f# - 多个序列的惰性笛卡尔积(序列的序列)

Question

你能建议更简单、更清晰的方法来编写这个函数吗？

let cartesian_product sequences = 
    let step acc sequence = seq { 
        for x in acc do 
        for y in sequence do 
        yield Seq.append x [y] }
    Seq.fold step (Seq.singleton Seq.empty) sequences 

Answer 1

我对链接到的 Juliet 函数进行了基准测试:

let items = List.init 6 (fun _ -> [0..9])
cart1 items |> Seq.length |> ignore

Real: 00:00:03.324, CPU: 00:00:03.322, GC gen0: 80, gen1: 0, gen2: 0

和你的一个稍微修改(使其成为苹果对苹果的比较)版本:

let cartesian items =
  items |> Seq.fold (fun acc s ->
    seq { for x in acc do for y in s do yield x @ [y] }) (Seq.singleton [])

cartesian items |> Seq.length |> ignore

Real: 00:00:00.763, CPU: 00:00:00.780, GC gen0: 37, gen1: 2, gen2: 1

你的速度明显更快(并且导致更少的GC)。在我看来，你拥有的东西很好。