I'll give it shot, but I'm also learning Haskell, so take this with a grain of salt :) randWhile :: ([Int]->Bool) -> StdGen -> [Int] randWhile predicate = head . filter predicate . blocks 10 . randomRs (0,9) where blocks n xs = let (y,ys) = splitAt n xs in y : blocks n ys main = newStdGen >>= print . randWhile (all even) or if you prefer non point free notation (point full?) randWhile :: ([Int]->Bool) -> StdGen -> [Int] randWhile predicate rndGen = head $ filter predicate $ blocks 10 $ randomRs (0,9) rndGen where blocks n xs = let (y,ys) = splitAt n xs in y : blocks n ys main = do rndGen <- newStdGen print $ randWhile (all even) rndGen 2008/12/28 Luke Palmer

On Sun, Dec 28, 2008 at 2:39 PM, Nicholas O. Andrews wrote:

...

Hi all,

What's the best way to implement the following Python code in Haskell? It is purposefully written in a functional style (and as a result will kill your recursion stack every other run).

Here's my solution, using MonadRandom (from Hackage). There may be more infinite-listy ways of doing it, but I wasn't able to make it come out clean.

import Control.Monad.Random

many n = sequence . replicate n

untilM p m = do x <- m if p x then return x else untilM p m

getList :: MonadRandom m => m [Int] getList = many 10 $ getRandomR (0,9)

main = print =<< evalRandIO (untilM (all even) getList)

# begin Python

...

from random import *

def genList (): return [randint(0,9) for x in range(10)]

def randWhile (predicate): result = genList () if predicate(result): return result else: return randWhile (predicate)

def allEven (list): return reduce(lambda x,y: x and y, [x%2 == 0 for x in list])

print randWhile (allEven) # End Python

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