system.reactive - 如何在另一个可观察对象发出 true 时缓冲项目，并在 false 时释放它们

Question

我有一个源流，通常希望在项目到达时发出它们。但是还有另一个可观察到的——我们称之为“门”。当门关闭时，源项目应该缓冲并仅在门打开时释放。

我已经能够编写一个函数来执行此操作，但它似乎比需要的更复杂。我必须使用 Observable.Create 方法。我假设有一种方法可以使用 Delay 或 Buffer 方法仅使用几行功能更强大的代码来实现我的目标，但我不知道如何实现。 Delay 似乎特别有前途，但我不知道如何有时延迟，有时允许一切立即通过(零延迟)。同样，我认为我可以使用 Buffer，然后使用 SelectMany；当门打开时，我会有长度为 1 的缓冲区，而当门关闭时，我会有更长的缓冲区，但我还是不知道如何让它工作。

这是我构建的适用于我所有测试的内容:

/// <summary>
/// Returns every item in <paramref name="source"/> in the order it was emitted, but starts
/// caching/buffering items when <paramref name="delay"/> emits true, and releases them when
/// <paramref name="delay"/> emits false.
/// </summary>
/// <param name="delay">
/// Functions as "gate" to start and stop the emitting of items. The gate is opened when true
/// and closed when false. The gate is open by default.
/// </param>

public static IObservable<T> DelayWhile<T>(this IObservable<T> source, IObservable<bool> delay) =>
    Observable.Create<T>(obs =>
    {
        ImmutableList<T> buffer = ImmutableList<T>.Empty;
        bool isDelayed = false;
        var conditionSubscription =
            delay
            .DistinctUntilChanged()
            .Subscribe(i =>
            {
                isDelayed = i;
                if (isDelayed == false)
                {
                    foreach (var j in buffer)
                    {
                        obs.OnNext(j);
                    }
                    buffer = ImmutableList<T>.Empty;
                }
            });
        var sourceSubscription =
            source
            .Subscribe(i =>
            {
                if (isDelayed)
                {
                    buffer = buffer.Add(i);
                }
                else
                {
                    obs.OnNext(i);
                }
            });
        return new CompositeDisposable(sourceSubscription, conditionSubscription);
    });

这是另一个通过测试的选项。它非常简洁，但没有使用 Delay 或 Buffer 方法；我需要手动进行延迟/缓冲。

public static IObservable<T> DelayWhile<T>(this IObservable<T> source, IObservable<bool> delay) =>
    delay
    .StartWith(false)
    .DistinctUntilChanged()
    .CombineLatest(source, (d, i) => new { IsDelayed = d, Item = i })
    .Scan(
        seed: new { Items = ImmutableList<T>.Empty, IsDelayed = false },
        accumulator: (sum, next) => new
        {
            Items = (next.IsDelayed != sum.IsDelayed) ?
                    (next.IsDelayed ? sum.Items.Clear() : sum.Items) :
                    (sum.IsDelayed ? sum.Items.Add(next.Item) : sum.Items.Clear().Add(next.Item)),
            IsDelayed = next.IsDelayed
        })
    .Where(i => !i.IsDelayed)
    .SelectMany(i => i.Items);

这些是我的测试:

[DataTestMethod]
[DataRow("3-a 6-b 9-c", "1-f", "3-a 6-b 9-c", DisplayName = "Start with explicit no_delay, emit all future items")]
[DataRow("3-a 6-b 9-c", "1-f 2-f", "3-a 6-b 9-c", DisplayName = "Start with explicit no_delay+no_delay, emit all future items")]
[DataRow("3-a 6-b 9-c", "1-t", "", DisplayName = "Start with explicit delay, emit nothing")]
[DataRow("3-a 6-b 9-c", "1-t 2-t", "", DisplayName = "Start with explicit delay+delay, emit nothing")]
[DataRow("3-a 6-b 9-c", "5-t 10-f", "3-a 10-b 10-c", DisplayName = "When delay is removed, all cached items are emitted in order")]
[DataRow("3-a 6-b 9-c 12-d", "5-t 10-f", "3-a 10-b 10-c 12-d", DisplayName = "When delay is removed, all cached items are emitted in order")]
public void DelayWhile(string source, string isDelayed, string expectedOutput)
{
    (long time, string value) ParseEvent(string e)
    {
        var parts = e.Split('-');
        long time = long.Parse(parts[0]);
        string val = parts[1];
        return (time, val);
    }
    IEnumerable<(long time, string value)> ParseEvents(string s) => s.Split(new char[] { ' ' }, StringSplitOptions.RemoveEmptyEntries).Select(ParseEvent);
    var scheduler = new TestScheduler();
    var sourceEvents = ParseEvents(source).Select(i => OnNext(i.time, i.value)).ToArray();
    var sourceStream = scheduler.CreateHotObservable(sourceEvents);
    var isDelayedEvents = ParseEvents(isDelayed).Select(i => OnNext(i.time, i.value == "t")).ToArray();
    var isDelayedStream = scheduler.CreateHotObservable(isDelayedEvents);
    var expected = ParseEvents(expectedOutput).Select(i => OnNext(i.time, i.value)).ToArray();
    var obs = scheduler.CreateObserver<string>();
    var result = sourceStream.DelayWhile(isDelayedStream);
    result.Subscribe(obs);
    scheduler.AdvanceTo(long.MaxValue);
    ReactiveAssert.AreElementsEqual(expected, obs.Messages);
}

[TestMethod]
public void DelayWhile_SubscribeToSourceObservablesOnlyOnce()
{
    var scheduler = new TestScheduler();
    var source = scheduler.CreateHotObservable<int>();
    var delay = scheduler.CreateHotObservable<bool>();

    // No subscriptions until subscribe
    var result = source.DelayWhile(delay);
    Assert.AreEqual(0, source.ActiveSubscriptions());
    Assert.AreEqual(0, delay.ActiveSubscriptions());

    // Subscribe once to each
    var obs = scheduler.CreateObserver<int>();
    var sub = result.Subscribe(obs);
    Assert.AreEqual(1, source.ActiveSubscriptions());
    Assert.AreEqual(1, delay.ActiveSubscriptions());

    // Dispose subscriptions when subscription is disposed
    sub.Dispose();
    Assert.AreEqual(0, source.ActiveSubscriptions());
    Assert.AreEqual(0, delay.ActiveSubscriptions());
}

[TestMethod]
public void DelayWhile_WhenSubscribeWithNoDelay_EmitCurrentValue()
{
    var source = new BehaviorSubject<int>(1);
    var emittedValues = new List<int>();
    source.DelayWhile(Observable.Return(false)).Subscribe(i => emittedValues.Add(i));
    Assert.AreEqual(1, emittedValues.Single());
}

// Subscription timing issue?
[TestMethod]
public void DelayWhile_WhenSubscribeWithDelay_EmitNothing()
{
    var source = new BehaviorSubject<int>(1);
    var emittedValues = new List<int>();
    source.DelayWhile(Observable.Return(true)).Subscribe(i => emittedValues.Add(i));
    Assert.AreEqual(0, emittedValues.Count);
}

[TestMethod]
public void DelayWhile_CoreScenario()
{
    var source = new BehaviorSubject<int>(1);
    var delay = new BehaviorSubject<bool>(false);
    var emittedValues = new List<int>();

    // Since no delay when subscribing, emit value
    source.DelayWhile(delay).Subscribe(i => emittedValues.Add(i));
    Assert.AreEqual(1, emittedValues.Single());

    // Turn on delay and buffer up a few; nothing emitted
    delay.OnNext(true);
    source.OnNext(2);
    source.OnNext(3);
    Assert.AreEqual(1, emittedValues.Single());

    // Turn off delay; should release the buffered items
    delay.OnNext(false);
    Assert.IsTrue(emittedValues.SequenceEqual(new int[] { 1, 2, 3 }));
}

Answer 1

编辑:我忘记了使用基于Join 和Join 的运算符(如WithLatestFrom) 当有两个冷的可观察对象时。不用说，下面提到的关于缺乏交易的批评比以往任何时候都更加明显。

我会推荐这个，它更像是我原来的解决方案，但使用了 Delay 重载。它通过了除 DelayWhile_WhenSubscribeWithDelay_EmitNothing 之外的所有测试。为了解决这个问题，我将创建一个接受起始默认值的重载:

public static IObservable<T> DelayWhile<T>(this IObservable<T> source, IObservable<bool> delay, bool isGateClosedToStart)
{
    return source.Publish(_source => delay
        .DistinctUntilChanged()
        .StartWith(isGateClosedToStart)
        .Publish(_delay => _delay
            .Select(isGateClosed => isGateClosed
                ? _source.TakeUntil(_delay).Delay(_ => _delay)
                : _source.TakeUntil(_delay)
            )
            .Merge()
        )
    );
}

public static IObservable<T> DelayWhile<T>(this IObservable<T> source, IObservable<bool> delay)
{
    return DelayWhile(source, delay, false);
}

---

旧答案:

我最近读了一本书，批评 Rx 不支持事务，我第一次尝试解决这个问题就是一个很好的例子，为什么:

public static IObservable<T> DelayWhile<T>(this IObservable<T> source, IObservable<bool> delay)
{
    return source.Publish(_source => delay
        .DistinctUntilChanged()
        .StartWith(false)
        .Publish(_delay => _delay
            .Select(isGateClosed => isGateClosed 
                ? _source.Buffer(_delay).SelectMany(l => l) 
                : _source)
            .Switch()
        )
    );
}

那应该可以工作，除了有太多东西依赖于 delay observable，而且订阅顺序很重要:在这种情况下，Switch在 Buffer 结束之前切换，因此当延迟门关闭时不会有任何结果。

这可以通过以下方式解决:

public static IObservable<T> DelayWhile<T>(this IObservable<T> source, IObservable<bool> delay)
{
    return source.Publish(_source => delay
        .DistinctUntilChanged()
        .StartWith(false)
        .Publish(_delay => _delay
            .Select(isGateClosed => isGateClosed 
                ? _source.TakeUntil(_delay).Buffer(_delay).SelectMany(l => l) 
                : _source.TakeUntil(_delay)
            )
            .Merge()
        )
    );
}

我的下一次尝试通过了你所有的测试，并且使用了你想要的 Observable.Delay 重载:

public static IObservable<T> DelayWhile<T>(this IObservable<T> source, IObservable<bool> delay)
{
    return delay
        .DistinctUntilChanged()
        .StartWith(false)
        .Publish(_delay => source
            .Join(_delay,
                s => Observable.Empty<Unit>(),
                d => _delay,
                (item, isGateClosed) => isGateClosed 
                    ? Observable.Return(item).Delay(, _ => _delay) 
                    : Observable.Return(item)
            )
            .Merge()
    );
}

Join 可以简化为 WithLatestFrom，如下所示:

public static IObservable<T> DelayWhile<T>(this IObservable<T> source, IObservable<bool> delay)
{
    return delay
        .DistinctUntilChanged()
        .StartWith(false)
        .Publish(_delay => source
            .WithLatestFrom(_delay,
                (item, isGateClosed) => isGateClosed 
                    ? Observable.Return(item).Delay(_ => _delay) 
                    : Observable.Return(item)
            )
            .Merge()
    );
}