similar to: frequency table

Using platform x86_64-pc-linux-gnu arch x86_64 os linux-gnu system x86_64, linux-gnu status major 2 minor 13.0 year 2011 month 04

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

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

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

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

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 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 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 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)

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

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

Hi I recently found a bug that was isomorphic to the following: ll1 <- 2 increment <- function(x) { l11 <- 1 return(x+ll1) #bug here } Of course, R is obeying the scoping rules just fine, but I'm evidently not setting the do.what.I.mean.not.what.I.say variable correctly. Now, how do I avoid making this type of error? ... and what is the best tool for tracking it down? --

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]

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 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 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: