f# - 懒惰地生成powerset

Question

我想计算一组的幂集。因为我一次不需要整个powerset，所以最好懒惰地生成它。

例如:

powerset (set ["a"; "b"; "c"]) =
seq {
  set [];
  set ["a"];
  set ["b"];
  set ["c"];
  set ["a"; "b"];
  set ["a"; "c"];
  set ["b"; "c"];
  set ["a";"b"; "c"];
}

由于结果是一个序列，我更喜欢上面的顺序。我怎样才能在 F# 中以惯用的方式做到这一点？

编辑:

这就是我将要使用的(基于 BLUEPIXY 的回答):

let powerset s =
    let rec loop n l =
        seq {
              match n, l with
              | 0, _  -> yield []
              | _, [] -> ()
              | n, x::xs -> yield! Seq.map (fun l -> x::l) (loop (n-1) xs)
                            yield! loop n xs
        }   
    let xs = s |> Set.toList     
    seq {
        for i = 0 to List.length xs do
            for x in loop i xs -> set x
    }

感谢大家的出色投入。

Answer 1

let rec comb n l =
  match n, l with
  | 0, _  -> [[]]
  | _, [] -> []
  | n, x::xs -> List.map (fun l -> x ::l) (comb (n - 1) xs) @ (comb n xs)

let powerset xs = seq {
    for i = 0 to List.length xs do
      for x in comb i xs -> set x
  }

演示

> powerset ["a";"b";"c"] |> Seq.iter (printfn "%A");;
set []
set ["a"]
set ["b"]
set ["c"]
set ["a"; "b"]
set ["a"; "c"]
set ["b"; "c"]
set ["a"; "b"; "c"]
val it : unit = ()