similar to: replace

I can set up an entry widget (thanks to an old post by Barry Rowlingson) that gets a password and exits when the user clicks on the "OK" button. Anyone have any clever ideas for returning/ destroying the window when the user types a carriage return/ENTER in the text window? I've messed around a little with validate, validatecommand, but don't see any obvious way to do it ...

From: Vitalie S. <vitosmail <at> rambler.ru> Subject: Bug in cut.POSIXt Newsgroups: gmane.comp.lang.r.general Date: 2009-10-15 15:47:48 GMT (1 hour and 29 minutes ago) Hello Everyone, Before reporting decided to post here first: tt <- structure(c(1254238817, 1254238859, 1254238969, 1254239080), class = c("POSIXt",

When I was studying the function cut I found this example: > x <- rep(0:8, tx0) > x [1] 0 0 0 0 0 0 0 0 0 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 4 4 4 5 5 5 5 5 5 5 5 5 5 6 [39] 6 6 6 6 7 7 7 8 8 8 8 8 > cut(x, b = 8) [1] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] [6] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (0.994,2] [11] (0.994,2]

Dear R users, I am trying to fill in arrays (5 different according to distinct "id") from objects produced from arbitrary data set below. a <-

Hi, Forgive a very basic question... I need to take two lists-of-lists, and apply a function to each pair of elements in the lists to return a single list... For example l1 <- list(1:5,6:10,2:15) l2 <- list(1:8,4:12,1:19,4:20) I could easily do an lapply across each of them, but is there a function that does a sort-of pairwise-apply across both together? Does anybody know of a good

Sweave does something clever with warnings, which I have so far been unable to figure out. There are a couple of threads on the list about this, but the best in here is a hack to redirect all the output and stick it back in. http://www.nabble.com/-R--Sweave-and-warning-messages-td7759353.html#a7759353

Hi, I have a binned data that looks like this. (8.048,18.05] (-21.95,-11.95] (-31.95,-21.95] (18.05,28.05] (-41.95,-31.95] 81 76 18 18 12 (-132,-122] (-122,-112] (-112,-102] (-162,-152] (-102,-91.95] 6 6 6 5 5 (-91.95,-81.95] (-192,-182]

I feel like a complete dolt, as I know this question has been asked by others on a fairly regular basis, but I'm going in circles trying to get the following to work: id.prob<-function (tt) { library(mvtnorm) #============================ Makeham<-function(tt) { a2=0.030386513 a3=0.006688287 b3=0.039047537 t<-tt-20 h.t<-a2+a3*exp(b3*t) S.t<-exp(-a2*t+a3/b3*(1-exp(b3*t)))

Hi! I use simtest fonction of multcomp package to compile a Dunnett's test. I have 10 treatments and one control group, so i create a matrix with: m<-matrix(0,10,11) m[1,1]<--1 m[1,2]<-1 m[2,1]<--1 m[2,3]<-1 m[3,1]<--1 m[3,4]<-1 m[4,1]<--1 m[4,5]<-1 m[5,1]<--1 m[5,6]<-1 m[6,1]<--1 m[6,7]<-1 m[7,1]<--1 m[7,8]<-1 m[8,1]<--1 m[8,9]<-1

Dear R users, I have a data frame (test) including five columns of upper (numeric), lower (numeric), observed (numeric), best_sim (numeric) and stname (factor with 80 levels, each level with different length). Now I would like to write a short program to draw one graph as follow for each level of stname but I would like also to draw each time 12 graphs for the 12 levels of stname in the same

Dear R people, The header practically says it all. I was wanting to include r code in a Latex file. Since R code using{ and }, which are interpreted by Latex as control characters, I would expect it to get upset. I believe that \{ would probably print as {, but I hoping I don't have to go through the code and add lots of \. I would rather use some global commannd, along the lines of

Dear all I have a dataset that looks like: fr.wt site 1 4400 glen 2 235 glen 3 225 glen ' ' ' ' ' ' ' ' ' 82 550 glen 83 550 kom 84 550 kom ' ' ' ' ' ' ' ' ' 191 820 kom 192 2000 soet ' ' ' ' ' ' I need to do a series of histograms for each of the codes, levels or factors in

Perhaps this is a common question but I haven't been able to find the answer. I have data with many factors, each taking many values. However, only relatively few combinations appear in the data, ie have nonzero counts, in other words the resulting table is sparse. Say we have 10 factors each with 10 levels. The result of table() would exceed the memory space (on a 32bit machine). Is there

hi - in version 2.1 the command >-2^2 gives -4 as the answer. (-2)^2 is evaluated correctly. Cheers, George Casella -- George Casella Phone: (352) 392-1941 Ext. 204 Distinguished Professor and Chair Cell: (352) 682-7210 Department of Statistics Fax: (352) 392-5175 University of Florida Email: casella at stat.ufl.edu P.O. Box 118545 Gainesville, FL

hi - in version 2.1 the command >-2^2 gives -4 as the answer. (-2)^2 is evaluated correctly. Cheers, George Casella -- George Casella Phone: (352) 392-1941 Ext. 204 Distinguished Professor and Chair Cell: (352) 682-7210 Department of Statistics Fax: (352) 392-5175 University of Florida Email: casella at stat.ufl.edu P.O. Box 118545 Gainesville, FL

If so, I''ve had trouble finding it. Can anyone help?

Dear List, I'd like to run a discriminate analysis on a data set, but have no idea how to go about this in R. I have attempted to locate info in the manuals, but may not be consulting the right sections or documents. Can anyone point me to appropriate documentation if such exists. Many thanks, David S. David White sdavidwhite at bigfoot.com Columbus, Ohio

Hi -- I''m going to be giving a free (but limited enrollment) one-day Rails workshop in Malmo, Sweden, on September 17. The event is presented by Polar Rose (http://www.polarrose.com). For more info: http://www.polarrose.com/about/ruby-on-rails-workshop-by-david-black Hop on over from London if you''re at RailsConf Europe! (Which [London] is where I am now.) David --

I've just spent today trying to fix a Heisenbug... this function returns a linear interpolator function: interpOne <- function(xl,yl){ f = function(data){ t = (data-min(xl))/(max(xl)-min(xl)) return(min(yl)+t*(max(yl)-min(yl))) } return(f) } > k=interpOne(c(0,1),c(4,5)) > k(0.5) [1] 4.5 and this function uses the above to return a function that returns a piece-wise

Dear R people, This is simple, nay trivial question. I do the following. I just want to know what is the slickest/most elegant method. I have an expression containing the term (y)*log(y) where y can take the value either 0 or 1. If y is 0 then the expression is undefined, but in this case I want to replace the result by 0 in the expression. I don't want to write a separate function for