如何等待RxPy并行线程完成

基于这个出色的SO答案，我可以让多个任务在RxPy中并行工作，我的问题是如何等待所有任务完成？我知道我可以使用线程，.join()但是Rx Schedulers似乎没有这种选择。.to_blocking()两者都不起作用，MainThread在所有通知都已触发并调用了完整的处理程序之前完成。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)

预期产量

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

实际产量

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

All done, thread: MainThread

如果我取消注释睡眠调用，则为实际输出

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