创建直方图 OCaml

我的任务是创建一个直方图，输出某个元素在列表中出现的次数。

Input:[2;2;2;3;4;4;1]
Output[(2, 3); (2, 2); (2, 1); (3, 1); (4, 2); (4, 1); (1, 1)]  
Expected output : [(2, 3); (3, 1); (4, 2); (1, 1)]

My code:

let rec count a ls = match ls with
  |[]              -> 0
  |x::xs  when x=a -> 1 + count a xs
  |_::xs           -> count a xs

let rec count a = function
  |[]              -> 0
  |x::xs  when x=a -> 1 + count a xs
  |_::xs           -> count a xs

let rec histo l = match l with
|[] -> []
|x :: xs ->  [(x, count x l)] @ histo xs ;;

我做错了什么？

问题是 xs 可能包含等于 x 的元素。这就是您在输出中看到的内容：(2,3) 表示列表中有 3 乘以 2； xs 则等于 [2;2;3;4;4;1]... 等等。

另外（不影响结论）：您有 2 个计数定义，但它们是相同的。

要实现直方图，请使用 Hashtbl ：

let h = Hashtbl.create 1000;;    
List.iter (fun x -> let c = try Hashtbl.find h x with Not_found -> 0 in Hashtbl.replace h x (c+1)) your_list;;
Hashtbl.fold (fun x y acc ->  (x,y)::acc) h [];;