构建已提交future的列表，然后使用 as_completed（） 来了解线程何时完成，其结果何时可用。

from concurrent.futures import ThreadPoolExecutor, as_completed
import time

def fnc(x, y):
    time.sleep(3)
    return x*y

futures = []

with ThreadPoolExecutor() as executor:
    for i in range(0, 4):
        print(f"Submitting {i}")
        futures += [executor.submit(fnc, i, i+1)]
    for future in as_completed(futures):
        print(future.result())

你在ThreadPoolExecutor上下文管理器外调用result方法，当你退出时，它调用__exit__方法：

def __exit__(self, exc_type, exc_val, exc_tb):
    self.shutdown(wait=True)
    return False

shutdown方法签名为：

shutdown(self, wait=True, *, cancel_futures=False)

医生说：

Args:
     wait: If True then shutdown will not return until all running
           futures have finished executing and the resources used by the
           executor have been reclaimed.
     cancel_futures: If True then shutdown will cancel all pending
           futures. Futures that are completed or running will not be
           cancelled.

我们可以看到，默认情况下，它会等到所有正在运行的future和它们的资源都停止运行，并且cancel_futures默认情况下会得到值False，因此我们不是取消挂起的future。
我们可以通过修改fnc来打印值而不是返回值，并且在ThreadPoolExecutor上下文管理器代码块之后什么也不做来证明这一点：

def fnc(x, y):
    time.sleep(3)
    print(x * y)

futures = []
with ThreadPoolExecutor(max_workers=1) as executor:
    for i in range(0, 4):
        print(f"Submitting {i}")
        futures += [executor.submit(fnc, i, i + 1)]

print("Blah!")

仍然打印值0, 2, 6, 12！，即使我们只将函数提交到执行列表...
通过在上下文管理器中移动for循环块来修复它：

from concurrent.futures import ThreadPoolExecutor
import time

def fnc(x, y):
    time.sleep(3)
    return x*y

futures = []
with ThreadPoolExecutor(max_workers=1) as executor:
    for i in range(0, 4):
        print(f"Submitting {i}")
        futures += [executor.submit(fnc, i, i+1)]
    for f in futures:
        print(f.result())

请注意，设置max_workers=1本质上是强制程序连续运行而不是并发运行，在此程序中，设置max_workers=X将一次打印fnc的X返回结果。
如果你想在两个结果之间等待3秒钟，那么可以将max_workers设置为1或者完全删除它。如果你想每3秒钟打印两个结果-设置max_workers=2等等。