Question

我是函数式语言编程的新手。我正在尝试实现F＃collect for list。

let rec collect func list =
     match list with
     | [] -> []
     | hd::tl -> let tlResult = collect func tl 
                 func hd::tlResult;;

collect (fun x -> [for i in 1..3 -> x * i]) [1;2;3];;

应打印：

val it : int list = [1; 2; 3; 2; 4; 6; 3; 6; 9]

但我得到了：

val it : int list = [[1; 2; 3;], [2; 4; 6;], [3; 6; 9]]

Answer 1

这是一个尾递归collect，它不会为大型列表堆栈溢出。

let collect f xs =
    let rec prepend res xs = function
        | [] -> loop res xs
        | y::ys -> prepend (y::res) xs ys
    and loop res = function
        | [] -> List.rev res
        | x::xs -> prepend res xs (f x)
    loop [] xs

一个更简单的版本，有点作弊，是：

let collect (f: _ -> list<_>) (xs: list<_>) = [ for x in xs do yield! f x ]

Answer 2

collect函数在函数式样式中有效实现很棘手，但您可以使用连接列表的@运算符轻松实现它：

let rec collect f input =
  match input with
  | [] -> []
  | x::xs -> (f x) @ (collect f xs)

在F＃中实现列表收集

2 个答案: