Question

我正在尝试在OCaml中创建一个函数，它可以提供＆＃34; k-average＆＃34;列表中的连续元素。例如：

 average 4 [1; 2; 3; 4; 5; 6] = [2; 3; 4]

由于1,2,3,4的平均值为2，因此2,3,4,5的平均值为3,3,4,5,6的平均值为4.

我创建了一个平均列表的函数，但是每两个元素都有一个：

 let rec average2 xs = match xs with
 | [] -> []
 | x :: [] -> [x]
 | x :: x' :: xs -> if xs = [] then [(x + x') / 2] else [(x + x') / 2] @ 
 (average2 (x'::xs))

如何修改它以允许我平均k元素？

Answer 1

你应该做的只是验证列表是否具有适当的长度，然后两个递归函数将轻松完成：

let average n l =
  if List.length l < n then failwith "List is too small"
  else
    (* this function computes one k-average and returns the result *)
    let rec aux2 acc i = function
      | hd :: tl when i < n -> aux2 (acc + hd) (i + 1) tl 
      | _ -> acc / n

    in 
    let rec aux acc l = match l with
      (* the resulting list is reversed *) 
      | [] -> List.rev acc
      | _ :: tl -> 
        (* Get the k-average of the k first elements of the list *)
        let avgn = aux2 0 0 l in
        (* if the rest of the list is too small, we reached the
           end for sure, end *)
        if List.length tl < n then List.rev (avgn :: acc)
        (* recursive call on the rest of the list (without the head) *)
        else aux (avgn :: acc) tl
    in aux [] l

计算OCaml中数组的k运行平均值

1 个答案: