Convert list op tuples into two lists of one-elemnt tuple

Question

In haskell I'm trying to acheive to convert a list like [(1,2),(3,4),(5,6)] into two lists: [1,3,5] and [2,4,6] using a recursive function only

I tried this so far, but it doesn't work

unpair :: [(a,b)] -> ([a], [b]) unpair [] = [] unpair ((x,y):xs) = x : unpair(xs) y : unpair (xs)

however, when I only try to make a list of the first value in the tuples, it does work:

unpair [] = [] unpair ((x,y):xs) = x : unpair(xs)

Any idea what I'm doing wrong?

unpair returns a pair, not a list. Hence, both your examples are wrong. Use case .. of (x,y) -> or let (x,y)=... in to access the components of a pair. — chi
– chi, Commented Dec 2, 2014 at 18:51

Sassa NF · Accepted Answer · 2014-12-02 17:55:10Z

The type of unpair xs is ([a], [b]), so you can't append an element to it like to a list. But you can use let statement to pattern-match the result from unpair xs, then construct the tuple with the new head element added to each.

I won't quote the actual answer because this looks like home work.

The reason the second implementation works is because in that one the type of unpair is [(a,b)]->[a].

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