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I have data on every grade in all elementary schools in Chicago over 5 years. I would like to estimate a trend over time for each grade in each school. There are 17,600 data all together (about 460 schools, nearly 8 grades each, over 5 years). Is there a not-so-hard way to do this in R (I was thinking of using rlm)? ______________________________________________________________________ Stuart

Is the width= parameter in barplot() supposed to work? I couldn't get it to work in my plot, and even in the example, data(VADeaths, package = "base") barplot(VADeaths, width=rep(0.1, 4)) the plot looks identical regardless of what I put in for the width. I looked at the source for barplot() and it looks like it SHOULD work (but what do I know). Am I doing something wrong?

Is there an efficient way to do linear model selection by choosing the model with the highest BIC from all possible models? ______________________________________________________________________ Stuart Luppescu -=-=- University of Chicago $(B:MJ8$HCRF`H~$NIc(B -=-=- s-luppescu at uchicago.edu http://www.consortium-chicago.org/people/sl.html http://musuko.uchicago.edu/pubkey.asc

I would like to make my horizontal barplot vertical axis labels perpendicular to the axis. I tried las=1, srt=90; I even tried yaxt='n' thinking I'd put the labels in using mtext, but the axis continued to be drawn. Can anyone help me with this? My barplot() statement looks like this: barplot(height=foo$rebint[o1], names=foo$Unit[o1], horiz=TRUE, col=mycolors[foo$type[o1]],

I would like to make the print format of the axis value labels appear as ``nn%'' instead of just the numbers. Is there a way to do this? Also, the x axis value labels on my horizontal barplot are rotated 90 degrees (perpendicular to the axis) no matter what I do. I tried las=1 but that didn't work. Is this doable? Thanks. $platform [1] "i686-pc-linux-gnu" $arch [1]

I seem to remember this coming up before, but I can't find it any messages I've saved or in the archives (searching by subject). I want to rotate mtext so that it's perpendicular to the right side. I tried srt=90 and lots of other values, but it seems to be ignored. Is there a way to do this? ______________________________________________________________________ Stuart Luppescu

I have plain-text data in lower triangular form that I want to read in. Does anyone know of an easy way to do this? ______________________________________________________________________ Stuart Luppescu -=-=- University of Chicago $(B:MJ8$HCRF`H~$NIc(B -=-=- s-luppescu at uchicago.edu http://www.consortium-chicago.org/people/sl.html http://musuko.uchicago.edu/pubkey.asc for PGP

Hello, I'm trying to read data in from a file using scan(). The last field is a character string that contains blanks. I had read it in in S-Plus using this code: ifile <- list(entry=0,measure=0,st=0,count=0,score=0, error=0,inmsq=0,instd=0,outms=0,outstd=0,displc=0,ptbis=0,a=0, r="",name="") if.widths <- c(1, 5, 8, 3, 6, 7, 7, 7, 7, 7, 7, 7, 7, 2,

QueryParser fails badly allocating enormous amount of memory when processing query strings with special/accented characters. See: irb(main):002:0> require ''rubygems'' irb(main):003:0> require ''ferret'' irb(main):004:0> include Ferret irb(main):005:0> index = Index::Index.new irb(main):008:0> index << "something" # Now the error

Hi, All. I'd have a set of data in an array: process <- c( 5 , 7 , 4 , 1 , 4 , 1 , 4 , 1 , 4 , 4 , 1 , 5 , 4 , ...) and I'd like to know the number of transitions in this data. I calculate transitions as the number of times a number follows another number. thus, something like this would be a 1 deep transition: 1 --> 1 : 10% (and actual number of 1 --> 1 occurrences) 1

Hi, I was wondering how a linux box configured as a firewall stacked up against some of the commercial products like checkpoint-1 and gauntlet. Can someone direct me to a good book or online doc that compares linux to some other firewall methods? Mind you, I''m not talking about a firewall in the classical sense, ie ip forwarding turned off and used as a proxy, but the typical Linux box

*************************************************************************************************** CISTI'2013 DOCTORAL SYMPOSIUM 8th Iberian Conference on Information Systems and Technologies Lisbn, Portugal, June 19 - 23, 2013 http://www.aisti.eu/cisti2013/index.php?option=com_content&view=article&id=64&Itemid=68&lang=en

*************************************************************************************************** CISTI'2013 DOCTORAL SYMPOSIUM 8th Iberian Conference on Information Systems and Technologies Lisbn, Portugal, June 19 - 23, 2013 http://www.aisti.eu/cisti2013/index.php?option=com_content&view=article&id=64&Itemid=68&lang=en

I just installed the 0.64 rpm for Linux and noticed that the links to function.html and packages.html on .../doc/html/index.html don't go anywhere. In fact, those pages don't seem to exist. Do I have to do something to generate them? Also, will the rpm's for the contrib packages be updated to 0.64 level, or just R-base?

I would like to force a specific variable to always be the first split criterion in rpart, somewhat analagous to forcing a variable into a stepwise regression. I looked in V & R, and in the help files for the rpart package. If it's there, it's pretty well hidden. Can anyone help me with this? ______________________________________________________________________ Stuart Luppescu

I would like to make pretty trees such as the one on page 383 of Chambers and Hastie, or page 346 of Venables and Ripley. I modified the post.rpart function, adding splits=T to text(), but the split criteria still don't print. Would anyone have any advice on this? Thanks. I'm running version 0.63 on Linux (intel). ______________________________________________________________________

I'm having trouble with coplot. I have two conditioning variables, and when I try to specify two sets of given.values, no matter what I do I get the message, Error: Object "b.levels" not found. Here is the section of code: c.ranges <- co.intervals(schses, number=4, overlap=0) a.ranges <- co.intervals(as.factor(aa94), number=2, overlap=0) PredAA <- as.factor(aa94)

I want to estimate the center of a distribution with lots of outliers in one tail, and thought I would use a function such as S-plus's location.m() with psi.fun=bisquare (as per MASS 3 p. 131). However, R seems not have such a function, so my questions are: 1) Is there an R equivalent to location.m()? 2) Would huber() give me results that are similar (i.e., close enough)? Thanks.

I need to calculate an index of predictive association, similar to Goodman and Kuskal's (1954) for a 2x2x2 array. I have two questions: 1) Is there an easy way to do this in R? (Of course, the answer to this kind of question is nearly always ``yes.'' Perhaps I should have asked, ``How can I do this in R?'') 2) Is it reasonable to try this with a 3-dimensional array? The way I

I want to leave a little extra room at the bottom for a legend in a box plot, but I find that the boxplot function ignores ylim. Is there any way around this? Thanks. ______________________________________________________________________ Stuart Luppescu -=-=- University of Chicago $(B:MJ8$HCRF`H~$NIc(B -=-=- s-luppescu at uchicago.edu