similar to: A couple of questions

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 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'm encountering a couple problems configuring and making R 0.64 under linux, redhat 5.2 These problems didn't occur when I build 0.63.2 Any suggestions would be greatly appreciated. (1) configure generates the warning: "configure: warning: Cannot determine how to create shared libraries. Please set CPICFLAGS, FPICFLAGS and SHLIBLDFLAGS in `config.site'." I'm not

Hi, all. I'm trying to use some linear regression models in which both the dependent and independent variables are measured with some error. To make things worse, while the errors in the dependent variable are uniform, the errors in the independent (or explanatory, or "x") variables can be heteroskedastic. I've been looking at the book _Measurement Error Models_ by Fuller

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...")