计算F#中的移动平均线

Question

我还在努力研究F#的东西 - 试图弄清楚如何在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.

其中一个例子是移动平均线,通常在平均n个项目时,长度为len的列表会产生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捏的队列)？

//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,这里有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.