Question

我有一个外部服务（ExternalDummyService），我在其中注册了一个回调。我想从该回调创建一个observable并订阅多个异步进程。

pyfiddle中的完整代码：https://pyfiddle.io/fiddle/da1e1d53-2e34-4742-a0b9-07838f2c13df *请注意，在pyfiddle版本中，“sleeps”被替换为“for i in range（10000）：foo + = i”，因为睡眠不能正常工作。

主要代码是：

thread = ExternalDummyService()
external_obs = thread.subject.publish()

external_obs.subscribe(slow_process)
external_obs.subscribe(fast_process)
external_obs.connect()

thread.start()

class ExternalDummyService(Thread):
    def __init__(self):
        self.subject = Subject()

    def run(self):
        for i in range(5):
            dummy_msg = { ... }
            self.subject.on_next(dummy_msg)

def fast_process(msg):
    print("FAST {0} {1}".format(msg["counter"], 1000*(time() - msg["timestamp"])))
    sleep(0.1)

def slow_process(msg):
    print("SLOW {0} {1}".format(msg["counter"], 1000*(time() - msg["timestamp"])))
    sleep(1)

我得到的输出是这一个，两个进程同步运行，而ExternalDummyService在两个进程完成每次执行之前都不会发出新值：

emitting 0
STARTED
SLOW 0 1.0008811950683594
FAST 0 2.0122528076171875
emitting 1
SLOW 1 1.5070438385009766
FAST 1 1.5070438385009766
emitting 2
SLOW 2 0.5052089691162109
FAST 2 0.9891986846923828
emitting 3
SLOW 3 1.0006427764892578
FAST 3 1.0006427764892578
emitting 4
SLOW 4 1.0013580322265625
FAST 4 1.0013580322265625

FINISHED

我想得到这样的东西，服务发出消息而不等待进程运行和进程异步运行：

STARTED
emitting 0
emitting 1
emitting 2
FAST 0 2.0122528076171875
FAST 1 1.5070438385009766
emitting 3
SLOW 0 1.0008811950683594
FAST 2 0.9891986846923828
emitting 4
FAST 3 1.0006427764892578
SLOW 1 1.5070438385009766
FAST 4 1.0013580322265625
SLOW 2 0.5052089691162109
SLOW 3 1.0006427764892578
SLOW 4 1.0013580322265625

FINISHED

我尝试过使用share（），ThreadPoolScheduler和其他我不知道我在做什么。

谢谢！

Answer 1

使用此问题的答案：RxJava concurrency with multiple subscribers and events

...我已用此代码获得了所需的结果：

optimal_thread_count = cpu_count()
pool_scheduler = ThreadPoolScheduler(optimal_thread_count)

thread = ExternalDummyService()
external_obs = thread.subject.publish()

external_obs \
    .flat_map(lambda msg: Observable.just(msg).subscribe_on(pool_scheduler)) \
    .subscribe(fast_process)

external_obs \
    .flat_map(lambda msg: Observable.just(msg).subscribe_on(pool_scheduler)) \
    .subscribe(slow_process)

external_obs.connect()

thread.start()

完整版：https://pyfiddle.io/fiddle/20f8871c-48d6-4d6b-b1a4-fdd0a4aa6f95/?m=Saved%20fiddle

输出结果为：

emitting 0
emitting 1
emitting 2
emitting 3
emitting 4
FAST 0 52.629709243774414
FAST 1 51.12814903259277
FAST 2 100.2051830291748
FAST 3 151.2434482574463
SLOW 0 503.0245780944824
SLOW 1 502.0263195037842
FAST 4 548.7725734710693
SLOW 2 551.4400005340576
SLOW 3 652.1098613739014
SLOW 4 1000.3445148468018

请随意提出任何改进建议。

RxPy：如何从外部回调创建热observable并订阅多个异步进程？

1 个答案: