我仍在研究 F# 的东西 - 试图弄清楚如何在 F# 中“思考”，而不仅仅是从我知道的其他语言翻译。
我最近一直在考虑在之前和之后之间没有 1:1 映射的情况。List.map 崩溃的情况。
其中一个示例是移动平均线，通常在对 n 个项目进行平均时，对于长度为 len 的列表，您将得到 len-n+1 个结果。
对于那里的大师来说，这是一个好方法吗（使用从Jomo Fisher捏出来的队列）？
//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从 Fifo 继承来实现 MovingAverageQueue？）