similar to: termplot - changes in defaults

Does anyone have, or has anyone ever considered making, a version of 'termplot' that allows the user to specify that all plots should have the same y-limits? This seems a natural thing to ask for, as the plots share a y-scale. If you don't have the same y-axes you can easily misread the comparative contributions of the different components. Notes: the current version of termplot

Does anyone have, or has anyone ever considered making, a version of 'termplot' that allows the user to specify that all plots should have the same y-limits? This seems a natural thing to ask for, as the plots share a y-scale. If you don't have the same y-axes you can easily misread the comparative contributions of the different components. Notes: the current version of termplot

I found writing the following default method the for the generic function "julian" causes R to crash. julian.default <- function(x, ...) { x <- as.Date(x) NextMethod("julian", x, ...) } Here is a test example > m <- as.Date("1972-09-27") + 0:10 > m [1] "1972-09-27" "1972-09-28" "1972-09-29"

I have just become painfully aware that objects of class "difftime", generated by the difference of two POSIXct objects, carry a "units" attribute, which flashes by when the object is printed, for example. The pain was occasioned when I tried to turn these objects into numberic objects for use elsewhere as a covariate. as.numeric(difftime object) simply turns off the units

I've noticed an increasing tendency for people to use '=' rather than the older '<-' symbol. When '=' became available as an assignment operator in S-PLUS in the late '90s my first reaction was to switch to it as well. Brian Ripley warned me that it was not a good idea. As usual he was right, but it took a couple of pretty serious finger-burning episodes

> I think we should get rid of source attributes completely, > since they are no longer needed, but your comment still > applies to source references. We should strip them when code > gets modified. > > Duncan Murdoch I would be very concerned about losing source attributes-- it would break a lot of my code :(! It's very useful to have a single portable R object that

The function "substitute" seems to fail to make a genuine substitution, although the printed verision seems fine. Here is an example. > m <- substitute(Y <- function(x) FUN(x+1), + list(Y = as.name("y"), FUN = as.name("sin"))) > m y <- function(x) sin(x + 1) > eval(m) > y function(x) FUN(x+1) However the story doesn't end there. The

This is not so much a bug as an infelicity in the code that can easily be fixed. The initialize expression in the quasi family function is, (uniformly for all links and all variance functions): initialize <- expression({ n <- rep.int(1, nobs) mustart <- y + 0.1 * (y == 0) }) This is inappropriate (and often fails) for variance function "mu(1-mu)".

If you would like a 10 second R puzzle, you might like to think about this one: > g <- expand.grid(long = 130:140, lat = -(10:25)) > gp <- transform(g, x = long, y = lat) Error in transform(g, x = long, y = lat) : object "long" not found I don't expect this hasn't come up before, but I can't find mention of it. I suggest that to minimise this little

My apologies for he very short time before things close, but it's not my fault. If necessary get your hat in the ring by Friday and see to the niceties later, may I suggest. This is a vacancy in my group in Cleveland. The work is interesting and the challenges large. Bill Venables. ---------------------------------------------------------------------------- Division of Marine Research

Peter, There is a clear and simple bug in glm() that I have noticed in 0.64.2 (Windows and Unix) but may have been present in earlier versions. The function starts as follows: > glm function (formula, family = gaussian, data = list(), weights = NULL, subset = NULL, na.action = na.fail, start = NULL, offset = NULL, control = glm.control(...), model = TRUE, method =

At 14:35 02/10/00 +0800, mohd zamri wrote: >new to R and starting to learn to program R. The underscore ("_") did some >suprising result. e.g > >> c <- c(1,2,3,4,5) >> mean(c) >[1] 3 >> c_mean <- mean(c) >> c >[1] 3 > >having some experience in C, I thought the underscore is "always" valid in >variable name. totally confuse

I have a big contingency table, approximately of size 60*2*500*500, and I need to count the number of cells containing a count of 1 for each of the factors values defining the first dimension. Here is my attempt: tab1<-with(pisa1,table(CNT,GENDER,ISCOF,ISCOM)) tab2<-apply(tab1,1:4,function(x)ifelse(sum(x)==1,1,0)) tab3<-apply(tab2,1,sum) Computing tab2 is very slow. Is there a faster

> -----Original Message----- > From: Cobalt [mailto:grosso at mail.ru] > Sent: Friday, 29 December 2000 7:23 > To: r-help at r-project.org > Subject: [R] anova > > > > ..Hello, Well Hello, whoever you are. (In this list anonymous messages are considered rather impolite. Next time please submit your message under your real name as a gesture of good faith. For this

I see Peter has just announced the release of 1.0.0, on time, even here in Australia, in accordance with a timetable privately announced about 6 months ago. At that time it seemed optimistic to put it mildly. I am not a member of the core team, but as an ordinary user of R with a more privileged inside view than most I'd like to offer my personal congratulations and thanks. This is the

Brian, Thanks for doing this. As the primary author, could I have a desk copy, please? Bill Venables. Bill Venables, CMIS, CSIRO Marine Laboratories, PO Box 120, Cleveland, Qld. 4163 AUSTRALIA Phone: +61 7 3826 7251 Fax: +61 7 3826 7304 Mobile: +61 419 634 642 <mailto: Bill.Venables at csiro.au> http://www.cmis.csiro.au/bill.venables/ -----Original Message----- From: Brian

Hi R, I wanted to know how do we access the elements of a list. In particular, v=list(c(1,2,3,4,5),c(1,2,33,4,5),c(1,2,333,4,5),c(1,2,3333,4,5)) I want to access all the thirds items of the elements of the list. i.e., I want to access the elements, 3,33,333,3333. This can be done through sapply as: sapply(v,function(x) x[3]) But I need to access this without using

I seem to have unwittingly stirred up a hornets' nest here, so let me have one more say and leave it. Firstly, congratulations to Renaud Launcelot, who seems to have a better feel for wryness and irony in English than some of my native speaker colleagues. Of course R A Fisher is a seminal writer, indeed close to the founder of modern statistics and every aspiring statistician should read

I recently took "." off my PATH for security reasons and now find that R does not work in my home directory where the .Renviron file resides. The fix is simple. In the R shell startup section replace ". .Renviron" by ". ./.Renviron". The relevant section is: # Startup if ${USE_R_ENVIRON}; then # use the one in current dir, or default if [ -r .Renviron ] then .

I have a set data={A,B,C,D,E,F,G} I want to choose 2 letter from 8 letters, i.e. generate the combination set for choose 2 letters from 8 letters. I want to get the liking: combination set={AB,AC,AD,....} Does anyone konw how to do in R. thanks, Aimin