Haskell组预定义的长度没有显式递归

Question

How can I make a function that groups element of a list into lists of predefined length without using explicit recursion.

For example, for 2:

[1, 2, 3, 4, 5, 6, 7, 8]
[[1, 2], [3, 4], [5, 6], [7, 8]]

Thank you!

Answer 1

The following would work (though you may consider it cheating to use groupBy from Data.List)

import Data.Function (on)
import Data.List     (groupBy)

group :: Int -> [a] -> [[a]]
group n = map (map snd)
        . groupBy ((==) `on` fst)
        . zip (enumFrom 1 >>= replicate n)

Answer 2

This is a one-liner if you set your wrap at 100 characters :). I'm guessing you are looking for something that uses a fold.

group :: Int -> [a] -> [[a]]
group n = uncurry (:) . foldr (\x (l,r) -> if length l == n then ([x],l:r) 
                                                            else (x:l,r))
                              ([],[])