F# 多序列的惰性笛卡尔积(序列序列)
你能推荐一种更简单更清晰的方法来编写这个函数吗F# 多序列的惰性笛卡尔积(序列序列),f#,sequences,F#,Sequences,你能推荐一种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 我对朱丽叶的功能进行了基
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
我对朱丽叶的功能进行了基准测试:
let items = List.init 6 (fun _ -> [0..9])
cart1 items |> Seq.length |> ignore
Real:00:03.324,CPU:00:00:03.322,GC gen0:80,gen1:0,gen2:0
还有一个稍加修改的(使之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:00:00.763,CPU:00:00.780,GC gen0:37,gen1:2,gen2:1
你的速度明显更快(并且导致地面军事系统更少)。在我看来,您拥有的很好。不那么优雅,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
)
请参阅和,对于基本上与此F#解决方案相同的C#解决方案,请参阅这似乎更快,因为您返回的是
seq
,而不是seq
,即您的内部序列没有被评估。试试看cartesian_product2 items |>Seq.map Seq.toList |>Seq.length
@Daniel:好的观点。但是它仍然比作者的解决方案快2倍(同时仍然是懒惰的,根据需要)。这个解决方案在构建序列和实现序列方面是否更快?或者两者都没有,因为测试没有实现子序列?老问题,但是你有一个gen 2集合。这意味着地面军事系统的数量减少了多少?
> 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