On Fri, Apr 17, 2009 at 12:19 PM, jim holtman <jholt...@gmail.com> wrote:
> try this:
>
>> matrixx<-function(A){
> + B=matrix(NaN,nrow=(A+1),ncol=4)
> + k <- 1
> + for (i in 3:A){
> + for (j in i:A) {
> + B[k,] <- c(NaN, i-2, i-1, j)
> + k <- k + 1
> + }
> + }
> + B
> + }
>> matrixx(5)
> [,1] [,2] [,3] [,4]
> [1,] NaN 1 2 3
> [2,] NaN 1 2 4
> [3,] NaN 1 2 5
> [4,] NaN 2 3 4
> [5,] NaN 2 3 5
> [6,] NaN 3 4 5
Here's a solution without the loop. I think it illustrates the intent
of the algorithm more clearly.
candidates <- t(combn(5,3))
firstdiff <- candidates[,2] - candidates[, 1]
cbind(NaN, candidates[firstdiff == 1, ])
Hadley
--
http://had.co.nz/
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