list - Nested chunksOf in Haskell? -

say want this:

nestedchunksof [3, 2] [1,1,1,2,2,2,3,3,3,4,4,4] == [[[1,1,1], [2,2,2]], [[3,3,3], [4,4,4]]] 

in python, can this

def group(a, *ns):     n in ns:         = [a[i:i+n] in xrange(0, len(a), n)]     return  group([1,1,1,2,2,2,3,3,3,4,4,4], 3, 2) == [[[1,1,1],[2,2,2]],[[3,3,3],[4,4,4]]] 

but in haskell, can't say

nestedchunksof :: [int] -> [a] -> [[a]] 

or

nestedchunksof :: [int] -> [a] -> [[[a]]] 

so how can achieve same thing in haskell?

it can done dependent types.

we'd express length of [int] argument determines type of result. need 2 things that: list type fixed length, , type-level function computes return type length:

{-# language datakinds, gadts, typefamilies #-}  import data.list.split  data nat = z | s nat -- natural numbers (zero, successor)  data vec n -- "n" length lists of "a" elements   nil  :: vec z   (:>) :: -> vec n -> vec (s n) infixr 5 :>  type family iterate n f   iterate z     f =   iterate (s n) f = f (iterate n f a) 

iterate n f a applies type constructor f n times argument. example, iterate (s (s z)) [] int reduces [[int]]. nestedchunksof can written directly now:

nestedchunksof :: vec n int -> [a] -> iterate (s n) [] nestedchunksof nil       = nestedchunksof (n :> ns) = chunksof n $ nestedchunksof ns 

usage:

> nestedchunksof (2 :> 3 :> nil) [1,1,1,2,2,2,3,3,3,4,4,4] [[[1,1,1],[2,2,2]],[[3,3,3],[4,4,4]]]