ocaml-取消可选列表:有没有更简单的方法? [英] ocaml - deoptionalize a list: is there a simpler way?

问题描述

我写了一个函数来取消整数列表的可选操作，我想知道是否有更好的写方法.

I have written a function to deoptionalize an integer list and I would like to know if there is a better way to write it.

let deoptionalize (lst:'a option list) : 'a list =
    List.map ~f:(fun x -> match x with Some x -> x | None -> assert false)
                    (List.filter ~f:(fun x -> x <> None) lst)
;;

在作业中，我目前正在使用地图，必须使用过滤器.

In the assignment I am currently working its using map and filter is a must.

推荐答案

我认为手动编码"解决方案(即没有map和filter的)更易于阅读，但是如果您确实需要使用他们，你去了:

I suppose that a "hand-coded" solution (i.e. without map and filter) is easier to read, but if you really need to use them, here you go:

似乎您正在使用Core库.如果是这样，我认为您的解决方案还不错，但是可以写得更紧凑一些:

It seems that you are using the Core library. If so, I think your solution is not so bad, but can be written a bit more compact:

let deoptionalize lst = 
    List.filter ~f:(is_some) lst
    |> List.map ~f:(function | Some x -> x | None -> assert false)

如果您不介意警告(我不鼓励您这样做)，您甚至可以省略其他内容:

If you don't mind warnings (which I discourage you to do), you can even leave out some more:

let deoptionalize lst = 
    List.filter ~f:(is_some) lst
    |> List.map ~f:(fun (Some x) -> x)

实际上，Core提供了将两者结合的filter_map(感谢@Ramon Snir作为提示)，因此您可以使用:

Actually, Core provides filter_map (thanks @Ramon Snir for the hint) which combines both, so you can use:

let deopt lst = 
    List.filter_map ~f:(fun x -> x) lst;;

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