我在搜索单元测试的超时调用时遇到了这个线程。我没有在答案或第三方包中找到任何简单的东西，所以我写了下面的装饰器，你可以直接放入代码中:

import multiprocessing.pool
import functools

def timeout(max_timeout):
    """Timeout decorator, parameter in seconds."""
    def timeout_decorator(item):
        """Wrap the original function."""
        @functools.wraps(item)
        def func_wrapper(*args, **kwargs):
            """Closure for function."""
            pool = multiprocessing.pool.ThreadPool(processes=1)
            async_result = pool.apply_async(item, args, kwargs)
            # raises a TimeoutError if execution exceeds max_timeout
            return async_result.get(max_timeout)
        return func_wrapper
    return timeout_decorator

然后就像这样简单地超时测试或任何你喜欢的函数:

@timeout(5.0)  # if execution takes longer than 5 seconds, raise a TimeoutError
def test_base_regression(self):
    ...

在@piro答案的基础上，您可以构建一个contextmanager。这允许非常易读的代码，将在成功运行后禁用警报信号(sets signal.alarm(0))

from contextlib import contextmanager
import signal
import time

@contextmanager
def timeout(duration):
    def timeout_handler(signum, frame):
        raise TimeoutError(f'block timedout after {duration} seconds')
    signal.signal(signal.SIGALRM, timeout_handler)
    signal.alarm(duration)
    try:
        yield
    finally:
        signal.alarm(0)

def sleeper(duration):
    time.sleep(duration)
    print('finished')

使用示例:

In [19]: with timeout(2):
    ...:     sleeper(1)
    ...:     
finished

In [20]: with timeout(2):
    ...:     sleeper(3)
    ...:         
---------------------------------------------------------------------------
Exception                                 Traceback (most recent call last)
<ipython-input-20-66c78858116f> in <module>()
      1 with timeout(2):
----> 2     sleeper(3)
      3 

<ipython-input-7-a75b966bf7ac> in sleeper(t)
      1 def sleeper(t):
----> 2     time.sleep(t)
      3     print('finished')
      4 

<ipython-input-18-533b9e684466> in timeout_handler(signum, frame)
      2 def timeout(duration):
      3     def timeout_handler(signum, frame):
----> 4         raise Exception(f'block timedout after {duration} seconds')
      5     signal.signal(signal.SIGALRM, timeout_handler)
      6     signal.alarm(duration)

Exception: block timedout after 2 seconds

下面是一个简单的例子，运行一个带有timeout的方法，并在成功时检索它的值。

import multiprocessing
import time

ret = {"foo": False}


def worker(queue):
    """worker function"""

    ret = queue.get()

    time.sleep(1)

    ret["foo"] = True
    queue.put(ret)


if __name__ == "__main__":
    queue = multiprocessing.Queue()
    queue.put(ret)

    p = multiprocessing.Process(target=worker, args=(queue,))
    p.start()
    p.join(timeout=10)

    if p.exitcode is None:
        print("The worker timed out.")
    else:
        print(f"The worker completed and returned: {queue.get()}")

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

我有一个不同的建议，这是一个纯函数(与线程建议相同的API)，似乎工作得很好(基于这个线程的建议)

def timeout(func, args=(), kwargs={}, timeout_duration=1, default=None):
    import signal

    class TimeoutError(Exception):
        pass

    def handler(signum, frame):
        raise TimeoutError()

    # set the timeout handler
    signal.signal(signal.SIGALRM, handler) 
    signal.alarm(timeout_duration)
    try:
        result = func(*args, **kwargs)
    except TimeoutError as exc:
        result = default
    finally:
        signal.alarm(0)

    return result

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"