12

In haskell, given a list of elements, xs, the simplest way to iterate over all pair permutations with repetitions is:

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

I wish to be able to do the same, but only on combinations. If x and y were comparable, I could do

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

But I'd prefer a solution that is more generic and more efficient (I know that asympotitic complexity will remain squared, but we can halve actual runtime complexity by avoiding the usage of a filtering condition)

duplode
  • 33,731
  • 7
  • 79
  • 150
tohava
  • 5,344
  • 1
  • 25
  • 47

2 Answers2

25

What about:

[ (x,y) | (x:rest) <- tails xs , y <- rest ]
chi
  • 111,837
  • 3
  • 133
  • 218
2

If I copy and simplify Math.Combinat.Sets.choose for the particular case k=2, this gives 

pairs :: [a] -> [[a]]
pairs []     = []
pairs (x:xs) = map (\y -> x:[y]) xs ++ pairs xs -- replace with \y -> (x,y) to get 2-tuples

>>> pairs [1,2,3]
[[1,2],[1,3],[2,3]]
Stéphane Laurent
  • 75,186
  • 15
  • 119
  • 225