On 30/07/07, Chad Scherrer
I'm trying to do something I thought would be pretty simple, but it's giving me trouble.
Given a list, say [1,2,3], I'd like to be able to generate an infinite list of random elements from that list, in this case maybe [1,2,1,3,2,1,3,2,3,1,2,...]. I'm using IO for random purely due to laziness (my own, not Haskell's).
I was thinking the best way to do this might be to first write this function:
randomElts :: [a] -> [IO a] randomElts [] = [] randomElts [x] = repeat (return x) randomElts xs = repeat r where bds = (1, length xs) xArr = listArray bds xs r = do i <- randomRIO bds return (xArr ! i)
Then I should be able to do this in ghci:
sequence . take 5 $ randomElts [1,2,3] [*** Exception: stack overflow
Any idea what's going on? I thought laziness (Haskell's, not my own) would save me on this one.
I don't get that result. However, you can't compute an infinite random list in IO without using something like unsafeInterleaveIO. However, you will probably be interested in randoms/randomRs, which take a random generator, and give an infinite list of results. Using that, we could write something like: randomElts :: [a] -> IO [a] randomElts [] = return [] randomElts xs = do g <- newStdGen return (map (xArr !) (randomRs bds g)) where bds = (1, length xs) xArr = listArray bds xs which for a nonempty input list, gives an infinite list of pseudorandom elements of that input list. - Cale