Question

申请要求：

订阅两个事件流A和B
对于每个A事件，一段时间后应该有相应的B事件
如果没有相应的B到达（及时）
B事件可以与A事件的不同顺序（但仅在之后），迟到或根本不到达

传统方法很简单：

记录集合中的每个事件
在相应的B事件到达时删除A事件
监控在超时内未收到B的A事件的集合，并生成警报

我想尝试一种反应性解决方案，但不知道哪些运营商会表达最佳效果。我可视化事件流（Observables）：

A
B流
也许也是一个计时器滴答流（，除非计时操作员满足此）

最终输出是一组警报：

someObservable.
    ...incantations, other observables...
.subscribe ((EventA a) -> raiseAlertForMissingB (a));

是否有优雅的反应方法，或者这只是不适合rx ？

（原始问题的后续补充）

ascii艺术大理石图可能看起来像这样

A stream A1----A2----------A3------------------A4--------------A5---------
B stream ------------B1------------B3---------------B4------------B5-----
                                            (A2 TIMEOUT) 
merged   ------------A1B1----------A3B3------A2??---A4B4----------A5B5---

订阅者从合并流中接收元组。如果元组是匹配的A和B事件，则会记录它，但如果元组是没有匹配B事件的A事件（在大理石图中显示为A2 ??），则订户会发出警报。中提琴！

但是如何为每个A事件触发定时等待匹配的B事件？

（另外一个）

为了说明第4个要点点数部分＆＃34; B事件可以从A事件的不同顺序到达（但仅在之后）......＆＃34;

A stream A6----A7----------A8----------------------
B stream ------------B7------------B6-------B8-----
merged   ------------A7B7----------A6B6-----A8B8---

2016年2月23日更新

我正在测试移植到Java的建议解决方案。

解决方案：电子化

public class Test06Enigmativity {
    private static final long A_PERIOD = 400;
    private static final long B_PERIOD = 500;
    private static final int TIMEOUT = 4_000;
    private static final int[] bOrder = {
        0, 1, 2,
        4,
        3,  // out of order
        6,
        5,  // out of order
        7, 8,
        10, 11, 12, 13, 14,
        9,  // out of order
        15
    };

    private final long startTime = System.currentTimeMillis ();

    public static void main (final String[] args) {
        Test06Enigmativity app = new Test06Enigmativity ();
        app.runEnigmativity ();
    }

    private void runEnigmativity () {
        Observable<Long> aStream =
            Observable.interval (A_PERIOD, TimeUnit.MILLISECONDS)
                .doOnNext (seq -> {
                    output ("       A%s", seq);
                }).take (bOrder.length);

        Observable<Long> bStream =
            Observable.interval (B_PERIOD, TimeUnit.MILLISECONDS)
                .map (seq -> {
                    long aId = (long) bOrder[seq.intValue ()];
                    output ("           B%s", aId);
                    return aId;
                })
                .take (bOrder.length);

        monitorEnigmativity (aStream, bStream, TIMEOUT)
            .subscribe (this::output);

        try {
            Thread.sleep (60_000);
        } catch (InterruptedException e) {
        }
    }

    private Observable<String> monitorEnigmativity (Observable<Long> aStream, Observable<Long> bStream, int thresholdMsec) {
        return Observable.create (subscriber ->
            bStream.publish (pb ->
                Observable.merge (
                    aStream.map (ax ->
                        pb
                            .filter (pbx -> pbx.equals (ax))
                            .take (1)
                            .timeout (thresholdMsec, TimeUnit.MILLISECONDS, Observable.defer (
                                () -> {
                                    output ("           timeout on B%s", ax);
                                    return Observable.just (-1L);
                                }
                            )).map (pbx -> String.format ("%s,%s", ax, pbx))
                    )
                )
            ).subscribe (subscriber::onNext)
        );
    }

    private void output (String format, Object... args) {
        System.out.printf ("tid:%3d %4dms %s%n", Thread.currentThread ().getId (), System.currentTimeMillis () - startTime,
            String.format (format, args));
    }
}

这适用于：

private static final long A_PERIOD = 400;
private static final long B_PERIOD = 500;
private static final int TIMEOUT = 4_000;

tid: 12  511ms        A0
tid: 15  626ms            B0
tid: 15  626ms 0,0
tid: 12  907ms        A1
tid: 15 1126ms            B1
tid: 15 1126ms 1,1
tid: 12 1307ms        A2
tid: 15 1626ms            B2
tid: 15 1626ms 2,2
tid: 12 1707ms        A3
tid: 12 2107ms        A4
tid: 15 2126ms            B4
tid: 15 2126ms 4,4
tid: 12 2507ms        A5
tid: 15 2625ms            B3
tid: 15 2625ms 3,3
tid: 12 2907ms        A6
tid: 15 3126ms            B6
tid: 15 3126ms 6,6
tid: 12 3307ms        A7
tid: 15 3626ms            B5
tid: 15 3626ms 5,5
tid: 12 3707ms        A8
tid: 12 4107ms        A9
tid: 15 4126ms            B7
tid: 15 4126ms 7,7
tid: 12 4507ms        A10
tid: 15 4626ms            B8
tid: 15 4626ms 8,8
tid: 12 4908ms        A11
tid: 15 5127ms            B10
tid: 15 5128ms 10,10
tid: 12 5307ms        A12
tid: 15 5626ms            B11
tid: 15 5626ms 11,11
tid: 12 5707ms        A13
tid: 12 6107ms        A14
tid: 15 6126ms            B12
tid: 15 6126ms 12,12
tid: 12 6507ms        A15
tid: 15 6626ms            B13
tid: 15 6626ms 13,13
tid: 13 7109ms            timeout on B9
tid: 13 7114ms 9,-1
tid: 15 7126ms            B14
tid: 15 7126ms 14,14
tid: 15 7625ms            B9
tid: 15 8126ms            B15
tid: 15 8126ms 15,15

但是随着B＆B接近A，事件可能会错过：

private static final long A_PERIOD = 400;
private static final long B_PERIOD = 410;
private static final int TIMEOUT = 2_000;

tid: 12  509ms        A0
tid: 15  538ms            B0
tid: 12  905ms        A1
tid: 15  948ms            B1
tid: 15  948ms 1,1
tid: 12 1305ms        A2
tid: 15 1358ms            B2
tid: 15 1358ms 2,2
tid: 12 1706ms        A3
tid: 15 1768ms            B4
tid: 12 2105ms        A4
tid: 15 2178ms            B3
tid: 15 2178ms 3,3
tid: 12 2505ms        A5
tid: 16 2538ms            timeout on B0
tid: 16 2544ms 0,-1
tid: 15 2588ms            B6
tid: 12 2905ms        A6
tid: 15 2998ms            B5
tid: 15 2998ms 5,5
tid: 12 3305ms        A7
tid: 15 3408ms            B7
tid: 15 3408ms 7,7
tid: 12 3705ms        A8
tid: 15 3817ms            B8
tid: 15 3817ms 8,8
tid: 12 4105ms        A9
tid: 14 4106ms            timeout on B4
tid: 14 4106ms 4,-1
tid: 15 4228ms            B10
tid: 12 4505ms        A10
tid: 15 4638ms            B11
tid: 12 4905ms        A11
tid: 16 4906ms            timeout on B6
tid: 16 4906ms 6,-1
tid: 15 5048ms            B12
tid: 12 5305ms        A12
tid: 15 5457ms            B13
tid: 12 5705ms        A13
tid: 15 5868ms            B14
tid: 12 6106ms        A14
tid: 13 6107ms            timeout on B9
tid: 13 6107ms 9,-1
tid: 15 6279ms            B9
tid: 14 6510ms            timeout on B10
tid: 12 6510ms        A15
tid: 14 6510ms 10,-1
tid: 15 6688ms            B15
tid: 15 6688ms 15,15

B0到达但是产生了超时而不是匹配。这是因为B0事件在A-stream观察者改变订阅B流之前到达。我认为对于热门B流，嵌套订阅的流行方法存在缺陷。我们需要的是某种有限的近期重播流 - 一个主题的应用程序？

解决方案：Supertopi

package test;

import rx.Observable;
import java.util.concurrent.TimeUnit;

public class Test06Supertopi {
    private static final long A_PERIOD = 400;
    private static final long B_PERIOD = 500;
    private static final int TIMEOUT = 3_000;
    private static final int[] bOrder = {
        0, 1, 2,
        4,
        3,  // out of order
        6,
        5,  // out of order
        7, 8,
        10, 11, 12, 13, 14,
        9,  // out of order
        15
    };

    private final long startTime = System.currentTimeMillis ();

    public static void main (final String[] args) {
        Test06Supertopi app = new Test06Supertopi ();
        app.runSupertopi ();
    }

    private void runSupertopi () {
        Observable<Long> aStream =
            Observable.interval (A_PERIOD, TimeUnit.MILLISECONDS)
                .doOnNext (seq -> {
                    output ("       A%s", seq);
                }).take (bOrder.length);

        Observable<Long> bStream =
            Observable.interval (B_PERIOD, TimeUnit.MILLISECONDS)
                .map (seq -> {
                    long aId = (long) bOrder[seq.intValue ()];
                    output ("           B%s", aId);
                    return aId;
                })
                .take (bOrder.length);

        monitorSupertopi (aStream, bStream, TIMEOUT)
            .subscribe (this::output);

        try {
            Thread.sleep (60_000);
        } catch (InterruptedException e) {
        }
    }

    private Observable<String> monitorSupertopi (Observable<Long> aStream, Observable<Long> bStream, int thresholdMsec) {
        return Observable.create (subscriber -> {
            Observable<Long> a = aStream.publish ().refCount ();
            Observable<Long> b = bStream.publish ().refCount ();

            a.subscribe ((Long aId) -> {
                Observable.merge (
                    Observable.timer (thresholdMsec, TimeUnit.MILLISECONDS)
                        .doOnNext (x -> {
                            output ("           timeout on B%s", aId);
                        })
                        .map (x -> String.format ("%s,%s", aId, -1L)),
                    b.filter ((Long j) -> j.equals (aId))
                        .map ((Long pbx) -> String.format ("%s,%s", aId, pbx))
                ).take (1)
                .subscribe (subscriber::onNext);
            });
        });
    }

    private void output (String format, Object... args) {
        System.out.printf ("tid:%3d %4dms %s%n", Thread.currentThread ().getId (), System.currentTimeMillis () - startTime,
            String.format (format, args));
    }
}

工作正常：

private static final long A_PERIOD = 400;
private static final long B_PERIOD = 500;
private static final int TIMEOUT = 4_000;

tid: 14  522ms        A0
tid: 14  922ms        A1
tid: 16 1054ms            B0
tid: 16 1055ms 0,0
tid: 14 1322ms        A2
tid: 16 1555ms            B1
tid: 16 1555ms 1,1
tid: 14 1721ms        A3
tid: 16 2055ms            B2
tid: 16 2055ms 2,2
tid: 14 2122ms        A4
tid: 14 2522ms        A5
tid: 16 2555ms            B4
tid: 16 2555ms 4,4
tid: 14 2922ms        A6
tid: 16 3055ms            B3
tid: 16 3055ms 3,3
tid: 14 3322ms        A7
tid: 16 3555ms            B6
tid: 16 3555ms 6,6
tid: 14 3722ms        A8
tid: 16 4055ms            B5
tid: 16 4055ms 5,5
tid: 14 4122ms        A9
tid: 14 4522ms        A10
tid: 16 4555ms            B7
tid: 16 4555ms 7,7
tid: 14 4922ms        A11
tid: 16 5055ms            B8
tid: 16 5055ms 8,8
tid: 14 5322ms        A12
tid: 16 5555ms            B10
tid: 16 5556ms 10,10
tid: 14 5723ms        A13
tid: 16 6056ms            B11
tid: 16 6057ms 11,11
tid: 14 6122ms        A14
tid: 14 6522ms        A15
tid: 16 6555ms            B12
tid: 16 6555ms 12,12
tid: 16 7055ms            B13
tid: 16 7055ms 13,13
tid: 13 7125ms            timeout on B9
tid: 13 7125ms 9,-1
tid: 16 7555ms            B14
tid: 16 7555ms 14,14
tid: 16 8055ms            B9
tid: 16 8555ms            B15
tid: 16 8555ms 15,15

但是由于时间更接近，最初的B事件会被收到两次。我还在调查这个。

private static final long A_PERIOD = 400;
private static final long B_PERIOD = 410;
private static final int TIMEOUT = 2_000;

tid: 14  539ms        A0
tid: 14  939ms        A1
tid: 16  983ms            B0
tid: 16  983ms 0,0
tid: 14 1339ms        A2
tid: 16 1393ms            B1
tid: 16 1393ms 1,1
tid: 14 1739ms        A3
tid: 16 1803ms            B2
tid: 16 1803ms 2,2
tid: 14 2139ms        A4
tid: 16 2213ms            B4
tid: 16 2213ms 4,4
tid: 14 2539ms        A5
tid: 16 2623ms            B3
tid: 16 2623ms 3,3
tid: 14 2939ms        A6
tid: 16 3032ms            B6
tid: 16 3032ms 6,6
tid: 14 3339ms        A7
tid: 16 3443ms            B5
tid: 16 3443ms 5,5
tid: 14 3739ms        A8
tid: 16 3852ms            B7
tid: 16 3852ms 7,7
tid: 14 4139ms        A9
tid: 16 4263ms            B8
tid: 16 4263ms 8,8
tid: 14 4539ms        A10
tid: 16 4672ms            B10
tid: 16 4672ms 10,10
tid: 14 4939ms        A11
tid: 16 5083ms            B11
tid: 16 5083ms 11,11
tid: 14 5339ms        A12
tid: 16 5493ms            B12
tid: 16 5493ms 12,12
tid: 14 5739ms        A13
tid: 16 5903ms            B13
tid: 16 5903ms 13,13
tid: 14 6139ms        A14
tid: 13 6140ms            timeout on B9
tid: 13 6140ms 9,-1
tid: 16 6313ms            B14
tid: 16 6313ms 14,14
tid: 14 6539ms        A15
tid: 14 6950ms            B0
tid: 14 7360ms            B1
tid: 14 7770ms            B2
tid: 14 8180ms            B4
tid: 13 8540ms            timeout on B15
tid: 13 8540ms 15,-1

2016年2月24日

解决方案：Lee Campbell

package test;

import rx.Observable;
import rx.subjects.PublishSubject;

import java.util.concurrent.TimeUnit;
import java.util.concurrent.TimeoutException;

public class Test06LeeCampbell {
    private static final int TIMEOUT = 500;

    class ScheduledEvent {
        final String type;
        final long aId;
        final long atMsec;
        volatile boolean expired;

        public ScheduledEvent (long atMsec, String type, long aId) {
            this.atMsec = atMsec;
            this.type = type;
            this.aId = aId;
        }
    }

    ScheduledEvent[] scheduledEvents = {
        new ScheduledEvent (10, "A", 0),
        new ScheduledEvent (90, "B", 0),
        new ScheduledEvent (110, "A", 1),
        new ScheduledEvent (140, "B", 1),
        new ScheduledEvent (190, "A", 2),
        new ScheduledEvent (270, "B", 2),
        new ScheduledEvent (310, "A", 3),
        new ScheduledEvent (410, "A", 4),
        new ScheduledEvent (440, "B", 4),
        new ScheduledEvent (480, "B", 3),
        new ScheduledEvent (510, "A", 5),
        new ScheduledEvent (610, "A", 6),
        //new ScheduledEvent (670, "B", 6),
        new ScheduledEvent (710, "A", 7),
        new ScheduledEvent (810, "A", 8),
        new ScheduledEvent (860, "B", 7),
        new ScheduledEvent (880, "B", 8),
        new ScheduledEvent (910, "A", 9),
        new ScheduledEvent (1100, "A", 10),
        new ScheduledEvent (1110, "A", 11),
        new ScheduledEvent (1120, "A", 12),
        new ScheduledEvent (1130, "A", 13),
        new ScheduledEvent (1140, "A", 14),
        //new ScheduledEvent (1200, "B", 10),
        //new ScheduledEvent (1210, "B", 11),
        //new ScheduledEvent (1220, "B", 12),
        new ScheduledEvent (1230, "B", 13),
        new ScheduledEvent (1240, "B", 14),
        new ScheduledEvent (1390, "B", 9),
        new ScheduledEvent (1450, "A", 15),
        new ScheduledEvent (3290, "B", 5),
        new ScheduledEvent (3350, "B", 15)
    };

    private final long startTime = System.currentTimeMillis ();

    public static void main (final String[] args) {
        Test06LeeCampbell app = new Test06LeeCampbell ();
        app.runLeeCampbell ();
    }

    private void runLeeCampbell () {
        Observable<Long> aStream =
            getCrudeSequencer ("A")
                .doOnNext (seq -> {
                    output ("       A%s", seq);
                });

        Observable<Long> bStreamCold =
            getCrudeSequencer ("B")
                .doOnNext (seq -> {
                    output ("           B%s", seq);
                });
        PublishSubject<Long> bStream = PublishSubject.create ();
        bStreamCold.subscribe (bStream);

        monitorLeeCampbell (aStream, bStream, TIMEOUT)
            .subscribe (this::output);

        pause (10_000);
    }

    private Observable<String> monitorLeeCampbell (Observable<Long> aStream, Observable<Long> bStream, int thresholdMsec) {
        return aStream.flatMap (a ->
            bStream.filter (b -> b.equals (a))
                .map (b -> String.format ("%s,%s", a, b))
                .take (1)
                .timeout (thresholdMsec, TimeUnit.MILLISECONDS)
                .onErrorResumeNext (
                    throwable -> {
                        output ("           timeout on B%s", a);
                        if (!(throwable instanceof TimeoutException)) {
                            throw new RuntimeException (throwable);
                        }
                        return Observable.just (String.format ("%s,%s", a, -1L));
                    }
                )
        );
    }

    private void output (String format, Object... args) {
        System.out.printf ("tid:%3d %4dms %s%n", Thread.currentThread ().getId (), getElapsedMsec (),
            String.format (format, args));
    }

    private long getElapsedMsec () {
        return System.currentTimeMillis () - startTime;
    }

    private Observable<Long> getCrudeSequencer (String name) {
        return Observable.create (subscriber ->
            new Thread (() -> {
                for (ScheduledEvent se : scheduledEvents) {
                    if (se.type.equals (name)) {
                        while (getElapsedMsec () < se.atMsec) {
                            pause (1);
                        }
                        subscriber.onNext (Long.valueOf (se.aId));
                        se.expired = true;
                    } else {
                        // Timing is not reliable for sequencing two threads
                        while (!se.expired) {
                            pause (1);
                        }
                    }
                }
                subscriber.onCompleted ();
            }).start ()
        );
    }

    private static void pause (final int millis) {
        try {
            Thread.sleep (millis);
        } catch (InterruptedException e) {
        }
    }
}

有效

tid: 12   90ms        A0
tid: 11  157ms            B0
tid: 11  157ms 0,0
tid: 12  159ms        A1
tid: 11  160ms            B1
tid: 11  160ms 1,1
tid: 12  190ms        A2
tid: 11  270ms            B2
tid: 11  270ms 2,2
tid: 12  310ms        A3
tid: 12  410ms        A4
tid: 11  440ms            B4
tid: 11  440ms 4,4
tid: 11  480ms            B3
tid: 11  480ms 3,3
tid: 12  510ms        A5
tid: 12  610ms        A6
tid: 12  710ms        A7
tid: 12  810ms        A8
tid: 11  860ms            B7
tid: 11  860ms 7,7
tid: 11  880ms            B8
tid: 11  880ms 8,8
tid: 12  910ms        A9
tid: 15 1011ms            timeout on B5
tid: 15 1012ms 5,-1
tid: 12 1100ms        A10
tid: 12 1110ms        A11
tid: 16 1111ms            timeout on B6
tid: 16 1111ms 6,-1
tid: 12 1120ms        A12
tid: 12 1130ms        A13
tid: 12 1140ms        A14
tid: 11 1230ms            B13
tid: 11 1230ms 13,13
tid: 11 1240ms            B14
tid: 11 1240ms 14,14
tid: 11 1390ms            B9
tid: 11 1390ms 9,9
tid: 12 1450ms        A15
tid: 14 1601ms            timeout on B10
tid: 14 1601ms 10,-1
tid: 15 1611ms            timeout on B11
tid: 15 1611ms 11,-1
tid: 16 1621ms            timeout on B12
tid: 16 1621ms 12,-1
tid: 19 1951ms            timeout on B15
tid: 19 1951ms 15,-1
tid: 11 3290ms            B5
tid: 11 3350ms            B15

Answer 1

假设：

B流很热，
您有一个可以表示匹配或不匹配的返回类型（而不是使用将终止订阅的OnError / Timeout）

然后这没关系。

AStream.SelectMany(a =>
    BStream.Where(b => b == a)
        .Select(b => new MatchMade(a, b))
        .Take(1)
        .Timeout(matchTimeout)
        .Catch<TimeoutException>(ex=>Observable.Return(new NoMatchMade(a)))
)

Answer 2

这应该有效。类型不同，我不想猜测你可能抽象的数据类型。您可以非常轻松地应用它们（函数参数，键比较和Select语句）

想法是，对于来自a的每个已发出的值，我们将第一个值按b.Where(keys match)或超时（由Observable.Timer显示）发出，并使我们的Select为基础关于这个信息。

我认为在超时情况下，您还希望通过一些错误配置机制进行OnNext通知。

private IObservable<string> MonitorAB(IObservable<long> a, IObservable<long> b,
                                      TimeSpan threshold)
{
    return Observable.Create<string>((obs) =>
    {
        a = a.Publish().RefCount();
        b = b.Publish().RefCount();
        return a.Subscribe(i =>
        {
            Observable.Merge(Observable.Timer(threshold).Select(_ => $"Timeout for A{i}"),
                             b.Where(j => j == i).Select(_ => $"Got matching B for A{i}"))
                      .Take(1)
                      .Subscribe(obs.OnNext);
        });
    });
}

我测试了这个。

private void Test()
{
    var a = Observable.Interval(TimeSpan.FromSeconds(2)).Take(5);
    var b = Observable.Interval(TimeSpan.FromSeconds(5)).Take(5);
    MonitorAB( a, b, TimeSpan.FromSeconds(13)).Subscribe(Console.WriteLine);
}

编辑：为了测试乱序情况，你可以翻转B流，例如

var b = Observable.Interval(TimeSpan.FromSeconds(2)).Select(i  => 4 - i).Take(5);

Answer 3

我认为这可以做你想要的，虽然它是C＃而不是Java - 我相信你可以轻松转换。

private IObservable<string> MonitorAB(
    IObservable<long> a, IObservable<long> b, TimeSpan threshold)
{
    return Observable.Create<string>(o =>
        b.Publish(pb =>
            a.Select(ax =>
                pb.Where(pbx => pbx == ax)
                    .Take(1)
                    .Timeout(threshold, Observable.Return(-1L))
                    .Select(pbx => String.Format("{0},{1}", ax, pbx)))
            .Merge())
            .Subscribe(o));
}

因此，这只是使用内联.Publish来确保b在查询中很热门。外部。Select过滤已发布的pb可观察对象，以匹配a（匹配a＆amp; b）的值，只有一个原因我们只需要一个，然后为.Timeout时间执行threshold并在达到超时时返回-1L（long）。内部.Select只会将两个long值转换为单个string。此时查询为IObservable<IObservable<string>>，因此.Merge将其展平。

对简单命令性任务的反应式方法

3 个答案: