你怎么能在RX中做一个简单的,有状态的序列变换？
假设我们想要对IObservable noisySequence进行指数移动平均变换.
每当noisySequence滴答时,emaSequence应该勾选并返回该值
(prevIoUsEmaSequenceValue *(1-lambda)latestNoisySequenceValue * lambda)
我猜我们使用的是主题,但究竟是怎么回事？
public static void Main()
{
var rand = new Random();
IObservable<double> sequence = Observable
.Interval(TimeSpan.FromMilliseconds(1000))
.Select(value => value + rand.NextDouble());
Func<double,double> addNoise = x => x + 10*(rand.NextDouble() - 0.5);
IObservable<double> noisySequence = sequence.Select(addNoise);
Subject<double> exponentialMovingAverage = new Subject<double>(); // ???
sequence.Subscribe(value => Console.WriteLine("original sequence "+value));
noisySequence.Subscribe(value => Console.WriteLine("noisy sequence " + value));
exponentialMovingAverage.Subscribe(value => Console.WriteLine("ema sequence " + value));
Console.ReadLine();
}
解决方法
这是您可以将状态附加到序列的方法.在这种情况下,它计算最后10个值的平均值.
var movingAvg = noisySequence.Scan(new List<double>(),(buffer,value)=>
{
buffer.Add(value);
if(buffer.Count>MaxSize)
{
buffer.RemoveAt(0);
}
return buffer;
}).Select(buffer=>buffer.Average());
但你可以使用Window(这是一种泛化的方式)来获得你的平均值.
noisySequence.Window(10) .Select(window=>window.Average()) .SelectMany(averageSequence=>averageSequence);
