CODEWITHSUNDEEP
haskell
demathieu
demathieu

Reputation: 479

alternative permutation of two lists

I'm trying to get a alternative permutation of two list with no fixed length for example:

x = ["a","b","c"]
y = [1,2,3]

should return:

[(a,1),(b,2),(c,3)] , [(a,2),(b,3),(c,1)], [(a,3),(b,1),(c,2)]

with list comprehensions I was able to get this:

[(x,y) | x<-x, y<-y ]

[("a",1),("a",2),("a",3),("b",1),("b",2),("b",3),("c",1),("c",2),("c",3)]

which is not exactly what I want

Upvotes: 0

Views: 426

Answers (1)

Cactus
Cactus

Reputation: 27626

You only need to permute one of your two lists, then pair up each permutation with the other list in the original order: 

import Data.List (permutations)

bijections :: [a] -> [b] -> [[(a, b)]]
bijections xs ys = map (zip xs) (permutations ys)

What this does is it generates all permutations of ys (so that would be [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1] for your example) and then for each of them, it pairs them up with xs in its original order.

Upvotes: 4

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