99 questions/Solutions/23
From HaskellWiki
Remi.berson (Talk  contribs) m (remove one useless applicative in the last solution) 

(One intermediate revision by one user not shown) 
Latest revision as of 13:03, 10 May 2014
Extract a given number of randomly selected elements from a list.
import System.Random import Control.Monad (replicateM) rnd_select :: [a] > Int > IO [a] rnd_select [] _ = return [] rnd_select l n  n<0 = error "N must be greater than zero."  otherwise = do pos < replicateM n $ getStdRandom $ randomR (0, (length l)1) return [l!!p  p < pos]
In order to use getStdRandom and randomR here, we need import module System.Random.
A more elegant solution using randomRs:
rnd_select xs n = do gen < getStdGen return $ take n [ xs !! x  x < randomRs (0, (length xs)  1) gen]
or using sequence all the way:
import Control.Monad (replicateM) rnd_select xs n  n < 0 = error "N must be greater than zero."  otherwise = replicateM n rand where rand = do r < randomRIO (0, (length xs)  1) return (xs!!r)
Alternative solution:
The original Lisp problem suggested we use our solution from problem 20. I believe that each item from the list should only appear once, whereas the above solution can reuse items.
Therefore here is an alternative which uses the "removeAt" function from problem 20:
rnd_select :: RandomGen g => [a] > Int > g > ([a], g) rnd_select _ 0 gen = ([], gen) rnd_select [] _ gen = ([], gen) rnd_select l count gen  count == (length l) = (l, gen)  otherwise = rnd_select (removeAt l (k+1)) count gen' where (k, gen') = randomR (0, (length l)  1) gen rnd_selectIO :: [a] > Int > IO [a] rnd_selectIO l count = getStdRandom $ rnd_select l count
If the number of items we want is the same as the number of items in the list, then we just return the list. Otherwise we remove a random item from the list and then recurse.
Another Alternative Solution:
Since the above Alternative Solution works by removing things to create the target list, it's most efficient when the target list length is > (orig list / 2). Here's another solution that's efficient in the other way (target < (orig list / 2)) by constructing an accumulator list of selected random elements. (This one also uses removeAt from problem 20)
rnd_select :: RandomGen g => [a] > Int > g > ([a], g) rnd_select ol ocount ogen = rnd_select' ol [] ocount ogen where rnd_select' l acc count gen  count == 0 = (acc, gen)  otherwise = rnd_select' (removeAt l (k+1)) ((l !! k) : acc) (count  1) gen' where (k, gen') = randomR (0, (length l)  1) gen rnd_selectIO :: [a] > Int > IO [a] rnd_selectIO l count = getStdRandom $ rnd_select l count
An O(N) algorithm:
import System.Random (randomRIO) rnd_select :: [a] > Int > IO [a] rnd_select _ 0 = return [] rnd_select (x:xs) n = do r < randomRIO (0, (length xs)) if r < n then do rest < rnd_select xs (n1) return (x : rest) else rnd_select xs n
A solution returns random results even when the number of items we want is the same as the number of items in the list:
import System.Random(randomRIO) rnd_select :: [a] > Int > IO [a] rnd_select _ 0 = return [] rnd_select [] _ = return [] rnd_select xs count = do r < randomRIO (0, (length xs)1) rest < rnd_select (removeAt xs (r+1)) (count1) return ((xs!!r) : rest)
A very simple and elegant solution which uses "nub" function from Data.List.
rnd_select :: Int > [a] > [a] rnd_select n x = map (x!!) is where is = take n . nub $ randomRs (0, length x  1) (mkStdGen 100)
IO a
The same solution using the global number generator and returning IO [a] (thanks to applicatives, $ is simply replaced by <$>):
rnd_select :: [a] > Int > IO [a] rnd_select lst n = map (lst !!) <$> indices where indices = take n . nub . randomRs (0, length lst  1) <$> getStdGen