如何制作一个 Python 并发 Future 以在 Futures 列表完成时发出信号？

Question

我有一个函数可以将多个任务提交给 ThreadPoolExecutor 和 return 每次提交创建的 Futures 列表：

def submit_tasks() -> [Future]:
    futures = []

    for i in range(10):
        future = executor.submit(Task())
        futures.append(future)

    return futures

def submit() -> Future:
    futures = submit_tasks()
    
    # I would like this function to return a single Future that clients can use to check
    # whether all futures in the list have completed. How to do that?

我在 Python 3.8

我想要这个函数 return 一个单一的未来，客户可以用它来检查列表中的所有未来是否已经完成。怎么做？

Answer 1

使用concurrent.futures.wait：https://docs.python.org/3/library/concurrent.futures.html#module-functions

from concurrent.futures import Future, wait

...

def submit() -> Future:
    return wait(submit_tasks())

如何制作一个 Python 并发 Future 以在 Futures 列表完成时发出信号？

How to make a Python concurrent Future that signals when a list of Futures has completed?

python

python-3.x

concurrency

concurrent.futures