Question

以下是我在标准库中找到的内容吗？

split transformer1 ([], []) [("Foo", 1); ("Bar", 2); ("FooBar", 3)];;
> val it : string list * int list = (["FooBar"; "Bar"; "Foo"], [3; 2; 1])

这是我的实现

let split transformer defval stream =
    let rec split' s acc =
        match s with
            | [] -> acc
            | x::xs -> split' xs (transformer x acc)
    split' stream defval

let transformer1 (key, item) (akey, aitem) = (key::akey, item::aitem)
let transformer2 (key, item) (akey) = (key::akey)
let transformer3 (key, item) (a1, a2, a3) = (key::a1, item::a2, (item + 1)::a3)

split transformer1 ([], []) [("Foo", 1); ("Bar", 2); ("FooBar", 3)];;
split transformer2 ([]) [("Foo", 1); ("Bar", 2); ("FooBar", 3)];;
split transformer3 ([], [], []) [("Foo", 1); ("Bar", 2); ("FooBar", 3)];;

Answer 1

据我所知，你已经彻底改造了List.fold。以下是fold：

重现结果的方法

[("Foo", 1); ("Bar", 2); ("FooBar", 3)]
|> List.fold (fun (akey, aitem) (key, item) -> key::akey, item::aitem) ([], [])

[("Foo", 1); ("Bar", 2); ("FooBar", 3)]
|> List.fold (fun (akey) (key, item) -> key::akey) []

[("Foo", 1); ("Bar", 2); ("FooBar", 3)]
|> List.fold (fun (a1, a2, a3) (key, item) -> key::a1, item::a2, (item + 1)::a3) ([], [], [])

还有Seq.fold函数（可能是懒惰评估的）序列。

是否有列表模块功能，就像使用Transformer解压缩一样

1 个答案: