list - 整数的Ocaml列表到整数列表的列表(与展平相反)

Question

带有整数列表，例如:

[1;2;3;4;5;6;7;8;9]

如何从上面创建一个整数列表，所有新列表都具有相同的指定长度？

例如，我需要从:

[1;2;3;4;5;6;7;8;9] to [[1;2;3];[4;5;6];[7;8;9]]

要拆分的数字为3？

谢谢你的时间。

Answer 1

所以你真正想要的是一个类型函数

val split : int list -> int -> int list list

需要一个整数列表和一个子列表大小。哪一个更普遍呢？

val split : 'a list -> int -> 'a list list

这里是实现:

let split xs size =
  let (_, r, rs) =
    (* fold over the list, keeping track of how many elements are still
       missing in the current list (csize), the current list (ys) and
       the result list (zss) *) 
    List.fold_left (fun (csize, ys, zss) elt ->
      (* if target size is 0, add the current list to the target list and
         start a new empty current list of target-size size *)
      if csize = 0 then (size - 1, [elt], zss @ [ys])
      (* otherwise decrement the target size and append the current element
         elt to the current list ys *)
      else (csize - 1, ys @ [elt], zss))
      (* start the accumulator with target-size=size, an empty current list and
         an empty target-list *)
        (size, [], []) xs
  in
  (* add the "left-overs" to the back of the target-list *)
  rs @ [r]

如果您为此获得加分，请告诉我! ;)