similar to: reshape()

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]

Dear helplist Many many thanks to everyone who helped me. The trick was to use tabulate() or, better, tab <- rep(0,50) names(tab) <- 1:50 tab[names(table(sleeps))] <- table(c) My original dataset was a list of 50 trees and a length 12 vector recording which tree a certain possum slept in on 12 nights. As Professor Ripley points out, a Monte-Carlo simulation is easy to set up, and it

dear helplist, my student has fifty trees, numbered one to fifty, and a vector recording which tree a certain possum slept in on 12 nights. R> c [1] 3 14 17 22 26 26 17 40 43 25 46 46 R> Thus it slept in tree #3 on Monday, then tree #14 on Tues, and so on. I wish to test the null hypothesis that the animal chooses trees randomly; try R> table(c) c 3 14 17 22 25 26 40 43 46 1 1

Dear R community A few weeks ago, I uploaded a small text file called R-and-octave.txt to the contributed docs section of CRAN. This file details octave/matlab commands and their (near) equivalents in R (Matlab is a widely-used high-level graphics/mathematics tool and octave a free clone). Someone has just pointed out to me that I never announced its existence to anyone, hence this email (I

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 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"

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 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.

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

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

Robin, the initial output from apply() is always in the form you have below, but if it can be 'simplified' into a structure like the matrix, it does so. The same thing happens with sapply(). If you want to produce a nice matrix as the out put you have to ensure that the simplification is possible. Here is one way. > apply(x2, 2, function(x, v) table(factor(x, levels=v)),

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) --

I am trying to do post-hoc tests associated with a repeated measures analysis with on factor nested within respondents. The factor (SOI) has 17 levels. The overall testing is working fine, but I can''t seem to get the multiple comparisons to work. The first step is to "stack" the data. Then I used "lme" to specify and test the overall model. Finally

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]

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)

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

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

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

Hi everyone. What's going on here? > a <- matrix(1:4,2,2) > a [,1] [,2] [1,] 1 3 [2,] 2 4 > apply(a,2,sum) [1] 3 7 > apply(a,2,"+") [,1] [,2] [1,] 1 3 [2,] 2 4 > apply(a,1,"+") [,1] [,2] [1,] 1 2 [2,] 3 4 > help(apply) says that "+" should be quoted but is otherwise silent on unary