Question

我还在努力研究F＃的事情 - 试图找出如何在F＃中“思考”，而不仅仅是翻译我所知道的其他语言。

我最近一直在考虑你之前和之后没有1：1地图的情况。 List.map崩溃的情况。

这方面的一个例子是移动平均线，当平均超过n个项目时，通常你会得到长度为len的len-n + 1个结果。

对于那里的大师来说，这是一个很好的方法吗（使用从Jomo Fisher收集的队列）？

//Immutable queue, with added Length member
type Fifo<'a> =
    new()={xs=[];rxs=[]}
    new(xs,rxs)={xs=xs;rxs=rxs}

    val xs: 'a list;
    val rxs: 'a list;

    static member Empty() = new Fifo<'a>()
    member q.IsEmpty = (q.xs = []) && (q.rxs = [])
    member q.Enqueue(x) = Fifo(q.xs,x::q.rxs)
    member q.Length() = (List.length q.xs) + (List.length q.rxs)
    member q.Take() =
        if q.IsEmpty then failwith "fifo.Take: empty queue"
        else match q.xs with
                | [] -> (Fifo(List.rev q.rxs,[])).Take()
                | y::ys -> (Fifo(ys, q.rxs)),y

//List module, add function to split one list into two parts (not safe if n > lst length)
module List =
    let splitat n lst =
        let rec loop acc n lst =
            if List.length acc = n then
                (List.rev acc, lst)
            else
                loop (List.hd lst :: acc) n (List.tl lst)
        loop [] n lst

//Return list with moving average accross len elements of lst
let MovingAverage (len:int) (lst:float list) = 
    //ugly mean - including this in Fifo kills genericity
    let qMean (q:Fifo<float>) = ((List.sum q.xs) + (List.sum q.rxs))/(float (q.Length()))

    //get first part of list to initialise queue
    let (init, rest) = List.splitat len lst

    //initialise queue with first n items
    let q = new Fifo<float>([], init)

    //loop through input list, use fifo to push/pull values as they come
    let rec loop (acc:float list) ls (q:Fifo<float>) =
        match ls with
        | [] -> List.rev acc
        | h::t -> 
            let nq = q.Enqueue(h) //enqueue new value
            let (nq, _) = nq.Take() //drop old value
            loop ((qMean nq)::acc) t nq //tail recursion

    loop [qMean q] rest q

//Example usage    
MovingAverage 3 [1.;1.;1.;1.;1.;2.;2.;2.;2.;2.]

（也许更好的方法是通过继承Fifo来实现MovingAverageQueue？）

Answer 1

如果您不太关心性能，这是一个非常简单的解决方案：

#light

let MovingAverage n s =
   Seq.windowed n s
   |> Seq.map Array.average

let avgs = MovingAverage 5000 (Seq.map float [|1..999999|])

for avg in avgs do
    printfn "%f" avg
    System.Console.ReadKey() |> ignore

这会从头开始重新计算每个“窗口”的平均值，因此如果窗口很大，则会很差。

无论如何，请查看Seq.windowed：

http://research.microsoft.com/projects/cambridge/fsharp/manual/FSharp.Core/Microsoft.FSharp.Collections.Seq.html

因为在你的后兜里放这些东西很方便。

Answer 2

如果你做关心性能，那么你可以使用这样的东西有效地计算移动平均线（假设我们在3天的窗口内计算移动平均线）

Numbers[n]    Running Total[n]
---------     ---------------
n[0] = 7       7 = Numbers[0]
n[1] = 1       8 = RunningTotal[1-1] + Numbers[1]
n[2] = 2      10 = RunningTotal[2-1] + Numbers[2]
n[3] = 8      11 = RunningTotal[3-1] + Numbers[3] - Numbers[3-3]
n[4] = 4      14 = RunningTotal[4-1] + Numbers[4] - Numbers[4-3]
n[5] = 1      13 = RunningTotal[5-1] + Numbers[5] - Numbers[5-3] 
n[6] = 9      14 = RunningTotal[6-1] + Numbers[6] - Numbers[6-3]
...
N             RunningTotal[N] = RunningTotal[N-1] + Numbers[N] - Numbers[N-3]

关于此问题的难点在于您先前的运行总数和Number _N-window。我想出了以下代码：

let movingAverage days l =
    seq {
        let queue = new Queue<_>(days : int)
        let divisor = decimal days

        let total = ref 0m
        for cur in l do
            queue.Enqueue(cur)
            total := !total + cur
            if queue.Count < days then
                yield (cur, 0m)
            else
                yield (cur, !total / divisor)
                total := !total - (queue.Dequeue())
    }

这个版本不像Haskell代码那样漂亮，但它应该避免与每次运行时重新计算“窗口”相关的性能问题。它保持一个运行总计并将以前使用过的数字保存在队列中，所以它应该非常快。

为了好玩，我写了一个简单的基准：

#light
open System
open System.Collections.Generic
open System.Diagnostics;

let windowAverage days (l : #seq<decimal>) = Seq.windowed days l |> Seq.map (Seq.average)

let princessAverage days l =
    seq {
        let queue = new Queue<_>(days : int)
        let divisor = decimal days

        let total = ref 0m
        for cur in l do
            queue.Enqueue(cur)
            total := !total + cur
            if queue.Count < days then
                yield (cur, 0m)
            else
                yield (cur, !total / divisor)
                total := !total - (queue.Dequeue())
    }

let testData =
    let rnd = new System.Random()
    seq { for a in 1 .. 1000000 -> decimal (rnd.Next(1000)) }

let benchmark msg f iterations =
    let rec loop = function
        | 0 -> ()
        | n -> f 3 testData |> ignore; loop (n - 1)

    let stopWatch = Stopwatch.StartNew()
    loop iterations
    stopWatch.Stop()
    printfn "%s: %f" msg stopWatch.Elapsed.TotalMilliseconds

let _ =
    let iterations = 10000000
    benchmark "princessAverage" princessAverage iterations
    benchmark "windowAverage" windowAverage iterations
    printfn "Done"

结果：

princessAverage: 1670.791800
windowAverage: 2986.146900

我的版本快了1.79倍。

Answer 3

这是一个（更正，我希望）F＃版本的Haskell解决方案提议here。

编辑：现在尾递归，不是更快，但是在n = 50000时不会爆炸。（请参阅编辑非尾递归版本的历史记录）

let LimitedAverage n ls = 
    let rec loop acc i n ls = 
        match i with
        | 0 -> acc //i counts down from n to 0, when we hit 0 we stop
        | _ -> match ls with
               | [] -> acc //if we hit empty list before end of n, we stop too
               | x::xs -> (loop (acc + (x / float n)) (i - 1) n xs) //divide this value by n, perform average on 'rest' of list
    loop 0. n n ls

LimitedAverage 50000 (List.map float [1..9999999])
//>val it : float = 25000.5

let rec MovingAverage3 n ls = 
    let rec acc loop i n ls = 
        match i with 
        | 0 -> List.rev acc //i counts down from n to 0, when we hit 0 we stop
        | _ -> match ls with
                | [] -> List.rev acc //if we hit empty list before end of n, we stop too
                | x::xs -> loop (LimitedAverage2 n ls :: acc) (i - 1) n xs // limited average on whole list, then moving average on tail
    loop [] (n + 1) n ls 

MovingAverage3 50000 (List.map float [1..9999999])
//>val it : float list = [25000.5; 25001.5; 25002.5; ...]

Answer 4

如果您关心性能并喜欢优雅代码，请尝试

module MovingAverage = 
    let selfZip n l =
        Seq.skip n l |> Seq.zip l 

    let runTotal i z =
        Seq.scan ( fun sum (s, e) -> sum - s + e ) i z

    let average n l:seq<'a> =
        Seq.skip n l
        |> selfZip n
        |> runTotal (l |> Seq.take n |> Seq.sum)
        |> Seq.map ( fun sum -> decimal sum / decimal n ) 

 let ma = MovingAverage.average 2 myseq

使用FSUnit我们可以测试它

 let myseq = seq { for i in 0..10 do yield i }

 Seq.nth 0 ma |> should equal 0.5
    Seq.nth 1 ma |> should equal 1.5
    Seq.nth 2 ma |> should equal 2.5
    Seq.nth 3 ma |> should equal 3.5

算法的技巧是前n个数和第一个和然后通过添加窗口的头部来保持运行总计并减去窗口的尾部。滑动窗口是通过对序列进行自我压缩而实现第二个按窗口大小拉高级的参数。

在管道的末尾，我们只是按窗口划分运行总计尺寸。

注意扫描就像折叠一样，但会将状态的每个版本都生成一个序列。

一个更加优雅的解决方案虽然可能会受到性能影响观察一下，如果我们填零我们不需要的序列计算初始总和。

namespace Utils

module MovingAverage = 
    let selfZip n l =
        Seq.skip n l |> Seq.zip l 

    let rec zeros = 
        seq { yield 0.0; yield! zeros} 

    // Create a running total given
    let runTotal z =
        Seq.scan (fun sum (s,e) -> sum - s + e ) 0.0 z

    let average n l =
        Seq.concat [(Seq.take n zeros); l]
        |> selfZip n
        |> runTotal
        |> Seq.map ( fun sum -> sum / float n ) 
        |> Seq.skip n

由于与第二个间接相关，可能会导致性能下降两个序列的包装，但也许并不重要关于窗口的大小

Answer 5

这是我的版本。

let sma list n =
    let len = List.length list
    let rec loop acc sum cnt =
        if cnt >= len then List.rev acc
        else if cnt < n-1 then loop (0.0::acc) (sum + List.nth list cnt) (cnt+1)
        else loop (((sum + List.nth list cnt)/(float n))::acc) (sum + (List.nth list cnt) - (List.nth list (cnt-n+1))) (cnt+1)
    loop [] 0.0 0

示例：

sma (List.map float [5..50]) 5
[0, 0, 0, 0, 7, 8, 9, ...]

Answer 6

据我所知，您的代码中充满了let个语句。我不熟悉F＃，但确实做了一些Haskell。功能范式意味着不考虑“如何”而是关于“什么”：你认为Fifo，但你实际上应该只指定移动平均线的语义。

-- the limited average of a list
limitedaverage 0 _ = 0
limited verage n (x:xs) = (x/n) + ( limited average (n-1) xs )

-- a list, transformed into a sequence of moving averages of 
movingaverages n [] = []
movingaverages n (x:xs) = ( movingaverage n (x:xs) : movingaverages n xs )

计算F＃中的移动平均线

6 个答案: