F＃将参数配置为高级函数或声明自己的函数

Question

我有一个将输入列表处理为不同类型的函数，但其​​本身并不是有趣的事情。

let testList = [(1,"c");(2,"a");(1,"b")] 

let rec toRel xs = 
    let rec insert (a, b) ys = 
     match ys with 
     | []      -> [(a, [b])] 
     | (a', b')::ys' when a' = a -> (a', b::b')::ys' 
     | y::ys'     -> y::insert (a, b) ys' 
    match xs with 
    | []   -> [] 
    | (a,b)::rest -> insert (a, b) (toRel rest) 

toRel testList //Expected [(1, ["c";"b"]); (2, ["a"])] 

这是很好，并且能够被重构到：

testList |> List.groupBy xs |> List.map (fun (k, v) -> (k, list.map snd v)) 

可以得到相同的结果。

当我试图将这个管道过程封装到函数中时，我遇到了问题。

let toRelHigherOrder xs = List.groupBy xs |> List.map (fun (k, v) -> (k, list.map snd v)) 

toRelHigherOrder testList 

This expression was expected to have type ''a -> 'b' but here has type '(int * string) list。

什么给？

Answer 1

我觉得你管是错误的，它应该是：

testList |> List.groupBy fst |> List.map (fun (k, v) -> (k, List.map snd v)) 

所以你的功能应该是：

let f x = x |> List.groupBy fst |> List.map (fun (k, v) -> (k, List.map snd v))