Question

我正在尝试从两个不同的列表创建一个元组列表，其中两个列表中的每个头都将成为一个元组（例如：headList1和headList2成为一个元组），并且每个元素都将继续），我在F＃中出现类型错误，我不确定自己在做什么错。我试着去掉括号内的括号，但仍然无济于事。

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 )

ex：（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

What about List.zip function?

> 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

或者您可以在F＃惯用语中将其写得更短：

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

如果您的问题是练习，并且您不想使用List.map2，则：

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

从两种不同类型的两个列表中创建元组

2 个答案: