Question

我有一个带有某些值的流：

Flux<Integer> stream = getStream();

我正试图实现每秒N个项目的功能

stream.bufferTimeout(MAX_SIZE_TWO, _1_SECOND).subscribe(val => {
  System.out.println(val);
});

我正在尝试找到与预期结果接近的运算符。

预期结果：

time: 15:00:00, stream_next_value: 1, output: {1}
time: 15:00:00, stream_next_value: 2, output: {2}
time: 15:00:00, stream_next_value: 3, no output => buffer
time: 15:00:00, stream_next_value: 4, no output => buffer
time: 15:00:00, stream_next_value: 5, no output => buffer
time: 15:00:01, stream_no_next_value, output: {3,4}
time: 15:00:01, stream_next_value: 6, no output => buffer
time: 15:00:02, stream_no_next_value, output: {5,6}

但是看起来缓冲区操作符的重载版本不支持此行为。

如何使用缓冲区运算符实现预期的行为？

Answer 1

也许您可以这样做：

Flowable<Long> stream = Flowable.generate(() -> 0L, (next, emitter) -> {
        emitter.onNext(next);
        return next + 1;
});

// Flowable<Long> stream = Flowable.interval(100, MILLISECONDS);
//                                 .onBackpressureDrop(); // to make it works otherwise get a MissingBackPressureException

stream.buffer(2)
      .zipWith(Flowable.interval(1, SECONDS), (first, second) -> first)
      .flatMap(Flowable::fromIterable)
      .subscribe(s -> LOGGER.info("received: " + s),
                 Throwable::printStackTrace);

当心stream必须承受背压，否则您必须添加一个onBackpressureXXX()运算符（例如，如果流为interval()（请参见注释代码）就是这种情况）。您将获得以下输出：

14:39:59.538 | INFO  | RxComputationThreadPool-1 | received: 0
14:39:59.540 | INFO  | RxComputationThreadPool-1 | received: 1
14:40:00.527 | INFO  | RxComputationThreadPool-1 | received: 2
14:40:00.528 | INFO  | RxComputationThreadPool-1 | received: 3
14:40:01.528 | INFO  | RxComputationThreadPool-1 | received: 4
14:40:01.528 | INFO  | RxComputationThreadPool-1 | received: 5
14:40:02.528 | INFO  | RxComputationThreadPool-1 | received: 6
14:40:02.528 | INFO  | RxComputationThreadPool-1 | received: 7
14:40:03.528 | INFO  | RxComputationThreadPool-1 | received: 8
14:40:03.528 | INFO  | RxComputationThreadPool-1 | received: 9

每秒缓冲N个值RxJava，Project Reactor

1 个答案: