list - F#中列表的所有选择-更优雅，更简单

Question

有人可以提出更好和/或更优雅的实现方式:

让每个xs =
让REC的每个acc左，右=
与之匹配
| []-> acc
|右-> let new_left =左@ [List.hd右]
让next = List.tl对
让结果=(List.hd右)，左@下一个
每个'(结果:: acc)new_left下一个
每个'[] [] xs

它这样做:

>每个[1..3] ;;
val it:(int * int list)list = [(3，[1; 2]); (2，[1; 3]); (1，[2; 3])]

此函数也可以反向返回结果。想法是将所有元素作为元组包含元素和其余元素列表。

Answer 1

此处的语义略有不同，但是从您提供的Set的示例可能很合适:

let each xs =
    let X = set xs                           
    [ for x in xs -> (x, X - set [x]) ]


> fsi.AddPrinter( fun (x:Set<int>) -> sprintf "%A" (Set.to_list x))
> each [1..3];;
> val it : (int * Set<int>) list = [(1, [2; 3]); (2, [1; 3]); (3, [1; 2])]

// Edited as per comments.