从两个不同类型的列表创建元组

Question

我试着去掉圆括号，把它们放在大括号里，等等，但还是没有用。

let rec pairToTuple letter count = // assume that each list is the same, will return a list with (letter, count) in itself
  match letter with
  | [] -> [()]
  | e1::rest1, e2::rest2 -> let tup = (e1, e2) 
                            tup::(pairToTuple rest1 rest2 )

例如：(a，b，c)和(10,20,30)将变成(a,10);(b,20);(c,30)

/home/codio/workspace/program/Program.fs(180,5): error FS0001: This expression was expected to have type    ''a list'    but here has type    ''b * 'c'
[/home/codio/workspace/program/program.fsproj]

Answer 1

那么List.zip函数呢？

> List.zip ["a"; "b"; "c"; "d"] [1; 2; 3; 4];;
val it : (string * int) list = [("a", 1); ("b", 2); ("c", 3); ("d", 4)]

Answer 2

您可以简单地使用List.map2 function

let pairToTuple letterList countList =
    List.map2 (fun letter count -> (letter, count)) letterList countList

或者你可以用F#的习惯用法把它写得更短：

let pairToTuple = List.map2 (fun letter count -> (letter, count))

如果你的问题是一个练习，并且你不想使用List.map2，那么：

let rec pairToTuple letterList countList =
    match letterList, countList with
    | [], _ -> []
    | _, [] -> []
    | letter :: res1, count :: res2 ->
        (letter, count) :: pairToTuple res1 res2