Question

当我运行使用多处理的python脚本时，我发现很难让它在收到Ctrl-C时干净利落地停止。必须多次按Ctrl-C，屏幕上才会显示各种错误消息。

如何制作使用多处理和退出的python脚本收到Ctrl-C时干净利落吗？

以此脚本为例

import numpy as np, time
from multiprocessing import Pool

def countconvolve(N):
    np.random.seed() # ensure seed is random

    count = 0
    iters = 1000000 # 1million

    l=12
    k=12
    l0=l+k-1

    for n in range(N):
        t = np.random.choice(np.array([-1,1], dtype=np.int8), size=l0 * iters)
        v = np.random.choice(np.array([-1,1], dtype=np.int8), size = l * iters)
        for i in xrange(iters):
            if (not np.convolve(v[(l*i):(l*(i+1))],
t[(l0*i):(l0*(i+1))], 'valid').any()):
                count += 1
    return count

if __name__ == '__main__':
    start = time.clock()

    num_processes = 8
    N = 13

    pool = Pool(processes=num_processes)
    res = pool.map(countconvolve, [N] * num_processes)
    print res, sum(res)

    print (time.clock() - start)

Answer 1

Jon的解决方案可能更好，但这里使用的是信号处理程序。我在一台VBox VM中尝试过，它非常慢，但很有效。我希望它会有所帮助。

import numpy as np, time
from multiprocessing import Pool
import signal

# define pool as global
pool = None


def term_signal_handler(signum, frame):
    global pool

    print 'CTRL-C pressed'
    try:
        pool.close()
        pool.join()
    except AttributeError:
        print 'Pool has been already closed'


def countconvolve(N):
    np.random.seed() # ensure seed is random

    count = 0
    iters = 1000000 # 1million

    l=12
    k=12
    l0=l+k-1

    for n in range(N):
        t = np.random.choice(np.array([-1,1], dtype=np.int8), size=l0 * iters)
        v = np.random.choice(np.array([-1,1], dtype=np.int8), size = l * iters)
        for i in xrange(iters):
            if (not np.convolve(v[(l*i):(l*(i+1))],t[(l0*i):(l0*(i+1))], 'valid').any()):
                count += 1
    return count


if __name__ == '__main__':
    # Register the signal handler
    signal.signal(signal.SIGINT, term_signal_handler)

    start = time.clock()

    num_processes = 8
    N = 13

    pool = Pool(processes=num_processes)
    res = pool.map(countconvolve, [N] * num_processes)
    print res, sum(res)

    print (time.clock() - start)

Answer 2

我相信a similar post here on SO中提到的try-catch可以用来覆盖它。

如果你在try-catch中包装pool.map调用然后调用terminate和join我认为会这样做。

<强> [编辑]

一些实验表明这些方面的效果很好：

from multiprocessing import Pool
import random
import time

def countconvolve(N):
    try:
        sleepTime = random.randint(0,5)
        time.sleep(sleepTime)
        count = sleepTime
    except KeyboardInterrupt as e:
        pass
    return count

if __name__ == '__main__':
    random.seed(0)
    start = time.clock()

    num_processes = 8
    N = 13

    pool = Pool(processes=num_processes)
    try:
        res = pool.map(countconvolve, [N] * num_processes)
        print res, sum(res)
        print (time.clock() - start)
    except KeyboardInterrupt as e:
        print 'Stopping..'

我简化了你的例子，以避免在我的机器上加载numpy进行测试，但关键部分是两个try-except调用来处理 CTRL + C 按键。

如何让多处理python应用程序干净利落

2 个答案: