计算移动平均线在F＃ [英] Calculating a moving average in F#

问题描述

我还在上groking F＃的东西 - 试图找出如何去思考F＃中，而不是从其他语言我只知道翻译

I'm still working on groking the F# thing - trying to work out how to 'think' in F# rather than just translating from other languages I know.

我最近一直在思考，你没有1例：前后1的地图。情况下，List.map倒下。

I've recently been thinking about the cases where you don't have a 1:1 map between before and after. Cases where List.map falls down.

这方面的一个例子是移动平均线，在那里通常你会在n个项目均在+ 1结果长度LEN列表len个-N。

One example of this is moving averages, where typically you will have len-n+1 results for a list of length len when averaging over n items.

对于大师在那里，这是一个很好的方式（使用队列从<一个捏​​做href="http://blogs.msdn.com/jomo_fisher/archive/2007/12/12/strange-confluence-an-immutable-queue-in-f.aspx"相对=nofollow>乔莫·费舍尔）？

For the gurus out there, is this a good way to do it (using queue pinched from Jomo Fisher)?

//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.]

（可能是更好的方式是由先进先出继承来实现MovingAverageQueue？）

(Maybe a better way would be to implement a MovingAverageQueue by inheriting from Fifo?)

推荐答案

如果你没有太在意性能，这里是一个非常简单的解决办法：

If you don't care too much about performance, here is a very simple solution:

#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

该重新计算每个从头开始'窗口'的平均值，因此它是可怜的，如果窗户都大。

This recomputes the average of each 'window' from scratch, so it is poor if the windows are large.

在任何情况下，请Seq.windowed：

In any case, check out Seq.windowed:

<一个href="http://research.microsoft.com/projects/cambridge/fsharp/manual/FSharp.Core/Microsoft.FSharp.Collections.Seq.html">http://research.microsoft.com/projects/cambridge/fsharp/manual/FSharp.Core/Microsoft.FSharp.Collections.Seq.html

，因为它很方便的在你的后面的口袋里这样的事情。

as it's handy to have in your back pocket for such things.

这篇关于计算移动平均线在F＃的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！