similar to: Curious about nomenclature: random deviates

Greetings all, I'm trying to figure out how to calculate the inverse CDF (i.e. a quantile) for a non-central F distribution. I could put together a quick numerical solver routine using the CDF, but I wonder if there's a function that I've missed that would be more efficient? Thank-you, Andrew Andrew Robinson Ph: 208 885 7115 Department of Forest Resources Fa: 208 885

Hello, has anyone tried to get R running on a Sharp Zaurus 5500? Andrew Andrew Robinson Phone: 208-885-7115 Department of Forest Resources Fax: 208-885-6226 University of Idaho E: andrewr at uidaho.edu Po Box 441133 WWW: http://www.uidaho.edu/~andrewr Moscow, ID 83843 and: http://www.biometrics.uidaho.edu/ No statement above

I don't get the same result, do you have a package loaded that would change the default behavior (such as Hmisc)? > list(grp.1) [[1]] [1] 1 2 Levels: 1 2 > list(grp.1[mask]) [[1]] [1] 1 Levels: 1 2 > library(Hmisc) <<snip>> > list(grp.1[mask]) [[1]] [1] 1 Levels: 1 --Matt > version _ platform i386-pc-mingw32 arch i386 os

Dear R community, In the excellent nlme package the default diagnostic plot graphs the innermost residuals against innermost fitted values. I recently fit a mixed-effects model in which there was a very clear positive linear trend in this plot. I inferred that this trend occurred because my fixed effect was a two-level factor, and my random effect was a 12-level factor. The negative residuals

If I place par(new=T) before I create a plot in a script that is sent to a pdf device, the pdf is unopenable and reports itself as having no pages. The postscript device seems to ignore the par instruction. I guess one of these is a bug, but I don't know which one! --please do not edit the information below-- Version: platform = i386-pc-linux-gnu arch = i386 os = linux-gnu system =

Hello all, Here's my query: Running R 1.6.2 on FreeBSD 5.0, and on WinXP, and I find that the following hangs the process: dchisq(alpha=0.01, df=1, ncp=295) but it does work for ncp < 294.92. Is this general? Best wishes to all, Andrew Andrew Robinson Ph: 208 885 7115 Department of Forest Resources Fa: 208 885 6226 University of Idaho E : andrewr at uidaho.edu PO

Dear R-community, I'm using Sweave to produce reports. The reports require the "axis" command. When I run axis the program returns NULL as well as creating the axis. > plot(1:4, rnorm(4), axes=FALSE) > axis(1, 1:4, LETTERS[1:4]) NULL > So, my Sweave tex files have \begin{Schunk} \begin{Soutput} NULL \end{Soutput} \end{Schunk} in front of each graphic that requires

Hi all, I'm using R 1.5.1 on Windows 2000. The following snippet of code doesn't seem to do anything - no error is reported, and there is no name change. names(myFrame[,c(1:3)]) <- c("name1", "name2", "name3") This code however works nicely: names(myFrame)[c(1:3)] <- c("name1", "name2", "name3") Can anyone suggest why

Hi dear R-community, I wonder if anyone has written code that will post-process the results of a tapply to create a dataframe that includes the outcome, the factor by which the function has been applied, and, if the factor is also an interaction, then also the levels of contributing factors? >From this outcome <- tapply(data, A.B, function) I'd like to arrange something like this:

Dear R community, I'm trying to understand this behavior of TukeyHSD. My goal is to obtain defensible, labelled multiple comparisons of an interaction term. Firstly, if I plot the TukeyHSD from the model that calculates its own interactions, then the y-axis labels appear to be reflected on their median when compared to the text output of the TukeyHSD statement. The labels are integers.

Greetings R-community, I'm running simulations within R that I wrote in C. The simulations require fitting that occasionally fails to finish. I was wondering if there is any kind of tool for process control in R, such that after e.g. 15 minutes I could kill the process, record the state for post-hoc analysis, and move to the next simulation? I'm running FreeBSD and could almost surely

Greetings R community, I am curious about the following behaviour: if I define a factor, and then store a subset of it in a list, the stored version seems to drop levels that were not included in the subset. E.g. .. > mask <- c(T, F) > grp.1 <- factor(c(1,2)) > list(grp.1) [[1]] [1] 1 2 Levels: 1 2 > list(grp.1[mask]) [[1]] [1] 1 Levels: 1 It is as though the list were

Greetings, R community. I am trying to create a multi-dimensional contingency table suitable for analysis by glm() using the poisson family. I have three factors, each with four levels, with some observed zeros. I'm trying to use aggregate to construct my contingency table, but it drops empty subsets, so the zeros get lost. I also tried tapply() but it doesn't carry over the main

Hi all, I'm trying to find a way to get xyplot to produce a second set of axes outside the right hand side of the graph. This is my progress so far: EE <- equal.count(ethanol$E, number=9, overlap=1/4) xyplot(NOx ~ C | EE, data = ethanol, prepanel = function(x, y) prepanel.loess(x, y, span = 1), xlab = "Compression Ratio", ylab = "NOx (micrograms/J)",

Dear R friends, I know that this topic has been mulled over before, and that there is a substantial difference between the convergence criteria for JMP and those for R. I apologize that this is somwehat raking cold coals. Summary: A model/data combination achieves convergence in JMP, and survives a reasonably rigorous examination (sensible parameter estimates, well-behaved surface,

Hello, I have been searching for ways to obtain these for combinations of fixed factors and levels other than the 'baseline' group (contrasts coded all 0's) from a mixed-effects model in lme. I've modelled the continuous variable y as a function of a continuous covariate x, and fixed factors A, B, and C. The fixed factors have two levels each and I'd like to know whether

I'm trying to do a sunflowerplot of Galton's data, with both regression lines and data ellipses, and I must be doing something wrong, because the lines do not intersect at \bar{x}, \bar{y}. The problem is likely in the line for x.y, but I don't know how to specify that correctly. The data is read in grouped form( galton), and then ungrouped (galton2): galton <-

I'm sure this is probably simple, but I can't find an answer... I have a data frame (Galton's data on heights of parents and children), in grouped form, parent child frequency 73.5 72.2 1 73.5 73.2 3 72.5 68.2 1 72.5 69.2 2 72.5 70.2 1 72.5 71.2 2 72.5 72.2 7 72.5 73.2 2 72.5 74.2 4 71.5 65.2 1 71.5 66.2 3 71.5 67.2 4 71.5 68.2 3 71.5 69.2 5 71.5 70.2 10 ... and need the ungrouped

I would appreciate some thoughts on using the bootstrap functions in the library "bootstrap" to estimate confidence intervals of ICC values calculated in lme. In lme, the ICC is calculated as tau/(tau+sigma-squared). So, for instance the ICC in the following example is 0.116: > tmod<-lme(CINISMO~1,random=~1|IDGRUP,data=TDAT) > VarCorr(tmod) IDGRUP = pdLogChol(1)

I have two lists: xa <- list( X=c(1,2,3), Y=c(4,5,6), Z=c(7,8,9) ) xb <- with( barley, tapply( X=seq(1:nrow(barley)), INDEX=site , FUN=function(z)yield[z])) I can convert xa to a dataframe easily with: as.data.frame(xa) But if i try the same with xb I get: as.data.frame(xb) Error in as.data.frame.default(xb) : can't coerce array into a data.frame What