[R] confusion matrix in randomForest
I have a question on the output generated by randomForest in classification mode, specifically, the confusion matrix. The confusion matrix lists the various classes and how the forest classified each one, plus the classification error. Are these numbers essentially averages over all the trees in the forest? If so, is there a way I can get the standard deviation values out of the randomForest, or do I have to evaluate each tree individually? By way of illustration, let me show the confusion matrix using the iris data. The output below shows that the forest correctly classified 47 versicolor irises, but this is the result for the entire forest. I'd like to know if every tree will have 47 correctly classified versicolor irises, but I don't think it will. Same for the class.error value. Not every tree will have those exact same values, right? But this raises another question. For this example, I used the entire data set to generate the forest, and so I assume that the confusion matrix is based on OOB data, so if I created a training set and evaluated trees individually in the test set I could get averages and standard deviations on the error rate. Any thoughts? Thanks in advance. -Miklos Z. Kiss print(iris.rf) Call: randomForest(formula = Species ~ ., data = iris, importance = TRUE, keep.forest = TRUE) Type of random forest: classification Number of trees: 500 No. of variables tried at each split: 2 OOB estimate of error rate: 5.33% Confusion matrix: setosa versicolor virginica class.error setosa 50 0 00.00 versicolor 0 47 30.06 virginica 0 5450.10 -- View this message in context: http://www.nabble.com/confusion-matrix-in-randomForest-tp18550873p18550873.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problem with read.table
Hi, Although I set check.names to FALSE in read.table, the duplicate names get modified. What should be done in this case? the text file to be read by read.table AM2 AM2 AM2 FAL 2 3 4 5 1 -1 -3 -2 t = read.table (my_file, check.names = FALSE, header = T) t AM2 AM2.1 AM2.2 FAL 2 3 4 5 1 -1 -3 -2 instead of AM2 AM2 AM2 FAL 2 3 4 5 1 -1 -3 -2 Best, carol [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R version 2.7.1 (warnings)
Thanks, Mark, for the response. The problem is vith SciViews. It is not stable under the latest version of R. I found a solution by downloading the latest version of Tinn-R, which communicates with the latest version of R, and now I can carry on like normal (with both windows tiled horizontally). Jacob Jacob L van Wyk Department of Statistics University of Johannesburg, APK Box 524 Auckland Park 2006 South Africa Office Tel: +27 11 559 3080 Fax: +27 11 559 2832 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sweave add code \201
Peter Dalgaard a écrit : Christophe Genolini wrote: Hi the list, I do not understand what change in my configuration, but Sweave add the code \201 before each special characters é è à ç ... Does someone know when it come from ? MULE (multilingal environment) in Emacs used to be prone to do this, but I haven't seen it for a while, so I'm afraid I have forgotten the details (if I ever understood them). Thanks ! I thaught I had a it was Sweave and thanks to you I start to search in emacs. Christophe __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R version 2.7.1 (warnings)
Hi Jaap, Great stuff! As the old adage went, Go well, go Bye, Mark. Van Wyk, Jaap wrote: Thanks, Mark, for the response. The problem is vith SciViews. It is not stable under the latest version of R. I found a solution by downloading the latest version of Tinn-R, which communicates with the latest version of R, and now I can carry on like normal (with both windows tiled horizontally). Jacob Jacob L van Wyk Department of Statistics University of Johannesburg, APK Box 524 Auckland Park 2006 South Africa Office Tel: +27 11 559 3080 Fax: +27 11 559 2832 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/R-version-2.7.1-%28warnings%29-tp18542181p18552685.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Discretize continous variables....
Frank E Harrell Jr [EMAIL PROTECTED] [Sun, Jul 20, 2008 at 12:20:28AM CEST]: Johannes Huesing wrote: Because regulatory bodies demand it? [...] And how anyway does this relate to predictors in a model? Not at all; you're correct. I was mixing the topic of this discussion up with another kind of silliness. I had a discussion with a biometrician in a pharmaceutical company though who stated that when you have only one df to spend it will be better to dichotomise it at a clinically meaningful point than to include it as a linear term. He kept the discussion on the ground of laboratory measurements like sodium, where a deviation from normal ranges is very significant (and unlike, say, cholesterol, where you have a gradual interpretation of the value). He has a point there, but in general the reason for sacrificing information is a mixture of laziness, the preference for presenting data in tables and to keep the modelling consistent with the tables (for instance to assign an odds ratio to each cell). -- Johannes Hüsing There is something fascinating about science. One gets such wholesale returns of conjecture mailto:[EMAIL PROTECTED] from such a trifling investment of fact. http://derwisch.wikidot.com (Mark Twain, Life on the Mississippi) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problem with read.table
Hi, Although I set check.names to FALSE in read.table, the duplicate names get modified. What should be done in this case? the text file to be read by read.table AM2 AM2 AM2 FAL 2 3 4 5 1 -1 -3 -2 t = read.table (my_file, check.names = FALSE, header = T) t AM2 AM2.1 AM2.2 FAL 2 3 4 5 1 -1 -3 -2 instead of AM2 AM2 AM2 FAL 2 3 4 5 1 -1 -3 -2 Best, carol [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] estimating volume from xyz points
On Sat, Jul 19, 2008 at 5:21 PM, milton ruser [EMAIL PROTECTED] wrote: Dear all, I have several sets of x-y-z points and I need to estimate the volume that encompass all my points. Recently I got some adivice to show the convex hull of my points using geometry package (see code below). But now I need to calculate the volume of my set of points. ?convhulln also in geometry package. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R interprets symbol as q()
R2.7.1, WinXP Hi, My question is probably easy to answer, but I can't seem to solve it myself. I need to read in a large number of datasets that were generated by another (old) software program. Each data file contains a bunch of additional code generated by the external software. It lists the data per respondent. So first the data for respondent 1, then the data for respondent 2, et cetera. After the data for the final respondent, the file contains a , followed by some zeroes. R now needs to read the data until the symbol. However, whenever it encounters the symbol it interprets it as a quit command! So, rather than continuing, R wants to quit on me. How can I make R see the symbol as just a particular symbol and NOT as the command to q()? Thanks, Roger ps. I hope the symbol is readable in the various email programs __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with read.table
If you really want to do it, here is a way, but don't expect to be able to reference the columns by name; this is just so it 'prints pretty': x - textConnection(AM2 AM2 AM2 FAL + 2 3 4 5 + 1 -1 -3-2) # read the header row header - read.table(x, as.is=TRUE, nrows=1) # read in the rest of the data myData - read.table(x, header=FALSE) # put header on the table colnames(myData) - header myData AM2 AM2 AM2 FAL 1 2 3 4 5 2 1 -1 -3 -2 myData$AM2 # only gives the first column [1] 2 1 On Sun, Jul 20, 2008 at 3:45 AM, carol white [EMAIL PROTECTED] wrote: Hi, Although I set check.names to FALSE in read.table, the duplicate names get modified. What should be done in this case? the text file to be read by read.table AM2 AM2 AM2 FAL 2 3 4 5 1 -1 -3-2 t = read.table (my_file, check.names = FALSE, header = T) t AM2 AM2.1 AM2.2 FAL 2 3 4 5 1 -1 -3-2 instead of AM2 AM2 AM2FAL 2 3 4 5 1 -1 -3-2 Best, carol [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R interprets symbol as q()
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. It would be nice to see at least a subset of the data you are reading. I would assume that if you are using 'read.table' to read in the data, then there is no reason for interpreting 'q()' as a quit command. Are you 'source'ing the data? So an example of both the script and the data are required to understand the problem you are trying to solve. On Sun, Jul 20, 2008 at 6:41 AM, Roger Leenders [EMAIL PROTECTED] wrote: R2.7.1, WinXP Hi, My question is probably easy to answer, but I can't seem to solve it myself. I need to read in a large number of datasets that were generated by another (old) software program. Each data file contains a bunch of additional code generated by the external software. It lists the data per respondent. So first the data for respondent 1, then the data for respondent 2, et cetera. After the data for the final respondent, the file contains a , followed by some zeroes. R now needs to read the data until the symbol. However, whenever it encounters the symbol it interprets it as a quit command! So, rather than continuing, R wants to quit on me. How can I make R see the symbol as just a particular symbol and NOT as the command to q()? Thanks, Roger ps. I hope the symbol is readable in the various email programs __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Parametric survival models with left truncated, right censored data
I uploaded eha-1.2-0 to CRAN a day ago. This version can handle
accelerated failure time and proportional hazards parametric models
for right censored, left truncated, and stratified data. Included
distributions are: Weibull, lognormal, loglogistic, Gompertz, and
extreme value. The new inner structure of eha makes it very simple
(for me) to add families of distributions of a shape-scale type (i.e.
the distribution of the log of the baseline survival time is a
location-scale family). Suggestions are welcome. The Gompertz and
Gompertz-Makeham families of distributions are unfortunately not of
this type. While the Gompertz case could be solved by a trick
(always include an intercept in the regression model), the
Gompertz-Makeham case is harder. Needs special treatment; hints are
welcome.
With the possibility of keeping the shape parameter(s) fixed, special
cases of the abovementioned distributions are also possible to fit,
e.g., the exponential (Weibull with shape = 1).
Göran
On Wed, Jan 23, 2008 at 5:46 PM, Beth Ireland [EMAIL PROTECTED] wrote:
Dear All,
I would like to fit some parametric survival models using left
truncated, right censored data in R. However I am having problems
finding a function to fit parametric survival models which can handle
left truncated data.
I have tested both the survreg function in package survival:
fit1 - survreg(Surv(start, stop, status) ~ X + Y + Z, data=data1)
and the psm function in package Design:
fit2 - psm(Surv(start, stop, status) ~ X + Y + Z, data=data1)
But neither function appears to work with left truncated data. The error
message Invalid survival type is received for both functions when left
truncated data is specified.
The function weibreg in the eha package fits Weibull survival models and
works with left truncated data. This function is useful, but I need to
fit parametric models other than Weibull.
Any suggestions of functions that can fit parametric survival models
other than Weibull on left truncated, right censored data would be
greatly appreciated.
Kind Regards,
Beth
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OC310836. A list of members of Hymans Robertson Financial Services
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--
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[R] Error in edit(name,file,title,editor)
Can anyone help me with the following attempt to use an external editor from within R vi(file=p286.R) Error in edit(name, file, title, editor) : unable to open file to read edit(file=p286.R) Error in edit(name, file, title, editor) : unable to open file to read I have only recently re-started trying to learn R. (I tried before but failed.) I am working with a Mac MacOsX 10.4.11. I have R 2.7.1 GUI 1.25 (5166) which is the Cocoa version of R. I have been reading Peter Dalgaard's book, which is very helpful, and I have not met any errors so far, so I don't understand why I should suddenly hit this one. Since edit and vi appear to be in the utils package, I did library(utils) to which I got a null response. I have searched the R-help-archives for the error message, and have not found it. I suppose I could use the function source(), but I haven't tried this. Thanks for any help David Epstein __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Order of columns(variables) in dataframe
Dear R experts,  I have a dataframe with 4 columns (variables). I want to redorder (or reposition) these columns on the basis of a value in its last row. e.g.  df1-data.frame( v1= c(2,3,1,9,5), v2=c(8,5,12,4,11), v3=c(7,8,2,6,9), v4=c(1,4,6,3,6))  df1   v1 v2 v3 v4 1 2 8 7 1 2 3 5 8 4 3 1 12 2 6 4 9 4 6 3 5 5 11 9 6 I wanto to get the order of df1 on the basis of value in last row (descending order) like    v2 v3 v4 v1 1 8 7 1 2 2 5 8 4 3 3 12 2 6 1 4 4 6 3 9 5 11 9 6 5  Could somebody help me?  Daniel Amsterdam     Send instant messages to your online friends http://uk.messenger.yahoo.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Order of columns(variables) in dataframe
Is this what you want: x - read.table(textConnection( v1 v2 v3 v4 + 1 2 8 7 1 + 2 3 5 8 4 + 3 1 12 2 6 + 4 9 4 6 3 + 5 5 11 9 6), header=TRUE) closeAllConnections() # order by the last row x[, order(unlist(x[5,]), decreasing=TRUE)] v2 v3 v4 v1 1 8 7 1 2 2 5 8 4 3 3 12 2 6 1 4 4 6 3 9 5 11 9 6 5 On Sun, Jul 20, 2008 at 7:32 AM, Daniel Wagner [EMAIL PROTECTED] wrote: Dear R experts, I have a dataframe with 4 columns (variables). I want to redorder (or reposition) these columns on the basis of a value in its last row. e.g. df1-data.frame( v1= c(2,3,1,9,5), v2=c(8,5,12,4,11), v3=c(7,8,2,6,9), v4=c(1,4,6,3,6)) df1 v1 v2 v3 v4 1 2 8 7 1 2 3 5 8 4 3 1 12 2 6 4 9 4 6 3 5 5 11 9 6 I wanto to get the order of df1 on the basis of value in last row (descending order) like v2 v3 v4 v1 1 8 7 1 2 2 5 8 4 3 3 12 2 6 1 4 4 6 3 9 5 11 9 6 5 Could somebody help me? Daniel Amsterdam Send instant messages to your online friends http://uk.messenger.yahoo.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error in edit(name,file,title,editor)
Make sure you are in the right directory; you can also use the full path names. It is basically saying it can not find the file. Do 'list.files()' to see what is addressable in whatever directory you are in. On Sun, Jul 20, 2008 at 7:19 AM, David Epstein [EMAIL PROTECTED] wrote: Can anyone help me with the following attempt to use an external editor from within R vi(file=p286.R) Error in edit(name, file, title, editor) : unable to open file to read edit(file=p286.R) Error in edit(name, file, title, editor) : unable to open file to read I have only recently re-started trying to learn R. (I tried before but failed.) I am working with a Mac MacOsX 10.4.11. I have R 2.7.1 GUI 1.25 (5166) which is the Cocoa version of R. I have been reading Peter Dalgaard's book, which is very helpful, and I have not met any errors so far, so I don't understand why I should suddenly hit this one. Since edit and vi appear to be in the utils package, I did library(utils) to which I got a null response. I have searched the R-help-archives for the error message, and have not found it. I suppose I could use the function source(), but I haven't tried this. Thanks for any help David Epstein __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Order of columns(variables) in dataframe
try this: df1 - data.frame(v1 = c(2,3,1,9,5), v2 = c(8,5,12,4,11), v3 = c(7,8,2,6,9), v4 = c(1,4,6,3,6)) vals - unlist(df1[5, ]) df1[order(vals, decreasing = TRUE)] I hope it helps. Best, Dimitris Dimitris Rizopoulos Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://perswww.kuleuven.be/dimitris_rizopoulos/ Quoting Daniel Wagner [EMAIL PROTECTED]: Dear R experts,  I have a dataframe with 4 columns (variables). I want to redorder (or reposition) these columns on the basis of a value in its last row. e.g.  df1-data.frame( v1= c(2,3,1,9,5), v2=c(8,5,12,4,11), v3=c(7,8,2,6,9), v4=c(1,4,6,3,6))  df1   v1 v2 v3 v4 1 2 8 7 1 2 3 5 8 4 3 1 12 2 6 4 9 4 6 3 5 5 11 9 6 I wanto to get the order of df1 on the basis of value in last row (descending order) like    v2 v3 v4 v1 1 8 7 1 2 2 5 8 4 3 3 12 2 6 1 4 4 6 3 9 5 11 9 6 5  Could somebody help me?  Daniel Amsterdam     Send instant messages to your online friends http://uk.messenger.yahoo.com [[alternative HTML version deleted]] Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Order of columns(variables) in dataframe
On 20-Jul-08 11:32:14, Daniel Wagner wrote: Dear R experts, I have a dataframe withÂ_4 columns (variables). I want to redorder (or reposition) these columns on the basis of a value in its last row. e.g. df1-data.frame( v1= c(2,3,1,9,5), v2=c(8,5,12,4,11), v3=c(7,8,2,6,9), v4=c(1,4,6,3,6)) I wanto to get the order of df1 on the basis of value in last row (descending order) like [Sorry, had to delete your examples because of intrusive special characters, but reproduced below anyway] Could somebody help me? Daniel Amsterdam Try the following. It seems one needs to pass from dataframe to matrix, since sort() does not like lists! Maybe others know better ... df1-data.frame( v1= c(2,3,1,9,5), v2=c(8,5,12,4,11), v3=c(7,8,2,6,9), v4=c(1,4,6,3,6)) df1 # v1 v2 v3 v4 # 1 2 8 7 1 # 2 3 5 8 4 # 3 1 12 2 6 # 4 9 4 6 3 # 5 5 11 9 6 M-as.matrix(df1) L-M[nrow(M),] ix-sort(L,decreasing=TRUE,index.return=TRUE)$ix df2-as.data.frame(M[,ix]) df2 # v2 v3 v4 v1 # 1 8 7 1 2 # 2 5 8 4 3 # 3 12 2 6 1 # 4 4 6 3 9 # 5 11 9 6 5 Hoping this helps, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 20-Jul-08 Time: 12:52:12 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] calculate differences - strange outcome
Em Qui, 2008-07-17 às 11:47 +0200, Kunzler, Andreas escreveu: Dear List, I ran into some trouble by calculating differences. For me it is important that differences are either 0 or not. So I don't understand the outcome of this calculation 865.56-(782.86+0+63.85+18.85+0) [1] -1.136868e-13 I run R version 2.71 on WinXP Hi Andreas Kunzler, Your problem is cause by numeric represntation in computer (Floating-Point numbers) and this topic is explain in R FAQ 7.31 But the solution for your calculations is possible using the guard digits approach, so if you needd solve: 865.56-(782.86+0+63.85+18.85+0) Using 865.56/100 -(782.86+0+63.85+18.85+0)/100 More details in http://docs.sun.com/source/806-3568/ncg_goldberg.html (This links is indicated in R FAQ 7.31) -- Bernardo Rangel Tura, M.D,MPH,Ph.D National Institute of Cardiology Brazil __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error in edit(name,file,title,editor)
Thanks. The file didn't exist at the time I tried edit, and that was my error. David On 20 Jul, 2008, at 12:44, jim holtman wrote: Make sure you are in the right directory; you can also use the full path names. It is basically saying it can not find the file. Do 'list.files()' to see what is addressable in whatever directory you are in. On Sun, Jul 20, 2008 at 7:19 AM, David Epstein [EMAIL PROTECTED] wrote: Can anyone help me with the following attempt to use an external editor from within R vi(file=p286.R) Error in edit(name, file, title, editor) : unable to open file to read edit(file=p286.R) Error in edit(name, file, title, editor) : unable to open file to read I have only recently re-started trying to learn R. (I tried before but failed.) I am working with a Mac MacOsX 10.4.11. I have R 2.7.1 GUI 1.25 (5166) which is the Cocoa version of R. I have been reading Peter Dalgaard's book, which is very helpful, and I have not met any errors so far, so I don't understand why I should suddenly hit this one. Since edit and vi appear to be in the utils package, I did library(utils) to which I got a null response. I have searched the R-help-archives for the error message, and have not found it. I suppose I could use the function source(), but I haven't tried this. Thanks for any help David Epstein __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] drawing segments through points with pch=1
Please excuse me for asking such basic questions: Here is my code y=c(1.21,0.51,0.14,1.62,-0.8,0.72,-1.71,0.84,0.02,-0.12) ybar=mean(y) ll=length(y); ybarv=rep(ybar,ll) x=1:ll plot(x,ybarv,pch=1) segments(x[1],ybar,x[ll],ybar) What I get is a collection of small circles, with a segment on top of the circles, which is almost what I want. But I don't want the segment to be visible inside any small circle. Is there an easy way to arrange for the segment to lie behind the pch=1 markers, as in hidden line removal, so that the circles remain with nothing inside them? I tried putting the segments command first, but then no segment appeared at all. In general, is there a method of laying a drawing on top of another. I tried inserting add=T as an argument to plot, and R objected strongly. Thanks for any help David Epstein __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] drawing segments through points with pch=1
On Sun, Jul 20, 2008 at 1:44 PM, David Epstein [EMAIL PROTECTED] wrote: Please excuse me for asking such basic questions: Here is my code y=c(1.21,0.51,0.14,1.62,-0.8,0.72,-1.71,0.84,0.02,-0.12) ybar=mean(y) ll=length(y); ybarv=rep(ybar,ll) x=1:ll plot(x,ybarv,pch=1) segments(x[1],ybar,x[ll],ybar) What I get is a collection of small circles, with a segment on top of the circles, which is almost what I want. But I don't want the segment to be visible inside any small circle. Is there an easy way to arrange for the segment to lie behind the pch=1 markers, as in hidden line removal, so that the circles remain with nothing inside them? I tried putting the segments command first, but then no segment appeared at all. In general, is there a method of laying a drawing on top of another. I tried inserting add=T as an argument to plot, and R objected strongly. What about replacing segments(x[1],ybar,x[ll],ybar) plot(x,ybarv,pch=1) with plot(x,ybarv,type=b) ? Paul __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Discretize continous variables....
Johannes Huesing wrote: Frank E Harrell Jr [EMAIL PROTECTED] [Sun, Jul 20, 2008 at 12:20:28AM CEST]: Johannes Huesing wrote: Because regulatory bodies demand it? [...] And how anyway does this relate to predictors in a model? Not at all; you're correct. I was mixing the topic of this discussion up with another kind of silliness. I had a discussion with a biometrician in a pharmaceutical company though who stated that when you have only one df to spend it will be better to dichotomise it at a clinically meaningful point than to include it as a linear term. He kept the discussion on the ground of laboratory measurements like sodium, where a deviation from normal ranges is very significant (and unlike, say, cholesterol, where you have a gradual interpretation of the value). He has a point there, but in general the reason for sacrificing information is a mixture of laziness, the preference for presenting data in tables and to keep the modelling consistent with the tables (for instance to assign an odds ratio to each cell). Nice points. I think the desire to be able to present things in tables is a major reason. The biometrician's idea that a piecewise flat line with one jump will fit a dataset better than a linear effect is quite a leap in logic. If I only have one d.f. to spend I'll take linear any day, but better to spend a little more and fit a smooth nonlinear relationship. A coherent approach is to shrink the fit down to the effective number of parameters the dataset will support estimating. There is no clinical laboratory measure that has a jump discontinuity in its effect on mortality or other patient outcomes. The fact that reference ranges exist (which are based only on supposedly normal subjects and don't related to the risk of an outcome) doesn't mean we should use them in formulated independent or dependent variables. It is common but distorted logic to want to make an odds ratio in a model be comparable to one in a table from which regression coefficients were just anti-logged (so that 1-unit changes could be used). The tabled odds ratio is a kind of crude population averaged odds ratio that may not apply to a single subject in the study. My book has many examples where laboratory measurements are related to risk using restricted cubic splines. Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] drawing segments through points with pch=1
On Sun, 20 Jul 2008, Uwe Ligges wrote: You probably want to make your code readable, read ?points and go ahead by making the plot without points (plot(., type=n)), drawing segments and at the end paint points with white background colour in order to overwrite the segments: Except that the background is not necessarily white (and you may want it to be transparent or translucent). It looks to me like lines(type=b) might be what was wanted. y - c(1.21, 0.51, 0.14, 1.62, -0.8, 0.72, -1.71, 0.84, 0.02, -0.12) ybar - mean(y) ll - length(y) ybarv - rep(ybar, ll) x - 1:ll plot(x, ybarv, type=n) segments(x[1], ybar, x[ll], ybar) points(x, ybarv, pch=21, bg=white) Uwe Ligges David Epstein wrote: Please excuse me for asking such basic questions: Here is my code y=c(1.21,0.51,0.14,1.62,-0.8,0.72,-1.71,0.84,0.02,-0.12) ybar=mean(y) ll=length(y); ybarv=rep(ybar,ll) x=1:ll plot(x,ybarv,pch=1) segments(x[1],ybar,x[ll],ybar) What I get is a collection of small circles, with a segment on top of the circles, which is almost what I want. But I don't want the segment to be visible inside any small circle. Is there an easy way to arrange for the segment to lie behind the pch=1 markers, as in hidden line removal, so that the circles remain with nothing inside them? I tried putting the segments command first, but then no segment appeared at all. In general, is there a method of laying a drawing on top of another. I tried inserting add=T as an argument to plot, and R objected strongly. Thanks for any help David Epstein __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] drawing segments through points with pch=1
What I don't like about type=b, also suggested by Paul Smith, is that the segments do not go right up to the little circles---a gap is left, which I don't like. So far, Uwes' solution is what suits me best. However, I understand Brian's objection, though it doesn't apply in my case. The discussion makes me fear that it's a very long road ahead before I can get fine control of R graphics. Thanks David On 20 Jul, 2008, at 14:54, Prof Brian Ripley wrote: On Sun, 20 Jul 2008, Uwe Ligges wrote: You probably want to make your code readable, read ?points and go ahead by making the plot without points (plot(., type=n)), drawing segments and at the end paint points with white background colour in order to overwrite the segments: Except that the background is not necessarily white (and you may want it to be transparent or translucent). It looks to me like lines(type=b) might be what was wanted. y - c(1.21, 0.51, 0.14, 1.62, -0.8, 0.72, -1.71, 0.84, 0.02, -0.12) ybar - mean(y) ll - length(y) ybarv - rep(ybar, ll) x - 1:ll plot(x, ybarv, type=n) segments(x[1], ybar, x[ll], ybar) points(x, ybarv, pch=21, bg=white) Uwe Ligges David Epstein wrote: Please excuse me for asking such basic questions: Here is my code y=c(1.21,0.51,0.14,1.62,-0.8,0.72,-1.71,0.84,0.02,-0.12) ybar=mean(y) ll=length(y); ybarv=rep(ybar,ll) x=1:ll plot(x,ybarv,pch=1) segments(x[1],ybar,x[ll],ybar) What I get is a collection of small circles, with a segment on top of the circles, which is almost what I want. But I don't want the segment to be visible inside any small circle. Is there an easy way to arrange for the segment to lie behind the pch=1 markers, as in hidden line removal, so that the circles remain with nothing inside them? I tried putting the segments command first, but then no segment appeared at all. In general, is there a method of laying a drawing on top of another. I tried inserting add=T as an argument to plot, and R objected strongly. Thanks for any help David Epstein __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R interprets symbol as q()
Roger Leenders r.t.a.j.leenders at rug.nl writes: After the data for the final respondent, the file contains a , followed by some zeroes. R now needs to read the data until the symbol. The symbol is not readable in my viewer, but try the following: put it into the clipboard, and paste it into the GUI (assuming Windows). In an earlier version, I had a similar case with Chinese characters that closed the application when pasted. The error was corrected within hours by B.Ripley. Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] smooth.spline
Are you aware that there are many different kinds of splines? With spline and splinefun, you can use method = fmm (Forsyth, Malcolm and Moler), natural, or periodic. I'm not familiar with fmm, but it seems to be adequately explained by the Manual spline evaluation you quoted from the documentation. Natural splines are perhaps the simplest: I(x-x0)*(x-x0)^j, where x0 is a knot, and I(z) = 1 if z0 and 0 otherwise. However, computations using natural splines are numerically unstable. The standard solution to this problem is to use B-splines, which are 0 outside a finite interval. Let's look at your example: n - 9 x - 1:n y - rnorm(n) plot(x, y, main = paste(spline[fun](.) through, n, points)) spl - smooth.spline(x,y) lines(spl) The 'smooth.spline' function uses B-splines. To see what they look like, let's do the following: library(fda) Bspl.basis - create.bspline.basis(unique(spl$fit$knot)) # Check to make sure: all.equal(knots(Bspl.basis, interior=FALSE), spl$fit$knot) # TRUE # What do B-splines look like? plot(Bspl.basis) abline(v=knots(Bspl.basis), lty='dotted', col='red') # 7 interior knots, 2 end knots replicated 4 times each, for a spline of order 4, degree 3 (cubic splines) # total of 15 knots # Each spline uses 5 consecutive knots, which means there will be 11 basis functions. # NOTE: 'smooth.spline' rescaled the interval [1, 9] to [0, 1]. # Evaluate the 11 B-splines at 'x' Bspl.basis.x - eval.basis((x-1)/8, Bspl.basis) round(Bspl.basis.x, 4) # Now the manual computation: y.spl - Bspl.basis.x %*% spl$fit$coef # Plot to confirm: plot(x, y, main = paste(spline[fun](.) through, n, points)) spl.xy - spline(x, y) lines(spl.xy) points(x, y.spl, pch=2, col='red') Hope this helps. Spencer [EMAIL PROTECTED] wrote: Fair enough. FOr a spline interpolation I can do the following: n - 9 x - 1:n y - rnorm(n) plot(x, y, main = paste(spline[fun](.) through, n, points)) lines(spline(x, y)) Then look at the coefficients generated as: f - splinefun(x, y) ls(envir = environment(f)) [1] ties ux z splinecoef - get(z, envir = environment(f)) slinecoef $method [1] 3 $n [1] 9 $x [1] 1 2 3 4 5 6 7 8 9 $y [1] 0.93571604 0.44240485 0.45451903 -0.96207396 -1.13246522 -0.60032698 [7] -1.77506105 -0.09171419 -0.23262573 $b [1] -1.53673409 0.22775629 -0.81788209 -1.16966436 0.73558677 -0.68744178 [7] 0.08639287 1.86770869 -2.92992167 $c [1] 1.3657783 0.3987121 -1.4443504 1.0925682 0.8126830 -2.2357115 3.0095462 [8] -1.2282303 -3.5694000 $d [1] -0.32235542 -0.61435416 0.84563953 -0.09329507 -1.01613149 1.74841922 [7] -1.41259217 -0.78038989 -0.78038989 WHen I look at ?spline there is even an example of manually using these coefficeients: ## Manual spline evaluation --- demo the coefficients : .x - get(ux, envir = environment(f)) u - seq(3,6, by = 0.25) (ii - findInterval(u, .x)) dx - u - .x[ii] f.u - with(splinecoef, y[ii] + dx*(b[ii] + dx*(c[ii] + dx* d[ii]))) stopifnot(all.equal(f(u), f.u)) For the smooth.spline as spl - smooth.spline(x,y) I can also look at the coefficients: spl$fit $knot [1] 0.000 0.000 0.000 0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000 [13] 1.000 1.000 1.000 $nk [1] 11 $min [1] 1 $range [1] 8 $coef [1] 0.90345898 0.73823276 0.40777431 -0.08046715 -0.54625461 -0.85205147 [7] -0.96233408 -0.91373830 -0.66529714 -0.47674774 -0.38246971 attr(,class) [1] smooth.spline.fit But there isn't an example on how to manual use these coefficients. This is what I was asking about. Once I hae the coefficients how do I manually interpolate using the coefficients given and x. Thank you. Kevin Spencer Graves [EMAIL PROTECTED] wrote: PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. I do NOT know how to do what you want, but with a self-contained example, I suspect many people on this list -- probably including me -- could easily solve the problem. Without such an example, there is a high probability that any answer might (a) not respond to your need, and (b) take more time to develop, just because we don't know enough of what you are asking. Spencer [EMAIL PROTECTED] wrote: Like I indicated. I understand the coefficients in a B-spline context. If I use the the 'spline' or 'splinefun' I can get the coefficients and they are grouped as 'a', 'b', 'c', and 'd' coefficients. But the coefficients for smooth.spline is just an array. I basically want to take these coefficients and outside of 'R' use them to form an interpolation. In other words I want 'R' to do the hard work and then export the results so they can be used else where. Thank you. Kevin Spencer Graves wrote: I believe that a short answer to your question is that the smooth is a linear combination of B-spline basis
Re: [R] principal factor analysis
Jinsong Zhao jszhao at mail.hzau.edu.cn writes: Hi, Is there a function to do principal factor analysis in R? Do a 'RSiteSearch(factor analysis)' and you will find several packages for Factor Analysis, such as: FAiRFactor Analysis in R FactoMineR Factor Analysis and Data Mining with R or factor.pa(), Principal Axis Factor Analysis, in package 'psych'. It seems, 'common factors' is a more colloquial term used in PCA books, or did you mean to compare 'principal components' with 'principal factors'? I am reading through ``A user's guide to principal components'' by J. E. Jackson. In Table 17.1, a comparison between principal components and common factors, and there is obvious difference between them. However, I don't know how to get the common factors. Any suggestions? Thanks! Jinsong __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] smooth.spline
On 20/07/2008 11:11 AM, Spencer Graves wrote: Are you aware that there are many different kinds of splines? With spline and splinefun, you can use method = fmm (Forsyth, Malcolm and Moler), natural, or periodic. I'm not familiar with fmm, but it seems to be adequately explained by the Manual spline evaluation you quoted from the documentation. Natural splines are perhaps the simplest: I(x-x0)*(x-x0)^j, where x0 is a knot, and I(z) = 1 if z0 and 0 otherwise. That's not what R means by natural spline in this context. Here it means that the function becomes linear outside the range of the knots. I would call the I(x-x0)*(x-x0)^j splines the truncated power basis for polynomial splines; B-splines are a different basis for the same set of splines (assuming the knots and degrees match). Natural splines are a subspace of these (since linear functions are a subspace of polynomials). I don't know of a simple basis for them. Duncan Murdoch However, computations using natural splines are numerically unstable. The standard solution to this problem is to use B-splines, which are 0 outside a finite interval. Let's look at your example: n - 9 x - 1:n y - rnorm(n) plot(x, y, main = paste(spline[fun](.) through, n, points)) spl - smooth.spline(x,y) lines(spl) The 'smooth.spline' function uses B-splines. To see what they look like, let's do the following: library(fda) Bspl.basis - create.bspline.basis(unique(spl$fit$knot)) # Check to make sure: all.equal(knots(Bspl.basis, interior=FALSE), spl$fit$knot) # TRUE # What do B-splines look like? plot(Bspl.basis) abline(v=knots(Bspl.basis), lty='dotted', col='red') # 7 interior knots, 2 end knots replicated 4 times each, for a spline of order 4, degree 3 (cubic splines) # total of 15 knots # Each spline uses 5 consecutive knots, which means there will be 11 basis functions. # NOTE: 'smooth.spline' rescaled the interval [1, 9] to [0, 1]. # Evaluate the 11 B-splines at 'x' Bspl.basis.x - eval.basis((x-1)/8, Bspl.basis) round(Bspl.basis.x, 4) # Now the manual computation: y.spl - Bspl.basis.x %*% spl$fit$coef # Plot to confirm: plot(x, y, main = paste(spline[fun](.) through, n, points)) spl.xy - spline(x, y) lines(spl.xy) points(x, y.spl, pch=2, col='red') Hope this helps. Spencer [EMAIL PROTECTED] wrote: Fair enough. FOr a spline interpolation I can do the following: n - 9 x - 1:n y - rnorm(n) plot(x, y, main = paste(spline[fun](.) through, n, points)) lines(spline(x, y)) Then look at the coefficients generated as: f - splinefun(x, y) ls(envir = environment(f)) [1] ties ux z splinecoef - get(z, envir = environment(f)) slinecoef $method [1] 3 $n [1] 9 $x [1] 1 2 3 4 5 6 7 8 9 $y [1] 0.93571604 0.44240485 0.45451903 -0.96207396 -1.13246522 -0.60032698 [7] -1.77506105 -0.09171419 -0.23262573 $b [1] -1.53673409 0.22775629 -0.81788209 -1.16966436 0.73558677 -0.68744178 [7] 0.08639287 1.86770869 -2.92992167 $c [1] 1.3657783 0.3987121 -1.4443504 1.0925682 0.8126830 -2.2357115 3.0095462 [8] -1.2282303 -3.5694000 $d [1] -0.32235542 -0.61435416 0.84563953 -0.09329507 -1.01613149 1.74841922 [7] -1.41259217 -0.78038989 -0.78038989 WHen I look at ?spline there is even an example of manually using these coefficeients: ## Manual spline evaluation --- demo the coefficients : .x - get(ux, envir = environment(f)) u - seq(3,6, by = 0.25) (ii - findInterval(u, .x)) dx - u - .x[ii] f.u - with(splinecoef, y[ii] + dx*(b[ii] + dx*(c[ii] + dx* d[ii]))) stopifnot(all.equal(f(u), f.u)) For the smooth.spline as spl - smooth.spline(x,y) I can also look at the coefficients: spl$fit $knot [1] 0.000 0.000 0.000 0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000 [13] 1.000 1.000 1.000 $nk [1] 11 $min [1] 1 $range [1] 8 $coef [1] 0.90345898 0.73823276 0.40777431 -0.08046715 -0.54625461 -0.85205147 [7] -0.96233408 -0.91373830 -0.66529714 -0.47674774 -0.38246971 attr(,class) [1] smooth.spline.fit But there isn't an example on how to manual use these coefficients. This is what I was asking about. Once I hae the coefficients how do I manually interpolate using the coefficients given and x. Thank you. Kevin Spencer Graves [EMAIL PROTECTED] wrote: PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. I do NOT know how to do what you want, but with a self-contained example, I suspect many people on this list -- probably including me -- could easily solve the problem. Without such an example, there is a high probability that any answer might (a) not respond to your need, and (b) take more time to develop, just because we don't know enough of what you are asking. Spencer [EMAIL PROTECTED] wrote: Like I indicated. I understand the coefficients in a B-spline context. If I use the
[R] Access to values of function arguments
Does anyone know of good reading material about the following? The R language definition does not appear to explicitly address my problem (maybe I misread that document?) I have a function definition: func(a) cat(Anova for variable ,a) What I wish to achieve is to call func with a value such as: func(Age) and then obtain: Anova for variable Age Using names(formals()) inside function func yields a. That is not what I need. I need the name contained in a, which in this case is Age. Thanks for your time. Willemf -- View this message in context: http://www.nabble.com/Access-to-values-of-function-arguments-tp18554602p18554602.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How do I install Joomla! 1.5?
Installing of Joomla! 1.5 is pretty easy. We assume you have set-up your Web site, and it is accessible with your browser. Download Joomla! 1.5, unzip it and upload/copy the files into the directory you Web site points to, fire up your browser and enter your Web site address and the installation will start. For full details on the installation processes check out the Installation Manual on the Joomla! Help Site where you will also find download instructions for a PDF version too. http://www.thecodeone.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] smooth.spline
in line Duncan Murdoch wrote: On 20/07/2008 11:11 AM, Spencer Graves wrote: Are you aware that there are many different kinds of splines? With spline and splinefun, you can use method = fmm (Forsyth, Malcolm and Moler), natural, or periodic. I'm not familiar with fmm, but it seems to be adequately explained by the Manual spline evaluation you quoted from the documentation. Natural splines are perhaps the simplest: I(x-x0)*(x-x0)^j, where x0 is a knot, and I(z) = 1 if z0 and 0 otherwise. That's not what R means by natural spline in this context. Here it means that the function becomes linear outside the range of the knots. I would call the I(x-x0)*(x-x0)^j splines the truncated power basis for polynomial splines; B-splines are a different basis for the same set of splines (assuming the knots and degrees match). Natural splines are a subspace of these (since linear functions are a subspace of polynomials). I don't know of a simple basis for them. Thanks for the correction. I erred by writing this from memory. Dierckx (1993, p. 4) says, A natural spline function is a spline of odd degree k = 2*m-1 (m=2) which satisfies the additional constraints (D^(m+j))s(a) = (D^(m+j))s(b) = 0, j = 0, 1, ..., m-2. He further (p. 5) defines the truncated power functions, which is what I mistakenly called natural splines. Thanks again for the correction. Spencer Duncan Murdoch However, computations using natural splines are numerically unstable. The standard solution to this problem is to use B-splines, which are 0 outside a finite interval. Let's look at your example: n - 9 x - 1:n y - rnorm(n) plot(x, y, main = paste(spline[fun](.) through, n, points)) spl - smooth.spline(x,y) lines(spl) The 'smooth.spline' function uses B-splines. To see what they look like, let's do the following: library(fda) Bspl.basis - create.bspline.basis(unique(spl$fit$knot)) # Check to make sure: all.equal(knots(Bspl.basis, interior=FALSE), spl$fit$knot) # TRUE # What do B-splines look like? plot(Bspl.basis) abline(v=knots(Bspl.basis), lty='dotted', col='red') # 7 interior knots, 2 end knots replicated 4 times each, for a spline of order 4, degree 3 (cubic splines) # total of 15 knots # Each spline uses 5 consecutive knots, which means there will be 11 basis functions. # NOTE: 'smooth.spline' rescaled the interval [1, 9] to [0, 1]. # Evaluate the 11 B-splines at 'x' Bspl.basis.x - eval.basis((x-1)/8, Bspl.basis) round(Bspl.basis.x, 4) # Now the manual computation: y.spl - Bspl.basis.x %*% spl$fit$coef # Plot to confirm: plot(x, y, main = paste(spline[fun](.) through, n, points)) spl.xy - spline(x, y) lines(spl.xy) points(x, y.spl, pch=2, col='red') Hope this helps. Spencer [EMAIL PROTECTED] wrote: Fair enough. FOr a spline interpolation I can do the following: n - 9 x - 1:n y - rnorm(n) plot(x, y, main = paste(spline[fun](.) through, n, points)) lines(spline(x, y)) Then look at the coefficients generated as: f - splinefun(x, y) ls(envir = environment(f)) [1] ties ux z splinecoef - get(z, envir = environment(f)) slinecoef $method [1] 3 $n [1] 9 $x [1] 1 2 3 4 5 6 7 8 9 $y [1] 0.93571604 0.44240485 0.45451903 -0.96207396 -1.13246522 -0.60032698 [7] -1.77506105 -0.09171419 -0.23262573 $b [1] -1.53673409 0.22775629 -0.81788209 -1.16966436 0.73558677 -0.68744178 [7] 0.08639287 1.86770869 -2.92992167 $c [1] 1.3657783 0.3987121 -1.4443504 1.0925682 0.8126830 -2.2357115 3.0095462 [8] -1.2282303 -3.5694000 $d [1] -0.32235542 -0.61435416 0.84563953 -0.09329507 -1.01613149 1.74841922 [7] -1.41259217 -0.78038989 -0.78038989 WHen I look at ?spline there is even an example of manually using these coefficeients: ## Manual spline evaluation --- demo the coefficients : .x - get(ux, envir = environment(f)) u - seq(3,6, by = 0.25) (ii - findInterval(u, .x)) dx - u - .x[ii] f.u - with(splinecoef, y[ii] + dx*(b[ii] + dx*(c[ii] + dx* d[ii]))) stopifnot(all.equal(f(u), f.u)) For the smooth.spline as spl - smooth.spline(x,y) I can also look at the coefficients: spl$fit $knot [1] 0.000 0.000 0.000 0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000 [13] 1.000 1.000 1.000 $nk [1] 11 $min [1] 1 $range [1] 8 $coef [1] 0.90345898 0.73823276 0.40777431 -0.08046715 -0.54625461 -0.85205147 [7] -0.96233408 -0.91373830 -0.66529714 -0.47674774 -0.38246971 attr(,class) [1] smooth.spline.fit But there isn't an example on how to manual use these coefficients. This is what I was asking about. Once I hae the coefficients how do I manually interpolate using the coefficients given and x. Thank you. Kevin Spencer Graves [EMAIL PROTECTED] wrote: PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. I do NOT know how to do what you want, but
Re: [R] Access to values of function arguments
See ?deparse . E.g. (with several stylistic improvements) func - function(a) cat(sprintf(Anova for variable %s\n, sQuote(deparse(substitute(a) func(Age) Note that you expliciltly do not want the *vaiue* of 'a', but the symbol passed. That is non-standard behaviour, and you might want to consider why func(Age) is not a better way to do this. On Sun, 20 Jul 2008, willemf wrote: Does anyone know of good reading material about the following? The R language definition does not appear to explicitly address my problem (maybe I misread that document?) 'S Programming' (see the R FAQ). I have a function definition: Well, that is not a function definition: please do use correct and reproducible examples as the posting guide asks. func(a) cat(Anova for variable ,a) What I wish to achieve is to call func with a value such as: func(Age) and then obtain: Anova for variable Age Using names(formals()) inside function func yields a. That is not what I need. I need the name contained in a, which in this case is Age. Thanks for your time. Willemf -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Access to values of function arguments
on 07/20/2008 08:42 AM willemf wrote:
Does anyone know of good reading material about the following? The R language
definition does not appear to explicitly address my problem (maybe I misread
that document?)
I have a function definition:
func(a)
cat(Anova for variable ,a)
What I wish to achieve is to call func with a value such as:
func(Age)
and then obtain:
Anova for variable Age
Using names(formals()) inside function func yields a. That is not what I
need. I need the name contained in a, which in this case is Age.
Thanks for your time.
Willemf
MyFun - function(a) {
cat(Anova for variable, deparse(substitute(a)), \n)}
Age - 50:60
MyFun(Age)
Anova for variable Age
The deparse(substitute(...)) idiom is what you are looking for.
See ?deparse and ?substitute
HTH,
Marc Schwartz
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[R] Indicator Kriging?
Hello All! I like to do some indicator kriging with R. So far I used geoR for simple and ordinary kriging. Does anybody know which package I should use? Thanks for your help! Sascha! -- Jetzt dabei sein: http://www.shortview.de/[EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] fill in area between 2 lines with a color
Hi - I'd like to have the area between 2 lines on a x-y plot be filled with grey, but I haven't had any luck using polygon or rect. (In reality, I'd like to do this for twice - once for a low group and once for a high group - and then I'd like to plot a set of data points for a 'normal' group together with these 2 grey areas.) Here's a simple example of the 2 lines: age=1:10 y.low=rnorm(length(age),150,25)+10*age y.high=rnorm(length(age),250,25)+10*age plot(age,y.high,type='n',ylim=c(100,400),ylab='Y Range',xlab='Age (years)') lines(age,y.low,col='grey') lines(age,y.high,col='grey') Is it possible to fill the area between the 2 lines filled with, for example, 'grey30' ? thanks very much in advance, David Freedman - David Freedman Atlanta -- View this message in context: http://www.nabble.com/fill-in-area-between-2-lines-with-a-color-tp18556096p18556096.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Conditionally Updating Lattice Plots
Hi All...
I can¹t seem to find an answer to this in the help pages, archives, or
Deepayan¹s Lattice Book.
I want to do a Lattice plot, and then update it, possibly more than once,
depending upon some logical options. Code below; it produces a second plot
page when the second update is called, from which I would infer that you
can't update the update or I'm not calling it correctly. I have a nagging
sense too that the real way to do this is with a non-standard use of
panel.superpose but I don't quite see how to do that from available
examples.
TIF for any suggestions, Bryan
Example: a function then, the call to the function
fancy.lm - function(x, y, fit = TRUE, resid = TRUE){
model - lm(y ~ x)
y.pred - predict(model) # Compute residuals for plotting
res.x - as.vector(rbind(x, x, rep(NA,length(x # NAs induce breaks in
line
res.y - as.vector(rbind(y, y.pred, rep(NA,length(x # after Fig 5.1 of
DAAG (clever!)
p - xyplot(y ~ x, pch = 20,
panel = function(...) {
panel.xyplot(...) # not strictly necessary if I understand correctly
})
plot(p, more = TRUE)
if (fit) {
plot(update(p, more = TRUE,
panel = function(...){
panel.xyplot(...)
panel.abline(model, col = red)
}))}
if (resid) {
plot(update(p, more = TRUE,
panel = function(...){
panel.xyplot(res.x, res.y, col = lightblue, type = l)
}))}
}
x - jitter(c(1:10), factor = 5)
y - jitter(c(1:10), factor = 10)
fancy.lm(x, y, fit = TRUE, resid = TRUE)
Session Info
sessionInfo()
R version 2.7.1 (2008-06-23)
i386-apple-darwin8.10.1
locale:
en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] datasets grid grDevices graphics stats utils methods
[8] base
other attached packages:
[1] fastICA_1.1-9 DescribeDisplay_0.1.3 ggplot_0.4.2
[4] RColorBrewer_1.0-2reshape_0.8.0 MASS_7.2-42
[7] pcaPP_1.5 mvtnorm_0.9-0 hints_1.0.1-1
[10] mvoutlier_1.3 robustbase_0.2-8 lattice_0.17-8
[13] rggobi_2.1.9 RGtk2_2.12.5-3
loaded via a namespace (and not attached):
[1] tools_2.7.1
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Erro updating HTML package descriptions in packages.html
Dear all, I have just installed the new version of R 2.7.1 and when i first instaled some packages it asked me to create a personal directory. It installed the packages, but not the html help page (packages.html) The downloaded packages are in C:\Users\jamaj\AppData\Local\Temp\Rtmp4MTuXN\downloaded_packages updating HTML package descriptions Warning message: In file.create(f.tg) : cannot create file 'C:\PROGRA~1\R\R-27~1.1/doc/html/packages.html', reason 'Permission denied' How can i fix the packages.html file to reflect the packages' instalation? Thanks in advance. José Augusto M. de Andrade Jr. (jamaj) Quant. Finance Researcher FEA-RP - University of São Paulo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] fill in area between 2 lines with a color
try the following: age - 1:10 y.low - rnorm(length(age), 150, 25) + 10*age y.high - rnorm(length(age), 250, 25) + 10*age plot(age,y.high,type = 'n', ylim = c(100, 400), ylab = 'Y Range', xlab = 'Age (years)') lines(age, y.low, col = 'grey') lines(age, y.high, col = 'grey') polygon(c(age, rev(age)), c(y.high, rev(y.low)), col = grey30, border = NA) I hope it helps. Best, Dimitris Dimitris Rizopoulos Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://perswww.kuleuven.be/dimitris_rizopoulos/ Quoting David Freedman [EMAIL PROTECTED]: Hi - I'd like to have the area between 2 lines on a x-y plot be filled with grey, but I haven't had any luck using polygon or rect. (In reality, I'd like to do this for twice - once for a low group and once for a high group - and then I'd like to plot a set of data points for a 'normal' group together with these 2 grey areas.) Here's a simple example of the 2 lines: age=1:10 y.low=rnorm(length(age),150,25)+10*age y.high=rnorm(length(age),250,25)+10*age plot(age,y.high,type='n',ylim=c(100,400),ylab='Y Range',xlab='Age (years)') lines(age,y.low,col='grey') lines(age,y.high,col='grey') Is it possible to fill the area between the 2 lines filled with, for example, 'grey30' ? thanks very much in advance, David Freedman - David Freedman Atlanta -- View this message in context: http://www.nabble.com/fill-in-area-between-2-lines-with-a-color-tp18556096p18556096.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] fill in area between 2 lines with a color
on 07/20/2008 11:34 AM David Freedman wrote: Hi - I'd like to have the area between 2 lines on a x-y plot be filled with grey, but I haven't had any luck using polygon or rect. (In reality, I'd like to do this for twice - once for a low group and once for a high group - and then I'd like to plot a set of data points for a 'normal' group together with these 2 grey areas.) Here's a simple example of the 2 lines: age=1:10 y.low=rnorm(length(age),150,25)+10*age y.high=rnorm(length(age),250,25)+10*age plot(age,y.high,type='n',ylim=c(100,400),ylab='Y Range',xlab='Age (years)') lines(age,y.low,col='grey') lines(age,y.high,col='grey') Is it possible to fill the area between the 2 lines filled with, for example, 'grey30' ? thanks very much in advance, David Freedman Try this: age=1:10 y.low=rnorm(length(age),150,25)+10*age y.high=rnorm(length(age),250,25)+10*age plot(age,y.high,type='n',ylim=c(100,400),ylab='Y Range', xlab='Age (years)') lines(age,y.low,col='grey') lines(age,y.high,col='grey') polygon(c(age, rev(age), age[1]), c(y.low, rev(y.high), y.low[1]), col = grey30) What you essentially need to do is to 'walk' the boundary of the polygon, be sure to 'close' the open end by returning to the starting point, and then fill it. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Conditionally Updating Lattice Plots
Try this:
library(lattice)
fancy.lm - function(x, y, fit = TRUE, resid = TRUE){
model - lm(y ~ x)
y.pred - predict(model) # Compute residuals for plotting
res.x - as.vector(rbind(x, x, rep(NA,length(x # NAs induce breaks
in line
res.y - as.vector(rbind(y, y.pred, rep(NA,length(x # after Fig
5.1 of DAAG (clever!)
plot(xyplot(y ~ x, pch = 20))
trellis.focus()
if (fit) panel.abline(model, col = red)
if (resid) panel.xyplot(res.x, res.y, col = lightblue, type = l)
trellis.unfocus()
}
x - jitter(c(1:10), factor = 5)
y - jitter(c(1:10), factor = 10)
fancy.lm(x, y, fit = TRUE, resid = TRUE)
On Sun, Jul 20, 2008 at 12:44 PM, Bryan Hanson [EMAIL PROTECTED] wrote:
Hi All...
I can¹t seem to find an answer to this in the help pages, archives, or
Deepayan¹s Lattice Book.
I want to do a Lattice plot, and then update it, possibly more than once,
depending upon some logical options. Code below; it produces a second plot
page when the second update is called, from which I would infer that you
can't update the update or I'm not calling it correctly. I have a nagging
sense too that the real way to do this is with a non-standard use of
panel.superpose but I don't quite see how to do that from available
examples.
TIF for any suggestions, Bryan
Example: a function then, the call to the function
fancy.lm - function(x, y, fit = TRUE, resid = TRUE){
model - lm(y ~ x)
y.pred - predict(model) # Compute residuals for plotting
res.x - as.vector(rbind(x, x, rep(NA,length(x # NAs induce breaks in
line
res.y - as.vector(rbind(y, y.pred, rep(NA,length(x # after Fig 5.1 of
DAAG (clever!)
p - xyplot(y ~ x, pch = 20,
panel = function(...) {
panel.xyplot(...) # not strictly necessary if I understand correctly
})
plot(p, more = TRUE)
if (fit) {
plot(update(p, more = TRUE,
panel = function(...){
panel.xyplot(...)
panel.abline(model, col = red)
}))}
if (resid) {
plot(update(p, more = TRUE,
panel = function(...){
panel.xyplot(res.x, res.y, col = lightblue, type = l)
}))}
}
x - jitter(c(1:10), factor = 5)
y - jitter(c(1:10), factor = 10)
fancy.lm(x, y, fit = TRUE, resid = TRUE)
Session Info
sessionInfo()
R version 2.7.1 (2008-06-23)
i386-apple-darwin8.10.1
locale:
en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] datasets grid grDevices graphics stats utils methods
[8] base
other attached packages:
[1] fastICA_1.1-9 DescribeDisplay_0.1.3 ggplot_0.4.2
[4] RColorBrewer_1.0-2reshape_0.8.0 MASS_7.2-42
[7] pcaPP_1.5 mvtnorm_0.9-0 hints_1.0.1-1
[10] mvoutlier_1.3 robustbase_0.2-8 lattice_0.17-8
[13] rggobi_2.1.9 RGtk2_2.12.5-3
loaded via a namespace (and not attached):
[1] tools_2.7.1
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Erro updating HTML package descriptions in packages.html
José Augusto Jr. wrote: Dear all, I have just installed the new version of R 2.7.1 and when i first instaled some packages it asked me to create a personal directory. It installed the packages, but not the html help page (packages.html) The downloaded packages are in C:\Users\jamaj\AppData\Local\Temp\Rtmp4MTuXN\downloaded_packages updating HTML package descriptions Warning message: In file.create(f.tg) : cannot create file 'C:\PROGRA~1\R\R-27~1.1/doc/html/packages.html', reason 'Permission denied' How can i fix the packages.html file to reflect the packages' instalation? If you need to, run R with privileges that permit write access to that file... Uwe Ligges Thanks in advance. José Augusto M. de Andrade Jr. (jamaj) Quant. Finance Researcher FEA-RP - University of São Paulo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Erro updating HTML package descriptions in packages.html
Read the VISTA HOW TO section of this page: http://batchfiles.googlecode.com On Sun, Jul 20, 2008 at 12:48 PM, José Augusto Jr. [EMAIL PROTECTED] wrote: Dear all, I have just installed the new version of R 2.7.1 and when i first instaled some packages it asked me to create a personal directory. It installed the packages, but not the html help page (packages.html) The downloaded packages are in C:\Users\jamaj\AppData\Local\Temp\Rtmp4MTuXN\downloaded_packages updating HTML package descriptions Warning message: In file.create(f.tg) : cannot create file 'C:\PROGRA~1\R\R-27~1.1/doc/html/packages.html', reason 'Permission denied' How can i fix the packages.html file to reflect the packages' instalation? Thanks in advance. José Augusto M. de Andrade Jr. (jamaj) Quant. Finance Researcher FEA-RP - University of São Paulo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Erro updating HTML package descriptions in packages.html
Dear all, Unfortunately, i´m running as administrator. So, i cannot run R in an upper level than this one i am already running. jamaj 2008/7/20 Gabor Grothendieck [EMAIL PROTECTED]: Read the VISTA HOW TO section of this page: http://batchfiles.googlecode.com On Sun, Jul 20, 2008 at 12:48 PM, José Augusto Jr. [EMAIL PROTECTED] wrote: Dear all, I have just installed the new version of R 2.7.1 and when i first instaled some packages it asked me to create a personal directory. It installed the packages, but not the html help page (packages.html) The downloaded packages are in C:\Users\jamaj\AppData\Local\Temp\Rtmp4MTuXN\downloaded_packages updating HTML package descriptions Warning message: In file.create(f.tg) : cannot create file 'C:\PROGRA~1\R\R-27~1.1/doc/html/packages.html', reason 'Permission denied' How can i fix the packages.html file to reflect the packages' instalation? Thanks in advance. José Augusto M. de Andrade Jr. (jamaj) Quant. Finance Researcher FEA-RP - University of São Paulo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matching Up Values
Hmm, I'm having a fair few difficulties using 'merge' now. I managed to get it to work successfully before, but in this case I'm trying to shorten (as oppose to lengthen as before) a file in relation to a 'master' file. These are the commands I've been using, followed by the dimensions of the files in question - as you can see, the row numbers of the merged file don't correlate to that of the 'coordinates' file (which is what I'm aiming to get 'merged' equal to): merge(PopDens.long, coordinates, by=c(Latitude,Longitude), all = TRUE) - merged dim(PopDens.long); dim(coordinates); dim(merged) [1] 67870 3 [1] 67420 2 [1] 69849 3 One thing I tried was swapping the order of the files in the merge command, but this causes 'merged' to have the same number of rows (69849). Something else I tried was to leave out the 'all = TRUE' command, as I'm essentially attempting the shorten the file, but this makes the output file *too* short! (65441 as opposed to the intended 67420). Again, the same applies when the order of the input files are swapped. merge(PopDens.long, coordinates, by=c(Latitude,Longitude)) - merged dim(PopDens.long); dim(coordinates); dim(merged) [1] 67870 3 [1] 67420 2 [1] 65441 3 Am I doing something obviously wrong? I'm pretty certain that 'coordinates' is a subset of 'PopDens.long' - so there should be equal numbers of common values when merged. Is there perhaps a more suitable function I could use, or a way of performing checks to see where I might be going wrong?! Many thanks, Steve _ 100’s of Nikon cameras to be won with Live Search __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] asp and ylim
#See David Williams' book Weighing the odds, p286 y - c(1.21, 0.51, 0.14, 1.62, -0.8, 0.72, -1.71, 0.84, 0.02, -0.12) ybar - mean(y) ylength - length(y) ybarv - rep(ybar, ylength) x - 1:ylength plot(x,y,asp=1,xlab=position,ylab=ybar,type=n,ylim=c(-1,1)) segments(x[1], ybar, x[ylength], ybar) segments(x,ybarv,x,y) points(x, ybarv, pch=21, bg=white) points(x,y,pch=19,col=black) With asp=1, the value of ylim seems to be totally ignored, as in the above code. With asp not set, R plays close attention to the value of ylim. This is not intuitive behaviour, or is it? How can I set the aspect ratio, and simultaneously set the plot region? The aspect ratio is one number and the plot region is given by four numbers (xleft, xright, yleft, yright). Logically, these 5 numbers are independent of each other and arbitrary, provided xleftxright and yleftyright. This should give a one-to-one bijection between 5-tuples and plots, determined up to a change of scale that is uniform in the x- and y-dirctions. My code above shows the (to me) obvious attempt, which fails. Thanks David __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Erro updating HTML package descriptions in packages.html
Just in case you are mistaken try the suggestion given already. If you determine that that does not work (and are not just guessing that it won't work) then in Windows Explorer, right click packages.html Properties Security and set the security levels appropriately. Not sure but you may have to do the same thing on its folder as well. On Sun, Jul 20, 2008 at 1:26 PM, José Augusto Jr. [EMAIL PROTECTED] wrote: Dear all, Unfortunately, i´m running as administrator. So, i cannot run R in an upper level than this one i am already running. jamaj 2008/7/20 Gabor Grothendieck [EMAIL PROTECTED]: Read the VISTA HOW TO section of this page: http://batchfiles.googlecode.com On Sun, Jul 20, 2008 at 12:48 PM, José Augusto Jr. [EMAIL PROTECTED] wrote: Dear all, I have just installed the new version of R 2.7.1 and when i first instaled some packages it asked me to create a personal directory. It installed the packages, but not the html help page (packages.html) The downloaded packages are in C:\Users\jamaj\AppData\Local\Temp\Rtmp4MTuXN\downloaded_packages updating HTML package descriptions Warning message: In file.create(f.tg) : cannot create file 'C:\PROGRA~1\R\R-27~1.1/doc/html/packages.html', reason 'Permission denied' How can i fix the packages.html file to reflect the packages' instalation? Thanks in advance. José Augusto M. de Andrade Jr. (jamaj) Quant. Finance Researcher FEA-RP - University of São Paulo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Off topic: SS formulae for 3-way repeated measure anova (for when aov() fails)
Pursuant to a prior on topic thread (http://tolstoy.newcastle.edu.au/R/e4/help/08/07/17192.html ) where I found I could not use AOV to perform an anova on my large data set, I'm now trying to code the analysis by hand so to speak. However, as demonstrated below, when comparing my attempt to aov() using a smaller data set, I seem to betray some sort of misunderstanding when I try to compute SSerr for the first interaction. Obviously I have missed something and although I have looked around for the explicit SSerr formulas for this design (my work thus far was extrapolated from understanding of a 2-way design), I can't seem to find any. Any help would be much obliged. N = 20 levs.a = 2 levs.b = 2 levs.d = 10 temp.sub = factor(1:N) temp.a = factor(1:levs.a) temp.b = factor(1:levs.b) temp.d = factor(1:levs.d) temp = expand.grid(sub=temp.sub, a=temp.a, b=temp.b, d=temp.d) temp$x = rnorm(length(temp[, 1])) #generate some random DV data sub=temp$sub a=temp$a b=temp$b d=temp$d x=temp$x this_aov = aov( x~a*b*d+Error(sub/(a*b*d)) ) summary(this_aov) #now let's try by hand, checking each sum-of-squares # ss against the analogous aov() produced ss (rounding # each to avoid small computational differences) #Get ss.subs sub.means = aggregate(x,list(sub=sub), mean) grand.mean = mean(sub.means$x) ss.total = sum((x-grand.mean)^2) ss.subs = levs.a*levs.b*levs.d*sum((sub.means$x-grand.mean)^2) round(ss.subs, 10)==round(summary(this_aov)[[1]][[1]]$Sum, 10) #Get ss.a a.means = aggregate(x, list(a=a), mean) ss.a = N*levs.b*levs.d*sum((a.means$x-grand.mean)^2) round(ss.a, 10)==round(summary(this_aov)[[2]][[1]]$Sum[1], 10) #ok! #Get ss.a.error a.cells = aggregate(x, list(a=a, sub=sub), mean) ss.a.cells = levs.b*levs.d*sum((a.cells$x-grand.mean)^2) ss.a.error = ss.a.cells - ss.a - ss.subs round(ss.a.error, 10)==round(summary(this_aov)[[2]][[1]]$Sum[2], 10) #ok! #Get ss.b b.means = aggregate(x, list(b=b), mean) ss.b = N*levs.a*levs.d*sum((b.means$x-grand.mean)^2) round(ss.b, 10)==round(summary(this_aov)[[3]][[1]]$Sum[1], 10) #ok! #Get ss.b.error b.cells = aggregate(x, list(b=b, sub=sub), mean) ss.b.cells = levs.a*levs.d*sum((b.cells$x-grand.mean)^2) ss.b.error = ss.b.cells - ss.b - ss.subs round(ss.b.error, 10)==round(summary(this_aov)[[3]][[1]]$Sum[2], 10) #ok! #Get ss.d d.means = aggregate(x, list(d=d), mean) ss.d = N*levs.a*levs.b*sum((d.means$x-grand.mean)^2) round(ss.d, 10)==round(summary(this_aov)[[4]][[1]]$Sum[1], 10) #ok! #Get ss.d.error d.cells = aggregate(x, list(d=d, sub=sub), mean) ss.d.cells = levs.a*levs.b*sum((d.cells$x-grand.mean)^2) ss.d.error = ss.d.cells - ss.d - ss.subs round(ss.d.error, 10)==round(summary(this_aov)[[4]][[1]]$Sum[2], 10) #ok! #Get ss.aBYb aBYb.means = aggregate(x, list(a=a, b=b), mean) ss.aBYb = N*levs.d*sum((aBYb.means$x-grand.mean)^2) - ss.a - ss.b round(ss.aBYb, 10)==round(summary(this_aov)[[5]][[1]]$Sum[1], 10) #ok! #Get ss.aBYb.error aBYb.cells = aggregate(x, list(a=a, b=b, sub=sub), mean) ss.aBYb.cells = levs.d*sum((aBYb.cells$x-grand.mean)^2) ss.aBYb.error = ss.aBYb.cells - ss.aBYb - ss.subs round(ss.aBYb.error, 10)==round(summary(this_aov)[[5]][[1]]$Sum[2], 10) #not ok :( -- Mike Lawrence Graduate Student Department of Psychology Dalhousie University www.thatmike.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Access to values of function arguments
Dear Mike and Brian, Thank you very much, this solves my immediate problem. Thanks for your time very much. I will do the reading on those two functions. I have looked for books that deal with the more programmatic aspects of R instead of the statistical side (e.g. dealing with R objects, manipulating environments, lists in R and the like). I have not seen a huge lot. The books that I have seen that deal with Programming in R are actually statistically-focused and, as far as I could find out, do not really cover the programmatic side. Any obvious suggestions for resources that I might have missed? I have several books on R but R programming (as opposed to performing statistical/graphical/mathematical manipulation) is one aspect that I am getting into now and for which I do not have a good reference. The language ref on CRAN is quite helpful but not a good tutorial at all. Kind regards, Willem -- View this message in context: http://www.nabble.com/Access-to-values-of-function-arguments-tp18554602p18557568.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Erro: cannot allocate vector of size 216.0 Mb
Please, I have a 2GB computer and a huge time-series to embedd, and i tried increasing memory.limit() and memory.size(max=TRUE), but nothing. Just before the command: memory.size(max=TRUE) [1] 13.4375 memory.limit() [1] 1535.875 gc() used (Mb) gc trigger (Mb) max used (Mb) Ncells 209552 5.6 407500 10.9 35 9.4 Vcells 125966 1.0 786432 6.0 496686 3.8 I increased the memory limit: memory.limit(3000) NULL memory.limit() [1] 3000 memory.size() [1] 11.33070 memory.size(max=TRUE) [1] 13.4375 gc() used (Mb) gc trigger (Mb) max used (Mb) Ncells 209552 5.6 407500 10.9 35 9.4 Vcells 125964 1.0 786432 6.0 496686 3.8 And even trying to increase the memory.limits, i still get and error. Any sugestions? Thanks in advance. jama __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Non-linearly constrained optimisation
Please have a look at the 'Rdonlp2' package at http://arumat.net/Rdonlp2/. It provides non-linear optinization with nonlinear constraints, so it may be exactly what you are looking for. Whether this algorithm is powerful enough for your task has to be seen. There are examples on the Web page that will enable a quick start. You did not give information about the complexity of your task, nor whether you are seeking local or global solutions. Hans Werner Borchers ABB Corporate Research Dear R Users, I am looking for some guidance on setting up an optimisation in R with non-linear constraints. Here is my simple problem: - I have a function h(inputs) whose value I would like to maximise - the 'inputs' are subject to lower and upper bounds - however, I have some further constraints: I would like to constrain the values for two other separate function f(inputs) and g(inputs) to be within certain bounds This means the 'inputs' must not only lie within the bounds specified by the 'upper' and 'lower' bounds, but they must also not take on values such that f(inputs) and g(inputs) take on values outside defined values. h, f and g are all non-linear. I believe constroptim would work if f and g were linear. Alas, they are not. Is there any other way I can achieve this in R ? Thanks in advance, Tolga - Hans W. Borchers ABB Corporate Research Germany -- View this message in context: http://www.nabble.com/Non-linearly-constrained-optimisation-tp18549310p18558334.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] enumerate subsets
Dear all, is there a R function that enumerate a partition of a vector of size n? (of course for n not very large). I would like enumerate all the (2 power n)-1 sub-sets. (2 power n)-1 since (2 power n) includes de empty subset. Best Regards. ps. It is not a homework. I never posted homework in this list. LL _ Cansado de espaço para só 50 fotos? Conheça o Spaces, o site de relac[[elided Hotmail spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Access to values of function arguments
Check out S Programming by Venables Ripley. On Sun, Jul 20, 2008 at 3:02 PM, willemf [EMAIL PROTECTED] wrote: Dear Mike and Brian, Thank you very much, this solves my immediate problem. Thanks for your time very much. I will do the reading on those two functions. I have looked for books that deal with the more programmatic aspects of R instead of the statistical side (e.g. dealing with R objects, manipulating environments, lists in R and the like). I have not seen a huge lot. The books that I have seen that deal with Programming in R are actually statistically-focused and, as far as I could find out, do not really cover the programmatic side. Any obvious suggestions for resources that I might have missed? I have several books on R but R programming (as opposed to performing statistical/graphical/mathematical manipulation) is one aspect that I am getting into now and for which I do not have a good reference. The language ref on CRAN is quite helpful but not a good tutorial at all. Kind regards, Willem -- View this message in context: http://www.nabble.com/Access-to-values-of-function-arguments-tp18554602p18557568.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] asp and ylim
On 21/07/2008, at 5:27 AM, David Epstein wrote:
#See David Williams' book Weighing the odds, p286
y - c(1.21, 0.51, 0.14, 1.62, -0.8,
0.72, -1.71, 0.84, 0.02, -0.12)
ybar - mean(y)
ylength - length(y)
ybarv - rep(ybar, ylength)
x - 1:ylength
plot(x,y,asp=1,xlab=position,ylab=ybar,type=n,ylim=c(-1,1))
segments(x[1], ybar, x[ylength], ybar)
segments(x,ybarv,x,y)
points(x, ybarv, pch=21, bg=white)
points(x,y,pch=19,col=black)
With asp=1, the value of ylim seems to be totally ignored, as in
the above code. With asp not set, R plays close attention to the
value of ylim. This is not intuitive behaviour, or is it?
How can I set the aspect ratio, and simultaneously set the plot
region? The aspect ratio is one number and the plot region is given
by four numbers (xleft, xright, yleft, yright). Logically, these 5
numbers are independent of each other and arbitrary, provided
xleftxright and yleftyright. This should give a one-to-one
bijection between 5-tuples and plots, determined up to a change of
scale that is uniform in the x- and y-dirctions. My code above
shows the (to me) obvious attempt, which fails.
I just tried
set.seed(42)
x - runif(10)
y - runif(10)
plot(x,y,ylim=c(-1,2),asp=1)
and it seemed to give results as expected/desired.
cheers,
Rolf Turner
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Re: [R] Erro: cannot allocate vector of size 216.0 Mb
What is huge? How big is it and what are the commands you are trying to use? Have you tried a smaller size to see how much memory is being used (do a 'gc' before and after). Depending on what you are trying to do, you may have memory fragmented. What are the sizes of all the other objects that you have in memory? Exactly what is your configuration. Some more information is need to understand why you are getting that error. It typically means what it says; you do not have enough memory to allocate the object of that size. On Sun, Jul 20, 2008 at 3:24 PM, José Augusto Jr. [EMAIL PROTECTED] wrote: Please, I have a 2GB computer and a huge time-series to embedd, and i tried increasing memory.limit() and memory.size(max=TRUE), but nothing. Just before the command: memory.size(max=TRUE) [1] 13.4375 memory.limit() [1] 1535.875 gc() used (Mb) gc trigger (Mb) max used (Mb) Ncells 209552 5.6 407500 10.9 35 9.4 Vcells 125966 1.0 786432 6.0 496686 3.8 I increased the memory limit: memory.limit(3000) NULL memory.limit() [1] 3000 memory.size() [1] 11.33070 memory.size(max=TRUE) [1] 13.4375 gc() used (Mb) gc trigger (Mb) max used (Mb) Ncells 209552 5.6 407500 10.9 35 9.4 Vcells 125964 1.0 786432 6.0 496686 3.8 And even trying to increase the memory.limits, i still get and error. Any sugestions? Thanks in advance. jama __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] enumerate subsets
lamack lamack wrote:
Dear all, is there a R function that enumerate a partition of a vector of size
n? (of course for n not very large).
I would like enumerate all the (2 power n)-1 sub-sets. (2 power n)-1 since (2
power n) includes de empty subset.
Best Regards.
ps. It is not a homework. I never posted homework in this list.
The easiest way is probably to generate the numbers 1:(2^n-1), convert
them to binary, and use the bits to indicate in/out status.
E.g.,
n - 4
x - 1:(2^n-1)
sapply(1:n, function(i) {r - as.logical(x%%2) ; x - x%/%2 ; r})
--
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c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
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Re: [R] enumerate subsets
Can you give a small example of what you data looks like and what you think the output should be. Have you looked at 'combn'? On Sun, Jul 20, 2008 at 4:33 PM, lamack lamack [EMAIL PROTECTED] wrote: Dear all, is there a R function that enumerate a partition of a vector of size n? (of course for n not very large). I would like enumerate all the (2 power n)-1 sub-sets. (2 power n)-1 since (2 power n) includes de empty subset. Best Regards. ps. It is not a homework. I never posted homework in this list. LL _ Cansado de espaço para só 50 fotos? Conheça o Spaces, o site de relac[[elided Hotmail spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] enumerate subsets
Try this replacing letters[1:4] with your set: do.call(expand.grid, as.data.frame(rbind(letters[1:4], NA))) V1 V2 V3 V4 1 abcd 2 NAbcd 3 a NAcd 4 NA NAcd 5 ab NAd 6 NAb NAd 7 a NA NAd 8 NA NA NAd 9 abc NA 10 NAbc NA 11a NAc NA 12 NA NAc NA 13ab NA NA 14 NAb NA NA 15a NA NA NA 16 NA NA NA NA On Sun, Jul 20, 2008 at 4:33 PM, lamack lamack [EMAIL PROTECTED] wrote: Dear all, is there a R function that enumerate a partition of a vector of size n? (of course for n not very large). I would like enumerate all the (2 power n)-1 sub-sets. (2 power n)-1 since (2 power n) includes de empty subset. Best Regards. ps. It is not a homework. I never posted homework in this list. LL _ Cansado de espaço para só 50 fotos? Conheça o Spaces, o site de relac[[elided Hotmail spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] asp and ylim
Try lattice: library(lattice) xyplot(y ~ x, ylim = c(0.5, 0.7), aspect = iso) On Sun, Jul 20, 2008 at 5:58 PM, David Epstein [EMAIL PROTECTED] wrote: On 20 Jul, 2008, at 22:00, Rolf Turner wrote: I just tried set.seed(42) x - runif(10) y - runif(10) plot(x,y,ylim=c(-1,2),asp=1) and it seemed to give results as expected/desired. cheers, Rolf Turner I tried plot(x,y,ylim=c(0.5,0.7),asp=1) on your data above and it failed to restrict the y-coordinates to between 0.5 and 0.7. In fact, the behaviour is independent of whether I insert the ylim=c(0.5,0.7) argument or omit it. So that's definitely not what I want to happen. Nor does the behaviour accord with the advertised behaviour seen from help(plot.window). I want the smallest y-coord seen on the plot to be .5 and the largest to be .7. I want points outside that range not to be displayed. This seems to be what help(plot.window) promises. When I added to your plot command the argument xlim=c(0,1), then that xlim instruction was also ignored. It seems that there is an unstated set of rules for how the plot should react to xlim and ylim settings when asp is set. I wonder if anyone knows what they are. I suppose one could always read the source. For example, one might guess from these examples discussed so far that asp=1 forces the forces the window to be square, but that is not the case. And the command plot(x,y,xlim=c(-6,-1),ylim=c(-1,2),asp=1) on your data suddenly decides that it ought to behave the way I think is reasonable---namely the plot is empty---in contrast to all the other examples where it has insisted on displaying ALL the points (x[i],y[i]), ignoring the values of xlim and ylim. It's fine by me to have xlim and ylim perform idiosyncratically. But can I achieve the degree of control I would like, perhaps using a different technique, or different tags? David __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] asp and ylim
Take a look at eqscplot() in package MASS for a different approach. You last para forgets that once you have set the device region and the margins the physical plot region and hence its aspect ratio is determined -- see the figures in 'An Introduction to R'. On Sun, 20 Jul 2008, David Epstein wrote: #See David Williams' book Weighing the odds, p286 y - c(1.21, 0.51, 0.14, 1.62, -0.8, 0.72, -1.71, 0.84, 0.02, -0.12) ybar - mean(y) ylength - length(y) ybarv - rep(ybar, ylength) x - 1:ylength plot(x,y,asp=1,xlab=position,ylab=ybar,type=n,ylim=c(-1,1)) segments(x[1], ybar, x[ylength], ybar) segments(x,ybarv,x,y) points(x, ybarv, pch=21, bg=white) points(x,y,pch=19,col=black) With asp=1, the value of ylim seems to be totally ignored, as in the above code. With asp not set, R plays close attention to the value of ylim. This is not intuitive behaviour, or is it? How can I set the aspect ratio, and simultaneously set the plot region? The aspect ratio is one number and the plot region is given by four numbers (xleft, xright, yleft, yright). Logically, these 5 numbers are independent of each other and arbitrary, provided xleftxright and yleftyright. This should give a one-to-one bijection between 5-tuples and plots, determined up to a change of scale that is uniform in the x- and y-dirctions. My code above shows the (to me) obvious attempt, which fails. Thanks David __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] asp and ylim
On Sun, Jul 20, 2008 at 11:14 PM, Gabor Grothendieck [EMAIL PROTECTED] wrote: Try lattice: library(lattice) xyplot(y ~ x, ylim = c(0.5, 0.7), aspect = iso) Alternatively, set.seed(42) x - runif(10) y - runif(10) plot(x[(y=0.5) (y=0.7)],y[(y=0.5) (y=0.7)],ylim=c(0.5,0.7),asp=1) Paul __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Sum efficiently from large matrix according to re-occuring levels of factor?
Hi, I am trying to calculate the sum for each occurrence of the level of a factor in a very large matrix. In addition, I want to save that sum together with the information of the level of the factor and the level of a second factor. My matrix looks like this: x-matrix(c(1,1,1,2,2,3,3,1,1,7,7,7,4,4,2,2,7,7,1,1,1,1,1,1,2,5,5),9,3) I want to sum according to the levels in the first column and save the sum with the information of the level in the first and the second column in a new matrix. That is, I want output in the matrix of form: 1 7 3 2 4 2 3 2 3 1 7 10 The important thing is, that a factor level such as 1 in the example can re-occur many times. There are no regularities on the number of re-occurences etc. How could I do this efficiently (the matrix is large:10^6 rows)? Many thanks!! -Ralph _ [[elided Hotmail spam]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] coin package (conditional inference / permutation): parameter teststat
Dear R-list members, This is in fact a question about statistics, not directly about the R software. The coin package, for conditional inference through permutation methods, has as it main function the function independence_test. One of its parameters is teststat, about which the package documentation says: teststat: a character, the type of test statistic to be applied: either a standardized scalar test statistic (scalar), or a maximum type statistic (max) or a quadratic form (quad). My current problem is a two-sample (two-group) one, with multivariate response consisting entirely of continuous numerical observations. Would someone please point me to references indicating the relative merits of the max and quad test statistics in this case? The pdf document Implementing a Class of Permutation Tests: The coin Package, by Hothorn et al., which comes with the package, says on page 5 that In the multivariate case (pq 1), a maximum-type statistic (...) is appropriate. But no reasons or references are provided. P. Good, Permutation, Parametric, and Bootstrap Tests of Hypotheses. 3rd. ed. (Springer), says on page 180 (Section 9.4.4, Which Statistic?) [my comments in brackets]: We've now considered three multivariate test statistics for testing hypothesis based on one or two samples. Which one should we use? To detect a simultaneous shift in the means of several variables, use Hotelling's T2 [a kind of 'quad' test statistics]; to detect a shift in any of several variables, use the maximum t [a kind of 'max' test statistics] (...) But again, no reasons or references are provided. So, I would like to know if anyone could indicate me references discussing the relative merits of the max and quad test statistics, in the case of a two-sample problem with continuous multivariate responses, when using permutation tests. Just for information, I currently use R 2.7.1 running on Windows XP. The coin package is version 0.6-9 (5 May 2008). Thank you very much. Regards, Paulo Barata --- Paulo Barata Fundacao Oswaldo Cruz Rua Leopoldo Bulhoes 1480 - 8A 21041-210 Rio de Janeiro - RJ Brazil E-mail: [EMAIL PROTECTED] Alternative e-mail: [EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sum efficiently from large matrix according to re-occuring levels of factor?
On Sun, Jul 20, 2008 at 4:16 PM, Ralph S. [EMAIL PROTECTED] wrote: Hi, I am trying to calculate the sum for each occurrence of the level of a factor in a very large matrix. In addition, I want to save that sum together with the information of the level of the factor and the level of a second factor. My matrix looks like this: x-matrix(c(1,1,1,2,2,3,3,1,1,7,7,7,4,4,2,2,7,7,1,1,1,1,1,1,2,5,5),9,3) I want to sum according to the levels in the first column and save the sum with the information of the level in the first and the second column in a new matrix. That is, I want output in the matrix of form: 1 7 3 2 4 2 3 2 3 1 7 10 Why that and not: 1 7 13 2 4 2 3 2 3 ? Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sum efficiently from large matrix according to re-occuring levels of factor?
On Sun, Jul 20, 2008 at 4:47 PM, hadley wickham [EMAIL PROTECTED] wrote: On Sun, Jul 20, 2008 at 4:16 PM, Ralph S. [EMAIL PROTECTED] wrote: Hi, I am trying to calculate the sum for each occurrence of the level of a factor in a very large matrix. In addition, I want to save that sum together with the information of the level of the factor and the level of a second factor. My matrix looks like this: x-matrix(c(1,1,1,2,2,3,3,1,1,7,7,7,4,4,2,2,7,7,1,1,1,1,1,1,2,5,5),9,3) I want to sum according to the levels in the first column and save the sum with the information of the level in the first and the second column in a new matrix. That is, I want output in the matrix of form: 1 7 3 2 4 2 3 2 3 1 7 10 Why that and not: 1 7 13 2 4 2 3 2 3 ? Here's a solution for that case: index - x[, 2] + x[, 1] * max(x[, 2]) cbind(x[!duplicated(index), 1:2], tapply(x[, 3], index, sum)) It takes about half a second for a million row matrix. Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] TeachingDemos question: my.symbols() alignment problems in complicated layout
Hi [EMAIL PROTECTED] wrote: Hello, After usefull suggestions by Paul Murrell, i have been trying to use my.symbols to plot arrows of varying angles on my plot in order to create a time series of wind direction... however,i have been unable to figure out how the allignment of symbols works... below i have included a simplified code that illustrates my problem: (i am creating a .ps file ... so i've included this in my mock code, in case it might be part of the problem...) # TEST CODE # postscript(test.ps, horizontal=FALSE, width=7.5, height=11.5,pointsize=10, paper = special ) library(TeachingDemos) opar - par(omi=c(0,0.1,0.7,0.1),xpd=T,mar=par()$mar+c(0,-1.5,-1,5)) layout_mat = matrix(c(1,2,3,4),nrow=4,ncol=1,byrow=TRUE) my_layout - layout(layout_mat,widths=c(1,1,1,1),heights=c(1.0,0.45,1.0,1.2),respect=FALSE) plot(1,1) plot(2,2) plot(3,3) plot(1:100,rep(c(9,1.5,2,8),25)) points(40,4,col=red) points(50,8,col=red) my.symbols(40,4,ms.polygon,n=3,inches=0.2,add=TRUE) my.symbols(40,4,ms.arrows,angle=pi/2,inches=0.2,add=TRUE) my.symbols(50,8,ms.arrows,angle=pi/4,inches=0.2,add=TRUE) dev.off() END of TEST CODE ### If i look at the output test.ps ... the first symbol is exactly where i want it to be i.e. at x=40,y=4 ... while the position of the two other ones is offset by some mysterious(!!) value then if i comment out the first my.symbols(...) command... the second symbol is at the right place, while the third is offset ...etc However if i simply do: plot(1:100,rep(c(9,1.5,2,8),25)) points(40,4,col=red) points(50,8,col=red) my.symbols(40,4,ms.polygon,n=3,inches=0.2,add=TRUE) my.symbols(40,4,ms.arrows,angle=pi/2,inches=0.2,add=TRUE) my.symbols(50,8,ms.arrows,angle=pi/4,inches=0.2,add=TRUE then everything is exactly where i want it ... suggesting there is a problem with the layout ??? I think the issue is that my.symbols() does a lot of this ... op - par(no.readonly = TRUE) on.exit(par(op)) ... which is not absolutely guaranteed to get you back to where you started (see the comment in the 'Value' section of the help page ?par). Fixing that will need the cooperation of the author of TeachingDemos. Here are a couple of options: (i) use the 'gridBase' package and do these arrow annotations using the 'grid' package, which allows you to control coordinate systems in a more rational manner. There's an example (perhaps slightly more complicated than you need) in: http://cran.r-project.org/doc/Rnews/Rnews_2003-2.pdf (ii) draw your main plot using 'lattice' and the annotations using 'grid' or possibly 'grImport'. There's a hint of an example of the latter on slide 18 of: http://www.stat.auckland.ac.nz/~paul/Talks/import.pdf (iii) draw the whole thing using 'grid'. You can start to get acquainted with grid here: http://www.stat.auckland.ac.nz/~paul/RGraphics/chapter5.pdf Unfortunately, all of these require a reasonable amount of extra learning on your part. Paul Any help would be really appreciated, Thank You a lot, maria __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dr 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/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] TeachingDemos question: my.symbols() alignment problems in complicated layout
Here are a couple of options: (i) use the 'gridBase' package and do these arrow annotations using the 'grid' package, which allows you to control coordinate systems in a more rational manner. There's an example (perhaps slightly more complicated than you need) in: http://cran.r-project.org/doc/Rnews/Rnews_2003-2.pdf (ii) draw your main plot using 'lattice' and the annotations using 'grid' or possibly 'grImport'. There's a hint of an example of the latter on slide 18 of: http://www.stat.auckland.ac.nz/~paul/Talks/import.pdf (iii) draw the whole thing using 'grid'. You can start to get acquainted with grid here: http://www.stat.auckland.ac.nz/~paul/RGraphics/chapter5.pdf (iv) Use ggplot2 - particularly geom_segment (http://had.co.nz/ggplot2/geom_segment.html) and stat_spoke (http://had.co.nz/ggplot2/stat_spoke.html) Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] smooth.spline
Actually that was my next question. From the books that I have I see a natural spline and a clamped spline. I am assuming that natural (Umerical Analysis, Burden, et. all.) cooresponds to 'R''s natural method. I am not clear on what a clamped spline cooresponds to (fmm or perodic). Or what the difference between fmm and periodic. Thank you. Kevin Spencer Graves [EMAIL PROTECTED] wrote: Are you aware that there are many different kinds of splines? With spline and splinefun, you can use method = fmm (Forsyth, Malcolm and Moler), natural, or periodic. I'm not familiar with fmm, but it seems to be adequately explained by the Manual spline evaluation you quoted from the documentation. Natural splines are perhaps the simplest: I(x-x0)*(x-x0)^j, where x0 is a knot, and I(z) = 1 if z0 and 0 otherwise. However, computations using natural splines are numerically unstable. The standard solution to this problem is to use B-splines, which are 0 outside a finite interval. Let's look at your example: n - 9 x - 1:n y - rnorm(n) plot(x, y, main = paste(spline[fun](.) through, n, points)) spl - smooth.spline(x,y) lines(spl) The 'smooth.spline' function uses B-splines. To see what they look like, let's do the following: library(fda) Bspl.basis - create.bspline.basis(unique(spl$fit$knot)) # Check to make sure: all.equal(knots(Bspl.basis, interior=FALSE), spl$fit$knot) # TRUE # What do B-splines look like? plot(Bspl.basis) abline(v=knots(Bspl.basis), lty='dotted', col='red') # 7 interior knots, 2 end knots replicated 4 times each, for a spline of order 4, degree 3 (cubic splines) # total of 15 knots # Each spline uses 5 consecutive knots, which means there will be 11 basis functions. # NOTE: 'smooth.spline' rescaled the interval [1, 9] to [0, 1]. # Evaluate the 11 B-splines at 'x' Bspl.basis.x - eval.basis((x-1)/8, Bspl.basis) round(Bspl.basis.x, 4) # Now the manual computation: y.spl - Bspl.basis.x %*% spl$fit$coef # Plot to confirm: plot(x, y, main = paste(spline[fun](.) through, n, points)) spl.xy - spline(x, y) lines(spl.xy) points(x, y.spl, pch=2, col='red') Hope this helps. Spencer [EMAIL PROTECTED] wrote: Fair enough. FOr a spline interpolation I can do the following: n - 9 x - 1:n y - rnorm(n) plot(x, y, main = paste(spline[fun](.) through, n, points)) lines(spline(x, y)) Then look at the coefficients generated as: f - splinefun(x, y) ls(envir = environment(f)) [1] ties ux z splinecoef - get(z, envir = environment(f)) slinecoef $method [1] 3 $n [1] 9 $x [1] 1 2 3 4 5 6 7 8 9 $y [1] 0.93571604 0.44240485 0.45451903 -0.96207396 -1.13246522 -0.60032698 [7] -1.77506105 -0.09171419 -0.23262573 $b [1] -1.53673409 0.22775629 -0.81788209 -1.16966436 0.73558677 -0.68744178 [7] 0.08639287 1.86770869 -2.92992167 $c [1] 1.3657783 0.3987121 -1.4443504 1.0925682 0.8126830 -2.2357115 3.0095462 [8] -1.2282303 -3.5694000 $d [1] -0.32235542 -0.61435416 0.84563953 -0.09329507 -1.01613149 1.74841922 [7] -1.41259217 -0.78038989 -0.78038989 WHen I look at ?spline there is even an example of manually using these coefficeients: ## Manual spline evaluation --- demo the coefficients : .x - get(ux, envir = environment(f)) u - seq(3,6, by = 0.25) (ii - findInterval(u, .x)) dx - u - .x[ii] f.u - with(splinecoef, y[ii] + dx*(b[ii] + dx*(c[ii] + dx* d[ii]))) stopifnot(all.equal(f(u), f.u)) For the smooth.spline as spl - smooth.spline(x,y) I can also look at the coefficients: spl$fit $knot [1] 0.000 0.000 0.000 0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000 [13] 1.000 1.000 1.000 $nk [1] 11 $min [1] 1 $range [1] 8 $coef [1] 0.90345898 0.73823276 0.40777431 -0.08046715 -0.54625461 -0.85205147 [7] -0.96233408 -0.91373830 -0.66529714 -0.47674774 -0.38246971 attr(,class) [1] smooth.spline.fit But there isn't an example on how to manual use these coefficients. This is what I was asking about. Once I hae the coefficients how do I manually interpolate using the coefficients given and x. Thank you. Kevin Spencer Graves [EMAIL PROTECTED] wrote: PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. I do NOT know how to do what you want, but with a self-contained example, I suspect many people on this list -- probably including me -- could easily solve the problem. Without such an example, there is a high probability that any answer might (a) not respond to your need, and (b) take more time to develop, just because we don't know enough of what you are asking. Spencer [EMAIL PROTECTED]
[R] Lattice Version of grconvertX or variant on panel.text?
Still playing with Lattice...
I want to use panel.text(x, y etc) but with x and y in plot coordinates
(0,1), not user coordinates.
I think if I had this problem with traditional graphics, I could use
grconvertX to make the change. I did come across convertX {grid} but this
doesn't seem to be what I need.
Is there a function like grconvertX in Lattice, or is there a flag or some
other method of making panel.text use plot coordinates?
Thanks, Bryan
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice Version of grconvertX or variant on panel.text?
On Sun, Jul 20, 2008 at 6:03 PM, Bryan Hanson [EMAIL PROTECTED] wrote:
Still playing with Lattice...
I want to use panel.text(x, y etc) but with x and y in plot coordinates
(0,1), not user coordinates.
I think if I had this problem with traditional graphics, I could use
grconvertX to make the change. I did come across convertX {grid} but this
doesn't seem to be what I need.
Looking at the code for panel.text (lattice:::ltext.default) it looks
like the easiest way to go would be to call grid.text directly,
specifying the units (npc) that you want to use.
Hadley
--
http://had.co.nz/
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sum efficiently from large matrix according to re-occuring levels of factor?
Does this do what you want:
# following up on another idea that was presented
# where are the breaks
dataBreaks - cumsum(c(0, (diff(x[, 2] + x[, 1] * max(x[, 2])) != 0)))
# sum up column 3 and output the first two columns with the indices
result - lapply(split(seq(nrow(x)), dataBreaks), function(.sect){
+ c(x[.sect[1], 1:2], sum(x[.sect, 3]))
+ })
do.call(rbind, result)
[,1] [,2] [,3]
0173
1242
2323
317 10
On Sun, Jul 20, 2008 at 7:57 PM, Ralph S. [EMAIL PROTECTED] wrote:
The first and second column are actually indices of another matrix (my
example may make this not sufficiently clear). I want to compare the sum with
that corresponding entry, and then record the result of that.
Any idea?
Best,
Ralph
Date: Sun, 20 Jul 2008 16:50:41 -0700
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: Re: [R] Sum efficiently from large matrix according to re-occuring
levels of factor?
CC: r-help@r-project.org
On Sun, Jul 20, 2008 at 4:47 PM, hadley wickham wrote:
On Sun, Jul 20, 2008 at 4:16 PM, Ralph S. wrote:
Hi,
I am trying to calculate the sum for each occurrence of the level of a
factor in a very large matrix. In addition, I want to save that sum
together with the information of the level of the factor and the level of
a second factor.
My matrix looks like this:
x-matrix(c(1,1,1,2,2,3,3,1,1,7,7,7,4,4,2,2,7,7,1,1,1,1,1,1,2,5,5),9,3)
I want to sum according to the levels in the first column and save the sum
with the information of the level in the first and the second column in a
new matrix.
That is, I want output in the matrix of form:
1 7 3
2 4 2
3 2 3
1 7 10
Why that and not:
1 7 13
2 4 2
3 2 3
?
Here's a solution for that case:
index - x[, 2] + x[, 1] * max(x[, 2])
cbind(x[!duplicated(index), 1:2], tapply(x[, 3], index, sum))
It takes about half a second for a million row matrix.
Hadley
--
http://had.co.nz/
_
With Windows Live for mobile, your contacts travel with you.
072008
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R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice Version of grconvertX or variant on panel.text?
Never mind, I just hard-coded it using ratios. Simpler than I thought.
Thanks, Bryan
On 7/20/08 9:03 PM, Bryan Hanson [EMAIL PROTECTED] wrote:
Still playing with Lattice...
I want to use panel.text(x, y etc) but with x and y in plot coordinates
(0,1), not user coordinates.
I think if I had this problem with traditional graphics, I could use
grconvertX to make the change. I did come across convertX {grid} but this
doesn't seem to be what I need.
Is there a function like grconvertX in Lattice, or is there a flag or some
other method of making panel.text use plot coordinates?
Thanks, Bryan
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sum efficiently from large matrix according to re-occuring levels of factor?
yes - thank you very much! slowly getting to the full power of R . . .
Date: Sun, 20 Jul 2008 21:21:35 -0400
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: Re: [R] Sum efficiently from large matrix according to re-occuring
levels of factor?
CC: [EMAIL PROTECTED]; r-help@r-project.org
Does this do what you want:
# following up on another idea that was presented
# where are the breaks
dataBreaks - cumsum(c(0, (diff(x[, 2] + x[, 1] * max(x[, 2])) != 0)))
# sum up column 3 and output the first two columns with the indices
result - lapply(split(seq(nrow(x)), dataBreaks), function(.sect){
+ c(x[.sect[1], 1:2], sum(x[.sect, 3]))
+ })
do.call(rbind, result)
[,1] [,2] [,3]
0173
1242
2323
317 10
On Sun, Jul 20, 2008 at 7:57 PM, Ralph S. wrote:
The first and second column are actually indices of another matrix (my
example may make this not sufficiently clear). I want to compare the sum
with that corresponding entry, and then record the result of that.
Any idea?
Best,
Ralph
Date: Sun, 20 Jul 2008 16:50:41 -0700
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: Re: [R] Sum efficiently from large matrix according to re-occuring
levels of factor?
CC: r-help@r-project.org
On Sun, Jul 20, 2008 at 4:47 PM, hadley wickham wrote:
On Sun, Jul 20, 2008 at 4:16 PM, Ralph S. wrote:
Hi,
I am trying to calculate the sum for each occurrence of the level of a
factor in a very large matrix. In addition, I want to save that sum
together with the information of the level of the factor and the level of
a second factor.
My matrix looks like this:
x-matrix(c(1,1,1,2,2,3,3,1,1,7,7,7,4,4,2,2,7,7,1,1,1,1,1,1,2,5,5),9,3)
I want to sum according to the levels in the first column and save the
sum with the information of the level in the first and the second column
in a new matrix.
That is, I want output in the matrix of form:
1 7 3
2 4 2
3 2 3
1 7 10
Why that and not:
1 7 13
2 4 2
3 2 3
?
Here's a solution for that case:
index - x[, 2] + x[, 1] * max(x[, 2])
cbind(x[!duplicated(index), 1:2], tapply(x[, 3], index, sum))
It takes about half a second for a million row matrix.
Hadley
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What is the problem you are trying to solve?
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Re: [R] Lattice Version of grconvertX or variant on panel.text?
Use grid.text instead:
library(lattice)
library(grid)
xyplot(demand ~ Time, BOD, panel = function(...) {
panel.xyplot(...)
grid.text(Some text, .1, .1) # near lower left
})
On Sun, Jul 20, 2008 at 9:03 PM, Bryan Hanson [EMAIL PROTECTED] wrote:
Still playing with Lattice...
I want to use panel.text(x, y etc) but with x and y in plot coordinates
(0,1), not user coordinates.
I think if I had this problem with traditional graphics, I could use
grconvertX to make the change. I did come across convertX {grid} but this
doesn't seem to be what I need.
Is there a function like grconvertX in Lattice, or is there a flag or some
other method of making panel.text use plot coordinates?
Thanks, Bryan
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] RODBC - problems using odbcDriverConnect without DSN
Hi,
I'm trying to use RODBC without having to set up a DSN, using hte
direct connection string in odbcDriverConnect. My connection attempt
looks something like:
odbcDriverConnect(connection =
SERVER=localhost;DRIVER={/usr/lib/odbc/libmyodbc.so};DATABASE=myDB;UID=reader;PASSWORD=insecure;)
And this returns the message:
Warning messages:
1: In odbcDriverConnect(connection = conn) :
[RODBC] ERROR: state IM002, code 0, message [unixODBC][Driver
Manager]Data source name not found, and no default driver specified
2: In odbcDriverConnect(connection = conn) : ODBC connection failed
I know this means it can't find x connection in the dsn... which means
my connection string isn't recognised as valid. I've used ODBC and
ODBC connection strings before, but I can't work out why this wouldn't
work here. I can successfully create the same connection using a user
DSN with the same settings as this, but this connection string won't
work. It's fairly important for this project that the code can
connect without a DSN needing to be set on every computer that runs
it. Does anyone know if I'm missing something in my connection
string, or can this not be done using RODBC?
Thanks,
Josiah.
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