I had the same thought, since my interval [1..10000] is rather large.
Thanks,
MIchael
--- On Tue, 6/14/11, Jonas Almström Duregård
Shuffle [1..20], then take 5? Yes, so simple, I'm embarrassed I didn't think of it.
That works well for small numbers, but I'm guessing it will evaluate the entire list so it should not be used for large inputs. If you have a large interval and use a relatively small part of it, the following function should be significantly faster (it builds a random permutation lazily):
import System.Random
randomOrder :: (Ord a, Num a, Random a, RandomGen g) => (a,a) -> g -> [a]randomOrder (low,high) g
| low > high = [] | otherwise = let
(a,g') = randomR (low,high) g (gl,gr) = split g'
in a : mergeRandom (a-1-low,randomOrder (low,a-1) gl) (high-a-1, randomOrder (a+1,high) gr) g'
where mergeRandom (_,[]) (_,xs) _ = xs
mergeRandom (_,xs) (_,[]) _ = xs mergeRandom (lx,x:xs) (ly,y:ys) g = let
(pick,g') = randomR (1,lx + ly) g in if pick <= lx
then x : mergeRandom (lx-1,xs) (ly,y:ys) g' else y : mergeRandom (lx,x:xs) (ly-1,ys) g'
On 14 June 2011 04:31, michael rice
Is there an (existing) way to select 5 Ints randomly (no duplicates) from a population, say 1-20 (inclusive)?
Yes, already implemented in the monte-carlo package as sampleSubset [1], sampleSubset :: MonadMC m => [a] -> Int -> m [a] Complete example code for your example: evalMC (sampleSubset [1..20] 5) (mt19937 0) Cheers! [1] http://hackage.haskell.org/packages/archive/monte-carlo/0.4.1/doc/html/Contr... -- Felipe. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe