@Enigmativity 的回答并不完全符合规范。它可能适用于你想要的东西。
他的答案定义了 1 秒窗口,并从每个窗口中获取第一个窗口。但是,这并不能保证您在项目之间保持一秒钟的沉默。考虑这种情况:
t : ---------1---------2---------3
source: ------1---2------3---4----5--|
window: ---------|---------|---------|
spec : ------1----------3-----------|
enigma: ------1---2----------4-------|
答案意味着您在第 1 项之后想要一秒钟什么都没有。之后的下一项是 3,然后一直保持沉默。这是测试代码:
var scheduler = new TestScheduler();
var source = scheduler.CreateColdObservable<int>(
RxTest.OnNext(700.MsTicks(), 1),
RxTest.OnNext(1100.MsTicks(), 2),
RxTest.OnNext(1800.MsTicks(), 3),
RxTest.OnNext(2200.MsTicks(), 4),
RxTest.OnNext(2600.MsTicks(), 5),
RxTest.OnCompleted<int>(3000.MsTicks())
);
var expectedResults = scheduler.CreateHotObservable<int>(
RxTest.OnNext(700.MsTicks(), 1),
RxTest.OnNext(1800.MsTicks(), 3),
RxTest.OnCompleted<int>(3000.MsTicks())
);
var target = source
.Window(() => Observable.Timer(TimeSpan.FromSeconds(1.0), scheduler))
.SelectMany(xs => xs.Take(1));
var observer = scheduler.CreateObserver<int>();
target.Subscribe(observer);
scheduler.Start();
ReactiveAssert.AreElementsEqual(expectedResults.Messages, observer.Messages);
我认为解决此问题的最佳方法是基于Scan 的解决方案,带有时间戳。您基本上将最后一条合法消息保存在内存中,并带有时间戳,如果新消息早一秒,则发出。否则,不要:
public static IObservable<T> TrueThrottle<T>(this IObservable<T> source, TimeSpan span)
{
return TrueThrottle<T>(source, span, Scheduler.Default);
}
public static IObservable<T> TrueThrottle<T>(this IObservable<T> source, TimeSpan span, IScheduler scheduler)
{
return source
.Timestamp(scheduler)
.Scan(default(Timestamped<T>), (state, item) => state == default(Timestamped<T>) || item.Timestamp - state.Timestamp > span
? item
: state
)
.DistinctUntilChanged()
.Select(t => t.Value);
}
注意:测试代码使用 Nuget Microsoft.Reactive.Testing 和以下帮助类:
public static class RxTest
{
public static long MsTicks(this int i)
{
return TimeSpan.FromMilliseconds(i).Ticks;
}
public static Recorded<Notification<T>> OnNext<T>(long msTicks, T t)
{
return new Recorded<Notification<T>>(msTicks, Notification.CreateOnNext(t));
}
public static Recorded<Notification<T>> OnCompleted<T>(long msTicks)
{
return new Recorded<Notification<T>>(msTicks, Notification.CreateOnCompleted<T>());
}
public static Recorded<Notification<T>> OnError<T>(long msTicks, Exception e)
{
return new Recorded<Notification<T>>(msTicks, Notification.CreateOnError<T>(e));
}
}