similar to: ylim doesn't work in boxplots?

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'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 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 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 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

Colors are real nice, but the publication I'm preparing these barplots for permits only black and white. The Splus plot options ``dbangle'' (or plain ``angle'') and ``density'' (as on p. 65 of MASS 1st ed.) don't seem to be available in R. Is there another way to do this? I'm running R 1.1.0 on Linux (intel). Thanks.

This is really a question of how to program this *BETTER*. It works as I have done it, but is quite ugly. I want to do a 3d scatterplot of the upper triangle of a matrix, where the z-values are the values in the matrix, and the row and column indices are the y- and x-values. The complete (11 by 11) matrix is mmtop94.2. Here is my awkward code: mmtop94.2[lower.tri(mmtop94.2)] <- NA # Here i

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,

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 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]

Hello, I want to add wireless capability to my Gentoo-linux based firewall/router at home, so I bought a Netgear MA311 PCI and installed the hostap package. I load the hostap_pci module and the wlan0 interface comes up fine. I can detect the signal from a wireless enabled laptop. Now I'm thinking I'm going to bridge the wlan0 interface and the eth1 interface, and run the firewall with br0

My mind has become corrupted by having to use SAS too much. I wanted to calculate the difference of elements of two vectors if the signs are the same. I tried it the corrupted way first: if (effects1[,3] < 0 & effects0[,3] < 0) { imprv1 <- effects0[,3] - effects1[,3] } else if (effects1[,3] > 0 & effects0[,3] > 0) { imprv1 <- effects1[,3] - effects0[,3] } else {

I'm trying to install packages in Gentoo Linux using install.packages(). The packages download fine, but when it comes to installation, I get this message: WARNING: ignoring environment value of R_HOME /var/tmp/portage/R-1.5.1/image//usr/lib/R/bin/Rcmd: /var/tmp/portage/R-1.5.1/image//usr/lib/R/bin/Rcmd: No such file or directory I have the R_HOME set to the correct value (/usr/lib/R), but I

I'm trying to shade part of a density plot. The code I'm trying (using the Old Faithful data as an example) is something like this: # The Old Faithful geyser data data(faithful) d <- density(faithful$eruptions, bw = "sj") plot(d) polygon(d[d$x>4], col = "wheat") I expected that the part of the curve to the right of 4 on the x axis should be shaded, but nothing

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 can't remember the programming term -- ``implicit declaration''? Has this changed between 1.4.1 and 1.5? This code used to run as is: for (i in 1:length(diff.time)) { if (diff.time[i] < 16) { total.time[i] <- ifelse(i==1, 7.5, total.time[i-1]) + diff.time[i] last.time[i] <- 0 } else { total.time[i] <- 7.5 last.time[i-1] <- 1 } } Now in 1.5 I