基于这个excellent SO answer我可以在RxPy中并行处理多个任务,我的问题是如何等待它们全部完成?我知道使用线程我可以做.join()
,但在Rx调度器中似乎没有这样的选择。.to_blocking()
也没有帮助,主线程在所有通知激发和完整处理程序被调用之前完成。下面是一个例子:
from __future__ import print_function
import os, sys
import time
import random
from rx import Observable
from rx.core import Scheduler
from threading import current_thread
def printthread(val):
print("{}, thread: {}".format(val, current_thread().name))
def intense_calculation(value):
printthread("calc {}".format(value))
time.sleep(random.randint(5, 20) * .1)
return value
if __name__ == "__main__":
Observable.range(1, 3) \
.select_many(lambda i: Observable.start(lambda: intense_calculation(i), scheduler=Scheduler.timeout)) \
.observe_on(Scheduler.event_loop) \
.subscribe(
on_next=lambda x: printthread("on_next: {}".format(x)),
on_completed=lambda: printthread("on_completed"),
on_error=lambda err: printthread("on_error: {}".format(err)))
printthread("\nAll done")
# time.sleep(2)
calc 1, thread: Thread-1
calc 2, thread: Thread-2
calc 3, thread: Thread-3
All done, thread: MainThread
calc 1, thread: Thread-1
calc 2, thread: Thread-2
calc 3, thread: Thread-3
All done, thread: MainThread
on_next: 2, thread: Thread-4
on_next: 3, thread: Thread-4
on_next: 1, thread: Thread-4
on_completed, thread: Thread-4
在此处发布完整的解决方案:
对于
ThreadPoolScheduler
,您可以:scheduler.executor.shutdown()
然后,你就可以得到所有的结果了。在
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