如何等待 RxPy 并行线程完成

Question

基于excellent SO answer，我可以让多个任务在 RxPy 中并行工作，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到.join()，但 Rx 调度程序似乎没有任何这样的选项。 .to_blocking() 也无济于事，主线程在所有通知触发并调用完整处理程序之前完成。这是一个例子：

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)

预期输出

calc 1, thread: Thread-1
calc 2, thread: Thread-2
calc 3, thread: Thread-3

on_next: 2, thread: Thread-4
on_next: 3, thread: Thread-4
on_next: 1, thread: Thread-4
on_completed, thread: Thread-4
All done, thread: MainThread

实际输出

calc 1, thread: Thread-1
calc 2, thread: Thread-2
calc 3, thread: Thread-3

All done, thread: MainThread

取消注释睡眠调用时的实际输出

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

Answer 1

在这里发布完整的解决方案：

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
from rx.concurrency import ThreadPoolScheduler

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__":
    scheduler = ThreadPoolScheduler(4)

    Observable.range(1, 3) \
        .select_many(lambda i: Observable.start(lambda: intense_calculation(i), scheduler=scheduler)) \
        .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")
    scheduler.executor.shutdown()
    # time.sleep(2)

Answer 2

对于ThreadPoolScheduler，您可以：

1. scheduler = ThreadPoolScheduler(pool_size)
2. 并行调用。
3. scheduler.executor.shutdown()

那么，完成后就可以得到所有结果了。

Answer 3

使用run() 等待 RxPy 并行线程完成。

BlockingObservables 已从 RxPY v3 中删除。

from threading import current_thread
import rx, random, multiprocessing, time
from rx import operators as ops

def intense_calculation(value):
   delay = random.randint(5, 20) * 0.2
   time.sleep(delay)
   print("From adding_delay: {0} Value : {1} {2}".format(current_thread(), value, delay))
   return (value[0], value[1]+ " processed")

thread_pool_scheduler = rx.scheduler.NewThreadScheduler()

my_dict={'A':'url1', 'B':'url2', 'C':'url3'}

new_dict = rx.from_iterable(my_dict.items()).pipe(
    ops.flat_map(lambda a: rx.of(a).pipe(
        ops.map(lambda a: intense_calculation(a)),
        ops.subscribe_on(thread_pool_scheduler)
    )),
    ops.to_dict(lambda x: x[0], lambda x: x[1])
).run()

print("From main: {0}".format(current_thread()))
print(str(new_dict))