Lazy cartesian product of multiple sequences (sequ

问题:

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 

回答1:

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

回答2:

I benchmarked the function Juliet linked to:

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

and a slightly modified (to make it an apples-to-apples comparison) version of yours:

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

Yours is significantly faster (and causes fewer GCs). Looks to me that what you have is good.