How to extract all unique pairs of a list in Haskell? [duplicate]

Question

I am pretty new to Haskell and I have a question. How can I write a function to return all the unique possible pairs of a list? Like: [1,2,3,4] -> [(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)]

Answer 1

Using a list comprehension is the easiest way.

import Data.List

pairs :: [a] -> [(a, a)]
pairs l = [(x,y) | (x:ys) <- tails l, y <- ys]

This only generates the relevant pairs so you don't have to filter out any duplicates as a separate step. tails [1,2,3,4] generates the lists of tails [[1,2,3,4], [2,3,4], [3,4], [4], []] and the list comprehension picks the first element from each tail and pairs it with the rest of the elements in that tail.

Answer 2

Looking at your example, this seems to be what you want

unique_pairs l = [(x,y) | x <- l, y <- l, x < y]

By the way, these are not just unique pairs, but unique pairs up to transposition.

Answer 3

You can use the nub function:

import Data.List
uniq_pairs l = nub [(x,y) | x <-l, y<-l, x < y]

Output:

*Main> uniq_pairs  [1,2,3,4]
[(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)]