python - 如何让多处理 python 应用程序干净地退出

Question

当我运行一个使用多处理的 python 脚本时，我发现很难让它在收到 Ctrl-C 时完全停止。必须多次按下 Ctrl-C，屏幕上会出现各种错误消息。

How can you make a python script that uses multiprocessing and quits cleanly when it receives a Ctrl-C ?

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 VBox VM 中尝试过它，它非常慢，但有效。希望对您有所帮助。

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)