similar to: How to get nice trees

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'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 have an old IBM Thinkpad 701 that has a 340MB disk and a 486 CPU that's not good for much anymore, so I thought I'd install R on it and make it a little useful. I went to *tremendous* pains to upgrade Linux on it without X so there would be some room to put R. But having done that I find that the binary is R.X11 and it's looking for (but can't find) libSM.so, which is ordinarily

Sorry to bother everyone with such a basic question again, but I can't seem to set my axis labels. When I use title(ylab="String") "String" just overwrites the variable name, producing a mess. When I try to put ylab="String" in the plot function, like this: plot(residuals(mschmod) ~ size94, ylab = "Productivity Adjusted for Context") I get this message:

Sorry to bother everyone with such a basic question again, but I can't seem to set my axis labels. When I use title(ylab="String") "String" just overwrites the variable name, producing a mess. When I try to put ylab="String" in the plot function, like this: plot(residuals(mschmod) ~ size94, ylab = "Productivity Adjusted for Context") I get this message:

I just installed 1.0 (great job, everyone!) and rebuilt all my packages. I have found an error in making VR_6.1-6.tar.gz: Installing package `MASS' ... libs gcc -I/usr/lib/R/include -mieee-fp -fPIC -O2 -m486 -fno-strength-reduce -g -c MASS.c -o MASS.o MASS.c:631: Arith.h: No such file or directory make: *** [MASS.o] Error 1 The file /usr/lib/R/include/Arith.h used to be there (locate

Can anyone explain to me what this error message means, why I'm getting it, and how to fix it? lines(lowess(xdat, ydat, f=.5), col=3) Error: NAs in foreign function call (arg 1) ______________________________________________________________________ Stuart Luppescu -=-=- University of Chicago ºÍʸ ¤ÈÃÒÆàÈþ¤ÎÉã(EUC) -=-=- s-luppescu at uchicago.edu

I am very confused about this. I want to convert a string to a name so I can use it to extract an element of a data frame using `$'. Here is my (non-working) code: do.graph <- function (meas) { fn <- paste("a", meas, ".dat", sep='') themeas <- read.table(fn, header=F) ameas <- as.name(paste("a", meas, sep=''))

I'm sorry if this is a basic question, but I'm stumped. I'm just trying to plot the residuals from a linear model against another variable in the data frame. Here are the lines I'm trying to execute: size <- read.table(file="/u67/abasl70/surveys/annenberg/mega/smschl.dat", header=T) sizef <- data.frame(size, row.names=size$unit) attach(sizef) mschmod <- lm

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 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 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

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'm stumped with this. When I execute the lines in the function singly, they run fine, but when I run the function, I get this error on the read.table() line: Error in count.fields(file, sep, quote, skip) : can't open file fspci1.dat Can anyone tell my why this should be so? Here is the program: library(rpart) wait <- function(str="Press a key when ready...")

Has the loess.smooth() function changed? It used to work, but now it causes R to abort with a segmentation fault. I stole the function points.lines() from V&R 1st ed. pp. 67--68, but now it only works if I remove the line with loess.smooth. Here's the function I'm using: points.lines <- function(x, y, ...) { cor1 <-round(cor(x, y, use="pairwise"), digits=2)

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