等待第一个子进程完成 [英] Wait for the first subprocess to finish

问题描述

我有一个 subprocess 进程列表.我不与他们沟通，只是等待.

I have a list of subprocess' processes. I do not communicate with them and just wait.

我想等待第一个进程完成(此解决方案有效):

I want to wait for the first process to finish (this solution works):

import subprocess

a = subprocess.Popen(['...'])
b = subprocess.Popen(['...'])

# wait for the first process to finish
while True:
    over = False
    for child in {a, b}:
        try:
            rst = child.wait(timeout=5)
        except subprocess.TimeoutExpired:
            continue  # this subprocess is still running

        if rst is not None:  # subprocess is no more running
            over = True
            break  # If either subprocess exits, so do we.
    if over:
        break

我不想使用 os.wait()，因为它可能从另一个 subprocess 返回，而不是我正在等待的列表的一部分.

I don't want use os.wait(), cause it could return from another subprocess not part of the list I'm waiting for.

一个好的和优雅的解决方案可能是使用 epoll 或 select 并且没有任何循环.

A nice and elegant solution would probably be with an epoll or select and without any loop.

推荐答案

这是一个使用 psutil 的解决方案 - 它正是针对这个用例:

Here's a solution using psutil - which is aimed exactly at this use-case:

import subprocess
import psutil

a = subprocess.Popen(['/bin/sleep', "2"])

b = subprocess.Popen(['/bin/sleep', "4"])

procs_list = [psutil.Process(a.pid), psutil.Process(b.pid)]

def on_terminate(proc):
     print("process {} terminated".format(proc))

# waits for multiple processes to terminate
gone, alive = psutil.wait_procs(procs_list, timeout=3, callback=on_terminate)

或者，如果您希望循环等待其中一个进程完成:

Or, if you'd like to have a loop waiting for one of the process to be done:

while True: 
    gone, alive = psutil.wait_procs(procs_list, timeout=3, callback=on_terminate) 
    if len(gone)>0: 
        break

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