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Ramosaurio
Asked: 2020-10-14 01:26:06 +0800 CST 2020-10-14 01:26:06 +0800 CST

Separate a Haskell list into an odd and even tuple

  • 772

I have the following function:

ParesImpares:: [Integer] -> ([Integer],[Integer])

I have to get even and odd numbers out of a list:

ParesImpares [3,2,4,5,7]

->([2,4],[3,5,7])

My code so far is as follows:

    paresImpares :: [Integer] -> ([Integer],[Integer])
    paresImpares [] = ([],[])
    paresImpares [x] 
    | x `mod` 2 == 0 = ([x],[])
    | otherwise = ([],[x])

I have defined a base case, but the truth is that I don't know how to proceed. I had in mind something like:

ParesImpares(x:xs) = ParesImpares [x]

But of course, it only evaluates the first element of the list.

haskell

2 Answers

  1. Best Answer

    There is a more direct solution:

    import Data.List (partition) 

paresImpares :: Integral a => [a] -> ([a],[a])
paresImpares = partition even
    • 1

  2. Well I found an answer:

    paresImpares :: Integral a => [a] -> ([a],[a])
paresImpares [] = ([],[])
paresImpares (x:xs)
| x `mod` 2 == 0 = (x:pares,impares)
| otherwise = (pares, x:impares)
where (pares,impares) = paresImpares xs

    This works perfectly xmod 2can be replaced byeven

    • 0