# 99 questions/Solutions/25

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< 99 questions | Solutions(Difference between revisions)

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rndPermutation xs = rndElem . permutations $ xs | rndPermutation xs = rndElem . permutations $ xs | ||

</haskell> | </haskell> | ||

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+ | WARNING: this may choke long lists |

## Revision as of 05:11, 12 January 2012

Generate a random permutation of the elements of a list.

rnd_permu xs = diff_select' (length xs) xs

Uses the solution for the previous problem. Choosing N distinct elements from a list of length N will yield a permutation.

Or we can generate the permutation recursively:

import System.Random (randomRIO) rnd_permu :: [a] -> IO [a] rnd_permu [] = return [] rnd_permu (x:xs) = do rand <- randomRIO (0, (length xs)) rest <- rnd_permu xs return $ let (ys,zs) = splitAt rand rest in ys++(x:zs) rnd_permu' [] = return [] rnd_permu' xs = do rand <- randomRIO (0, (length xs)-1) rest <- let (ys,(_:zs)) = splitAt rand xs in rnd_permu' $ ys ++ zs return $ (xs!!rand):rest

permutations

Data.List

import System.Random (getStdGen, randomRIO) import Data.List (permutations) rndElem :: [a] -> IO a rndElem xs = do gen <- getStdGen index <- randomRIO (0, length xs - 1) return $ xs !! index rndPermutation :: [a] -> IO [a] rndPermutation xs = rndElem . permutations $ xs

WARNING: this may choke long lists