My code is: [[ -- |Given a list of lists, construct a new list of lists where -- each member of the new list is the same length as the original -- list, and each member corresponds to a different choice of -- one element from each of the original members in the -- corresponding position. Thus: -- -- listProduct [[a1,a2],[b1],[c1,c2]] = -- [ [a1,b1,c1], [a1,b1,c2], [a2,b1,c1], [a2,b1,c2] ] -- -- Note: The length of the resulting list is the product of -- lengths of the components of the original list. Thus, if -- any member of the original list is empty then so is the -- resulting list: -- -- listProduct [[a1,a2],[],[c1,c2]] = [] -- listProduct :: [[a]] -> [[a]] listProduct [] = [[]] listProduct (as:ass) = concat [ map (a:) (listProduct ass) | a <- as ]
test1 = listProduct [["a1","a2"],["b1"],["c1","c2"]] test2 = listProduct [["a1","a2"],[],["c1","c2"]] ]]
And a supplementary point. Following a comment made to me previously, I can replace this list comprehension (not having any guard expressions) with another map function, thus:
[[ lp [] = [[]] lp (as:ass) = concatMap (\a -> (map (a:) (lp ass))) as ]]
I think I should also be able to eliminate the lambda-abstraction, but I can't see how. I prefer the list comprehension, as I find that easier to read than the \-expression.
#g
------------ Graham Klyne For email: http://www.ninebynine.org/#Contact
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