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] mapM takes a function which returns a computation for a given argument. In this case the function always returns the computation [-1,0,1] which you can think of as a non-deterministic computation resulting in either -1, or 0, or 1. This computation is executed for every value in the list [1..3] and because this list has three values the execution results in [x,y,z] where each of x, y, and z is either -1, or 0, or -1. This gives you all variations. You can also write: [[x,y,z] | x<-[-1,0,1], y<-[-1,0,1], z<-[-1,0,1]] Sincerely, Jan. -- Heriot-Watt University is a Scottish charity registered under charity number SC000278.