多序列的懒惰笛卡尔积(序列序列)

Question

你能建议更简单,更清晰的方式来编写这个函数吗？

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

不太优雅，但（似乎）更快的解决方案：

let cartesian_product2 sequences = 
    let step acc sequence = seq { 
        for x in acc do 
        for y in sequence do 
        yield seq { yield! x ; yield y } }
    Seq.fold step (Seq.singleton Seq.empty) sequences 

;

> cartesian items |> Seq.length;;
Real: 00:00:00.405, CPU: 00:00:00.405, GC gen0: 37, gen1: 1, gen2: 0
val it : int = 1000000
> cartesian_product2 items |> Seq.length;;
Real: 00:00:00.228, CPU: 00:00:00.234, GC gen0: 18, gen1: 0, gen2: 0
val it : int = 1000000