similar to: ellipsis question

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 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 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

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

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 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 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

Dear All, I tried to plot contour lines using R function contour, but got the results which are not expected. require(RTOMO) x <- seq(-1,1,0.1) y <- seq(-1,1,0.1) xy <- meshgrid(x,y) z <- xy$x^2+ 3*xy$y^2 contour(x,y,z,col="blue",xlab="x",ylab="y") The above code gave me the contour graph for z=3*x^2+y^2 rather than z=x^2+3*y^2. Is anyone know the

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) --

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 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)) {

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

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

Hello everybody. I have a dozen or so vectors (of different lengths) which I want to coalesce into a dataframe, as in the following toy example. R> foo <- c(1,2,3) R> bar <- c(7,8) R> data.frame(name =c(rep("foo",length(foo)),rep("bar",length(bar))), value=c(foo,bar)) name value 1 foo 1 2 foo 2 3 foo 3 4 bar 7 5 bar

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 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

Hi all, I have a 2D matrix, which has 100 rows, and 100 columns, I have a 2D matrix, with 100 rows and 100 columns, I want to display it using 3D plot, much like plot3d and mesh/surf functions in matlab. Specifically, in matlab, I just need to do the following: -------- [X, Y]=meshgrid([0:0.01:0.99, 0:0.01:0.99]); % Z is my 2D matrix, surf(X, Y, Z); -------- Note that X and Y are created so