如何根据一个可观测序列计算一个ETA?

IObservable<Timestamped<double>> estimatedCompletion =
    progressObservable
        .Timestamp()
        .Buffer(2, 1)
        .Where(x => x.Count() == 2)
        .Scan((a, b) => a.Take(1).Concat(b).Take(1).Concat(b.Skip(1)).ToList())
        .Select(x => new
        {
            current = x[1].Value,
            delta = x[1].Timestamp.Subtract(x[0].Timestamp),
        })
        .Select(x => new
        {
            x.current,
            rate = x.current / x.delta.TotalSeconds,
        })
        .Select(x => new
        {
            x.current,
            estimated = DateTimeOffset.Now.AddSeconds((1.0 - x.current) / x.rate),
        })
        .Select(x => new Timestamped<double>(x.current, x.estimated));

IObservable<Timestamped<double>> DateTimeOffset 1.0 100%

.Buffer(2, 1).Where(x => x.Count() == 2) .Scan((a, b) => a.Take(1).Concat(b).Take(1).Concat(b.Skip(1)).ToList())

您可以使用以下代码测试它:

var rnd = new Random();
var progressObservable = Observable.Generate(0, x => x <= 100, x => x + 1, x => x / 100.0, x => TimeSpan.FromSeconds(rnd.NextDouble()));

它可能看起来像这样:

var secondsRemaining = progressObservable
    .Timestamp()
    .Buffer(5, 1)
    .Select(l => ((l[4].Timestamp - l[0].Timestamp).TotalMilliseconds / (l[4].Value - l[0].Value)) * (100 - l[4].Value))
    .Select(msRemaining => msRemaining / 1000);

说明:

对于每个进度更新,

• .Buffer(5, 1) 发布最近5个进度更新的列表。
• .Timestamp() 将时间戳钉在每个时间戳上。
• 第一个 Select 运算符计算最远时间戳和最新时间戳之间的毫秒数,将其除以实现的进度,然后将其乘以剩余进度,输出剩余毫秒数。
• 最后 选择 除以1000得到剩余秒数。