Knuth shuffle: Difference between revisions

Line 3,956:

===Short version===

After accounting for R being 1-indexed rather than 0-indexed, it's not hard to implement the pseudo-code given in the task almost exactly:

<lang rsplus>knuth<-function(vec)

<lang rsplus>knuth <- function(vec)

{

last<-length(vec)

last <- length(vec)

if(last<2)~~{return(vec)}~~

if(last >= 2)

else for(i in last:2)

{

~~j<-sample~~(1:i,1)

for(i in last:2)

{

⚫

vec[c(i,j)]<-vec[c(j,i)]

j <- sample(seq_len(i), size = 1)

⚫

vec[c(i, j)] <- vec[c(j, i)]

}

vec

Line 3,970:

Line 3,972:

knuth(integer(0))

knuth(c(10))

replicate(10,knuth(c(10,20)))

replicate(10, knuth(c(10, 20)))

replicate(10,knuth(c(10,20,30)))

replicate(10, knuth(c(10, 20, 30)))

knuth(c("Also","works","for","strings"))</lang>

knuth(c("Also", "works", "for", "strings"))</lang>

<pre>> knuth(integer(0))

Line 3,978:

Line 3,980:

> knuth(c(10))

[1] 10

> replicate(10,knuth(c(10,20)))

> replicate(10, knuth(c(10, 20)))

[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]

[1,] 20 20 10 10 20 10 20 10 20 10

[2,] 10 10 20 20 10 20 10 20 10 20

> replicate(10,knuth(c(10,20,30)))

> replicate(10, knuth(c(10, 20, 30)))

[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]

[1,] 30 10 20 20 30 30 10 30 10 10

[2,] 10 20 30 10 10 10 20 20 20 20

[3,] 20 30 10 30 20 20 30 10 30 30

> knuth(c("Also","works","for","strings"))

> knuth(c("Also", "works", "for", "strings"))

[1] "strings" "Also" "for" "works"</pre>