列表列表，最多为 1 个偶数的列表的数量

Question

I'm new in Haskell and I tried this using recursion

f :: [[Integer]] -> Integer
f [] = 0
f [x] = 1
f ((x : xs) : xss)
    | even x = 1 + f (xs : xss)
    | otherwise = f (xs : xss)

but doesn't work and I think its totally wrong. How can I make this using recursion, list comprehension and map, filter functions? For exemple, f[[2,3,4], [1,0], [1,3,5,7]] = 2

Answer 1

If I get the tile right you want to count all inner-lists with at-most 1 even number in them.

So let's split this - first count all even numbers in a list:

evenNumberCount :: [Integer] -> Int
evenNumberCount ls = length (filter even ls)

having this we can do a test if a list has at most one:

atMostOneEvenNumber :: [Integer] -> Bool
atMostOneEvenNumber ls = evenNumberCount ls <= 1

I guess at this point it's easy enough to plug together the answer:

f :: [[Integer]] -> Int
f ls = length (filter atMostOneEvenNumber ls)

your example:

> f[[2,3,4], [1,0], [1,3,5,7]]
2

Note: this has the wrong signature (because length:: [a] -> Int ) - if you want Integer either add fromIntegral to the right side or switch to genericLength and adapt the signatures

(Note to Haskellers reading: I'm aware that length:: Foldable t => ta -> Int don't think this helps here;) )