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 <lrpal...@gmail.com>
> On Sun, Dec 28, 2008 at 2:39 PM, Nicholas O. Andrews <nandr...@vt.edu>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
>>
>> Thanks!
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>>
>
>
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