On Fri, Apr 24, 2009 at 11:36:44AM +0100, Jan Jakubuv wrote:
On Fri, Apr 24, 2009 at 12:16:06AM +0200, Erik Quaeghebeur wrote:
Aha. Great. Thanks, Jan. And now I realized that I don't really care about the domain, so I said:
Prelude> let m = mapM (\x -> [(x,-1),(x,0),(x,1)]) ['a','b','c'] Prelude> map (\x -> snd $ unzip x) m [[-1,-1,-1],[-1,-1,0],[-1,-1,1],[-1,0,-1],[-1,0,0],[-1,0,1],[-1,1,-1],[-1,1,0],[-1,1,1],[0,-1,-1],[0,-1,0],[0,-1,1],[0,0,-1],[0,0,0],[0,0,1],[0,1,-1],[0,1,0],[0,1,1],[1,-1,-1],[1,-1,0],[1,-1,1],[1,0,-1],[1,0,0],[1,0,1],[1,1,-1],[1,1,0],[1,1,1]]
Any more direct way of doing this? (Still trying to understand how the mapM works... I've found it's sequence.map, but that doesn't really help.)
Well, you can write:
mapM (const [-1,0,1]) [1..3]
Better yet (in my opinion), you can just write sequence (replicate 3 [-1,0,1]) which is really the same thing, since mapM = sequence . map. Mapping (const [-1,0,1]) over [1..3] yields [[-1,0,1], [-1,0,1], [-1,0,1]], that is, (replicate 3 [-1,0,1]). It's the 'sequence' that does the magic of selecting an item from each of the three lists in all possible ways. -Brent