Re: [R] Need help to estimate the Coef matrices in mAr
Have you tried 'RSiteSearch(multivariate autoregression, functions)'? This produced 14 hits for me just now, the first of which mentions a package 'MSBVAR'. Have you looked at that? If that failed, I don't think it would be too hard to modify 'mAr.est' to do what you want. If it were my problem, I might a local copy of the function, then add an argument accepting a 2 or 3-dimensional array with numbers for AR coefficients to be fixed and NAs for the coefficients. Then I'd use 'debug' to walk through the function line by line until I figured out how to modify the function to do what I wanted. I haven't checked all the details, so I don't know for sure if this would work, but the function contains a line 'R = qr.R(qr((rbind(K, diag(scale, complete = TRUE)' which I would start by decomposing, possibly starting as follows: Z - rbind(K, diag(scale) I'd figure out how the different columns of Z relate to my problem, then modify it appropriately to get what I wanted. Another alternative would be to program it from scratch using something like 'optim' to minimize the sum of squares of residuals over the free parameters in my AR matrices. I'm confident I could make this work, even if the I somehow could not get it with either of the other two. There may be something else better, e.g., a Kalman filter representation, but I can't think how to do that off the top if my head. Hope this helps. Spencer Graves Arun Kumar Saha wrote: Dear R users, I am using mAr package to fit a Vector autoregressive model to my data. But here I want to put some predetermined values for some elements in coefficient matrix that mAr.est going to estimate. For example if p=3 then I want to put A3[1,3] = 0 and keep rest of the elements of coefficient matrices to be determined by mAr.est. Can anyone please tell me how can I do that? Sincerely yours, Arun [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Subsetting vectors based on condition
Hello, I have a question regarding subsetting of vectors. Here's an example of what I'm trying to do: vect.1 - c(76,195, 290, 380) vect.2 - c(63, 95, 133, 170, 215, 253, 285, 299, 325, 375) I would like to subset vect.2 so that it has the same length as vect.1, and its numbers are the first corresponging higher value compared to vect.1. The output should be: final.output = (95, 215, 299, NA) What is the fastest/most eficient way to accompllish this in R? Thanks for the help in advance, Mahesh Krishnan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Command line cut-off
Hi, I noticed that, when working with R on a command line, I cannot enter anything into the command line which is longer than 1023 characters. If I input sth longer, it'l be just cut off at this lenght, and a + will be presented to me on the command line. It's not as bad since I can copy'n'paste commands longer than this with newline characters in between onto the prompt, but still kind of annoying. Does anybody know how I can make my command line accept lines longer than 1023 characters? Regards, Björn -- Q: How many lawyers does it take to change a light bulb? A: One. Only it's his light bulb when he's done. -- Important! Please recognize my new GPG Public Key! Björn Thalheim gpg fingerprint: 2F22 AAEB 1818 1548 EC78 1AE8 9D2E FCB4 0980 28CC download key: wget http://www.ifsr.de/~bjoern/gpg/public_key.asc See also: http://www.ifsr.de/~bjoern/gpg/key.html signature.asc Description: OpenPGP digital signature __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] New forecasting bundle of packages
v1.0 of the forecasting bundle of packages is now on CRAN and will propagate to mirrors shortly. The forecasting bundle of R packages provides new forecasting methods, and graphical tools for displaying and analysing forecasts. It comprises the following packages: * forecast: Functions and methods for forecasting. * fma: All data sets from Makridakis, Wheelwright and Hyndman (1998) Forecasting: methods and applications, Wiley Sons: New York. * Mcomp: All data from the M1 and M3 forecast competitions. Key features: * a forecast method and class which can be applied to Arima, StructTS, HoltWinters and other time series models. This is preferred to predict() as it provides output in a consistent format (the forecast class) that can be used by other functions. * automatic univariate time series forecasting based on exponential smoothing state space models. This is much more general and flexible than HoltWinters(). * automatic ARIMA forecasting based on minimizing the AIC or BIC. * several new forecasting methods and time series graphics. Some features of the forecast package were the subject of my talk at UseR! in Vienna in June. Slides of the talk are at http://www.robhyndman.info/talks/Hyndman_UseR.pdf Anyone who has been using earlier versions of the packages from my web pages should check out the list of changes at http://www.robhyndman.info/Rlibrary/forecast/ __ Professor Rob J Hyndman Department of Econometrics Business Statistics, Monash University, VIC 3800, Australia http://www.robhyndman.info/ ___ R-packages mailing list R-packages@stat.math.ethz.ch https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] generating loglogistic distribution in R
Hi Dear, Can someone please inform me on genreating random variables of Loglogistic, and PERT beta distributions? Thanks for your time and help, in advance. Regards, Murthy. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Command line cut-off
On Mon, 4 Sep 2006, Björn Thalheim wrote: Hi, I noticed that, when working with R on a command line, I cannot enter anything into the command line which is longer than 1023 characters. Actually, 1000 or 1022 or 1023 bytes, and it depends on your 'command line' (and you have not even told us your OS). You will find details in the R-devel list archives, e.g. https://stat.ethz.ch/pipermail/r-devel/2006-August/038985.html If I input sth longer, it'l be just cut off at this lenght, and a + will be presented to me on the command line. It's not as bad since I can copy'n'paste commands longer than this with newline characters in between onto the prompt, but still kind of annoying. How do you think thousands users of R for a decade have managed? What are you doing that needs 'commands' (R has function calls) longer than 1000 or so characters that need to be on one line? Does anybody know how I can make my command line accept lines longer than 1023 characters? See the R-devel version of R for a full description: on some consoles it works there. -- 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__ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] xlsReadWrite 1.0
The first version of xlsReadWrite has been uploaded to CRAN. -- WHAT IS IT? / R COMMANDS The package allows you to read and write Excel files natively. The supported Format is BIFF 8, i.e. Excel from version 97 up to 2003. read.xls( file, colNames = TRUE, sheet = 1, type = data.frame, from = 1 ) write.xls( x, file, colNames = TRUE, sheet = 1, from = 1 ) -- PLATFORM / LICENSE Currently the package is only available on Windows (see also below). The license is GPLv2 with an exception to allow the use of a third party library (flexcel). With this exception the package as a whole becomes a non-free package (according to the R license non-free packages are not encouraged but at least possible/tolerated). See also README. -- TECHNICAL BACKGROUND / PRO VERSION: The core of this package is written in Delphi (PASCAL) and uses the R headerfiles for Delphi which I have written earlier (see website below). For the actual read/write operation in excel a proprietary library (Flexcel) is used as I wouldn't be able to write this by myself and because for pascal there are not open source libraries available. (AFAIK. - I am aware of Apache POI (Java), a library for OpenOffice (C, too much work to translate to pascal) and also some older one which don't support BIFF8). I intend to make available (in 2 - 3 weeks) a pro version which has some more functionality and formal support. As an independant developer I am certainly happy if people will check this version out or consider a contribution, but it is not my idea to play silly shareware games, the open source version works well and I am commited to support it as far as I can. -- OTHER PLATFORMS (Linux/Mac) Technically xlsReadWrite could be crossplatform. Flexcel already has a Kylix version, which means that it could be made to run on Linux almost immediately. The IMO better way would be to port the code to *FreePascal*: this fits better in the open source world of R and this also would make it possible to supply a Mac version. - While I started to port it, I didn't succeed in the 2 days I have given me. - If there are any pascal-lovers out there I would be very interested to get in touch, it would be more fun to do it together than to try all by myself. -- MORE INFOS / CONTACT: At http://treetron.googlepages.com/ you will find an additional file with testdata/-scripts and more infos (contact, bugs (currently nothing), suggestions and todos). I hope you enjoy using this package (I certainly do...) and am looking forward to any feedback you might have. Best regards, Hans-Peter [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] merge files after cor.test
Dear All, Suppose I have 2 files: # first one : testid.csv A B C D E (id- read.table (testid.csv,col.name=c(id))) id 1 A 2 B 3 C 4 D 5 E # second file is the result file I calculate from cor.text, which shows the correlation coefficient. cor.value.t 1 2 3 4 5 1 1.000 0.2156213 0.31000492 0.22282154 0.1822277 2 0.2156213 1.000 -0.31183893 0.42681488 0.3421535 3 0.3100049 -0.3118389 1. -0.02801885 -0.1307732 4 0.2228215 0.4268149 -0.02801885 1. 0.1454049 5 0.1822277 0.3421535 -0.13077317 0.14540493 1.000 # I tried to merge these two files together. # what I expected is like this: A 1.000 0.2156213 0.31000492 0.22282154 0.1822277 B 0.2156213 1.000 -0.31183893 0.42681488 0.3421535 C 0.3100049 -0.3118389 1. -0.02801885 -0.1307732 D 0.2228215 0.4268149 -0.02801885 1. 0.1454049 E 0.1822277 0.3421535 -0.13077317 0.14540493 1.000 # but after I used (output-merge(id,cor.value.t)), which shows 25 lines (below): id 1 2 3 4 5 1 A 1.000 0.2156213 0.31000492 0.22282154 0.1822277 2 B 1.000 0.2156213 0.31000492 0.22282154 0.1822277 3 C 1.000 0.2156213 0.31000492 0.22282154 0.1822277 4 D 1.000 0.2156213 0.31000492 0.22282154 0.1822277 5 E 1.000 0.2156213 0.31000492 0.22282154 0.1822277 6 A 0.2156213 1.000 -0.31183893 0.42681488 0.3421535 7 B 0.2156213 1.000 -0.31183893 0.42681488 0.3421535 8 C 0.2156213 1.000 -0.31183893 0.42681488 0.3421535 9 D 0.2156213 1.000 -0.31183893 0.42681488 0.3421535 10 E 0.2156213 1.000 -0.31183893 0.42681488 0.3421535 11 A 0.3100049 -0.3118389 1. -0.02801885 -0.1307732 12 B 0.3100049 -0.3118389 1. -0.02801885 -0.1307732 13 C 0.3100049 -0.3118389 1. -0.02801885 -0.1307732 14 D 0.3100049 -0.3118389 1. -0.02801885 -0.1307732 15 E 0.3100049 -0.3118389 1. -0.02801885 -0.1307732 16 A 0.2228215 0.4268149 -0.02801885 1. 0.1454049 17 B 0.2228215 0.4268149 -0.02801885 1. 0.1454049 18 C 0.2228215 0.4268149 -0.02801885 1. 0.1454049 19 D 0.2228215 0.4268149 -0.02801885 1. 0.1454049 20 E 0.2228215 0.4268149 -0.02801885 1. 0.1454049 21 A 0.1822277 0.3421535 -0.13077317 0.14540493 1.000 22 B 0.1822277 0.3421535 -0.13077317 0.14540493 1.000 23 C 0.1822277 0.3421535 -0.13077317 0.14540493 1.000 24 D 0.1822277 0.3421535 -0.13077317 0.14540493 1.000 25 E 0.1822277 0.3421535 -0.13077317 0.14540493 1.000 Is anybody know why it outputs 5 times? Thanks a lot! Nina [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] my error with augPred
Hallo thank you for your response. I am not sure but maybe fixed effects cannot be set to be influenced by a factor to be able to use augPred. lob-Loblolly[Loblolly$Seed!=321,] set.seed(1) lob-data.frame(lob, x1=sample(letters[1:3], replace=T)) # add a #factor lob-groupedData(height~age|Seed, data=lob) fm1 - nlme(height ~ SSasymp(age, Asym, R0, lrc), data = lob, fixed = Asym + R0 + lrc ~ 1, random = Asym ~ 1, start = c(Asym = 103, R0 = -8.5, lrc = -3.3)) fm2-update(fm1, fixed=list(Asym~x1, R0+lrc~1), start=c(103,0,-8.5,- 3)) ^^^ and plot(augPred(fm2)) Throws an error. So it is not possible to use augPred with such constructions. Best regards. Petr Pikal On 2 Sep 2006 at 17:58, Spencer Graves wrote: Date sent: Sat, 02 Sep 2006 17:58:05 -0700 From: Spencer Graves [EMAIL PROTECTED] To: Petr Pikal [EMAIL PROTECTED] Copies to: r-help@stat.math.ethz.ch Subject:Re: [R] my error with augPred comments in line Petr Pikal wrote: Dear all I try to refine my nlme models and with partial success. The model is refined and fitted (using Pinheiro/Bates book as a tutorial) but when I try to plot plot(augPred(fit4)) I obtain Error in predict.nlme(object, value[1:(nrow(value)/nL), , drop = FALSE], : Levels (0,3.5],(3.5,5],(5,7],(7,Inf] not allowed for vykon.fac Is it due to the fact that I have unbalanced design with not all levels of vykon.fac present in all levels of other explanatory factor variable? I don't know, but I'm skeptical. I try to repeat 8.19 fig which is OK until I try: fit4 - update(fit2, fixed = list(A+B~1,xmid~vykon.fac, scal~1), start = c(57, 100, 700, rep(0,3), 13)) I know I should provide an example but maybe somebody will be clever enough to point me to an explanation without it. I'm not. To answer these questions without an example from you, I'd have to make up my own example and try to see if I could replicate the error messages you report, and I'm not sufficiently concerned about this right now to do that. Have you tried taking an example from the book and deleting certain rows from the data to see if you can force it to reproduce your error? Alternatively, have you tried using 'debug' to trace through the code line by line until you learn enough of what it's doing to answer your question? Spencer Graves nlme version 3.1-75 SSfpl model R 2.4.0dev (but is the same in 2.3.1), W2000. Thank you Best regards. Petr PikalPetr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] abline and plot(augPred) help
Dear all
as I did not get any response on my post about abline and
plot(augPred)) I try again. I hope I do not break some posting guide
rules. I would try to contact package maintainer directly but there
is stated to be R-core people, so I feel R-help list shall be OK.
I need to draw straight lines through augPred plotted panels
(vertical or horizontal) at specified point. I know I shall probably
use panel.abline but I am missing correct syntax. Below you can see
my attempts together with results. I hope somebody can point me to
right direction.
I am probably somewhere close but I have no clue, which parameter I
shall modify to get measured points, fitted lines and vertical lines
in panels together.
Please help
Thank you
Best regards.
Petr Pikal
fm1 - lme(Orthodont)
# standard plot
plot(augPred(fm1, level = 0:1, length.out = 2))
#plot with vertical but without points and fitted lines
plot(augPred(fm1, level = 0:1, length.out = 2),
panel=function(v,...) {
panel.abline(v=10)}
)
# plot with vertical but without fitted lines
plot(augPred(fm1, level = 0:1, length.out=2),
panel=function(x,y,...) {
panel.xyplot(x,y,...)
panel.abline(v=10)}
)
# plot with vertical and with all points (fitted lines are drawn as
points)
plot(augPred(fm1, level = 0:1),
panel=function(x,y,...) {
panel.xyplot(x,y,...)
panel.abline(v=10)}
)
Petr Pikal
[EMAIL PROTECTED]
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Re: [R] Subsetting vectors based on condition
vect.1 [1] 76 195 290 380 vect.2 [1] 63 95 133 170 215 253 285 299 325 375 unlist(lapply(vect.1, function(x)vect.2[which(vect.2 x)[1]])) [1] 95 215 299 NA On 9/3/06, Mahesh Krishnan [EMAIL PROTECTED] wrote: Hello, I have a question regarding subsetting of vectors. Here's an example of what I'm trying to do: vect.1 - c(76,195, 290, 380) vect.2 - c(63, 95, 133, 170, 215, 253, 285, 299, 325, 375) I would like to subset vect.2 so that it has the same length as vect.1, and its numbers are the first corresponging higher value compared to vect.1. The output should be: final.output = (95, 215, 299, NA) What is the fastest/most eficient way to accompllish this in R? Thanks for the help in advance, Mahesh Krishnan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fitting Pareto distribution to some data
On 9/3/06, Prof Brian Ripley [EMAIL PROTECTED] wrote:
I am trying to fit Pareto distribution to some data. MASS package does
not support Pareto distribution. Is there some alternative way?
Actually fitdistr{MASS} does if you supply the pdf for a Pareto.
That is not in base R, but easy to write for yourself.
It seems that Pareto and generalized Pareto is in several packages,
including POT SoPhy VaR evd evir fExtremes lmomco.
Thanks, Prof Brian Ripley.
Paul
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Re: [R] Question on Chi-square of null model in sem package
Dear Wei-Wei, As I explained to you in private email yesterday (perhaps you didn't receive my reply?), the problem that you point out is due to a bug in the sem function that I fixed some time ago and then inadvertently reintroduced. Yesterday, I sent a corrected version of the sem package (0.9-5) to CRAN; the source package is there now and I'm sure that the compiled Windows package will appear in due course. Thank you once more for bringing the problem to my attention. John John Fox Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Guo Wei-Wei Sent: Monday, September 04, 2006 12:34 AM To: r-help@stat.math.ethz.ch Subject: [R] Question on Chi-square of null model in sem package Dear all, I met a problem while doing SEM by sem package. I got a negative chi-square of null model. Because the theoretical value of chi-square cannot be negative, I checked the source code of sem.R in sem package and I found the Chi-square of null model was computed by the following expression: result$chisqNull - (N - 1) * (sum(diag(S %*% diag(1/diag(S + log(prod(diag(S I think the reason for negative Chi-square is the too small value of prod(diag(S)) of my data. I'm working on a data.frame named emc.data from a sample of a 16-item questioinnaire. The variance of items are diag(cov(emc.data)) EMC1 EMC2 EMC3 EMC4 EMC5 EMC6 EMC7 EMC8 0.364 0.2350041 0.2488009 0.2901653 0.3195399 0.3107343 0.3436622 0.2345912 EMC9 EMC10 EMC11 EMC12 EMC13 EMC14 EMC15 EMC16 0.2621680 0.3230400 0.4039245 0.3803105 0.2773370 0.4348342 0.2757216 0.3405252 The fit indices of RMSEA and GFI are good, so I think the problem might be solve by another way for computing the Chi-square of null model. I'm not well trained in maths, so I come for help. Any advise is appreciated. Best wishes, Wei-Wei __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] library(plgem)
Dear Users, library(plgem) doesn't exist directly in the list of available packages of R. Where could it be found? Thanks so much for help. Amir - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to fit gauss beam?
Hello,
I am having a hard time fitting a gauss beam using R. In
gnutplot I did something like
$ w(z) = w0 * sqrt(1+(z/z0)**2)
$ fit w(z) 'before_eom.txt' using 1:2 via w0, z0
to obtain w0 and z0. Now I want to do the same in R. I tried
a linear model like this (r = radius, z = distance):
beam - function(z) {
sum(sqrt(1 + z**2))
}
lm(r ~ I(beam(z)), data = before_eom)
Which gives nonsensical answers ...
Then I tried a nonlinear model:
d - read.table (before_eom.tab, header=T)
z - d$d
r - d$minor * 1e-6
beam- function(p) {
M - 1.1
sum((r-M*p[1]*sqrt(1 + (z/p[2])**2))^2)
}
out - nlm(beam, p=c(400, 0.1), hessian=TRUE)
out$estimate
This is very sensitive to the starting values and gives
nonsensical answers as well...
Is there a simple way to translate the gnuplot fit into R?
Thanks,
Thomas
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Re: [R] library(plgem)
Perhaps here? http://www.bioconductor.org/packages/bioc/1.6/src/contrib/html/plgem.htm l -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Amir Safari Sent: 04 September 2006 14:44 To: R-help@stat.math.ethz.ch Subject: [R] library(plgem) Dear Users, library(plgem) doesn't exist directly in the list of available packages of R. Where could it be found? Thanks so much for help. Amir - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] newbie question about index
Thanks for all these replies, all work perfectly. Sun --- Petr Pikal [EMAIL PROTECTED]写道: Hallo probably there are other options but outer(1:3,a, ==)*1 can do what you want. HTH Petr On 31 Aug 2006 at 22:41, z s wrote: Date sent:Thu, 31 Aug 2006 22:41:27 +0800 (CST) From: z s [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Subject: [R] newbie question about index Hi, I am trying to convert a variable a = sample(1:3,100,rep = T) represents choices into a 3X100 dummy varible b with corresponding element set to 1 otherwise 0. eg. a: 1 3 2 1 2 3 1 1 b: 1 0 0 1 0 0 1 1.. 0 0 1 0 1 0 0 0... 0 1 0 0 0 1 0 0... Is there something like b[a] =1 existing? I could not figure this out myself. - Mp3�杩袼�-新歌热歌高速下 [[alternative HTML version deleted]] Petr Pikal [EMAIL PROTECTED] ___ 抢注雅虎免费邮箱-3.5G容量，20M附件！ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] generating loglogistic distribution in R
nmi13 nmi13 at ext.canterbury.ac.nz writes: Hi Dear ??? Can someone please inform me on genreating random variables of Loglogistic, and PERT beta distributions? loglogistic: exp(n,rlogis(n,...)) I'm not sure about the PERT beta -- a few seconds of web browsing suggests that it's a shifted, scaled version of the beta. Hence a + (b-a)*rbeta(n,shape1,shape2) cheers Ben Bolker __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Questions about sort data
Dear R users, I am doing my project which I want to plot a piecewise function, I knew that I can use the command segments to plot. But the problem is I want to use my real data which needs me to sort of my data by using the 'if else'command, I use it If(t[i]36) lambda-0.5 Else lambda-0.2 The funny thing is when I look at my data set, it did not follow my command to sort data, also the final numbers do not change either. The other problem is I have a big data set, sometimes I want to know the number of some data, such as how many numbers are over 60, how to do this by using R? Does anybody got any ideas about my two pros? Thank you in advance Best wishes Nessie^_^ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Questions about sort data
Regarding your other questions On 9/4/06, Xiao Zhao [EMAIL PROTECTED] wrote: Dear R users, I am doing my project which I want to plot a piecewise function, I knew that I can use the command segments to plot. But the problem is I want to use my real data which needs me to sort of my data by using the 'if else'command, I use it If(t[i]36) lambda-0.5 Else lambda-0.2 The funny thing is when I look at my data set, it did not follow my command to sort data, also the final numbers do not change either. The other problem is I have a big data set, sometimes I want to know the number of some data, such as how many numbers are over 60, how to do this by using R? Does anybody got any ideas about my two pros? Using the built in data set rivers there are sum(rivers 1000) rivers greater than 1000 miles long. See the last two lines on every message to r-help. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problems with 2-D Kernel Density Estimation using MASS KernSmooth
Dear R-Users, I am using the two dimensional Kernel density estimation function in MASS package (specifically kde2d) and am having a recurrent problem. The problem is that when my data vectors have less then 3 entries then the functions gives me the density estimate. But when the vector size increases beyond 3 then I get an error. I have pasted below my steps and the error message. Pmin-2 Pmax-14 Mmin-100 Mmax-1000 N-200 dens-kde2d(data3[[3]],data3[[4]], h = c(width.SJ(data3[[3]],nb=100, +method=dpi), width.SJ(data3[[4]], nb=100,method=dpi)), n=N, +lims=c(Pmin,Pmax,log10(Mmin),log10(Mmax))) When data3[[3]] and data3[[4]] are each less then 3 entries then the function runs fine else it gives me following error message: Error in SDh(cnt, (2.394/(n * TD))^(1/7), n, d) : NA/NaN/Inf in foreign function call (arg 5) I also tried using the bkde2D from KernSmooth package and am getting the same error for longer vector sizes. I shall be thankful if you could suggest what is going wrong. Alternatively I will be glad to know about other possible ways of fast density estimation on bigger vectors. Thanks in advance! Best Regards, Chanchal === Chanchal Kumar, Ph.D. Candidate Dept. of Proteomics and Signal Transduction Max Planck Institute of Biochemistry Am Klopferspitz 18 82152 D-Martinsried (near Munich) Germany e-mail: [EMAIL PROTECTED] Phone: (Office) +49 (0) 89 8578 2296 Fax:(Office) +49 (0) 89 8578 2219 http://www.biochem.mpg.de/mann/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cross-correlation between two time series data
'ccf' provides a plot with red, dashed lines indicating an approximate 95% threshold for the correlation. Beyond that, with any particular model fit, you can get confidence intervals and anova tests for any particular parameter estimated. If neither of these are adequate, I suppose one might be able to try Markov Chain Monte Carlo, but I've never used that, so I can't comment further on that. If you would like more help from this listserve, please provide more detail of your application including commented, minimal, self-contained, reproducible code, explaining something you've tried and why it is not adequate (as suggested in the posting guide www.R-project.org/posting-guide.html). Hope this helps. Spencer Graves Juni Joshi wrote: Hi all, I have two time series data (say x and y). I am interested to calculate the correlation between them and its confidence interval (or to test no correlation). Function cor.test(x,y) does the test of no correlation. But this test probably is wrong because of autocorrelated data. ccf() calculates the correlation between two series data. But it does not provide the confidence intervals of cross correlation. Is there any function that calculates the confidence interval of correlation between two time series data or performs the test of no correlation between two time series data. Thanks. Jun __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] opening files in directory
Hi there
I want to be able to take all the files in a given directory, read them in one
at a time, calculate a distance matrix for them (the files are data matrices)
and then print them out to separate files. This is the code I thought I would
be able to use
(all files are in directory data_files)
for(i in 1:length(files))
+ {
+ x-read.table(data_files/files[[i]])
+ dist-dist(x, method=euclidean, diag=TRUE)
+ mat-as.matrix(dist)
+ write.table(mat, file=files[[i]])
+ }
But I get this error when I try to open the first file using read.table
Error in file(file, r) : unable to open connection
In addition: Warning message:
cannot open file 'data_files/files[[i]]'
if I try the read.table command without the quotation marks like so
x-read.table(data_matrix_files/files[[i]])
I get the error
Error in read.table(data_matrix_files/files[[i]]) :
Object data_matrix_files not found
But if I go to the directory where the files are kept before starting up R, the
read.table command without the quotation marks works.
I don't want to start up R in the same directory as the where the files I will
be using reside though so how do I rectify this?
Any help much appreciated
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Re: [R] opening files in directory
FAQ ...
Uwe Ligges
Ffenics wrote:
Hi there
I want to be able to take all the files in a given directory, read them in
one at a time, calculate a distance matrix for them (the files are data
matrices) and then print them out to separate files. This is the code I
thought I would be able to use
(all files are in directory data_files)
for(i in 1:length(files))
+ {
+ x-read.table(data_files/files[[i]])
+ dist-dist(x, method=euclidean, diag=TRUE)
+ mat-as.matrix(dist)
+ write.table(mat, file=files[[i]])
+ }
But I get this error when I try to open the first file using read.table
Error in file(file, r) : unable to open connection
In addition: Warning message:
cannot open file 'data_files/files[[i]]'
if I try the read.table command without the quotation marks like so
x-read.table(data_matrix_files/files[[i]])
I get the error
Error in read.table(data_matrix_files/files[[i]]) :
Object data_matrix_files not found
But if I go to the directory where the files are kept before starting up R,
the read.table command without the quotation marks works.
I don't want to start up R in the same directory as the where the files I
will be using reside though so how do I rectify this?
Any help much appreciated
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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[R] read csv
Hi, I have a csv file where the number of filled columns varies in the different rows: Sun 5-Feb-06,15,,,01:30:00,0:06:00, Mon 6-Feb-06,, Tue 7-Feb-06,7,,,00:41:00,0:05:51, Wed 8-Feb-06,, I would like to use read.table (or whatever is appropriate) to read in only those rows that have two or more columns filled. Any hint will be appreciated. Georg __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] opening files in directory
Yes, already looked at the FAQ. Thats how I have got where I am.
I didnt see a solution to this particular problem on there though
Thanks anyway.
Uwe Ligges [EMAIL PROTECTED] wrote: FAQ ...
Uwe Ligges
Ffenics wrote:
Hi there
I want to be able to take all the files in a given directory, read them in
one at a time, calculate a distance matrix for them (the files are data
matrices) and then print them out to separate files. This is the code I
thought I would be able to use
(all files are in directory data_files)
for(i in 1:length(files))
+ {
+ x-read.table(data_files/files[[i]])
+ dist-dist(x, method=euclidean, diag=TRUE)
+ mat-as.matrix(dist)
+ write.table(mat, file=files[[i]])
+ }
But I get this error when I try to open the first file using read.table
Error in file(file, r) : unable to open connection
In addition: Warning message:
cannot open file 'data_files/files[[i]]'
if I try the read.table command without the quotation marks like so
x-read.table(data_matrix_files/files[[i]])
I get the error
Error in read.table(data_matrix_files/files[[i]]) :
Object data_matrix_files not found
But if I go to the directory where the files are kept before starting up R,
the read.table command without the quotation marks works.
I don't want to start up R in the same directory as the where the files I
will be using reside though so how do I rectify this?
Any help much appreciated
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] read csv
On Mon, 4 Sep 2006, Georg Otto wrote: Hi, I have a csv file where the number of filled columns varies in the different rows: Sun 5-Feb-06,15,,,01:30:00,0:06:00, Mon 6-Feb-06,, Tue 7-Feb-06,7,,,00:41:00,0:05:51, Wed 8-Feb-06,, I would like to use read.table (or whatever is appropriate) to read in only those rows that have two or more columns filled. Any hint will be appreciated. See ?read.table is the main hint. Use read.table(fill=TRUE) and post-process, e.g. by A - A[rowSums(!is.na(A)) 2), ] -- 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 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] read csv
I have a csv file where the number of filled columns
varies in the different rows:
I would hack it like this, but then I am totally new to R,
which means you should not trust me.:
d - read.csv(testdata, header=F)
selection - apply(d, c(1),
function(x) {sum(!is.na(x) x != ) 2})
as.data.frame(t(as.data.frame(t(d))[selection]))
Tom
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Re: [R] Fitting Pareto distribution to some data
Paul, Package lmomco fits generalized pareto (three parameter) using method of L-moments. I suspect that other packages that Brian identified use method of moments or other. William On Sep 3, 2006, at 10:55 AM, Paul Smith wrote: Dear All I am trying to fit Pareto distribution to some data. MASS package does not support Pareto distribution. Is there some alternative way? Thanks in advance, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subsetting vectors based on condition
Mahesh Krishnan wrote: Hello, I have a question regarding subsetting of vectors. Here's an example of what I'm trying to do: vect.1 - c(76,195, 290, 380) vect.2 - c(63, 95, 133, 170, 215, 253, 285, 299, 325, 375) I would like to subset vect.2 so that it has the same length as vect.1, and its numbers are the first corresponging higher value compared to vect.1. The output should be: final.output = (95, 215, 299, NA) What is the fastest/most eficient way to accompllish this in R? Thanks for the help in advance, Mahesh Krishnan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. vect.1 - c(76,195, 290, 380) vect.2 - c(63, 95, 133, 170, 215, 253, 285, 299, 325, 375) vect.2[ findInterval(vect.1,vect.2) + 1 ] [1] 95 215 299 NA J. R. M. Hosking __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fitting Pareto distribution to some data
On 9/4/06, William Asquith [EMAIL PROTECTED] wrote: Package lmomco fits generalized pareto (three parameter) using method of L-moments. I suspect that other packages that Brian identified use method of moments or other. That is excellent to learn that, William. Thanks. Paul On Sep 3, 2006, at 10:55 AM, Paul Smith wrote: Dear All I am trying to fit Pareto distribution to some data. MASS package does not support Pareto distribution. Is there some alternative way? Thanks in advance, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Optimization
Have you considered talking logarithms of the expression you mentioned: log(Yield) = a1*log(A)+b1*log(B)+c2*log(C)+... where a1 = a/(a+b+...), etc. This model has two constraints not present in ordinary least squares: First, the intercept is assumed to be zero. Second, the coefficients in this log formulation must sum to 1. If I were you, I might use something like lm to test them both. To explain how, I'll modify the notation, replacing A by X1, B by X2, ..., up to Xkm1 (= X[k-1]) and Xk for k different environmental variables. Then I might try something like the following: fit0 - lm(log(Yield) ~ log(X1) + ... + log(Xk)-1 ) fit1 - lm(log(Yield) ~ log(X1) + ... + log(Xk) ) fit.1 - lm(log(Yield/Xk) ~ log(X1/Xk) + ... + log(Xkm1/Xk) ) fit.0 - lm(log(Yield/Xk) ~ log(X1/Xk) + ... + log(Xkm1/Xk)-1 ) anova(fit1, fit0) would test the no-constant model, and if I haven't made a mistake in this, anova(fit0, fit.0) and anova(fit1, fit.1) would test the constraint that all the coefficients should sum to 1. If you would like further help from this listserve, please provide commented, minimal, self-contained, reproducible code to help potential respondents understand your question and concerns (as suggested in the posting guide www.R-project.org/posting-guide.html). Hope this helps. Spencer Graves Simone Vincenzi wrote: Dear R-list, I'm trying to estimate the relative importance of 6 environmental variables in determining clam yield. To estimate clam yield a previous work used the function Yield = (A^a*B^b*C^c...)^1/(a+b+c+...) where A,B,C... are the values of the environmental variables and the weights a,b,c... have not been calibrated on data but taken from literature. Now I'd like to estimate the weights a,b,c... by using a dataset with 110 observations of yield and values of the environmental variables. I'm wondering if it is feasible or if the number of observation is too low, if some data transformation is needed and which R function is the most appropriate to try to estimate the weights. Any help would be greatly appreciated. Simone Vincenzi _ Simone Vincenzi, PhD Student Department of Environmental Sciences University of Parma Parco Area delle Scienze, 33/A, 43100 Parma, Italy Phone: +39 0521 905696 Fax: +39 0521 906611 e.mail: [EMAIL PROTECTED] -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] opening files in directory
R won't do variable interpolation inside quotation marks as perl does.
You could try amending your code with, for e.g.
file.name-paste(sep=/,data_files,files[[i]])
x-read.table(file.name)
Regards,
Mike
On 9/4/06, Ffenics [EMAIL PROTECTED] wrote:
Hi there
I want to be able to take all the files in a given directory, read them in
one at a time, calculate a distance matrix for them (the files are data
matrices) and then print them out to separate files. This is the code I
thought I would be able to use
(all files are in directory data_files)
for(i in 1:length(files))
+ {
+ x-read.table(data_files/files[[i]])
+ dist-dist(x, method=euclidean, diag=TRUE)
+ mat-as.matrix(dist)
+ write.table(mat, file=files[[i]])
+ }
But I get this error when I try to open the first file using read.table
Error in file(file, r) : unable to open connection
In addition: Warning message:
cannot open file 'data_files/files[[i]]'
if I try the read.table command without the quotation marks like so
x-read.table(data_matrix_files/files[[i]])
I get the error
Error in read.table(data_matrix_files/files[[i]]) :
Object data_matrix_files not found
But if I go to the directory where the files are kept before starting up R,
the read.table command without the quotation marks works.
I don't want to start up R in the same directory as the where the files I
will be using reside though so how do I rectify this?
Any help much appreciated
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Regards,
Mike Nielsen
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] opening files in directory
On 9/4/06, Mike Nielsen [EMAIL PROTECTED] wrote: R won't do variable interpolation inside quotation marks as perl does. Just as an aside, gsubfn in package gsubfn will do perl-style (well, sort of) string interpolation: library(gsubfn) i - 1 gsubfn(x = data_files/file$i) [1] data_files/file1 cati and cati0 in the same package provide for such interpolation within cat. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] abline and plot(augPred) help
Hi
Petr Pikal wrote:
Dear all
as I did not get any response on my post about abline and
plot(augPred)) I try again. I hope I do not break some posting guide
rules. I would try to contact package maintainer directly but there
is stated to be R-core people, so I feel R-help list shall be OK.
I need to draw straight lines through augPred plotted panels
(vertical or horizontal) at specified point. I know I shall probably
use panel.abline but I am missing correct syntax. Below you can see
my attempts together with results. I hope somebody can point me to
right direction.
I am probably somewhere close but I have no clue, which parameter I
shall modify to get measured points, fitted lines and vertical lines
in panels together.
The problem is that you do not know about the default panel function
that nlme:::plot.augPred() uses, so your panel functions are not
replicating all of the default behaviour as well as adding your vertical
lines. Some possible solutons suggested below ...
fm1 - lme(Orthodont)
# standard plot
plot(augPred(fm1, level = 0:1, length.out = 2))
#plot with vertical but without points and fitted lines
plot(augPred(fm1, level = 0:1, length.out = 2),
panel=function(v,...) {
panel.abline(v=10)}
)
# plot with vertical but without fitted lines
plot(augPred(fm1, level = 0:1, length.out=2),
panel=function(x,y,...) {
panel.xyplot(x,y,...)
panel.abline(v=10)}
)
# plot with vertical and with all points (fitted lines are drawn as
points)
plot(augPred(fm1, level = 0:1),
panel=function(x,y,...) {
panel.xyplot(x,y,...)
panel.abline(v=10)}
)
One option is to take a sneak a peek at nlme:::plot.augPred() to see
what the default panel function is doing. Here I have replicated the
default panel function and added a call to panel.abline().
plot(augPred(fm1, level = 0:1, length.out = 2),
panel=function(x, y, subscripts, groups, ...) {
orig - groups[subscripts] == original
panel.xyplot(x[orig], y[orig], ...)
panel.superpose(x[!orig], y[!orig], subscripts[!orig],
groups, ..., type = l)
panel.abline(v=10)
})
The problem with this approach is that you need to crawl around in the
code of nlme:::plot.augPred(). An alternative approach is to annotate
the plot after-the-fact. This is shown below.
plot(augPred(fm1, level = 0:1, length.out = 2))
for (i in 1:5) {
for (j in 1:6) {
if (i 5 || j 4) {
trellis.focus(panel, j, i, highlight=FALSE)
panel.abline(v=10)
}
}
}
This avoids crawling around in code, but the problem with this is
knowing how many rows and columns of panels there are. If you
explicitly controlled the 'layout' of the original plot, you could
guarantee that your annotation works properly.
Hope that helps.
Paul
--
Dr Paul Murrell
Department of Statistics
The University of Auckland
Private Bag 92019
Auckland
New Zealand
64 9 3737599 x85392
[EMAIL PROTECTED]
http://www.stat.auckland.ac.nz/~paul/
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Re: [R] abline and plot(augPred) help
On 9/4/06, Paul Murrell [EMAIL PROTECTED] wrote:
Hi
Petr Pikal wrote:
Dear all
as I did not get any response on my post about abline and
plot(augPred)) I try again. I hope I do not break some posting guide
rules. I would try to contact package maintainer directly but there
is stated to be R-core people, so I feel R-help list shall be OK.
I need to draw straight lines through augPred plotted panels
(vertical or horizontal) at specified point. I know I shall probably
use panel.abline but I am missing correct syntax. Below you can see
my attempts together with results. I hope somebody can point me to
right direction.
I am probably somewhere close but I have no clue, which parameter I
shall modify to get measured points, fitted lines and vertical lines
in panels together.
The problem is that you do not know about the default panel function
that nlme:::plot.augPred() uses, so your panel functions are not
replicating all of the default behaviour as well as adding your vertical
lines. Some possible solutons suggested below ...
fm1 - lme(Orthodont)
# standard plot
plot(augPred(fm1, level = 0:1, length.out = 2))
#plot with vertical but without points and fitted lines
plot(augPred(fm1, level = 0:1, length.out = 2),
panel=function(v,...) {
panel.abline(v=10)}
)
# plot with vertical but without fitted lines
plot(augPred(fm1, level = 0:1, length.out=2),
panel=function(x,y,...) {
panel.xyplot(x,y,...)
panel.abline(v=10)}
)
# plot with vertical and with all points (fitted lines are drawn as
points)
plot(augPred(fm1, level = 0:1),
panel=function(x,y,...) {
panel.xyplot(x,y,...)
panel.abline(v=10)}
)
One option is to take a sneak a peek at nlme:::plot.augPred() to see
what the default panel function is doing. Here I have replicated the
default panel function and added a call to panel.abline().
plot(augPred(fm1, level = 0:1, length.out = 2),
panel=function(x, y, subscripts, groups, ...) {
orig - groups[subscripts] == original
panel.xyplot(x[orig], y[orig], ...)
panel.superpose(x[!orig], y[!orig], subscripts[!orig],
groups, ..., type = l)
panel.abline(v=10)
})
The problem with this approach is that you need to crawl around in the
code of nlme:::plot.augPred(). An alternative approach is to annotate
the plot after-the-fact. This is shown below.
plot(augPred(fm1, level = 0:1, length.out = 2))
for (i in 1:5) {
for (j in 1:6) {
if (i 5 || j 4) {
trellis.focus(panel, j, i, highlight=FALSE)
panel.abline(v=10)
}
}
}
This avoids crawling around in code, but the problem with this is
knowing how many rows and columns of panels there are. If you
explicitly controlled the 'layout' of the original plot, you could
guarantee that your annotation works properly.
You can find that out with trellis.currentLayout:
tcL - trellis.currentLayout()
for(i in 1:nrow(tcL))
for(j in 1:ncol(tcL))
if (tcL[i,j] 0) {
trellis.focus(panel, j, i, highlight = FALSE)
panel.abline(v = 10)
trellis.unfocus()
}
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[R] Coercing elements of a matrix from integer to double
Ive been converting elements of matrices and arrays, e.g., from
Integers to double-precision, by vectorizing the matrix and then
remaking it. Alternatively, I can redefine one element as double
which then redefines them all. Both methods are quick, so I guess
I shouldn't complain, but I would have thought there'd be something
more obvious. Have I missed it?
Here's my redimensioning example:
## Matrix M...
M - 1:2e6 ; dim(Mi) - c(1e3,2e3)
dim(M)
class(M)
## ...has integer elements, e.g.,
class(M[1,1])
## The as.double() command changes
## these to double-precision, but it
## also strips away dimensions...
Md - as.double(Mi)
dim(Md)
class(Md)
## ...so I have to put them back.
dim(Md) - dim(Mi)
dim(Md)
class(Md)
class(Md[1,1])
Here's my tail wagging the dog example:
M[1,1] - as.double(M[1,1])
class(M[2,2])
Thanks,
-John Thaden
Confidentiality Notice: This e-mail message, including any a...{{dropped}}
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Re: [R] Coercing elements of a matrix from integer to double
Thaden, John J [EMAIL PROTECTED] writes: Ive been converting elements of matrices and arrays, e.g., from Integers to double-precision, by vectorizing the matrix and then remaking it. Alternatively, I can redefine one element as double which then redefines them all. Both methods are quick, so I guess I shouldn't complain, but I would have thought there'd be something more obvious. Have I missed it? storage.mode(M) - double -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to fit gauss beam?
Reply to self: I am having a hard time fitting a gauss beam using R. In gnutplot I did something like $ w(z) = w0 * sqrt(1+(z/z0)**2) $ fit w(z) 'before_eom.txt' using 1:2 via w0, z0 to obtain w0 and z0. Now I want to do the same in R. I tried a linear model like this (r = radius, z = This works fine: data - read.table (nach_eom.tab, header=T) M - 1.1 nls (major ~ M*w0*sqrt(1+(d/z0)^2), data = data, start = list(w0 = 460, z0=0.02)) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cross-correlation between two time series data
Jun, If your interest is to estimate the correlation and either a confidence interval or a test for no correlation, then you might try to proceed as follows. This is a Monte-Carlo significance test, and a useful strategy. 1) use ccf() to compute the cross-correlation between x and y. 2) repeat the following steps, say, 1000 times. 2a) randomly reorder the values of one of the time series, say x. Call the randomly reordered series x'. 2b) use ccf() to compute the cross-correlation between x' and y. Store that cross-correlation. 3) the 1000 cross-correlation estimates computed in step 2 are all estimating cross-correlation 0, conditional on the data. A two-tailed test then is: if the cross-correlation computed in step 1 is outside the (0.025, 0.975) quantiles of the empirical distribution of the cross-correlations computed in step 2, then, reject the null hypothesis that x and y are uncorrelated, with size 0.05. I hope that this helps. Andrew Juni Joshi wrote: Hi all, I have two time series data (say x and y). I am interested to calculate the correlation between them and its confidence interval (or to test no correlation). Function cor.test(x,y) does the test of no correlation. But this test probably is wrong because of autocorrelated data. ccf() calculates the correlation between two series data. But it does not provide the confidence intervals of cross correlation. Is there any function that calculates the confidence interval of correlation between two time series data or performs the test of no correlation between two time series data. Thanks. Jun __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Andrew Robinson Department of Mathematics and StatisticsTel: +61-3-8344-9763 University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599 Email: [EMAIL PROTECTED] http://www.ms.unimelb.edu.au __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subsetting vectors based on condition
Sincere thanks to Jim Holtman and J. Hosking for their suggestions. both their solutions work perfectly, in particular the findInterval function is what I was looking for. Cheers. Mahesh Krishnan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Fitting generalized additive models with constraints?
Hello, I am trying to fit a GAM for a simple model, a simple model, y ~ s(x0) + s(x1) ; with a constraint that the fitted smooth functions s(x0) and s(x1) have to each always be 0. From the library documentation and a search of the R-site and R-help archives I have not been able to decipher whether the following is possible using this, or other GAM libraries, or whether I will have to try to roll my own. I see from the mgcv docs that GAMs need to be constrained such that the smooth functions have zero mean. Is there a way around this? Is such a constraint possible? thanks very much for any advice or pointers. -David [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problem with Variance Components (and general glmm confusion)
Dear list, I am having some problems with extracting Variance Components from a random-effects model: I am running a simple random-effects model using lme: model-lme(y~1,random=~1|groupA/groupB) which returns the output for the StdDev of the Random effects, and model AIC etc as expected. Until yesterday I was using R v. 2.0, and had no problem in calling the variance components of the above model using VarCorr(model), together with their 95% confidence intervals using intervals() - although for some response variables a call to intervals() returns the error: Cannot get confidence intervals on var-cov components: Non-positive definite approximate variance-covariance. I have now installed R v. 2.3.1 and am now experiencing odd behaviour with VarCorr(lme.object), with an error message typically being returned: Error in VarCorr(model) : no direct or inherited method for function 'VarCorr' for this call Is this known to happen? For instance could it be due to the subsequent loading of new packages? (lme4 for instance?). To get around this problem I have tried running the same model using lmer: model2-lmer(y~1 + (1|groupA) + (1|groupB)) Should this not produce the same model? The variance components are very similar but not identical, making me think that I am doing something wrong. I am also correct in thinking that intervals() does not work with lmer? I get: Error in intervals(model2) : no applicable method for intervals GLMM I have a general application question - please excuse my ignorance, I am relatively new to this and trying to find a way through the maze. In short I need to compile generalized linear mixed models both for (a) Poisson data and (b) binonial data incorporating a two nested random factors, and I need to be able to extract AIC values as I am taking an information-theoretic approach to model selection. Prior to sending an email to the list I have spent quite a few days reading the background on a number of functions, all of which offer potential for this; glmmML, glmmPQL, lmer, and glmmADMB. I can understand that glmmPQL is unsuitable because there is no way of knowing the maximised likelihood, but is there much difference between the remaining three options? I have seen simulation comparisons published on this list between glmmADMB and glmmPQL and lmer, but it seems these are before the latest release of lmer, and also they do not evaluate glmmML. To a newcomer this myriad ! of options is bewildering, can anyone offer advice as to the most robust approach? Many thanks for your time and patience, Toby Gardner School of Environmental Sciences University of East Anglia Norwich, NR4 7TJ United Kingdom Email: [EMAIL PROTECTED] Website: www.uea.ac.uk/~e387495 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cross-correlation between two time series data
Hi, Andrew: This will produce a permutation distribution for the correlation under the null hypothesis of zero correlation between the variables. This is a reasonable thing to do, and would probably produce limits more accurate than the dashed red lines on the 'ccf' plot. However, they would NOT be confidence interval(s). For a confidence interval on cross correlation, you'd have to hypothesize some cross correlation pattern between x and y, preferably parameterized parsimoniously, then somehow determine an appropriate range of values consistent with the data. By the time you've done all that, you've effectively fit some model and constructed confidence intervals on the parameter(s). Best Wishes, Spencer Andrew Robinson wrote: Jun, If your interest is to estimate the correlation and either a confidence interval or a test for no correlation, then you might try to proceed as follows. This is a Monte-Carlo significance test, and a useful strategy. 1) use ccf() to compute the cross-correlation between x and y. 2) repeat the following steps, say, 1000 times. 2a) randomly reorder the values of one of the time series, say x. Call the randomly reordered series x'. 2b) use ccf() to compute the cross-correlation between x' and y. Store that cross-correlation. 3) the 1000 cross-correlation estimates computed in step 2 are all estimating cross-correlation 0, conditional on the data. A two-tailed test then is: if the cross-correlation computed in step 1 is outside the (0.025, 0.975) quantiles of the empirical distribution of the cross-correlations computed in step 2, then, reject the null hypothesis that x and y are uncorrelated, with size 0.05. I hope that this helps. Andrew Juni Joshi wrote: Hi all, I have two time series data (say x and y). I am interested to calculate the correlation between them and its confidence interval (or to test no correlation). Function cor.test(x,y) does the test of no correlation. But this test probably is wrong because of autocorrelated data. ccf() calculates the correlation between two series data. But it does not provide the confidence intervals of cross correlation. Is there any function that calculates the confidence interval of correlation between two time series data or performs the test of no correlation between two time series data. Thanks. Jun __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] plot a new picture against an old one to see the difference between them
Hello, useR:, Suppose I have two plots made by using contour() function, say Cont1 and Cont2 respectively. They have slightly difference because of the two slightly different data I used. I want to see the difference between them so I want to plot Cont2 on Cont1, are there any methods to plot it without filling the frame of Cont1 totally of Cont2. I mean, how I can integreate the two plots together that they kind of have weighted colors? Thanks very much in Advance! Leon [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot a new picture against an old one to see the difference between them
Dear Roger, Thanks, that's really helpful, do you know how to deal with it if the two plots are generated by plot(), not by contour(). Best, Leon - Original Message - From: roger koenker [EMAIL PROTECTED] To: Am Stat [EMAIL PROTECTED] Sent: Monday, September 04, 2006 8:06 PM Subject: Re: [R] plot a new picture against an old one to see the difference between them for the second call to contour use the argument add=TRUE. On Sep 4, 2006, at 6:42 PM, Am Stat wrote: Hello, useR:, Suppose I have two plots made by using contour() function, say Cont1 and Cont2 respectively. They have slightly difference because of the two slightly different data I used. I want to see the difference between them so I want to plot Cont2 on Cont1, are there any methods to plot it without filling the frame of Cont1 totally of Cont2. I mean, how I can integreate the two plots together that they kind of have weighted colors? Thanks very much in Advance! Leon [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help with plotmath
Hi R users:
How can I have several subscript number with a comma in a plot.
I would like to have the LaTeX equivalent of
x_{i,j}.
I try:
plot(1:10,1:10,type=n)
text(5,5,expression(x[i,j]))
but it doesn´t work.
Thank you for your help.
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[R] Sweave and the [ function
I am wanting to use the [ operator in an S-chunk, e.g. = str(women) women$height women[,1] [(women,1) @ to show the equivalence of three methods of extracting an element from a data.frame. However Sweave returns the last of these as women[1] in the S input chunk How can I force it not to do this and return [(women,1) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question on Chi-square of null model in sem package
Dear Pref. Fox Sorry, I didn't receive your reply. I try the new sem package. It's great. The following is the results that I got. The fit indices are fine. Model Chisquare = 208 Df = 98 Pr(Chisq) = 6.6e-10 Chisquare (null model) = 1741 Df = 120 Goodness-of-fit index = 0.9 Adjusted goodness-of-fit index = 0.87 RMSEA index = 0.066 90 % CI: (0.054, 0.079) Bentler-Bonnett NFI = 0.88 Tucker-Lewis NNFI = 0.92 Bentler CFI = 0.93 BIC = -336 Thank you very much. You help me out so many problems. Best wishes, Wei-Wei 2006/9/4, John Fox [EMAIL PROTECTED]: Dear Wei-Wei, As I explained to you in private email yesterday (perhaps you didn't receive my reply?), the problem that you point out is due to a bug in the sem function that I fixed some time ago and then inadvertently reintroduced. Yesterday, I sent a corrected version of the sem package (0.9-5) to CRAN; the source package is there now and I'm sure that the compiled Windows package will appear in due course. Thank you once more for bringing the problem to my attention. John __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cross-correlation between two time series data
Hi Spencer, you are quite right. I should have been careful to emphasize that the strategy I suggested was intended only to produce the test for no correlation clause of the either a confidence interval or a test for no correlation sentence. Cheers Andrew On Mon, Sep 04, 2006 at 04:00:44PM -0700, Spencer Graves wrote: Hi, Andrew: This will produce a permutation distribution for the correlation under the null hypothesis of zero correlation between the variables. This is a reasonable thing to do, and would probably produce limits more accurate than the dashed red lines on the 'ccf' plot. However, they would NOT be confidence interval(s). For a confidence interval on cross correlation, you'd have to hypothesize some cross correlation pattern between x and y, preferably parameterized parsimoniously, then somehow determine an appropriate range of values consistent with the data. By the time you've done all that, you've effectively fit some model and constructed confidence intervals on the parameter(s). Best Wishes, Spencer Andrew Robinson wrote: Jun, If your interest is to estimate the correlation and either a confidence interval or a test for no correlation, then you might try to proceed as follows. This is a Monte-Carlo significance test, and a useful strategy. 1) use ccf() to compute the cross-correlation between x and y. 2) repeat the following steps, say, 1000 times. 2a) randomly reorder the values of one of the time series, say x. Call the randomly reordered series x'. 2b) use ccf() to compute the cross-correlation between x' and y. Store that cross-correlation. 3) the 1000 cross-correlation estimates computed in step 2 are all estimating cross-correlation 0, conditional on the data. A two-tailed test then is: if the cross-correlation computed in step 1 is outside the (0.025, 0.975) quantiles of the empirical distribution of the cross-correlations computed in step 2, then, reject the null hypothesis that x and y are uncorrelated, with size 0.05. I hope that this helps. Andrew Juni Joshi wrote: Hi all, I have two time series data (say x and y). I am interested to calculate the correlation between them and its confidence interval (or to test no correlation). Function cor.test(x,y) does the test of no correlation. But this test probably is wrong because of autocorrelated data. ccf() calculates the correlation between two series data. But it does not provide the confidence intervals of cross correlation. Is there any function that calculates the confidence interval of correlation between two time series data or performs the test of no correlation between two time series data. Thanks. Jun __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Andrew Robinson Department of Mathematics and StatisticsTel: +61-3-8344-9763 University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599 Email: [EMAIL PROTECTED] http://www.ms.unimelb.edu.au __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with plotmath
Try:
text(5,5,expression(x[i * , * j]))
On 9/4/06, Kenneth Cabrera [EMAIL PROTECTED] wrote:
Hi R users:
How can I have several subscript number with a comma in a plot.
I would like to have the LaTeX equivalent of
x_{i,j}.
I try:
plot(1:10,1:10,type=n)
text(5,5,expression(x[i,j]))
but it doesn´t work.
Thank you for your help.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Sweave and the [ function
= str(women) women$height women[,1] [(women,1) @ to show the equivalence of three methods of extracting an element from a data.frame. However Sweave returns the last of these as women[1] in the S input chunk How can I force it not to do this and return [(women,1) I don't think you can. Sweave parses your R code and from then on uses the internal R representation. R normalises the parse tree in certain ways (eg. strips comments, formats source code, and clearly normalises some function calls). Since sweave uses this, and not the original text, I don't think there is anyway to get around this, unless there is some trick during parsing. (And don't forget women[[1]]) Hadley __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sweave and the [ function
Le Mardi 5 Septembre 2006 0:03, hadley wickham a écrit : = str(women) women$height women[,1] [(women,1) @ to show the equivalence of three methods of extracting an element from a data.frame. However Sweave returns the last of these as women[1] in the S input chunk How can I force it not to do this and return [(women,1) I don't think you can. Sweave parses your R code and from then on uses the internal R representation. R normalises the parse tree in certain ways (eg. strips comments, formats source code, and clearly normalises some function calls). Since sweave uses this, and not the original text, I don't think there is anyway to get around this, unless there is some trick during parsing. (And don't forget women[[1]]) Hadley So here's a workaround (untested): echo=TRUE, eval=TRUE= str(women) women$height women[,1] @ echo=TRUE, eval=FALSE= [(women,1) @ echo=FALSE, eval=TRUE= [(women,1) @ I often end up doing similar things. HTHVincent -- Vincent Goulet, Professeur agrégé École d'actuariat Université Laval, Québec [EMAIL PROTECTED] http://vgoulet.act.ulaval.ca __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] colorRamp
Hi,
I am using colorRamp in the following way. I am *sure* there is a
better way to do this, so if you'd be so kind to show me the true R way:
Step 0: Create a new variable, say, x, that maps some other
continuous variable I have onto the [0,1] line.
Step 1: Store the result from colorRamp (a function), into, say, test
test - colorRamp(mypalette)
Step 2: In my data frame, data
data$colorTemp - test(data$x)
Step 3: Write a new function
bob - function(temp) { rgb(temp[1],temp[2],temp[3],maxColorValue=255) }
Step 4:
for (i in 1:dim(data)[1]) data[i,color] - bob(data[i,
colorTemp])
Step 5:
map(states, region=data$region, fill=T, col=data$color)
Thanks in advance!
Rick
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[R] Quick question about lm()
Hi, Feel awkward to ask , but really couldn't find a answer anywhere, How could I extract the R^2 and t-stat. from the result of lm()? Thanks a lot. best __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Quick question about lm()
Say, my.lm - lm(y ~ x, data=my.data) Then if you try: names(summary(my.lm)) you will see the components of the summary.lm object. The coefficients and t-statistics can be extracted by summary(my.lm)$coefficients and similarly for the r-squared and other statistics provided in the summary report. -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Tong Wang Sent: Tuesday, September 05, 2006 1:35 AM To: r-help@stat.math.ethz.ch Subject: [R] Quick question about lm() Hi, Feel awkward to ask , but really couldn't find a answer anywhere, How could I extract the R^2 and t-stat. from the result of lm()? Thanks a lot. best __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
