多个序列的惰性笛卡尔积(序列的序列) [英] Lazy cartesian product of multiple sequences (sequence of sequences)

问题描述

你能建议更简单、更清晰的方法来编写这个函数吗?

Can you suggest simpler and clearer way to write this function?

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 

推荐答案

不太优雅，但(似乎)更快的解决方案:

Less elegant, but (seems to be) faster solution:

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

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