我们也可以用信号来表示。我认为下面的例子会对你有用。与线程相比，它非常简单。

import signal

def timeout(signum, frame):
    raise myException

#this is an infinite loop, never ending under normal circumstances
def main():
    print 'Starting Main ',
    while 1:
        print 'in main ',

#SIGALRM is only usable on a unix platform
signal.signal(signal.SIGALRM, timeout)

#change 5 to however many seconds you need
signal.alarm(5)

try:
    main()
except myException:
    print "whoops"

#!/usr/bin/python2
import sys, subprocess, threading
proc = subprocess.Popen(sys.argv[2:])
timer = threading.Timer(float(sys.argv[1]), proc.terminate)
timer.start()
proc.wait()
timer.cancel()
exit(proc.returncode)

我也遇到过同样的问题，但我的情况是需要在子线程上工作，信号不适合我，所以我写了一个python包:timeout-timer来解决这个问题，支持用作上下文或装饰器，使用信号或子线程模块来触发超时中断:

from timeout_timer import timeout, TimeoutInterrupt

class TimeoutInterruptNested(TimeoutInterrupt):
    pass

def test_timeout_nested_loop_both_timeout(timer="thread"):
    cnt = 0
    try:
        with timeout(5, timer=timer):
            try:
                with timeout(2, timer=timer, exception=TimeoutInterruptNested):
                    sleep(2)
            except TimeoutInterruptNested:
                cnt += 1
            time.sleep(10)
    except TimeoutInterrupt:
        cnt += 1
    assert cnt == 2

查看更多信息:https://github.com/dozysun/timeout-timer

Tim Savannah的func_timeout包对我来说工作得很好。

安装:

PIP安装func_timeout

用法:

import time
from func_timeout import func_timeout, FunctionTimedOut

def my_func(n):
    time.sleep(n)

time_to_sleep = 10

# time out after 2 seconds using kwargs
func_timeout(2, my_func, kwargs={'n' : time_to_sleep})

# time out after 2 seconds using args
func_timeout(2, my_func, args=(time_to_sleep,))

下面是一个POSIX版本，它结合了前面的许多答案来提供以下特性:

子进程阻塞执行。 timeout函数在类成员函数上的使用。 严格要求终止时间。

下面是代码和一些测试用例:

import threading
import signal
import os
import time

class TerminateExecution(Exception):
    """
    Exception to indicate that execution has exceeded the preset running time.
    """


def quit_function(pid):
    # Killing all subprocesses
    os.setpgrp()
    os.killpg(0, signal.SIGTERM)

    # Killing the main thread
    os.kill(pid, signal.SIGTERM)


def handle_term(signum, frame):
    raise TerminateExecution()


def invoke_with_timeout(timeout, fn, *args, **kwargs):
    # Setting a sigterm handler and initiating a timer
    old_handler = signal.signal(signal.SIGTERM, handle_term)
    timer = threading.Timer(timeout, quit_function, args=[os.getpid()])
    terminate = False

    # Executing the function
    timer.start()
    try:
        result = fn(*args, **kwargs)
    except TerminateExecution:
        terminate = True
    finally:
        # Restoring original handler and cancel timer
        signal.signal(signal.SIGTERM, old_handler)
        timer.cancel()

    if terminate:
        raise BaseException("xxx")

    return result

### Test cases
def countdown(n):
    print('countdown started', flush=True)
    for i in range(n, -1, -1):
        print(i, end=', ', flush=True)
        time.sleep(1)
    print('countdown finished')
    return 1337


def really_long_function():
    time.sleep(10)


def really_long_function2():
    os.system("sleep 787")


# Checking that we can run a function as expected.
assert invoke_with_timeout(3, countdown, 1) == 1337

# Testing various scenarios
t1 = time.time()
try:
    print(invoke_with_timeout(1, countdown, 3))
    assert(False)
except BaseException:
    assert(time.time() - t1 < 1.1)
    print("All good", time.time() - t1)

t1 = time.time()
try:
    print(invoke_with_timeout(1, really_long_function2))
    assert(False)
except BaseException:
    assert(time.time() - t1 < 1.1)
    print("All good", time.time() - t1)


t1 = time.time()
try:
    print(invoke_with_timeout(1, really_long_function))
    assert(False)
except BaseException:
    assert(time.time() - t1 < 1.1)
    print("All good", time.time() - t1)

# Checking that classes are referenced and not
# copied (as would be the case with multiprocessing)


class X:
    def __init__(self):
        self.value = 0

    def set(self, v):
        self.value = v


x = X()
invoke_with_timeout(2, x.set, 9)
assert x.value == 9

有很多建议，但没有一个是使用并发的。期货，我认为这是最清晰的处理方式。

from concurrent.futures import ProcessPoolExecutor

# Warning: this does not terminate function if timeout
def timeout_five(fnc, *args, **kwargs):
    with ProcessPoolExecutor() as p:
        f = p.submit(fnc, *args, **kwargs)
        return f.result(timeout=5)

超级简单的阅读和维护。

我们创建一个池，提交一个进程，然后等待5秒，然后引发一个TimeoutError，你可以根据需要捕获和处理它。

本机为python 3.2+，并反向移植到2.7 (pip install futures)。

线程和进程之间的切换非常简单，只需将ProcessPoolExecutor替换为ThreadPoolExecutor。

如果您想在超时时终止进程，我建议您查看Pebble。