similar to: new R documentation on CRAN

Hi everyone This must be a FAQ but I can't find it anywhere... I want a postscript image of a contour() plot, with axes of equal length. Try R> postscript(file="~/f.ps") R> contour(matrix(rnorm(100),10,10)) R> dev.off() This isn't what I want: the plotting region is, as documented, quarter of an inch shy of the paper edge and the axes appear to be different lengths.

Hello everyone. If I do f <- function(n){max(rnorm(n))} plot(sapply(rep(5000,4000),f)) #[this takes my PC about 30 seconds] then I get something quite unexpected: gaps in the distribution. For me, the most noticable one is at about 3.6. Do others get this? Is it an optical illusion? It can't be right, can it? Or maybe I just don't understand the good ol' Gaussian very

Hello R experten I have just written a little function to calculate all pairwise combinations of two vector arguments: > pair(c(1,2,3),c(7,8)) [,1] [,2] [1,] 1 7 [2,] 1 8 [3,] 2 7 [4,] 2 8 [5,] 3 7 [6,] 3 8 > I want to generalize this to any number of arguments, for example, <fantasy> > ntuple(c(1,2,3),c(7,8),c(14,15)) [,1] [,2]

Dear List the other day, I had my yearly staff interview with my Head of Department. Under my list of publications, I included a document which I wrote (R-and-octave.txt) that ended up in the "contributed docs" section of CRAN. Unfortunately, neither my HoD nor the personnel person were terribly impressed with it, even though its preparation time was commensurate with many of my (co

Hello everybody. I have a question that has stumped me and the usual "apply" tricks don't seem to work. I run a course where each student's performance is marked by one or more assessors. I have a data frame containing students' names, assessors' names and their marks, arranged as follows: ID student assessor Q1A Q1B Q1C Q2A Q2B Q3 1 2152833

Hello ALL: How do I get R to list all possible orderings of 2 x's and 3 y's? It should look like this (which rows appear first is unimportant): x x y y y x y x y y x y y x y x y y y x y x x y y y x y x y y x y y x y y x x y y y x y x y y y x x Thanks, ANDREW

Hello everybody I have a generic problem which the following toy function illustrates: f <- function(n) { if(abs(n) < pi) { return(TRUE) } else { return(FALSE) } } I want it to return TRUE if abs(n)<pi and FALSE otherwise. f() is fine as far as it goes, but does not deal well with NA or NaN or NULL (I want these to signal some problem with the

Hello List How do I turn R> simple.example.alphabetic [,1] [,2] [,3] [1,] "a" "b" "c" [2,] "d" "e" "f" [3,] "g" "h" "i" into R> simple.example.numeric [,1] [,2] [,3] [1,] 1 2 3 [2,] 4 5 6 [3,] 7 8 9 [ie "a" becomes 1, ..., "z"

Kia Ora everybody. There must be an obvious answer to this, but I can't see it.... I want four square plots in one postscript file. The canonical answer would be: postscript(file="~/f.ps",width=5,height=5) par(pty="s",mfrow=c(2,2)) plot(1:19,xlab="") plot(1:19,xlab="") plot(1:19,xlab="") plot(1:19,xlab="") dev.off() But this

Hello list (thanks for all the help on my data.frame() question, especially to Professor R for a working script...I was pleased to see the solution wasn't obvious!) Anyway, now I'm trying to run R in batch mode, but I'm getting extra output, which I don't want (RedHat 8.3, R-1.7.0): r:~% cat test.R options(echo=FALSE) write(rnorm(4),"") r:~% R CMD BATCH --vanilla

Hi everyone check this out [R-1.7.0]: R> f1 <- function(x){x^2} R> f1 -> f2 R> f2(4) [1] 16 R> R> function(x){x^2} -> f3 function(x){x^2} -> f3 R> f3(4) Error: couldn't find function "f3" Why does right assignment "->" work in the first but not the second case? Can anyone else reproduce this? -- Robin Hankin, Lecturer, School of

Hello everybody. Does anyone else find the last test in the following sequence odd? Can anyone else reproduce it or is it just me? > binom.test(100,200,0.13)$p.value [1] 2.357325e-36 > binom.test(100,200,0.013)$p.value [1] 6.146546e-131 > binom.test(100,200,0.0013)$p.value [1] 1.973702e-230 > binom.test(100,200,0.00013)$p.value [1] 0.9743334 (R 1.5.1, Linux RedHat 7.1) --

Hello everybody Is there a good way for a function to tell whether the caller used the defaults? I'm writing a little function that may take a pair of real arguments or a single complex argument (in which case I want the real and imaginary components). "e" <- function(first,second=first) { if (all(first == second) & is.complex(first)) {

Hi everyone another documented feature that was a bit unexpected for me: R> x <- 19 R> f <- function(t){t+x} R> f(100) [1] 119 --as expected: x is visible from within f() ..but... R> g <- function(a){x <- 1e99 ; return(f(a))} R> g(4) [1] 23 --the "x" that is visible from within g() is just 19, which is not the one I expected it to find. R> rm(x)

Dear R-ers I've recently upgraded to R-1.4 but I have noticed that it is slower to load datasets than R-1.3.1: R-1.3.1: > system.time(a <- read.table("~/people/academics/mitchell/nzsl02.27",header=T)) [1] 40.86 0.51 54.10 0.00 0.00 > R-1.4.0: > system.time(a <- read.table("~/people/academics/mitchell/nzsl02.27",header=T)) [1] 293.24 30.76 478.35

Dear Helplist Some time ago, Professor Ripley gave me a tip which I thought was very very useful for Monte Carlo simulation; I thought I'd pass it on to the list, and ask whether this or a similar example could be added to the sapply() manpage. Suppose I have ten N(0,1) random variables and I'm interested in the pair that are closest together: R> min(diff(sort(rnorm(10)))) [1]

Hi everyone thanks for the replies. The issue was NOT a font problem; I deliberately chose ll1 and l11 as examples of easily confused variable names (evidently these were too easily confused ;-). The code snippet was written as intended, and increment() contained a deliberate, highlighted, bug. I was asking for guidance on avoiding/finding this sort of coding error. That was why I wrote

Dear Helplist I have a dataframe that holds the Southern Oscillation Index over the last few years: R> soi[1:3,] Year Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 1 1993 -9 -10 -12 -24 -9 -18 -11 -18 -9 -15 -1 0 2 1994 -2 -1 -14 -26 -13 -12 -18 -20 -19 -16 -9 -15 3 1995 -4 -5 2 -19 -9 -3 4 -1 3 -2 0 -8 QUESTION: how do I coerce reshape() into giving me this:

Dear R-xperts I have just written a little hypergeometric function, included below [the hypergeometric function crops up when solving a common type of ODE]. It works fine on single values of the primary argument z, but vectorizing it is getting confusing. The best I have come up with so far just tests for z being longer than 1 and if so, uses sapply() recursively. This is fine, except that it

Hello everybody I'm trying to write a simple version of matlab's "quiver". The idea is that I have fluid with velocity defined on a grid. I have a matrix of x-components of velocity and a matrix of y-components and I want to see the overall flow pattern. (I work with 2D fluid mechanics problems). My first-stab function is below: quiver <- function(u,v,scale=1) # first