thanks a lot for your help. Beat [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] control of font size colour for title, subtitles, axis, and tick marks in LATTICE graph
Hi,
I very much appreciate any help on this fine tuning problem in a lattice
graph (I am new to LATTICE and could not find an example in the help files
that worked for me. My apologies if I missed it there).
I am running the following box plots to compare conditional distributions of
x at different levels of y under two treatment conditions ID=1 (upper
panel ) ID=0 (lower panel of the plot).
bwplot(HF.ELECYEAR ~ stparvotech | ID ,
data=data, aspect=1,
layout=c(1,2),
xlab=Changes in Party Vote Shares,
xlim=(-20:20),
ylab=Periods Following Last Federal Election,
main=Divided Government,
panel = function(x,y)
{
panel.bwplot(x,y)
panel.abline(v=0, col=red)
}
)
How can I:
1. Control the font size of the main title, the panel titles, the axis, and
the tick marks? The usual cex.main=1/3, cex.xlab etc. do not work.
2. Change the color of the boxes where the panel titles (ID=1 ID=0) are
located?
Thank you very much.
Best,
Jens
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Re: [R] pairs correlations colors
Valeria Edefonti wrote:
I have the following problem.
I want to use pairs function and get a matrix of scatterplots with the
correlations in the upper panel and the ordinary scatterplots in the
lower panel.
Moreover, I want to have points colored in five differet ways in the
lower panel, because I have five subgroups.
In order to do that I tried to combine examples on pairs function help.
I got a colored matrix using hints on iris dataset.
I got a black and white matrix with correlations using function
panel.cor, exactly as it is in the example.
Unfortunately, the line:
jpeg(filename=/home/valeria/Thesis/lung/fig/scatterplotcolnames.jpg)
pairs(aggiunta[,1:
6],labels=c(ALCAM,ITGB5,MSN,CSTB,DHCR24,TRIM29), main =
Scatterplots selected genes,pch=21,
bg = c(red, green3, blue,
brown,orange)[aggiunta[,7]],upper.panel=panel.cor)
dev.off()
doesn't allow me to get the desidered matrix with colors and correlations.
I also tried to create a function panel.col for the lower.panel:
## put colors on the lower panels
panel.col - function(datiepheno)
{
usr - par(usr); on.exit(par(usr))
par(bg = c(red, green3, blue,
brown,orange)[datiepheno[,7]],pch=21, usr = c(0, 1, 0, 1))
}
but it doesn't work as well.
Any idea?
I hope I'll be precise but not too much precise!
Thank you very much
Valeria
Looks like there was no answer yet:
Please specify am example with data available in R, so that we can
reproduce your example. This will save much time for those people who
are going to help.
Rearranging your examples with data we do not have available requires
quite a lot of effort.
What I guess is that you are going to specify
lower.panel = function(x, y)
points(x, y,
col = sapply(aggiunta[,7], switch, 1=red, 2=green3, .))
Uwe Ligges
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Re: [R] loading error of the Rcmdr library on Debian Sid
You can apt-get RGL in sid.
Thomas Schönhoff [EMAIL PROTECTED] writes:
Hello,
I just tried to get Rcmdr package working, resulting in:
--
library(Rcmdr)
Loading required package: tcltk
Loading required package: lattice
Loading required package: foreign
Loading required package: abind
Loading required package: lmtest
Loading required package: multcomp
Loading required package: relimp
Loading required package: effects
Loading required package: rgl
Error in dyn.load(x, as.logical(local), as.logical(now)) :
unable to load shared library
/usr/lib/R/site-library/rgl/libs/rgl.so:
libnvidia-tls.so.1: cannot handle TLS data
Error in .C(symbol.C(rgl_quit), success = FALSE, PACKAGE = rgl) :
C function name not in DLL for package rgl
Loading required package: mgcv
This is mgcv 1.1-1
Loading required package: car
Error: Missing packages: rgl
Error: .onLoad failed in loadNamespace
Error in library(Rcmdr) : package/namespace load failed
Missing rgl-package ?
---
thomas dpkg -l |grep r-
r-cran-abind 1.1.0-1
r-cran-car 1.0.13-1
r-cran-foreign 0.7-1
r-cran-lattice 0.9.16-1
r-cran-mgcv1.1.1.1-1
r-cran-rgl 0.64.13-1
r-cran-relimp 0.8.4-1
r-cran-rcmdr 0.9.11-1
r-cran-lmtest 0.9.6-2
r-cran-effects 1.0.5-1
r-cran-multcom 0.4.7-1
r-cran-mvtnorm 0.6.8-1
amongst other R related (basic and specific) packages.
Am I still missing some required packages to run RCommander smoothly?
My system:
Debian Sid (unstable)
GNU R 1.9.1
Xfree 4.3
Anyone else noticed this on Debian Sid whilst trying to run the R
Commander !? So far I didn't find a related bug report for
Linux. Maybe this is a Debian related problem, I really have no clue
at the moment.
Regards
Thomas
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Biomedical and Health Informatics University of Washington
Biostatistics, SCHARP/HVTN Fred Hutchinson Cancer Research Center
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RE: [R] Bessel function
Currently, I'm implementing the generalized hyperbolic distribution into Splus. Unfortunately the Bessel function is not implemented in Splus. In R the Bessel function does exist but it is an internal function and I'm not able to look at the code. Is there any possibility to see the code of the Bessel function in R or does anybody has an implementation of the Bessel function in Splus? Have a look at Press-Teukolsky-Vetterling-Flannery: Numerical Recipes in C, Cambridge University Press, Chapter 6 (2nd edition). I guess you are looking for modified Bessel functions (see, 6.6). Hannu Kahra Progetti Speciali Monte Paschi Asset Management SGR S.p.A. Via San Vittore, 37 IT-20123 Milano, Italia Tel.: +39 02 43828 754 Mobile: +39 333 876 1558 Fax: +39 02 43828 247 E-mail: [EMAIL PROTECTED] Web: www.mpsam.it __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Density Estimation
Try fitting it with a Johnson function -- see SuppDists. If you can fit it you will then be able to use the functions in SuppDists just as you can for any other distribution supported by R. Brian Mac Namee wrote: Hi there, Sorry if this is a rather loing post. I have a simple list of single feature data points from which I would like to generate a probability that an unseen point comes from the same distribution. To do this I am trying to estimate the probability density of the list of points and use this to generate a probability for the new unseen points. I have managed to use the R density function to generate the density estimate but have not been able to do anything with this - i.e. generate a rpobability that a new point comes from the same distribution. Is there a function to do this, or am I way off the mark using the density function at all? Thanks in advance, Brian. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Bob Wheeler --- http://www.bobwheeler.com/ ECHIP, Inc. --- Randomness comes in bunches. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Read.fwf
It might help to read that help file again: d - paste(c(rep(0, 80), 1), collapse=) d [1] 000 01 f - file(try.dat, w) writeLines(d, f) writeLines(d, f) writeLines(d, f) close(f) fw - c(rep(1, 80), 5) x - read.fwf(try.dat, fw) x V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21 V22 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 V23 V24 V25 V26 V27 V28 V29 V30 V31 V32 V33 V34 V35 V36 V37 V38 V39 V40 V41 V42 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 V43 V44 V45 V46 V47 V48 V49 V50 V51 V52 V53 V54 V55 V56 V57 V58 V59 V60 V61 V62 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 V63 V64 V65 V66 V67 V68 V69 V70 V71 V72 V73 V74 V75 V76 V77 V78 V79 V80 V81 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 Andy From: Doran, Harold Dear List I have a fixed width file with variables of varying width. The help is pretty transparent for this feature, but I can't seem to figure out how I can make effective use of the package with my data. In my dataset, the first 80 columns are of width 1 followed by other variables with width larger than 1. I think the correct way to do this, by brute force, would be read.fwf(dataset, 1,1,1,...,1,5,...). I'm sure I do not need to actually enter 80 1s, correct? How might I do this more effectively? Harold [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Density Estimation
Hi!
The function density returns you a object of class density.
This object has an x and an y attribute which you can access by x y,
Hi!
Use approx and runif.
eg.:
dd-density(rnorm(100,3,5))
plot(dd)
Using the function ?approx you can compute the density value for any x.
#the x is a dummy here.
mydist-function(x,dd)
{
while(1)
{
tmp - runif(1,min=min(dd$x),max=max(dd$x))
lev - approx(dd$x,dd$y,tmp)$y
if(runif(1,c(0,1)) = lev)
{
return(tmp)
}
}
}
x - 0
mydist(x,dd)
res-rep(0,500)
res-sapply(res,mydist,dd)
lines(density(res),col=2)
/E.
*** REPLY SEPARATOR ***
On 9/15/2004 at 12:36 PM Brian Mac Namee wrote:
Hi there,
Sorry if this is a rather loing post. I have a simple list of single
feature data points from which I would like to generate a probability
that an unseen point comes from the same distribution. To do this I am
trying to estimate the probability density of the list of points and
use this to generate a probability for the new unseen points. I have
managed to use the R density function to generate the density estimate
but have not been able to do anything with this - i.e. generate a
rpobability that a new point comes from the same distribution. Is
there a function to do this, or am I way off the mark using the
density function at all?
Thanks in advance,
Brian.
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Dipl. bio-chem. Witold Eryk Wolski @ MPI-Moleculare Genetic
Ihnestrasse 63-73 14195 Berlin'v'
tel: 0049-30-83875219/ \
mail: [EMAIL PROTECTED]---W-Whttp://www.molgen.mpg.de/~wolski
[EMAIL PROTECTED]
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Re: [R] loading error of the Rcmdr library on Debian Sid
Hello, A.J. Rossini schrieb: You can apt-get RGL in sid. An apt-cache search or (Synaptic search) RGL gives me r-cran-rgl 0.64.13-1. This package is already installed! regards Thomas __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] fractal dimension of an image
Hello, I have a problem that I think can be solved in R but I'm not sure how to tie things together. I have a digital image of a crystal growth in 2 dimensions. And my aim is to calculate the fractal dimension of the crystal. I was planning to use the box counting method. So I need to read in the image in R (for which I can use the pixmap or rimage packages) and then draw a grid over at a series of resolutions and for each grid, count how many cells of the grid are occupied by the crystal. Since the image can be converted to grayscale, I figured that specifying an intensity threshold would allow me to differentiate between crystal and surrounding solution. I know this can be done in Matlab, but since I'm more comfortable in R can it be done in R? I found the fdim package but that calculates the dimension for a data.frame - but I'm not sure whether this would be feasible for my image. Any pointers would be appreciated --- Rajarshi Guha [EMAIL PROTECTED] http://jijo.cjb.net GPG Fingerprint: 0CCA 8EE2 2EEB 25E2 AB04 06F7 1BB9 E634 9B87 56EE --- So the Zen master asked the hot-dog vendor, Can you make me one with everything? - TauZero on Slashdot __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Density Estimation
On 15-Sep-04 Brian Mac Namee wrote: Sorry if this is a rather loing post. I have a simple list of single feature data points from which I would like to generate a probability that an unseen point comes from the same distribution. To do this I am trying to estimate the probability density of the list of points and use this to generate a probability for the new unseen points. I have managed to use the R density function to generate the density estimate but have not been able to do anything with this - i.e. generate a rpobability that a new point comes from the same distribution. Is there a function to do this, or am I way off the mark using the density function at all? It's not clear what you're really after, but it looks as though you may be wanting to sample from the distribution estimated by 'density'. A possible approach, which you could refine, is exemplified by x-rnorm(1000) d-density(x,n=4096) y-sample(d$x,size=1000,prob=d$y) Check performance with hist(y) Looks OK to me! See ?density and ?sample. On an alternative interpretation, perhaps you want to first estimate the density based on data you already have, and then when you have got further data (but these would then be seen and not unseen) come to a judgement about whether these new points are compatible with coming from the distributikon you have estimated. A possible approach to this question (again susceptible to refinement) would be as follows. 1. Use a fine-grained grid for 'density', i.e. a large value for n. 2. Replace each of the points in the new data by the nearest point in this grid. Call these values z1, z2, ... , zk corresponding to index values i1, i2, ... , ik in d$x. 3. Evaluate the probability P(z1,...,zk) from the density as the product of d$y[i] where i-c(i1,...,ik). Better still, evaluated the logarithm of this. Call the result L. 4. Now simulate a large number of draws of k values from d on the lines of sample(d$x,size=k,prob=d$y) as above, and evaluate L for each of these. Where is the value of L from (3) situated in the distribution of these values of L from (4)? If (say) only 1 per cent of the simulated values of L from d are less than the value of L from (3), then you have a basis for a test that your new data did not come from the distribution you have estimated from your old data, in that the new data are from the low-density part of the estimated distribution. There are of course alternative ways to view this question. The value of k is relevant. In particular, if k is small (say 3 or 4) then the suggestion in (4) is probably the best way to approach it. However, if k is large then you can use a test on the lines of Kolmogorov-Smirnov with the reference distribution estimated as the cumulative distribution of d$y and the distribution being tested as the empirical cumulative distribution of your new data. Even sharper focus is available if you are in a position to make a paramatric model for your data, but your description does not suggest that this is the case. Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 167 1972 Date: 15-Sep-04 Time: 15:07:33 -- XFMail -- __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] control of font size colour for title, subtitles, axis, and tick marks in LATTICE graph
On Wednesday 15 September 2004 08:24, Jens Hainmueller wrote:
Hi,
I very much appreciate any help on this fine tuning problem in a
lattice graph (I am new to LATTICE and could not find an example in
the help files that worked for me. My apologies if I missed it
there).
Well, if the examples were to cover all possible uses, there would be so
many that they would become practically useless as reference. You are
also expected to read the documentation. At least for your first
question, the solution is described pretty much where you would expect
it to be (see below).
I am running the following box plots to compare conditional
distributions of x at different levels of y under two treatment
conditions ID=1 (upper panel ) ID=0 (lower panel of the plot).
bwplot(HF.ELECYEAR ~ stparvotech | ID ,
data=data, aspect=1,
layout=c(1,2),
xlab=Changes in Party Vote Shares,
xlim=(-20:20),
ylab=Periods Following Last Federal Election,
main=Divided Government,
panel = function(x,y)
{
panel.bwplot(x,y)
panel.abline(v=0, col=red)
}
)
How can I:
1. Control the font size of the main title, the panel titles, the
axis, and the tick marks? The usual cex.main=1/3, cex.xlab etc. do
not work.
Why do you expect them to? In help(bwplot), the entry for 'main' says:
main: character string or expression or list describing main title
to be placed on top of each page. Defaults to 'NULL'. Can be
a character string or expression, or a list with components
'label, cex, col, font'. The 'label' tag can be omitted if it
is the first element of the list. Expressions are treated as
specification of LaTeX-like markup as in 'plotmath'
which suggests that
main = list(label = Divided Government, cex = 1/3)
is what you want.
2. Change the color of the boxes where the panel titles
(ID=1 ID=0) are located?
See help(strip.default). You need to use something like
strip = function(..., bg) strip.default(..., bg = white)
(The next version of lattice will have a slightly easier way to do this,
along with an example.)
All this can also be done by changing the settings globally (see
help(trellis.par.get)), but that's probably not what you want.
Hope that helps,
Deepayan
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[R] adding observations to lm for fast recursive residuals?
dear R community: i have been looking but failed to find the
following: is there a function in R that updates a plain OLS lm()
model with one additional observation, so that I can write a function
that computes recursive residuals *quickly*?
PS: (I looked at package strucchange, but if I am not mistaken, the
recresid function there takes longer than iterating over the models
fresh from start to end.) I know the two functions do not do the same
thing, but the main part (OLS) is the same:
handrecurse.test - function( y, x ) { z- rep(NA, T); for (i in
2:T) { z[i] - coef(lm(y[1:i] ~ x[1:i]))[2]; }; return(z); }
system.time(handrecurse.test(y,x))
[1] 0.69 0.00 0.70 0.00 0.00
system.time(length(recresid( y~x )))
[1] 1.44 0.07 1.59 0.00 0.00
pointers appreciated. regards, /iaw
---
ivo welch
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Re: [R] adding observations to lm for fast recursive residuals?
In my quantreg package there is a function called lm.fit.recursive()
that, as the .Rd file
says:
Description:
This function fits a linear model by recursive least squares. It
is a utility routine for the 'khmaladzize' function of the
quantile regression package.
Usage:
lm.fit.recursive(X, y, int=TRUE)
Arguments:
X: Design Matrix
y: Response Variable
int: if TRUE then append intercept to X
Value:
return p by n matrix of fitted parameters, where p. The ith column
gives the solution up to time i.
It is written in fortran so it should be reasonably quick.
HTH
url:www.econ.uiuc.edu/~rogerRoger Koenker
email [EMAIL PROTECTED] Department of Economics
vox:217-333-4558University of Illinois
fax:217-244-6678Champaign, IL 61820
On Sep 15, 2004, at 9:53 AM, [EMAIL PROTECTED] wrote:
dear R community: i have been looking but failed to find the
following: is there a function in R that updates a plain OLS lm()
model with one additional observation, so that I can write a function
that computes recursive residuals *quickly*?
PS: (I looked at package strucchange, but if I am not mistaken, the
recresid function there takes longer than iterating over the models
fresh from start to end.) I know the two functions do not do the same
thing, but the main part (OLS) is the same:
handrecurse.test - function( y, x ) { z- rep(NA, T); for (i in
2:T) { z[i] - coef(lm(y[1:i] ~ x[1:i]))[2]; }; return(z); }
system.time(handrecurse.test(y,x))
[1] 0.69 0.00 0.70 0.00 0.00
system.time(length(recresid( y~x )))
[1] 1.44 0.07 1.59 0.00 0.00
pointers appreciated. regards, /iaw
---
ivo welch
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Re: [R] rolling n-step predictions for ARIMA models
There are functions in the dse bundle that do this. (See featherForecasts and horizonForecasts.) You might look through the users' guide to get an idea if they are exactly what you want. Paul Gilbert Michael Roberts wrote: Hello: I would like to generate rolling, multiperiod forecasts from an estimated ARIMA model, but the function predict.Arima seems only to generate forecasts from the last observation in the data set. To implement this, I was looking for an argument like 'newdata=' in predict.lm. I can write some code that does this for my particular problem, but might there exist a package/function that does this that I cannot find? Thanks, -Michael Michael J. Roberts Resource Economics Division Production, Management, and Technology USDA-ERS (202) 694-5557 (phone) (202) 694-5775 (fax) __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] [R-pkgs] RODBC 1.1-1
The first non-maintenance update of RODBC since January 2003 is now on
CRAN and will soon propagate to mirrors. From the ChangeLog:
* Select the decimal point from Sys.localeconv.
* Add an external reference and finalizer so open channels get
closed at the end of the session or when there is no R object
referring to them.
* There is no longer a restriction to 16 channels.
* Add NAMESPACE.
* odbcConnect{Access,Dbase,Excel} allow a missing file name
(and will bring up a dialog box to search for it).
* odbcGetInfo returns more information in a 8-element character
vector (based on an idea of Matthew Dowle).
* The C code calls SQLExecuteDirect rather than SQLExecute and
does not call SQLCloseCursor, based on a problem report from
Matthew Dowle using MS SQLServer. Repeated calls to
sqlGetResults now work.
* New function sqlFetchMore.
* Table names in Access with embedded spaces are mapped to the
[name space] form which Access requires.
* Table creation no longer removes _ from column names.
* New functions get/setSqlTypeInfo and the typeInfo argument to
sqlSave allow users to specify the mapping from R types to DBMS
datatypes. sqlSave also allows the specification of DBMS
datatypes by column.
* It is now possible to write more than 255 chars to a field with
sqlSave and sqlUpdate.
* Dates and timestamps are now read as 'Date' and 'POSIXct'
columns by sqlGetResults (unless as.is = TRUE for the column).
I have been able to test SQL Server reasonably extensively this time
around, as well as MySQL, PostgreSQL, Access, Excel and SQLite.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] cluster analysis and null hypothesis testing
Hi, testing the randomness of a cluster analysis is not a well defined problem, because it depends crucially on your null model. In fpc, there is nothing like this. Function prabtest in package prabclus performs such a test, but this is for a particular data structure, namely presence-absence data in biogeography. In principle, a Monte Carlo test can be constructed (and thus implemented in R) as follows: 1) You need a null model H_0, from which you generate data. 2) You need a test statistic T. 3) Compute T on your data (call it T_0). 4) Repeat k times: a) Generate data from H_0 b) Compute T on the generated data. 5) The p-value is (K+1)/(k+1), where K is the number of generated datasets for which T=T_0 (given that T small indicates the tendency of clustering). Standard choices for H_0 will be a normal or uniform distribution. (In prabtest, it is a complicated distribution on presence-absence data.) There are lots of possible choices of T. prabtest uses the ratio between the 25% smallest distances in the dataset and the 25% largest distances. This should be reasonable in fairly general settings. For a discussion of this and alternative choices (and references on them), you may take a look into C. Hennig and B. Hausdorf: Distance-based parametric bootstrap tests for clustering of species ranges, Computational Statistics and Data Analysis 45 (2004), 875-896. A preprint of this can be obtained from my web page. If you want to test the significance of a solution from a particular cluster analysis method, you should think about choosing T so that it is somehow connected to the method. (In the Hennig and Hausdorf paper, there are for example two alternatives discussed that are connected to Single Linkage.) Best, Christian On Wed, 15 Sep 2004, Patrick Giraudoux wrote: Hi, I am wondering if a Monte Carlo method (or equivalent) exist permitting to test the randomness of a cluster analysis (eg got by hclust(). I went through the package fpc (maybe too superficially) but dit not find such method. Thanks for any hint, Patrick Giraudoux __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html *** Christian Hennig Fachbereich Mathematik-SPST/ZMS, Universitaet Hamburg [EMAIL PROTECTED], http://www.math.uni-hamburg.de/home/hennig/ ### ich empfehle www.boag-online.de __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] fractal dimension of an image
From www.r-project.org - search - R site search - fractal dimension, I got 17 hits, the third of which discussed package RandomFields, which includes a function fractal. Are you familiar with this? hope this helps. spencer graves p.s. Have you read the posting guide! http://www.R-project.org/posting-guide.html;? Rajarshi Guha wrote: Hello, I have a problem that I think can be solved in R but I'm not sure how to tie things together. I have a digital image of a crystal growth in 2 dimensions. And my aim is to calculate the fractal dimension of the crystal. I was planning to use the box counting method. So I need to read in the image in R (for which I can use the pixmap or rimage packages) and then draw a grid over at a series of resolutions and for each grid, count how many cells of the grid are occupied by the crystal. Since the image can be converted to grayscale, I figured that specifying an intensity threshold would allow me to differentiate between crystal and surrounding solution. I know this can be done in Matlab, but since I'm more comfortable in R can it be done in R? I found the fdim package but that calculates the dimension for a data.frame - but I'm not sure whether this would be feasible for my image. Any pointers would be appreciated --- Rajarshi Guha [EMAIL PROTECTED] http://jijo.cjb.net GPG Fingerprint: 0CCA 8EE2 2EEB 25E2 AB04 06F7 1BB9 E634 9B87 56EE --- So the Zen master asked the hot-dog vendor, Can you make me one with everything? - TauZero on Slashdot __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Spencer Graves, PhD, Senior Development Engineer O: (408)938-4420; mobile: (408)655-4567 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] heatmap2
Can anybody tell me where to find a copy of heatmap2? I've seen it in my travels across the web but didn't bookmark it and can't find it again. Thanks in advance. Paul `-:-. ,-;`-:-. ,-;`-:-. ,-;`-:-. ,-;`-:-. ,-;`-:-. `=`,'=/ `=`,'=/ `=`,'=/ `=`,'=/ `=`,'=/ `=` ==/==/==/==/==/ ,=,-=`.,=,-=`.,=,-=`.,=,-=`.,=,-=`.,=, ,-'-' `-=_,-'-' `-=_,-'-' `-=_,-'-' `-=_,-'-' `-=_,-'-' Paul Lepp, Ph.D. Stanford School of Medicine VAPAHCS, 154T Dept. of Microbiology Immunology 3801 Miranda Ave Stanford University Palo Alto, CA 94304 Stanford, CA (650) 493-5000 x66762 fax: (650) 852-3291 http://cmgm.stanford.edu/~pwlepp [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] heatmap2
Paul, It is called heatmap.2 and it is in the gregmisc package. Sean On Sep 15, 2004, at 1:03 PM, Paul Lepp wrote: Can anybody tell me where to find a copy of heatmap2? I've seen it in my travels across the web but didn't bookmark it and can't find it again. Thanks in advance. Paul `-:-. ,-;`-:-. ,-;`-:-. ,-;`-:-. ,-;`-:-. ,-;`-:-. `=`,'=/ `=`,'=/ `=`,'=/ `=`,'=/ `=`,'=/ `=` ==/==/==/==/==/ ,=,-=`.,=,-=`.,=,-=`.,=,-=`.,=,-=`.,=, ,-'-' `-=_,-'-' `-=_,-'-' `-=_,-'-' `-=_,-'-' `-=_,-'-' Paul Lepp, Ph.D. Stanford School of Medicine VAPAHCS, 154T Dept. of Microbiology Immunology 3801 Miranda Ave Stanford University Palo Alto, CA 94304 Stanford, CA (650) 493-5000 x66762 fax: (650) 852-3291 http://cmgm.stanford.edu/~pwlepp [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] lomb periodogram package
Hi, Does anyone know the name of the package that includes a function for computing the lomb periodogram on irregular spaced ts data? I saw the package once ~ 1 month ago but cannot find it now ... , Rich __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Problem installing source packages on OS X
I am attempting to install the Hmisc, rreport and Design packages, but am not able to do so. I am running R v1.9.1 on Mac OS 10.3.5. I have tried installation using both the GUI version of R and also running R from sudo in a terminal session. In the terminal I receive the following error: * Installing *source* package 'Design' ... ** libs g77 -fno-common -g -O2 -c lrmfit.f -o lrmfit.o make: g77: Command not found make: *** [lrmfit.o] Error 127 ERROR: compilation failed for package 'Design' ** Removing '/Library/Frameworks/R.framework/Versions/1.9.1/Resources/library/Design ' I get the same error for Hmisc (rreport is not on CRAN). It looks like it is trying to use g77 to compile the source package. How can I change the default compiler? Will this solve the problem? I cannot find a binary version of either package. thanks, aric __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Problem installing source packages on OS X
On 15 Sep 2004, at 20:29, Aric Gregson wrote: I am attempting to install the Hmisc, rreport and Design packages, but am not able to do so. I am running R v1.9.1 on Mac OS 10.3.5. I get the same error for Hmisc (rreport is not on CRAN). It looks like it is trying to use g77 to compile the source package. How can I change the default compiler? Will this solve the problem? I cannot find a binary version of either package. R is trying to build a Fortran program, and it needs a Fortran compiler. Fortran compiler does not ship with MacOS X, but you got to get one. See the MacOS FAQ for R. If I remember correctly, it tells you to go http://hpc.sourceforge.net/ for the compiler. Normally I wouldn't remember addresses like this, but just today I had to make a visit there: I had installed g77 using fink, and that puts its stuff into /sw instead of /usr/local. Some R routines had hardcoded the g77 path to /usr/local/bin/g77 and so building a package failed in the false claim of missing g77 (yeah, it was in the path). cheers, jari oksanen -- Jari Oksanen, Oulu, Finland __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] BUGS and OS X
Is there a way of running BUGS on OS X (from R)? I only see Windows versions on their website. If not, what are the alternatives for Bayesian analysis? Thanks, Tamas __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Bessel function
You can find C++ source code for Bessel function and similar functions for example here: http://root.cern.ch/root/htmldoc/src/TMath.cxx.html Hope this helps Best regards Christian -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- C.h.r.i.s.t.i.a.n. .S.t.r.a.t.o.w.a V.i.e.n.n.a. .A.u.s.t.r.i.a -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- Ravi Varadhan wrote: Dear Beat, You might also want to look at the book by Zang and Jin - Computation of Special Functions, John Wiley. They have Fortran sources for all the special functions covered in there. Ravi. Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology, Johns Hopkins University, 2024 E. Monument Street, Suite 2-700 Baltimore, MD 21205. (410) 502 - 7806. From: [EMAIL PROTECTED] on behalf of [EMAIL PROTECTED] Sent: Wed 9/15/2004 4:24 AM To: [EMAIL PROTECTED] Subject: [R] Bessel function Dear all Currently, I'm implementing the generalized hyperbolic distribution into Splus. Unfortunately the Bessel function is not implemented in Splus. In R the Bessel function does exist but it is an internal function and I'm not able to look at the code. Is there any possibility to see the code of the Bessel function in R or does anybody has an implementation of the Bessel function in Splus? Thanks a lot for your help. Beat [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Slightly off-topic --- distribution name.
I've built R functions to ``effect'' a particular distribution, and
would like to find out if that distribution is already ``known'' by
an existing name. (I.e. suppose it were called the ``Melvin''
distribution --- I've built dmelvin, pmelvin, qmelvin, and rmelvin as
it were, but I need a real name to substitute for melvin.)
The distribution is really just a toy --- but it provides a nice (and
``non-obviouse'') example of a two parameter distribution where both
the moment and maximum likelihood equations for the parameter
estimators are readily solvable, but at the same time are
``interesting''. So it's good for exercises in an intro math-stats
course.
The distribution is simply that of the ***difference*** of two
independent exponential variates, with different parameters.
I.e. X = U - V where U ~ exp(beta) and V ~ exp(alpha) (where
E(U) = beta, E(V) = alpha).
This makes the distribution of X something like an asymetric Laplace
distribution, with its mode at 0. (One could shift the mode too, but
that would add a third parameter, which would be de trop.)
Anyhow: Is this a ``known'' distribution? Does it have a name?
(I've never seen it mentioned in any of the intro math-stat books
that I've looked into.) If not, can anyone suggest a good name for
it? (Don't be rude now!)
cheers,
Rolf Turner
[EMAIL PROTECTED]
P. S. To save you putting pen to paper and working it out,
the density function is
{ exp(x/alpha)/(alpha + beta) for x = 0
f(x) = {
{ exp(-x/beta)/(alpha + beta) for x = 0
The mean and variance are mu = beta - alpha and
sigma^2 = alpha^2 + beta^2 respectfully. :-)
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Re: [R] BUGS and OS X
On Wed, Sep 15, 2004 at 02:21:18PM -0400, Liaw, Andy wrote: That's more of a question for the BUGS developers. BUGS is not open source, so whatever binary is provided, that's all you can use. If I'm not mistaken, WinBUGS is the only version under development. I found something called JAGS, and I am still exploring it. It appears to be an open-source BUGS replacement, thought with limitations. I was asking what software people would recommend for the same functionality, not a drop-in replacement. I am just baffled by the bewildering array of R packages, and would be so happy if somebody told me what THEY use for Bayesian analysis, so I could read the docs and get started. MCMC? Boa? etc. Suggestions on how experienced users do bayesian analysis in R would be welcome. Thanks, Tamas __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Slightly off-topic --- distribution name.
Have you checked Johnson and Kotz? That's the obvious place to start
looking for distributions beyond the usual.
Rolf Turner [EMAIL PROTECTED] writes:
I've built R functions to ``effect'' a particular distribution, and
would like to find out if that distribution is already ``known'' by
an existing name. (I.e. suppose it were called the ``Melvin''
distribution --- I've built dmelvin, pmelvin, qmelvin, and rmelvin as
it were, but I need a real name to substitute for melvin.)
The distribution is really just a toy --- but it provides a nice (and
``non-obviouse'') example of a two parameter distribution where both
the moment and maximum likelihood equations for the parameter
estimators are readily solvable, but at the same time are
``interesting''. So it's good for exercises in an intro math-stats
course.
The distribution is simply that of the ***difference*** of two
independent exponential variates, with different parameters.
I.e. X = U - V where U ~ exp(beta) and V ~ exp(alpha) (where
E(U) = beta, E(V) = alpha).
This makes the distribution of X something like an asymetric Laplace
distribution, with its mode at 0. (One could shift the mode too, but
that would add a third parameter, which would be de trop.)
Anyhow: Is this a ``known'' distribution? Does it have a name?
(I've never seen it mentioned in any of the intro math-stat books
that I've looked into.) If not, can anyone suggest a good name for
it? (Don't be rude now!)
cheers,
Rolf Turner
[EMAIL PROTECTED]
P. S. To save you putting pen to paper and working it out,
the density function is
{ exp(x/alpha)/(alpha + beta) for x = 0
f(x) = {
{ exp(-x/beta)/(alpha + beta) for x = 0
The mean and variance are mu = beta - alpha and
sigma^2 = alpha^2 + beta^2 respectfully. :-)
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[R] replacing NA's with 0 in a dataframe for specified columns
I know that there must be a cool way of doing this, but I can't think of it. Let's say I have an dataframe with NA's. x - data.frame(a = c(0,1,2,NA), b = c(0,NA,1,2), c = c(NA, 0, 1, 2)) x a b c 1 0 0 NA 2 1 NA 0 3 2 1 1 4 NA 2 2 I know it is easy to replace all the NA's with zeroes. x[is.na(x)] - 0 x a b c 1 0 0 0 2 1 0 0 3 2 1 1 4 0 2 2 But how do I do this for just columns a and c, leaving the NA in column b alone? Thanks, Dave Kane R.version _ platform i686-pc-linux-gnu arch i686 os linux-gnu system i686, linux-gnu status major1 minor9.1 year 2004 month06 day 21 language R __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] lomb periodogram package
Does anyone know the name of the package that includes a function for computing the lomb periodogram on irregular spaced ts data? I saw the package once ~ 1 month ago but cannot find it now ... I have a LombScargleLibrary.R file that I will be talking about at CAMDA '04 in November: Searching for Periodic Microarray Expression Patterns Using Lomb-Scargle Periodograms http://research.stowers-institute.org/efg/2004/CAMDA/ I'll get the file online by the time of the conference on Nov 10: http://www.camda.duke.edu/camda04 I could E-mail you the preliminary version before then if you'd like. Just drop me an E-mail. efg Earl F. Glynn Scientific Programmer Bioinformatics Department Stowers Institute for Medical Research __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] [R-pkgs] Announcing snowFT 0.1
Parallel programming with snowFT
Our package snowFT is now available at CRAN. It is an extention of the
package snow, which adds fault tolerance (in the sense of recomputing
computational units when hardware/network failures occur on compute
nodes) and a tighter notion of reproducibility for computations
running on clusters. It additionally provides tools for flexible
management of cluster size as well as computation transparency.
snowFT is written by Hana Sevcikova and Tony Rossini.
(this tool currently requires rpvm for the SNOW backend, though we are
exploring the possibility of extensions for the Rmpi backend. It is
unlikely that these extensions will be implemented for the socket
backend.
best,
-tony
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Biomedical and Health Informatics University of Washington
Biostatistics, SCHARP/HVTN Fred Hutchinson Cancer Research Center
UW (Tu/Th/F): 206-616-7630 FAX=206-543-3461 | Voicemail is unreliable
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Re: [R] replacing NA's with 0 in a dataframe for specified columns
Have you considered the following: x - data.frame(a = c(0,1,2,NA), b = c(0,NA,1,2), c = c(NA, 0, 1, 2)) x$a[is.na(x$a)] - 0 x$c[is.na(x$c)] - 0 x a b c 1 0 0 0 2 1 NA 0 3 2 1 1 4 0 2 2 hope this helps. spencer graves David Kane wrote: I know that there must be a cool way of doing this, but I can't think of it. Let's say I have an dataframe with NA's. x - data.frame(a = c(0,1,2,NA), b = c(0,NA,1,2), c = c(NA, 0, 1, 2)) x a b c 1 0 0 NA 2 1 NA 0 3 2 1 1 4 NA 2 2 I know it is easy to replace all the NA's with zeroes. x[is.na(x)] - 0 x a b c 1 0 0 0 2 1 0 0 3 2 1 1 4 0 2 2 But how do I do this for just columns a and c, leaving the NA in column b alone? Thanks, Dave Kane R.version _ platform i686-pc-linux-gnu arch i686 os linux-gnu system i686, linux-gnu status major1 minor9.1 year 2004 month06 day 21 language R __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Spencer Graves, PhD, Senior Development Engineer O: (408)938-4420; mobile: (408)655-4567 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] replacing NA's with 0 in a dataframe for specified columns
try: x[is.na(x$a) | is.na(x$c),] - 0 At 02:44 PM 9/15/2004 -0400, David Kane wrote: I know that there must be a cool way of doing this, but I can't think of it. Let's say I have an dataframe with NA's. x - data.frame(a = c(0,1,2,NA), b = c(0,NA,1,2), c = c(NA, 0, 1, 2)) x a b c 1 0 0 NA 2 1 NA 0 3 2 1 1 4 NA 2 2 I know it is easy to replace all the NA's with zeroes. x[is.na(x)] - 0 x a b c 1 0 0 0 2 1 0 0 3 2 1 1 4 0 2 2 But how do I do this for just columns a and c, leaving the NA in column b alone? Thanks, Dave Kane R.version _ platform i686-pc-linux-gnu arch i686 os linux-gnu system i686, linux-gnu status major1 minor9.1 year 2004 month06 day 21 language R __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html With best wishes and kind regards I am Sincerely, Corey A. Moffet Rangeland Scientist ## USDA-ARS# Northwest Watershed Research Center # 800 Park Blvd, Plaza IV, Suite 105 ### Boise, ID 83712-7716 ## # # Voice: (208) 422-0718 ## FAX: (208) 334-1502 ## # # ### ## __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] replacing NA's with 0 in a dataframe for specified columns
But Spencer's solution would require looping to generalize to a large numbers of columns. In fact, you've given the answer already in your post: xsub-x[,somecols] xsub[is.na(xsub)]-0 x[,somecols]-xsub -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA The business of the statistician is to catalyze the scientific learning process. - George E. P. Box -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Spencer Graves Sent: Wednesday, September 15, 2004 12:01 PM To: David Kane Cc: [EMAIL PROTECTED] Subject: Re: [R] replacing NA's with 0 in a dataframe for specified columns Have you considered the following: x - data.frame(a = c(0,1,2,NA), b = c(0,NA,1,2), c = c(NA, 0, 1, 2)) x$a[is.na(x$a)] - 0 x$c[is.na(x$c)] - 0 x a b c 1 0 0 0 2 1 NA 0 3 2 1 1 4 0 2 2 hope this helps. spencer graves David Kane wrote: I know that there must be a cool way of doing this, but I can't think of it. Let's say I have an dataframe with NA's. x - data.frame(a = c(0,1,2,NA), b = c(0,NA,1,2), c = c(NA, 0, 1, 2)) x a b c 1 0 0 NA 2 1 NA 0 3 2 1 1 4 NA 2 2 I know it is easy to replace all the NA's with zeroes. x[is.na(x)] - 0 x a b c 1 0 0 0 2 1 0 0 3 2 1 1 4 0 2 2 But how do I do this for just columns a and c, leaving the NA in column b alone? Thanks, Dave Kane R.version _ platform i686-pc-linux-gnu arch i686 os linux-gnu system i686, linux-gnu status major1 minor9.1 year 2004 month06 day 21 language R __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Spencer Graves, PhD, Senior Development Engineer O: (408)938-4420; mobile: (408)655-4567 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] replacing NA's with 0 in a dataframe for specified columns
Dear David, How about the following? cols - c(1,3) x[,cols][is.na(x[,cols])] - 0 I hope that this helps, John On Wed, 15 Sep 2004 14:44:53 -0400 David Kane [EMAIL PROTECTED] wrote: I know that there must be a cool way of doing this, but I can't think of it. Let's say I have an dataframe with NA's. x - data.frame(a = c(0,1,2,NA), b = c(0,NA,1,2), c = c(NA, 0, 1, 2)) x a b c 1 0 0 NA 2 1 NA 0 3 2 1 1 4 NA 2 2 I know it is easy to replace all the NA's with zeroes. x[is.na(x)] - 0 x a b c 1 0 0 0 2 1 0 0 3 2 1 1 4 0 2 2 But how do I do this for just columns a and c, leaving the NA in column b alone? Thanks, Dave Kane R.version _ platform i686-pc-linux-gnu arch i686 os linux-gnu system i686, linux-gnu status major1 minor9.1 year 2004 month06 day 21 language R __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html John Fox Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] replacing NA's with 0 in a dataframe for specified columns
mydata - data.frame(a = c(0,1,2,NA), b = c(0,NA,1,2), c = c(NA, 0, 1, 2))
mydata
a b c
1 0 0 NA
2 1 NA 0
3 2 1 1
4 NA 2 2
mydata[,c(a, c)] -
apply(mydata[,c(a,c)], 2, function(x){replace(x, is.na(x), 0)})
mydata
a b c
1 0 0 0
2 1 NA 0
3 2 1 1
4 0 2 2
David Kane wrote:
I know that there must be a cool way of doing this, but I can't think
of it. Let's say I have an dataframe with NA's.
x - data.frame(a = c(0,1,2,NA), b = c(0,NA,1,2), c = c(NA, 0, 1, 2))
x
a b c
1 0 0 NA
2 1 NA 0
3 2 1 1
4 NA 2 2
I know it is easy to replace all the NA's with zeroes.
x[is.na(x)] - 0
x
a b c
1 0 0 0
2 1 0 0
3 2 1 1
4 0 2 2
But how do I do this for just columns a and c, leaving the NA in
column b alone?
Thanks,
Dave Kane
R.version
_
platform i686-pc-linux-gnu
arch i686
os linux-gnu
system i686, linux-gnu
status
major1
minor9.1
year 2004
month06
day 21
language R
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Re: [R] replacing NA's with 0 in a dataframe for specified columns
The col funtion can be helpful here. We want to satisfy two conditions: 1. the element is an NA 2. the element lies in one of the specified columns The first two lines below calculate logical vectors for these two, respectively, and the last line assigns 0 to those elements. isna - is.na(x) iscol - col(isna) %in% c(1,3) x[isna iscol] - 0 To specify the columns by name rather than number first calculate their column numbers and then proceed as before. That is, replace the second line by: cols - match(c(a, c), colnames(x)) iscol - col(isna) %in% cols David Kane dave at kanecap.com writes: : : I know that there must be a cool way of doing this, but I can't think : of it. Let's say I have an dataframe with NA's. : : x - data.frame(a = c(0,1,2,NA), b = c(0,NA,1,2), c = c(NA, 0, 1, 2)) : x :a b c : 1 0 0 NA : 2 1 NA 0 : 3 2 1 1 : 4 NA 2 2 : : : I know it is easy to replace all the NA's with zeroes. : : x[is.na(x)] - 0 : x : a b c : 1 0 0 0 : 2 1 0 0 : 3 2 1 1 : 4 0 2 2 : : : But how do I do this for just columns a and c, leaving the NA in : column b alone? : : Thanks, : : Dave Kane : : R.version : _ : platform i686-pc-linux-gnu : arch i686 : os linux-gnu : system i686, linux-gnu : status : major1 : minor9.1 : year 2004 : month06 : day 21 : language R : : : __ : R-help at stat.math.ethz.ch mailing list : https://stat.ethz.ch/mailman/listinfo/r-help : PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html : : __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Splitting vector into individual elements
do.call() is good for this, I believe:
offred.rgb - c(1, 0, 0) * 0.60
offred.col - do.call(rgb, c(as.list(offred.rgb), names=offred))
offred.col
[1] #99
HTH,
Andy
From: Paul Roebuck
Is there a means to split a vector into its individual
elements without going the brute-force route for arguments
to a predefined function call?
offred.rgb - c(1, 0, 0) * 0.60;
## Brute force style
offred.col - rgb(offred.rgb[1],
offred.rgb[2],
offred.rgb[3],
names = offred)
## Desired style
offred.col - rgb(silver.bullet(offred.rgb),
names = offred)
Neither of my attempts gets it right.
silver.bullet.try1 - function(x) {
expr - cat(x, sep = ,)
return(parse(text = expr))
}
silver.bullet.try2 - function(x) {
expr - expression(cat(x, sep = ,))
return(eval(expr))
}
--
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Re: [R] Splitting vector into individual elements
Have you considered do.call:
do.call(rgb, as.list((1:3)/10))
[1] #1A334C
same as:
rgb(.1, .2, .3)
[1] #1A334C
Hope this helps. spencer graves
Paul Roebuck wrote:
Is there a means to split a vector into its individual
elements without going the brute-force route for arguments
to a predefined function call?
offred.rgb - c(1, 0, 0) * 0.60;
## Brute force style
offred.col - rgb(offred.rgb[1],
offred.rgb[2],
offred.rgb[3],
names = offred)
## Desired style
offred.col - rgb(silver.bullet(offred.rgb),
names = offred)
Neither of my attempts gets it right.
silver.bullet.try1 - function(x) {
expr - cat(x, sep = ,)
return(parse(text = expr))
}
silver.bullet.try2 - function(x) {
expr - expression(cat(x, sep = ,))
return(eval(expr))
}
--
SIGSIG -- signature too long (core dumped)
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O: (408)938-4420; mobile: (408)655-4567
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[R] RWAVE axis notation
I am using Rwave wavelets and I need better axis notation. Does anyone have code similar to matlab's centfreq or scale2frq functions that turn a scale for a wavelet transform into a good looking scale to plot on my wavelet transfoms? I am using the rwave package to investigate seismic signals from an exploding volcano. Jonathan Lees -- == Prof. Jonathan M. Lees Department of Geological Sciences CB #3315, Mitchell Hall University of North Carolina Chapel Hill, NC 27599-3315 (919) 962-0695 FAX (919) 966-4519 [EMAIL PROTECTED] http://www.unc.edu/~leesj __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Splitting vector into individual elements
Paul Roebuck roebuck at odin.mdacc.tmc.edu writes: : Is there a means to split a vector into its individual : elements without going the brute-force route for arguments : to a predefined function call? : : offred.rgb - c(1, 0, 0) * 0.60; : : ## Brute force style : offred.col - rgb(offred.rgb[1], : offred.rgb[2], : offred.rgb[3], : names = offred) : ## Desired style : offred.col - rgb(silver.bullet(offred.rgb), : names = offred) See: http://maths.newcastle.edu.au/~rking/R/help/03a/7417.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Splitting vector into individual elements
Dear Paul,
How about do.call(rgb, as.list(offred.rgb)) ?
I hope that this helps,
John
On Wed, 15 Sep 2004 15:20:24 -0500 (CDT)
Paul Roebuck [EMAIL PROTECTED] wrote:
Is there a means to split a vector into its individual
elements without going the brute-force route for arguments
to a predefined function call?
offred.rgb - c(1, 0, 0) * 0.60;
## Brute force style
offred.col - rgb(offred.rgb[1],
offred.rgb[2],
offred.rgb[3],
names = offred)
## Desired style
offred.col - rgb(silver.bullet(offred.rgb),
names = offred)
Neither of my attempts gets it right.
silver.bullet.try1 - function(x) {
expr - cat(x, sep = ,)
return(parse(text = expr))
}
silver.bullet.try2 - function(x) {
expr - expression(cat(x, sep = ,))
return(eval(expr))
}
--
SIGSIG -- signature too long (core dumped)
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John Fox
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/
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Re: [R] Splitting vector into individual elements
Paul Roebuck [EMAIL PROTECTED] writes: Is there a means to split a vector into its individual elements without going the brute-force route for arguments to a predefined function call? offred.rgb - c(1, 0, 0) * 0.60; ## Brute force style offred.col - rgb(offred.rgb[1], offred.rgb[2], offred.rgb[3], names = offred) ## Desired style offred.col - rgb(silver.bullet(offred.rgb), names = offred) The closest is probably this: offred.col - do.call(rgb, c(as.list(offred.rgb), list(names=offred))) (ever read/seen The Handmaid's Tale, btw?) -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] efficient submatrix extraction
Hi,
I have a matrix of say 1024x1024 and I want to look at it in chunks.
That is I'd like to divide into a series of submatrices of order 2x2.
| 1 2 3 4 5 6 7 8 ... |
| 1 2 3 4 5 6 7 8 ... |
| 1 2 3 4 5 6 7 8 ... |
| 1 2 3 4 5 6 7 8 ... |
...
So the first submatrix would be
| 1 2 |
| 1 2 |
the second one would be
| 3 4 |
| 3 4 |
and so on. That is I want the matrix to be evenly divided into 2x2
submatrices. Now I'm also doing this subdivision into 4x4, 8x8 ...
256x256 submatrices.
Currently I'm using loops and I'm sure there is a mroe efficient way to
do it:
m - matrix(runif(1024*1024), nrow=1024)
boxsize - 2^(1:8)
for (b in boxsize) {
bcount - 0
bstart - seq(1,1024, by=b)
for (x in bstart) {
for (y in bstart) {
xend - x + b - 1
yend - y + b - 1
if (length(which( m[ x:xend, y:yend ] 0.7)) 0) {
bcount - bcount + 1
}
}
}
}
Is there any way to vectorize the two inner loops?
Thanks,
---
Rajarshi Guha [EMAIL PROTECTED] http://jijo.cjb.net
GPG Fingerprint: 0CCA 8EE2 2EEB 25E2 AB04 06F7 1BB9 E634 9B87 56EE
---
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Re: [R] Splitting vector into individual elements
On Wed, 15 Sep 2004, Peter Dalgaard wrote: Paul Roebuck [EMAIL PROTECTED] writes: Is there a means to split a vector into its individual elements without going the brute-force route for arguments to a predefined function call? offred.rgb - c(1, 0, 0) * 0.60; ## Brute force style offred.col - rgb(offred.rgb[1], offred.rgb[2], offred.rgb[3], names = offred) ## Desired style offred.col - rgb(silver.bullet(offred.rgb), names = offred) The closest is probably this: offred.col - do.call(rgb, c(as.list(offred.rgb), list(names=offred))) Everyone offered 'do.call' as the solution. While that works, is it to say that there is no means of expanding the expression as an argument to the original function? (ever read/seen The Handmaid's Tale, btw?) Not yet. Though renaming my sample variable 'off.red.col' would avoid future confusion with oppressed handmaids. -- SIGSIG -- signature too long (core dumped) __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] efficient submatrix extraction
I think you should be able to do something with reassigning the dim
attribute, and then using apply(), something along the lines of the
following (which doesn't do your computation on the data in the subarrays,
but merely illustrates how to create and access them):
x - matrix(1:64,ncol=8)
x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]19 17 25 33 41 49 57
[2,]2 10 18 26 34 42 50 58
[3,]3 11 19 27 35 43 51 59
[4,]4 12 20 28 36 44 52 60
[5,]5 13 21 29 37 45 53 61
[6,]6 14 22 30 38 46 54 62
[7,]7 15 23 31 39 47 55 63
[8,]8 16 24 32 40 48 56 64
x2 - x
dim(x2) - c(2,4,2,4)
x2[,1,,1]
[,1] [,2]
[1,]19
[2,]2 10
x2[,2,,1]
[,1] [,2]
[1,]3 11
[2,]4 12
x2[,1,,2]
[,1] [,2]
[1,] 17 25
[2,] 18 26
x4 - x
dim(x4) - c(4,2,4,2)
x4[,1,,1]
[,1] [,2] [,3] [,4]
[1,]19 17 25
[2,]2 10 18 26
[3,]3 11 19 27
[4,]4 12 20 28
invisible(apply(x4, c(2,4), print))
[,1] [,2] [,3] [,4]
[1,]19 17 25
[2,]2 10 18 26
[3,]3 11 19 27
[4,]4 12 20 28
[,1] [,2] [,3] [,4]
[1,]5 13 21 29
[2,]6 14 22 30
[3,]7 15 23 31
[4,]8 16 24 32
[,1] [,2] [,3] [,4]
[1,] 33 41 49 57
[2,] 34 42 50 58
[3,] 35 43 51 59
[4,] 36 44 52 60
[,1] [,2] [,3] [,4]
[1,] 37 45 53 61
[2,] 38 46 54 62
[3,] 39 47 55 63
[4,] 40 48 56 64
hope this helps,
Tony Plate
At Wednesday 03:10 PM 9/15/2004, Rajarshi Guha wrote:
Hi,
I have a matrix of say 1024x1024 and I want to look at it in chunks.
That is I'd like to divide into a series of submatrices of order 2x2.
| 1 2 3 4 5 6 7 8 ... |
| 1 2 3 4 5 6 7 8 ... |
| 1 2 3 4 5 6 7 8 ... |
| 1 2 3 4 5 6 7 8 ... |
...
So the first submatrix would be
| 1 2 |
| 1 2 |
the second one would be
| 3 4 |
| 3 4 |
and so on. That is I want the matrix to be evenly divided into 2x2
submatrices. Now I'm also doing this subdivision into 4x4, 8x8 ...
256x256 submatrices.
Currently I'm using loops and I'm sure there is a mroe efficient way to
do it:
m - matrix(runif(1024*1024), nrow=1024)
boxsize - 2^(1:8)
for (b in boxsize) {
bcount - 0
bstart - seq(1,1024, by=b)
for (x in bstart) {
for (y in bstart) {
xend - x + b - 1
yend - y + b - 1
if (length(which( m[ x:xend, y:yend ] 0.7)) 0) {
bcount - bcount + 1
}
}
}
}
Is there any way to vectorize the two inner loops?
Thanks,
---
Rajarshi Guha [EMAIL PROTECTED] http://jijo.cjb.net
GPG Fingerprint: 0CCA 8EE2 2EEB 25E2 AB04 06F7 1BB9 E634 9B87 56EE
---
The way to love anything is to realize that it might be lost.
__
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RE: [R] Splitting vector into individual elements
From: Paul Roebuck On Wed, 15 Sep 2004, Peter Dalgaard wrote: Paul Roebuck [EMAIL PROTECTED] writes: Is there a means to split a vector into its individual elements without going the brute-force route for arguments to a predefined function call? offred.rgb - c(1, 0, 0) * 0.60; ## Brute force style offred.col - rgb(offred.rgb[1], offred.rgb[2], offred.rgb[3], names = offred) ## Desired style offred.col - rgb(silver.bullet(offred.rgb), names = offred) The closest is probably this: offred.col - do.call(rgb, c(as.list(offred.rgb), list(names=offred))) Everyone offered 'do.call' as the solution. While that works, is it to say that there is no means of expanding the expression as an argument to the original function? What would be the point? That's what do.call() does for you internally. Andy (ever read/seen The Handmaid's Tale, btw?) Not yet. Though renaming my sample variable 'off.red.col' would avoid future confusion with oppressed handmaids. -- SIGSIG -- signature too long (core dumped) __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Splitting vector into individual elements
Paul Roebuck [EMAIL PROTECTED] writes: Everyone offered 'do.call' as the solution. While that works, is it to say that there is no means of expanding the expression as an argument to the original function? Not really. You need an explicit expansion of the argument to a list somehow, and there's no silver bullet that can convert one function argument to several. There are solutions without do.call, like offred.rgb - c(1, 0, 0) * 0.60 x - quote(rgb(.,.,.,names=offred)) x[2:4] - as.list(offred.rgb) eval(x) offred #99 but you might find it difficult to explain how it works a year later -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Splitting vector into individual elements
From: Liaw, Andy From: Paul Roebuck On Wed, 15 Sep 2004, Peter Dalgaard wrote: Paul Roebuck [EMAIL PROTECTED] writes: Is there a means to split a vector into its individual elements without going the brute-force route for arguments to a predefined function call? offred.rgb - c(1, 0, 0) * 0.60; ## Brute force style offred.col - rgb(offred.rgb[1], offred.rgb[2], offred.rgb[3], names = offred) ## Desired style offred.col - rgb(silver.bullet(offred.rgb), names = offred) The closest is probably this: offred.col - do.call(rgb, c(as.list(offred.rgb), list(names=offred))) Everyone offered 'do.call' as the solution. While that works, is it to say that there is no means of expanding the expression as an argument to the original function? What would be the point? That's what do.call() does for you internally. Is this what you're after? toCall - c(as.name(rgb), as.list(offred.rgb), names=offred) toCall [[1]] rgb [[2]] [1] 0.6 [[3]] [1] 0 [[4]] [1] 0 $names [1] offred toCall - as.call(toCall) toCall rgb(0.6, 0, 0, names = offred) eval(toCall) [1] #99 Andy Andy (ever read/seen The Handmaid's Tale, btw?) Not yet. Though renaming my sample variable 'off.red.col' would avoid future confusion with oppressed handmaids. -- SIGSIG -- signature too long (core dumped) __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Notice: This e-mail message, together with any attachments, contains information of Merck Co., Inc. (One Merck Drive, Whitehouse Station, New Jersey, USA 08889), and/or its affiliates (which may be known outside the United States as Merck Frosst, Merck Sharp Dohme or MSD and in Japan, as Banyu) that may be confidential, proprietary copyrighted and/or legally privileged. It is intended solely for the use of the individual or entity named on this message. If you are not the intended recipient, and have received this message in error, please notify us immediately by reply e-mail and then delete it from your system. -- __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Slightly off-topic --- distribution name.
I believe this is the skew-Laplace distribution, although the skew-Laplace
does allow for the location of the mode of the distribution to vary.
Have a look at the function dskewlap in HyperbolicDist. The help on that
function gives a reference to a paper by Feiller et al which describes the
distribution.
David Scott
On Wed, 15 Sep 2004, Rolf Turner wrote:
I've built R functions to ``effect'' a particular distribution, and
would like to find out if that distribution is already ``known'' by
an existing name. (I.e. suppose it were called the ``Melvin''
distribution --- I've built dmelvin, pmelvin, qmelvin, and rmelvin as
it were, but I need a real name to substitute for melvin.)
The distribution is really just a toy --- but it provides a nice (and
``non-obviouse'') example of a two parameter distribution where both
the moment and maximum likelihood equations for the parameter
estimators are readily solvable, but at the same time are
``interesting''. So it's good for exercises in an intro math-stats
course.
The distribution is simply that of the ***difference*** of two
independent exponential variates, with different parameters.
I.e. X = U - V where U ~ exp(beta) and V ~ exp(alpha) (where
E(U) = beta, E(V) = alpha).
This makes the distribution of X something like an asymetric Laplace
distribution, with its mode at 0. (One could shift the mode too, but
that would add a third parameter, which would be de trop.)
Anyhow: Is this a ``known'' distribution? Does it have a name?
(I've never seen it mentioned in any of the intro math-stat books
that I've looked into.) If not, can anyone suggest a good name for
it? (Don't be rude now!)
cheers,
Rolf Turner
[EMAIL PROTECTED]
P. S. To save you putting pen to paper and working it out,
the density function is
{ exp(x/alpha)/(alpha + beta) for x = 0
f(x) = {
{ exp(-x/beta)/(alpha + beta) for x = 0
The mean and variance are mu = beta - alpha and
sigma^2 = alpha^2 + beta^2 respectfully. :-)
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_
David Scott Department of Statistics, Tamaki Campus
The University of Auckland, PB 92019
AucklandNEW ZEALAND
Phone: +64 9 373 7599 ext 86830 Fax: +64 9 373 7000
Email: [EMAIL PROTECTED]
Graduate Officer, Department of Statistics
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Re: [R] Splitting vector into individual elements
Slightly more transparent but arguably uglier: offred.rgb - c(1, 0, 0) * 0.60 ofr - paste(offred.rgb, collapse=,) ofr. - paste(rgb(, ofr, ',names=offred)') ofr. [1] rgb( 0.6,0,0 ,names=\offred\) eval(parse(text=ofr.)) offred #99 As long as I can remember eval(parse(text=, this is for me the most transparent and works for constructing virtually any R command. hope this helps. spencer graves Peter Dalgaard wrote: Paul Roebuck [EMAIL PROTECTED] writes: Everyone offered 'do.call' as the solution. While that works, is it to say that there is no means of expanding the expression as an argument to the original function? Not really. You need an explicit expansion of the argument to a list somehow, and there's no silver bullet that can convert one function argument to several. There are solutions without do.call, like offred.rgb - c(1, 0, 0) * 0.60 x - quote(rgb(.,.,.,names=offred)) x[2:4] - as.list(offred.rgb) eval(x) offred #99 but you might find it difficult to explain how it works a year later -- Spencer Graves, PhD, Senior Development Engineer O: (408)938-4420; mobile: (408)655-4567 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Signs of loadings from princomp on Windows
I am sorry to insist, but we have three other people that were able to reproduce the behavior I mentioned. I have also installed R 1.9.1 from the CRAN binaries on a different Windows machine and again I see the differents signs as mentioned before. What would be causing the difference? -Francisco On Tue, 14 Sep 2004 11:04:29 -0600, Tony Plate [EMAIL PROTECTED] wrote: FWIW, I see the same behavior as Francisco on my Windows machine (also an installation of the windows binary without trying to install any special BLAS libraries): library(MASS) data(painters) pca.painters - princomp(painters[ ,1:4]) loadings(pca.painters) Loadings: Comp.1 Comp.2 Comp.3 Comp.4 Composition 0.484 -0.376 0.784 -0.101 Drawing 0.424 0.187 -0.280 -0.841 Colour -0.381 -0.845 -0.211 -0.310 Expression 0.664 -0.330 -0.513 0.432 Comp.1 Comp.2 Comp.3 Comp.4 SS loadings 1.00 1.00 1.00 1.00 Proportion Var 0.25 0.25 0.25 0.25 Cumulative Var 0.25 0.50 0.75 1.00 pca.painters - princomp(painters[ ,1:4]) loadings(pca.painters) Loadings: Comp.1 Comp.2 Comp.3 Comp.4 Composition -0.484 -0.376 0.784 -0.101 Drawing -0.424 0.187 -0.280 -0.841 Colour 0.381 -0.845 -0.211 -0.310 Expression -0.664 -0.330 -0.513 0.432 Comp.1 Comp.2 Comp.3 Comp.4 SS loadings 1.00 1.00 1.00 1.00 Proportion Var 0.25 0.25 0.25 0.25 Cumulative Var 0.25 0.50 0.75 1.00 R.version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major1 minor9.1 year 2004 month06 day 21 language R My machine is a dual-processor hp xw8000. I also get the same results with R 2.0.0 dev as in R.version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status Under development (unstable) major2 minor0.0 year 2004 month09 day 13 language R -- Tony Plate At Tuesday 10:25 AM 9/14/2004, Prof Brian Ripley wrote: On Tue, 14 Sep 2004, Francisco Chamu wrote: I have run this on both Windows 2000 and XP. All I did was install the binaries from CRAN so I think I am using the standard Rblas.dll. To reproduce what I see you must run the code at the beginning of the R session. We did, as you said `start a clean session'. I think to reproduce what you see we have to be using your account on your computer. After the second run, all subsequent runs give the same result as the second set. Thanks, Francisco On Tue, 14 Sep 2004 08:29:25 +0200, Uwe Ligges [EMAIL PROTECTED] wrote: Prof Brian Ripley wrote: I get the second set each time, on Windows, using the build from CRAN. Which BLAS are you using? Works also well for me with a self compiled R-1.9.1 (both with standard Rblas as well as with the Rblas.dll for Athlon CPU from CRAN). Is this a NT-based version of Windows (NT, 2k, XP)? Uwe On Tue, 14 Sep 2004, Francisco Chamu wrote: I start a clean session of R 1.9.1 on Windows and I run the following code: library(MASS) data(painters) pca.painters - princomp(painters[ ,1:4]) loadings(pca.painters) Loadings: Comp.1 Comp.2 Comp.3 Comp.4 Composition 0.484 -0.376 0.784 -0.101 Drawing 0.424 0.187 -0.280 -0.841 Colour -0.381 -0.845 -0.211 -0.310 Expression 0.664 -0.330 -0.513 0.432 Comp.1 Comp.2 Comp.3 Comp.4 SS loadings 1.00 1.00 1.00 1.00 Proportion Var 0.25 0.25 0.25 0.25 Cumulative Var 0.25 0.50 0.75 1.00 However, if I rerun the same analysis, the loadings of the first component have the opposite sign (see below), why is that? I have read the note in the princomp help that says The signs of the columns of the loadings and scores are arbitrary, and so may differ between different programs for PCA, and even between different builds of R. However, I still would expect the same signs for two runs in the same session. pca.painters - princomp(painters[ ,1:4]) loadings(pca.painters) Loadings: Comp.1 Comp.2 Comp.3 Comp.4 Composition -0.484 -0.376 0.784 -0.101 Drawing -0.424 0.187 -0.280 -0.841 Colour 0.381 -0.845 -0.211 -0.310 Expression -0.664 -0.330 -0.513 0.432 Comp.1 Comp.2 Comp.3 Comp.4 SS loadings 1.00 1.00 1.00 1.00 Proportion Var 0.25 0.25 0.25 0.25 Cumulative Var 0.25 0.50 0.75 1.00 R.version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major1 minor9.1 year 2004
[R] Cross-validation for Linear Discrimitant Analysis
Hello: I am new to R and statistics and I have two questions. First I need help to interpret the cross-validation result from the R linear discriminant analysis function lda. I did the following: lda (group ~ Var1 + Var2, CV=T) where CV=T tells the lda to do cross-validation. The output of lda are the posterior probabilities among other things, but I can't find an error term (like delta returned by cv.glm). My question is how to get such an error term from the output? Can I just simply calculate the prediction accuracy using the posterior probabilities from the cross-validation, and use that to measure the quality of the model? Another question is more basic: how to determine if a lda model is significant? (There is no p-value.) Thanks, Yu Shao Wadsworth Research Center Department of Health of New York State Albany, NY 12208 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Signs of loadings from princomp on Windows
You could investigate this yourself by looking at the code of princomp (try getAnywhere(princomp.default)). I'd suggest making a file that in-lines the body of princomp.default into the commands you had below. See if you still get the difference. (I'd be surprised if you didn't). Then try commenting out lines the second pass through the commands produces the same results as the first. The very last thing you commented out might help to answer your question What would be causing the difference? (The fact that various people chimed in to say they could reproduce the behavior that bothered you, but didn't bother dig deeper suggests it didn't bother them that much, which further suggests that you are the person most motivated by this and thus the best candidate for investigating it further...) -- Tony Plate At Wednesday 05:07 PM 9/15/2004, Francisco Chamu wrote: I am sorry to insist, but we have three other people that were able to reproduce the behavior I mentioned. I have also installed R 1.9.1 from the CRAN binaries on a different Windows machine and again I see the differents signs as mentioned before. What would be causing the difference? -Francisco On Tue, 14 Sep 2004 11:04:29 -0600, Tony Plate [EMAIL PROTECTED] wrote: FWIW, I see the same behavior as Francisco on my Windows machine (also an installation of the windows binary without trying to install any special BLAS libraries): library(MASS) data(painters) pca.painters - princomp(painters[ ,1:4]) loadings(pca.painters) Loadings: Comp.1 Comp.2 Comp.3 Comp.4 Composition 0.484 -0.376 0.784 -0.101 Drawing 0.424 0.187 -0.280 -0.841 Colour -0.381 -0.845 -0.211 -0.310 Expression 0.664 -0.330 -0.513 0.432 Comp.1 Comp.2 Comp.3 Comp.4 SS loadings 1.00 1.00 1.00 1.00 Proportion Var 0.25 0.25 0.25 0.25 Cumulative Var 0.25 0.50 0.75 1.00 pca.painters - princomp(painters[ ,1:4]) loadings(pca.painters) Loadings: Comp.1 Comp.2 Comp.3 Comp.4 Composition -0.484 -0.376 0.784 -0.101 Drawing -0.424 0.187 -0.280 -0.841 Colour 0.381 -0.845 -0.211 -0.310 Expression -0.664 -0.330 -0.513 0.432 Comp.1 Comp.2 Comp.3 Comp.4 SS loadings 1.00 1.00 1.00 1.00 Proportion Var 0.25 0.25 0.25 0.25 Cumulative Var 0.25 0.50 0.75 1.00 R.version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major1 minor9.1 year 2004 month06 day 21 language R My machine is a dual-processor hp xw8000. I also get the same results with R 2.0.0 dev as in R.version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status Under development (unstable) major2 minor0.0 year 2004 month09 day 13 language R -- Tony Plate At Tuesday 10:25 AM 9/14/2004, Prof Brian Ripley wrote: On Tue, 14 Sep 2004, Francisco Chamu wrote: I have run this on both Windows 2000 and XP. All I did was install the binaries from CRAN so I think I am using the standard Rblas.dll. To reproduce what I see you must run the code at the beginning of the R session. We did, as you said `start a clean session'. I think to reproduce what you see we have to be using your account on your computer. After the second run, all subsequent runs give the same result as the second set. Thanks, Francisco On Tue, 14 Sep 2004 08:29:25 +0200, Uwe Ligges [EMAIL PROTECTED] wrote: Prof Brian Ripley wrote: I get the second set each time, on Windows, using the build from CRAN. Which BLAS are you using? Works also well for me with a self compiled R-1.9.1 (both with standard Rblas as well as with the Rblas.dll for Athlon CPU from CRAN). Is this a NT-based version of Windows (NT, 2k, XP)? Uwe On Tue, 14 Sep 2004, Francisco Chamu wrote: I start a clean session of R 1.9.1 on Windows and I run the following code: library(MASS) data(painters) pca.painters - princomp(painters[ ,1:4]) loadings(pca.painters) Loadings: Comp.1 Comp.2 Comp.3 Comp.4 Composition 0.484 -0.376 0.784 -0.101 Drawing 0.424 0.187 -0.280 -0.841 Colour -0.381 -0.845 -0.211 -0.310 Expression 0.664 -0.330 -0.513 0.432 Comp.1 Comp.2 Comp.3 Comp.4 SS loadings 1.00 1.00 1.00 1.00 Proportion Var 0.25 0.25 0.25 0.25 Cumulative Var 0.25 0.50 0.75 1.00 However, if I rerun the same analysis, the loadings of the first component have the opposite sign (see below), why is that? I have read the note in the princomp help that says The signs of the columns of the loadings and scores are arbitrary, and so may differ between
[R] Newbie q. need some help understanding this code.
dear all.
Would someone be kind and willing to explain the code
below for a person who has never used R? ( that is if
one has enough time and inclination)
It implements gillepsie's stochastic algorithm for
Lotka Volterra model.
What would help me tremendously is to see the
breakdown of the line by line code into plain english.
thanks for any insights or other comments.
sean
library(stepfun)
lv - function(N=1000,cvec=c(1,0.005,0.6),x=c(50,100))
{
m-length(cvec)
n-length(x)
xmat-matrix(nrow=N+1,ncol=n)
tvec-vector(numeric,N)
h-vector(numeric,m)
t-0
xmat[1,]-x
for (i in 1:N) {
h[1]-cvec[1]*x[1]
h[2]-cvec[2]*x[1]*x[2]
h[3]-cvec[3]*x[2]
h0=sum(h)
tp-rexp(1,h0)
t-t+tp
u-runif(1,0,1)
if ( u h[1]/h0 ) {
x[1] - x[1]+1
} else if ( u (h[1]+h[2])/h0 ) {
x[1] - x[1]-1
x[2] - x[2]+1
} else {
x[2] - x[2]-1
}
xmat[i+1,]-x
tvec[i]-t
}
list(stepfun(tvec,xmat[,1]),stepfun(tvec,xmat[,2]))
}
results - lv(N=1)
op-par(mfrow=c(2,1))
plot(results[[1]],do.points=True)
plot(results[[2]],do.points=False)
par(op)
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Re: [R] Newbie q. need some help understanding this code.
Hi,
sean kim wrote:
thanks for any insights or other comments.
I would suggest that you run the code line by line, to see what it does
yourself. It is the best way to learn!
library(stepfun)
This just loads the package stepfun.
lv - function(N=1000,cvec=c(1,0.005,0.6),x=c(50,100))
{
Declaration of a function named lv, where N, cvec and x are the
parameters/arguments with default values.
m-length(cvec)
m is a new variable, with in this case is just a number, the length of
the vector cvec. In this case, 3.
n-length(x)
Ditto.
xmat-matrix(nrow=N+1,ncol=n)
Generate a matrix with N + 1 (1001) rows and n columns.
tvec-vector(numeric,N)
h-vector(numeric,m)
Generating two vectors.
t-0
xmat[1,]-x
Assign x to the first row of the xmat matrix.
for (i in 1:N) {
h[1]-cvec[1]*x[1]
The first element in cvec and x, multiply them together and the result
becomes the first element of h.
h[2]-cvec[2]*x[1]*x[2]
h[3]-cvec[3]*x[2]
Ditto.
h0=sum(h)
Get the sum of h.
tp-rexp(1,h0)
Generating an exponential random number with a rate of h0.
t-t+tp
u-runif(1,0,1)
Generating a uniform random number, [0, 1]
if ( u h[1]/h0 ) {
x[1] - x[1]+1
If the uniform random number, u, is smaller than h[1]/h0 then do...
otherwise do the following...
} else if ( u (h[1]+h[2])/h0 ) {
x[1] - x[1]-1
x[2] - x[2]+1
} else {
x[2] - x[2]-1
}
xmat[i+1,]-x
Increment to the next row.
tvec[i]-t
}
list(stepfun(tvec,xmat[,1]),stepfun(tvec,xmat[,2]))
}
Put things together into a list.
Again, check these line by line and you'll have a better understanding!
Kev
--
Ko-Kang Kevin Wang
PhD Student
Centre for Mathematics and its Applications
Building 27, Room 1004
Mathematical Sciences Institute (MSI)
Australian National University
Canberra, ACT 0200
Australia
Homepage: http://wwwmaths.anu.edu.au/~wangk/
Ph (W): +61-2-6125-2431
Ph (H): +61-2-6125-7407
Ph (M): +61-40-451-8301
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Re: [R] Splitting vector into individual elements
Gabor Grothendieck ggrothendieck at myway.com writes:
:
: Paul Roebuck roebuck at odin.mdacc.tmc.edu writes:
:
: :
: : On Wed, 15 Sep 2004, Peter Dalgaard wrote:
: :
: : Paul Roebuck roebuck at odin.mdacc.tmc.edu writes:
: :
: : Is there a means to split a vector into its individual
: : elements without going the brute-force route for arguments
: : to a predefined function call?
: :
: : offred.rgb - c(1, 0, 0) * 0.60;
: :
: : ## Brute force style
: : offred.col - rgb(offred.rgb[1],
: : offred.rgb[2],
: : offred.rgb[3],
: : names = offred)
: : ## Desired style
: : offred.col - rgb(silver.bullet(offred.rgb),
: : names = offred)
: :
: : The closest is probably this:
: :
: : offred.col - do.call(rgb, c(as.list(offred.rgb),
: : list(names=offred)))
: :
: :
: : Everyone offered 'do.call' as the solution. While that
: : works, is it to say that there is no means of expanding
: : the expression as an argument to the original function?
:
: This is not a true answer to the question of expanding a list
: into arguments without using do.call but it does allow you to
: carry out either syntax in this particular case using S3
: dispatch:
:
: R silver.bullet - as.list
: R rgb - function(x, ...) UseMethod(rgb)
: R rgb.list - function(x, ...) rgb(x[[1]],x[[2]],x[[3]],...)
: R rgb.default - graphics::rgb
:
: R offred.rgb - c(1, 0, 0) * 0.60;
: R # original syntax
: R rgb(offred.rgb[1], offred.rgb[2], offred.rgb[3], names = offred)
:offred
: #99
: R # list syntax
: R rgb(silver.bullet(offred.rgb), names = offred)
:offred
: #99
Here is a second, different approach.
Again, it is not exactly what you are asking for since silver.bullet,
here called flatten.args, is applied to the function rather than the
argument in question and operates on all arguments, not just one but
I think its closer in spirit to your query than my previous solution:
flatten.args - function(f)
function(...) {
L - list()
for(i in list(...)) L - c(L, unlist(i))
do.call(as.character(substitute(f)), L)
}
flatten.args(rgb)(0.6 * c(1,0,0), name = offred)
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Re: [R] Cross-validation for Linear Discrimitant Analysis
On Wed, 15 Sep 2004, Yu Shao wrote: I am new to R and statistics and I have two questions. Perhaps then you need to start by explaining why you are using LDA. Please take a good look at the posting guide. First I need help to interpret the cross-validation result from the R linear discriminant analysis function lda. You mean Professor Ripley's function lda in package MASS, I guess. I did the following: lda (group ~ Var1 + Var2, CV=T) R allows you to use meaningful names, so please do so. where CV=T tells the lda to do cross-validation. The output of lda are the posterior probabilities among other things, but I can't find an error term (like delta returned by cv.glm). My question is how to get such an error term from the output? Can I just simply calculate the prediction accuracy using the posterior probabilities from the cross-validation, and use that to measure the quality of the model? cv.glm as in Dr Canty's package boot? If you are trying to predict classifications, LDA is not the right tool, and LOO CV probably is not either. There is no unique definition of `error term' (true for cv.glm as well), and people have written whole books about how to assess classifiers. LDA is about `discrimination' not `allocation' in the jargon used ca 1960. Another question is more basic: how to determine if a lda model is significant? (There is no p-value.) Thanks, Please do read the references on the ?lda page. It's not a useful question, as LDA is about discriminating between populations and makes the unrealistic assumption of multivariate normality. (Analogously for linear regression, there are ways to test if that is (statistically) `significant', but knowledgable users almost never do so.) Perhaps more realistic advice is to suggest you seek some statistical consultancy. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] sapply() with cat()
Hi,
Suppose I've got a data frame:
inter.df
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12
1 3.3 NA NA NA NA NA NA NA NA NA NA NA
2 0.0 0.1 NA NA NA NA NA NA NA NA NA NA
3 1.0 0.9 0.2 NA NA NA NA NA NA NA NA NA
4 1.6 0.0 2.9 0.7 NA NA NA NA NA NA NA NA
5 0.0 0.1 2.9 0.1 0.1 NA NA NA NA NA NA NA
6 2.4 1.0 0.6 0.4 1.9 0.1 NA NA NA NA NA NA
7 2.8 1.4 1.2 7.5 0.0 0.0 4.2 NA NA NA NA NA
8 0.3 3.1 0.8 3.7 5.7 0.0 0.8 0.0 NA NA NA NA
9 0.1 2.9 0.3 1.3 0.2 0.2 0.5 1.4 0.9 NA NA NA
10 0.8 2.6 0.0 0.0 0.1 4.1 0.8 4.3 0.6 2.2 NA NA
11 0.0 4.0 0.0 0.3 0.5 0.9 0.0 1.5 0.2 0.7 0.8 NA
12 0.6 0.8 0.3 0.0 0.2 1.2 0.0 0.8 1.5 0.9 0.4 0
foo - function(x) {
+ ifelse(x 3.8, NA, x)
+ }
sapply(inter.df, foo)
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12
[1,] 3.3 NA NA NA NA NA NA NA NA NA NA NA
[2,] 0.0 0.1 NA NA NA NA NA NA NA NA NA NA
[3,] 1.0 0.9 0.2 NA NA NA NA NA NA NA NA NA
[4,] 1.6 0.0 2.9 0.7 NA NA NA NA NA NA NA NA
[5,] 0.0 0.1 2.9 0.1 0.1 NA NA NA NA NA NA NA
[6,] 2.4 1.0 0.6 0.4 1.9 0.1 NA NA NA NA NA NA
[7,] 2.8 1.4 1.2 NA 0.0 0.0 NA NA NA NA NA NA
[8,] 0.3 3.1 0.8 3.7 NA 0.0 0.8 0.0 NA NA NA NA
[9,] 0.1 2.9 0.3 1.3 0.2 0.2 0.5 1.4 0.9 NA NA NA
[10,] 0.8 2.6 0.0 0.0 0.1 NA 0.8 NA 0.6 2.2 NA NA
[11,] 0.0 NA 0.0 0.3 0.5 0.9 0.0 1.5 0.2 0.7 0.8 NA
[12,] 0.6 0.8 0.3 0.0 0.2 1.2 0.0 0.8 1.5 0.9 0.4 0
Up to here, sapply() does what I want, replacing values that are greater
than 3.8 to NA, as per my foo() function. But...
goo - function(x) {
+ ifelse(x 3.8, cat(\\textbf{, x, }, sep = ), x)
+ }
sapply(inter.df, goo)
\textbf{NA0.10.900.111.43.12.92.640.8}Error in [-(`*tmp*`, test,
value = rep(yes, length.out = length(ans))[test]) :
incompatible types
If instead whenever I get a value that's greater than 3.8 I want to
change it what goo() does, e.g. the value 4.0 should become:
\textbf{4.0}
but I got the above error.
What I'm intending is to then pass the inter.df (with the \textbf{}
markups) into xtable() so the values will be bolded in the resulting
LaTeX table (as so far, I cannot see how I can achieve bolding only
certain numbers with xtable() alone).
Any suggestions will be welcome!
Kevin
--
Ko-Kang Kevin Wang
PhD Student
Centre for Mathematics and its Applications
Building 27, Room 1004
Mathematical Sciences Institute (MSI)
Australian National University
Canberra, ACT 0200
Australia
Homepage: http://wwwmaths.anu.edu.au/~wangk/
Ph (W): +61-2-6125-2431
Ph (H): +61-2-6125-7407
Ph (M): +61-40-451-8301
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Re: [R] sapply() with cat()
On Thu, 16 Sep 2004, Kevin Wang wrote:
Hi,
Suppose I've got a data frame:
inter.df
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12
1 3.3 NA NA NA NA NA NA NA NA NA NA NA
2 0.0 0.1 NA NA NA NA NA NA NA NA NA NA
3 1.0 0.9 0.2 NA NA NA NA NA NA NA NA NA
4 1.6 0.0 2.9 0.7 NA NA NA NA NA NA NA NA
5 0.0 0.1 2.9 0.1 0.1 NA NA NA NA NA NA NA
6 2.4 1.0 0.6 0.4 1.9 0.1 NA NA NA NA NA NA
7 2.8 1.4 1.2 7.5 0.0 0.0 4.2 NA NA NA NA NA
8 0.3 3.1 0.8 3.7 5.7 0.0 0.8 0.0 NA NA NA NA
9 0.1 2.9 0.3 1.3 0.2 0.2 0.5 1.4 0.9 NA NA NA
10 0.8 2.6 0.0 0.0 0.1 4.1 0.8 4.3 0.6 2.2 NA NA
11 0.0 4.0 0.0 0.3 0.5 0.9 0.0 1.5 0.2 0.7 0.8 NA
12 0.6 0.8 0.3 0.0 0.2 1.2 0.0 0.8 1.5 0.9 0.4 0
foo - function(x) {
+ ifelse(x 3.8, NA, x)
+ }
sapply(inter.df, foo)
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12
[1,] 3.3 NA NA NA NA NA NA NA NA NA NA NA
[2,] 0.0 0.1 NA NA NA NA NA NA NA NA NA NA
[3,] 1.0 0.9 0.2 NA NA NA NA NA NA NA NA NA
[4,] 1.6 0.0 2.9 0.7 NA NA NA NA NA NA NA NA
[5,] 0.0 0.1 2.9 0.1 0.1 NA NA NA NA NA NA NA
[6,] 2.4 1.0 0.6 0.4 1.9 0.1 NA NA NA NA NA NA
[7,] 2.8 1.4 1.2 NA 0.0 0.0 NA NA NA NA NA NA
[8,] 0.3 3.1 0.8 3.7 NA 0.0 0.8 0.0 NA NA NA NA
[9,] 0.1 2.9 0.3 1.3 0.2 0.2 0.5 1.4 0.9 NA NA NA
[10,] 0.8 2.6 0.0 0.0 0.1 NA 0.8 NA 0.6 2.2 NA NA
[11,] 0.0 NA 0.0 0.3 0.5 0.9 0.0 1.5 0.2 0.7 0.8 NA
[12,] 0.6 0.8 0.3 0.0 0.2 1.2 0.0 0.8 1.5 0.9 0.4 0
Up to here, sapply() does what I want, replacing values that are greater
than 3.8 to NA, as per my foo() function. But...
goo - function(x) {
+ ifelse(x 3.8, cat(\\textbf{, x, }, sep = ), x)
+ }
sapply(inter.df, goo)
\textbf{NA0.10.900.111.43.12.92.640.8}Error in [-(`*tmp*`, test,
value = rep(yes, length.out = length(ans))[test]) :
incompatible types
If instead whenever I get a value that's greater than 3.8 I want to
change it what goo() does, e.g. the value 4.0 should become:
\textbf{4.0}
but I got the above error.
cat writes to connection (here stdout) and then returns NULL. I think you
meant to use paste():
goo - function(x)ifelse(x 3.8, paste(\\textbf{, x, }, sep = ), x)
works.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] sapply() with cat()
Prof Brian Ripley ripley at stats.ox.ac.uk writes:
:
: On Thu, 16 Sep 2004, Kevin Wang wrote:
:
: Hi,
:
: Suppose I've got a data frame:
: inter.df
: V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12
: 1 3.3 NA NA NA NA NA NA NA NA NA NA NA
: 2 0.0 0.1 NA NA NA NA NA NA NA NA NA NA
: 3 1.0 0.9 0.2 NA NA NA NA NA NA NA NA NA
: 4 1.6 0.0 2.9 0.7 NA NA NA NA NA NA NA NA
: 5 0.0 0.1 2.9 0.1 0.1 NA NA NA NA NA NA NA
: 6 2.4 1.0 0.6 0.4 1.9 0.1 NA NA NA NA NA NA
: 7 2.8 1.4 1.2 7.5 0.0 0.0 4.2 NA NA NA NA NA
: 8 0.3 3.1 0.8 3.7 5.7 0.0 0.8 0.0 NA NA NA NA
: 9 0.1 2.9 0.3 1.3 0.2 0.2 0.5 1.4 0.9 NA NA NA
: 10 0.8 2.6 0.0 0.0 0.1 4.1 0.8 4.3 0.6 2.2 NA NA
: 11 0.0 4.0 0.0 0.3 0.5 0.9 0.0 1.5 0.2 0.7 0.8 NA
: 12 0.6 0.8 0.3 0.0 0.2 1.2 0.0 0.8 1.5 0.9 0.4 0
:
: foo - function(x) {
: + ifelse(x 3.8, NA, x)
: + }
: sapply(inter.df, foo)
:V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12
: [1,] 3.3 NA NA NA NA NA NA NA NA NA NA NA
: [2,] 0.0 0.1 NA NA NA NA NA NA NA NA NA NA
: [3,] 1.0 0.9 0.2 NA NA NA NA NA NA NA NA NA
: [4,] 1.6 0.0 2.9 0.7 NA NA NA NA NA NA NA NA
: [5,] 0.0 0.1 2.9 0.1 0.1 NA NA NA NA NA NA NA
: [6,] 2.4 1.0 0.6 0.4 1.9 0.1 NA NA NA NA NA NA
: [7,] 2.8 1.4 1.2 NA 0.0 0.0 NA NA NA NA NA NA
: [8,] 0.3 3.1 0.8 3.7 NA 0.0 0.8 0.0 NA NA NA NA
: [9,] 0.1 2.9 0.3 1.3 0.2 0.2 0.5 1.4 0.9 NA NA NA
: [10,] 0.8 2.6 0.0 0.0 0.1 NA 0.8 NA 0.6 2.2 NA NA
: [11,] 0.0 NA 0.0 0.3 0.5 0.9 0.0 1.5 0.2 0.7 0.8 NA
: [12,] 0.6 0.8 0.3 0.0 0.2 1.2 0.0 0.8 1.5 0.9 0.4 0
:
: Up to here, sapply() does what I want, replacing values that are greater
: than 3.8 to NA, as per my foo() function. But...
:
: goo - function(x) {
: + ifelse(x 3.8, cat(\\textbf{, x, }, sep = ), x)
: + }
: sapply(inter.df, goo)
: \textbf{NA0.10.900.111.43.12.92.640.8}Error in [-(`*tmp*`, test,
: value = rep(yes, length.out = length(ans))[test]) :
: incompatible types
:
: If instead whenever I get a value that's greater than 3.8 I want to
: change it what goo() does, e.g. the value 4.0 should become:
: \textbf{4.0}
: but I got the above error.
:
: cat writes to connection (here stdout) and then returns NULL. I think you
: meant to use paste():
:
: goo - function(x)ifelse(x 3.8, paste(\\textbf{, x, }, sep = ), x)
:
: works.
:
Also, if you convert your data frame to a matrix you won't need sapply:
inter.mat - as.matrix(inter.df)
ifelse(inter.mat 3.8,
paste(\\textbf{, inter.mat, }, sep = ),
inter.mat)
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