更新
我正在寻找一种控制一个可观察到的一个人的流动的方法。例如,让我们有2个单独增加(重要(整数的可观察物:
source : 1----2-2---2--3--3--4----4--5---6----8---9---10--------11------
control : -1----3----------------5-----------6-------9-----------12------
我需要产生一个新的观察值,其元素与源完全匹配,但是它们的时间由可观察的控制方式控制的,以下方式:源值应始终较小或等于控制值。这意味着,只有所有比最近发布的控制大的源值才能等到Control
"释放"它们source : 1----2-2---2--3--3--4----4--5---6----8---9---10--------11------
control : -1----3----------------5-----------6-------9-----------12------
expected result: -1----2-2--2--3--3-----4-4--5------6-------8-9---------10-11---
请查看以下代码示例:
private static <T, C> Observable<T> combine(Observable<T> source, Observable<C> control, BiFunction<T, C, Boolean> predicate) {
// ???
}
@Test
public void testControl() throws InterruptedException {
Subject<Integer> control = PublishSubject.create();
Observable<Integer> source = Observable.fromArray(1, 2, 2, 2, 3, 3, 4, 4, 5, 6, 8, 10, 11);
Observable<Integer> combined = combine(source, control, (s, c) -> s <= c);
control.subscribe(val -> System.out.println("Control: " + val));
combined.observeOn(Schedulers.io()).subscribe(val -> System.out.println("Value: " + val));
control.onNext(3); // should release 1,2,2,2,3,3
Thread.sleep(1000);
control.onNext(6); // should release 4,4,5,6
Thread.sleep(1000);
control.onNext(11); // should release 8,10,11
Thread.sleep(1000);
}
由于我没有找到任何优雅的解决方案,因此我最终以自己的方式实现它。如果我有人建议更优雅的解决方案,我会很高兴(在这种情况下,我会不了解这个答案,并且会接受更好的答案(。以下是我的解决方案:
private static <T, C> Observable<T> combine(Observable<T> source, Observable<C> control, BiFunction<T, C, Boolean> predicate) {
return Observable.create(emitter -> {
Queue<T> buffer = new ArrayDeque<>();
AtomicReference<C> lastControl = new AtomicReference<>();
CompletableSubject sourceCompletable = CompletableSubject.create();
CompletableSubject controlCompletable = CompletableSubject.create();
Disposable disposable = new CompositeDisposable(
control.subscribe(
val -> {
lastControl.set(val);
synchronized (buffer) {
while (!buffer.isEmpty() && predicate.apply(buffer.peek(), val)) {
emitter.onNext(buffer.poll());
}
}
},
emitter::onError,
controlCompletable::onComplete),
source.subscribe(
val -> {
C lastControlVal = lastControl.get();
synchronized (buffer) {
if (lastControlVal != null && predicate.apply(val, lastControlVal)) {
emitter.onNext(val);
} else {
buffer.add(val);
}
}
},
emitter::onError,
sourceCompletable::onComplete),
controlCompletable.andThen(sourceCompletable).subscribe(emitter::onComplete));
emitter.setDisposable(disposable);
});
}
您需要将缓冲区功能(将为您为您的项目缓冲(与CombineLatest功能(这将确定何时可以释放缓冲区(。
Observable<Integer> source;
Observable<Integer> control;
Observable<Integer> canRelease = Observable.combineLatest(source, control, (s, c) -> s < c ? s : null).filter(val -> val != null);
Observable<Integer> result = source.buffer(canRelease).flatMap(Observable::from);
最终可观察的将缓冲源项目,直到canRelease
中有一个值。canRelease
每次源可观察的最新项目都会发出少于可观察的最新项目;