Is this what you want:
> swap <- function(z){
+ a <- sample(length(z), 2)
+ z[a] <- z[rev(a)]
+ z
+ }
> swap(1:10)
[1] 2 1 3 4 5 6 7 8 9 10
>
> swap(1:10)
[1] 5 2 3 4 1 6 7 8 9 10
> swap(1:10)
[1] 8 2 3 4 5 6 7 1 9 10
> swap(1:10)
[1] 2 1 3 4 5 6 7 8 9 10
> swap(1:10)
[1] 4 2 3 1 5 6 7 8 9 10
> swap(swap(swap(1:10)))
[1] 8 10 3 5 4 6 7 1 9 2
> swap(swap(swap(1:10)))
[1] 8 2 7 4 5 6 3 10 9 1
>
On Fri, Aug 22, 2008 at 8:57 AM, amor Gandhi <[EMAIL PROTECTED]> wrote:
> Hello Richie,
> I would like to do three (or k) swap steps in each step just 2 ID recursive
> swaping
> x <- 1:10
> swap <- function(x){
> a <- sample(x,2)
> x[x==a[1]] <- swap[2]
> x[x==a[2]] <- swap[1]
> return(x)
> }
> swap(swap(swap(x))) -> mix
>
> Is this possible?
> Thanks you in advance!
> Amor
>
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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