我目前正在学习函数式编程和F#,我想做一个循环控制直到n-2。例如:
Given a list of doubles, find the pairwise average,
e.g. pairwiseAverage [1.0; 2.0; 3.0; 4.0; 5.0] will give [1.5; 2.5; 3.5; 4.5]在做了一些实验和搜索之后,我有几种方法可以做到:
方法1:
let pairwiseAverage (data: List<double>) =
[for j in 0 .. data.Length-2 do
yield (data.[j]+data.[j+1])/2.0]方法2:
let pairwiseAverage (data: List<double>) =
let averageWithNone acc next =
match acc with
| (_,None) -> ([],Some(next))
| (result,Some prev) -> ((prev+next)/2.0)::result,Some(next))
let resultTuple = List.fold averageWithNone ([],None) data
match resultTuple with
| (x,_) -> List.rev x方法3:
let pairwiseAverage (data: List<double>) =
// Get elements from 1 .. n-1
let after = List.tail data
// Get elements from 0 .. n-2
let before =
data |> List.rev
|> List.tail
|> List.rev
List.map2 (fun x y -> (x+y)/2.0) before after我只想知道是否有其他方法来处理这个问题。谢谢。
发布于 2017-03-13 11:50:20
仅使用内置:
list |> Seq.windowed 2 |> Seq.map Array.averageSeq.windowed n为您提供n个元素的滑动窗口。
发布于 2017-03-13 10:46:44
另一种简单的方法是使用Seq.pairwise
就像这样
list |> Seq.pairwise |> Seq.map (fun (a,b) -> (a+b)/2.0)发布于 2017-03-13 15:17:11
以上建议的方法适用于短窗口,如问题中的方法。对于长度大于2的窗户,不能成对使用。hlo的答案概括为更宽的窗口,如果窗口长度不太大,这是一种干净而快速的方法。对于非常宽的窗口,下面的代码运行得更快,因为它只添加一个数字,并从前一个窗口的值中减去另一个数字。注意,Seq.map2 (和Seq.map)自动处理不同长度的序列。
let movingAverage (n: int) (xs: float List) =
let init = xs |> (Seq.take n) |> Seq.sum
let additions = Seq.map2 (fun x y -> x - y) (Seq.skip n xs) xs
Seq.fold (fun m x -> ((List.head m) + x)::m) [init] additions
|> List.rev
|> List.map (fun (x: float) -> x/(float n))
xs = [1.0..1000000.0]
movingAverage 1000 xs
// Real: 00:00:00.265, CPU: 00:00:00.265, GC gen0: 10, gen1: 10, gen2: 0作为比较,上面的函数执行上面的计算比加窗的等价物快60倍:
let windowedAverage (n: int) (xs: float List) =
xs
|> Seq.windowed n
|> Seq.map Array.average
|> Seq.toList
windowedAverage 1000 xs
// Real: 00:00:15.634, CPU: 00:00:15.500, GC gen0: 74, gen1: 74, gen2: 71我试图使用List.rev消除foldBack,但没有成功。
https://stackoverflow.com/questions/42761575
复制相似问题