函数调用超时

Question

我正在 Python 中调用一个函数，我知道该函数可能会停止并迫使我重新启动脚本。

如何调用该函数或者将其包装在什么中，以便如果花费时间超过 5 秒，脚本就会取消该函数并执行其他操作？

Answer 1

如果您在 UNIX 上运行，则可以使用 signal 包：

In [1]: import signal

# Register an handler for the timeout
In [2]: def handler(signum, frame):
   ...:     print("Forever is over!")
   ...:     raise Exception("end of time")
   ...: 

# This function *may* run for an indetermined time...
In [3]: def loop_forever():
   ...:     import time
   ...:     while 1:
   ...:         print("sec")
   ...:         time.sleep(1)
   ...:         
   ...:         

# Register the signal function handler
In [4]: signal.signal(signal.SIGALRM, handler)
Out[4]: 0

# Define a timeout for your function
In [5]: signal.alarm(10)
Out[5]: 0

In [6]: try:
   ...:     loop_forever()
   ...: except Exception, exc: 
   ...:     print(exc)
   ....: 
sec
sec
sec
sec
sec
sec
sec
sec
Forever is over!
end of time

# Cancel the timer if the function returned before timeout
# (ok, mine won't but yours maybe will :)
In [7]: signal.alarm(0)
Out[7]: 0

调用后 10 秒< code>signal.alarm(10)，调用处理程序。 这会引发一个异常，您可以从常规 Python 代码中拦截该异常。

这个模块不能很好地与线程配合使用（但是，谁呢？）

请注意，因为我们在超时发生时引发异常，所以它最终可能会在函数内部被捕获并忽略，例如一个这样的函数：

def loop_forever():
    while 1:
        print('sec')
        try:
            time.sleep(10)
        except:
            continue

Answer 2

您可以使用 multiprocessing.Process 来做到这一点。

代码

import multiprocessing
import time

# bar
def bar():
    for i in range(100):
        print "Tick"
        time.sleep(1)

if __name__ == '__main__':
    # Start bar as a process
    p = multiprocessing.Process(target=bar)
    p.start()

    # Wait for 10 seconds or until process finishes
    p.join(10)

    # If thread is still active
    if p.is_alive():
        print "running... let's kill it..."

        # Terminate - may not work if process is stuck for good
        p.terminate()
        # OR Kill - will work for sure, no chance for process to finish nicely however
        # p.kill()

        p.join()

Answer 3

如何调用该函数或者将其包装在什么中，以便如果花费时间超过 5 秒，脚本就会取消它？

我发布了一个 gist ，它用装饰器和 threading.Timer 解决了这个问题/问题。 这是有故障的。

兼容性的导入和设置

它已经使用 Python 2 和 3 进行了测试。它也应该在 Unix/Linux 和 Windows 下工作。

首先是进口。 这些尝试使代码保持一致，无论 Python 版本如何：

from __future__ import print_function
import sys
import threading
from time import sleep
try:
    import thread
except ImportError:
    import _thread as thread

使用与版本无关的代码：

try:
    range, _print = xrange, print
    def print(*args, **kwargs): 
        flush = kwargs.pop('flush', False)
        _print(*args, **kwargs)
        if flush:
            kwargs.get('file', sys.stdout).flush()            
except NameError:
    pass

现在我们已经从标准库导入了我们的功能。

exit_after 装饰器

接下来我们需要一个函数来终止子线程中的 main()：

def quit_function(fn_name):
    # print to stderr, unbuffered in Python 2.
    print('{0} took too long'.format(fn_name), file=sys.stderr)
    sys.stderr.flush() # Python 3 stderr is likely buffered.
    thread.interrupt_main() # raises KeyboardInterrupt

这是装饰器本身：

def exit_after(s):
    '''
    use as decorator to exit process if 
    function takes longer than s seconds
    '''
    def outer(fn):
        def inner(*args, **kwargs):
            timer = threading.Timer(s, quit_function, args=[fn.__name__])
            timer.start()
            try:
                result = fn(*args, **kwargs)
            finally:
                timer.cancel()
            return result
        return inner
    return outer

用法

这是直接回答您的问题的用法5 秒后退出！：

@exit_after(5)
def countdown(n):
    print('countdown started', flush=True)
    for i in range(n, -1, -1):
        print(i, end=', ', flush=True)
        sleep(1)
    print('countdown finished')

演示：

>>> countdown(3)
countdown started
3, 2, 1, 0, countdown finished
>>> countdown(10)
countdown started
10, 9, 8, 7, 6, countdown took too long
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 11, in inner
  File "<stdin>", line 6, in countdown
KeyboardInterrupt

第二个函数调用将不会完成，进程应该退出并带有回溯！

KeyboardInterrupt 并不总是停止睡眠线程

请注意，在 Windows 上的 Python 2 上，睡眠并不总是会被键盘中断打断，例如：

@exit_after(1)
def sleep10():
    sleep(10)
    print('slept 10 seconds')

>>> sleep10()
sleep10 took too long         # Note that it hangs here about 9 more seconds
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 11, in inner
  File "<stdin>", line 3, in sleep10
KeyboardInterrupt

也不可能中断扩展中运行的代码，除非它显式检查对于 PyErr_CheckSignals()，请参阅 Cython、Python 和 KeyboardInterrupt 被忽略

我会避免更多地休眠线程无论如何，不​​到一秒——这就是处理器时间的亿万年。

如何调用该函数或者将其包装在什么中，以便如果花费的时间超过 5 秒，脚本就会取消它并执行其他操作？

要捕获它并执行其他操作，您可以捕获键盘中断。

>>> try:
...     countdown(10)
... except KeyboardInterrupt:
...     print('do something else')
... 
countdown started
10, 9, 8, 7, 6, countdown took too long
do something else

Answer 4

我有一个不同的建议，它是一个纯函数（具有与线程建议相同的 API），并且似乎工作正常（基于此线程的建议）

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

Answer 5

我在单元测试中搜索超时调用时遇到了这个线程。 我在答案或第 3 方包中没有找到任何简单的内容，因此我在下面编写了装饰器，您可以直接将其放入代码中：

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

然后就可以像这样简单地使测试或您喜欢的任何函数超时：

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

Answer 6

在 pypi 上找到的 stopit 包似乎可以很好地处理超时。

我喜欢 @stopit.threading_timeoutable 装饰器，它向装饰函数添加一个 timeout 参数，它会执行您所期望的操作，它会停止该函数。

在 pypi 上查看：https://pypi.python.org/pypi/stopit

Answer 7

我是 wrapt_timeout_decorator 的作者。

乍一看，这里介绍的大多数解决方案在 Linux 下工作得很好 - 因为我们有 fork() 和 signals() - 但在 Windows 上，事情看起来有点不同。
当谈到 Linux 上的子线程时，你不能再使用信号了。

为了在 Windows 下生成进程，它需要是可挑选的 - 而许多修饰函数或类方法则不是。

因此，您需要使用更好的pickler，例如dill和multiprocess（不是pickle和multiprocessing） - 这就是为什么您不能使用ProcessPoolExecutor（或仅具有有限的功能）。

对于超时本身 - 您需要定义超时的含义 - 因为在 Windows 上，需要相当长的（且不可确定的）时间来生成进程。 这对于短超时来说可能会很棘手。 让我们假设，生成过程大约需要 0.5 秒（很容易！！！）。 如果你给出 0.2 秒的超时时间，会发生什么？
该函数是否应该在 0.5 + 0.2 秒后超时（因此让该方法运行 0.2 秒）？
或者被调用的进程应该在 0.2 秒后超时（在这种情况下，修饰函数将始终超时，因为在那段时间它甚至没有生成）？

嵌套装饰器也可能很糟糕，并且您不能在子线程中使用信号。 如果你想创建一个真正通用的、跨平台的装饰器，所有这些都需要考虑（并测试）。

其他问题是将异常传递回调用者，以及日志记录问题（如果在装饰函数中使用 - 不支持在另一个进程中记录到文件）

我试图涵盖所有边缘情况，您可能会查看包wrapt_timeout_decorator，或者至少测试受此处使用的单元测试启发的您自己的解决方案。

@Alexis Eggermont - 不幸的是我没有足够的观点来评论 - 也许其他人可以通知你 - 我想我解决了你的导入问题。

Answer 8

有很多建议，但没有使用并发。futures，我认为这是处理这个问题的最清晰的方法。

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，然后等待最多 5 秒，然后引发一个 TimeoutError，您可以根据需要捕获并处理该错误。

原生于 python 3.2+ 并向后移植到 2.7 (pip install futures)。

线程和进程之间的切换就像用 ThreadPoolExecutor 替换 ProcessPoolExecutor 一样简单。

如果您想在超时时终止进程，我建议您查看 Pebble。

Answer 9

在 @piro 的答案的基础上构建和增强，您可以构建一个上下文管理器。 这允许非常可读的代码，它将在成功运行后禁用警报信号（设置 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')

示例用法：

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

Answer 10

很棒、易于使用且可靠的 PyPi 项目 timeout-decorator (https://pypi.org/project/timeout-decorator/)

安装：

pip install timeout-decorator

使用：

import time
import timeout_decorator

@timeout_decorator.timeout(5)
def mytest():
    print "Start"
    for i in range(1,10):
        time.sleep(1)
        print "%d seconds have passed" % i

if __name__ == '__main__':
    mytest()

Answer 11

timeout-decorator 不适用于 Windows 系统，因为 Windows 不能很好地支持 signal。

如果您在 Windows 系统中使用 timeout-decorator，您将得到以下信息

AttributeError: module 'signal' has no attribute 'SIGALRM'

Some suggest to use use_signals=False but did not work for me.

作者 @bitranox 创建了以下包：

pip install https://github.com/bitranox/wrapt-timeout-decorator/archive/master.zip

代码示例：

import time
from wrapt_timeout_decorator import *

@timeout(5)
def mytest(message):
    print(message)
    for i in range(1,10):
        time.sleep(1)
        print('{} seconds have passed'.format(i))

def main():
    mytest('starting')


if __name__ == '__main__':
    main()

给出以下异常：

TimeoutError: Function mytest timed out after 5 seconds

Answer 12

为了防止它对任何人有帮助，在 @piro 的答案的基础上，我制作了一个函数装饰器：在

import time
import signal
from functools import wraps


def timeout(timeout_secs: int):
    def wrapper(func):
        @wraps(func)
        def time_limited(*args, **kwargs):
            # Register an handler for the timeout
            def handler(signum, frame):
                raise Exception(f"Timeout for function '{func.__name__}'")

            # Register the signal function handler
            signal.signal(signal.SIGALRM, handler)

            # Define a timeout for your function
            signal.alarm(timeout_secs)

            result = None
            try:
                result = func(*args, **kwargs)
            except Exception as exc:
                raise exc
            finally:
                # disable the signal alarm
                signal.alarm(0)

            return result

        return time_limited

    return wrapper

具有 20 秒 超时的函数上使用包装器将如下所示：

    @timeout(20)
    def my_slow_or_never_ending_function(name):
        while True:
            time.sleep(1)
            print(f"Yet another second passed {name}...")

    try:
        results = my_slow_or_never_ending_function("Yooo!")
    except Exception as e:
        print(f"ERROR: {e}")

Answer 13

亮点

• 引发TimeoutError使用异常来警告超时 - 可以轻松修改
• 跨平台：Windows和Windows Mac OS X
• 兼容性：Python 3.6+（我还在 python 2.7 上进行了测试，它可以进行小的语法调整）

有关并行映射的完整说明和扩展，请参阅此处 https://flipdazed.github.io/blog/quant%20dev/parallel-functions-with-timeouts

最小示例

>>> @killer_call(timeout=4)
... def bar(x):
...        import time
...        time.sleep(x)
...        return x
>>> bar(10)
Traceback (most recent call last):
  ...
__main__.TimeoutError: function 'bar' timed out after 4s

和预期

>>> bar(2)
2

完整代码

import multiprocessing as mp
import multiprocessing.queues as mpq
import functools
import dill

from typing import Tuple, Callable, Dict, Optional, Iterable, List, Any

class TimeoutError(Exception):

    def __init__(self, func: Callable, timeout: int):
        self.t = timeout
        self.fname = func.__name__

    def __str__(self):
            return f"function '{self.fname}' timed out after {self.t}s"


def _lemmiwinks(func: Callable, args: Tuple, kwargs: Dict[str, Any], q: mp.Queue):
    """lemmiwinks crawls into the unknown"""
    q.put(dill.loads(func)(*args, **kwargs))


def killer_call(func: Callable = None, timeout: int = 10) -> Callable:
    """
    Single function call with a timeout

    Args:
        func: the function
        timeout: The timeout in seconds
    """

    if not isinstance(timeout, int):
        raise ValueError(f'timeout needs to be an int. Got: {timeout}')

    if func is None:
        return functools.partial(killer_call, timeout=timeout)

    @functools.wraps(killer_call)
    def _inners(*args, **kwargs) -> Any:
        q_worker = mp.Queue()
        proc = mp.Process(target=_lemmiwinks, args=(dill.dumps(func), args, kwargs, q_worker))
        proc.start()
        try:
            return q_worker.get(timeout=timeout)
        except mpq.Empty:
            raise TimeoutError(func, timeout)
        finally:
            try:
                proc.terminate()
            except:
                pass
    return _inners

if __name__ == '__main__':
    @killer_call(timeout=4)
    def bar(x):
        import time
        time.sleep(x)
        return x

    print(bar(2))
    bar(10)

注释

由于 dill 的工作方式，您将需要在函数内部导入。

这也意味着如果目标函数中有导入，这些函数可能与 doctest 不兼容。 您将遇到找不到 __import__ 的问题。

Answer 14

我们可以使用信号来实现同样的目的。 我认为下面的例子对你有用。 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"

Answer 15

asyncio 的另一个解决方案：

如果您想取消后台任务，而不仅仅是在运行的主代码上超时，那么您需要主线程的显式通信来要求任务的代码取消，例如 threading.Event()

import asyncio
import functools
import multiprocessing
from concurrent.futures.thread import ThreadPoolExecutor


class SingletonTimeOut:
    pool = None

    @classmethod
    def run(cls, to_run: functools.partial, timeout: float):
        pool = cls.get_pool()
        loop = cls.get_loop()
        try:
            task = loop.run_in_executor(pool, to_run)
            return loop.run_until_complete(asyncio.wait_for(task, timeout=timeout))
        except asyncio.TimeoutError as e:
            error_type = type(e).__name__ #TODO
            raise e

    @classmethod
    def get_pool(cls):
        if cls.pool is None:
            cls.pool = ThreadPoolExecutor(multiprocessing.cpu_count())
        return cls.pool

    @classmethod
    def get_loop(cls):
        try:
            return asyncio.get_event_loop()
        except RuntimeError:
            asyncio.set_event_loop(asyncio.new_event_loop())
            # print("NEW LOOP" + str(threading.current_thread().ident))
            return asyncio.get_event_loop()

# ---------------

TIME_OUT = float('0.2')  # seconds

def toto(input_items,nb_predictions):
    return 1

to_run = functools.partial(toto,
                           input_items=1,
                           nb_predictions="a")

results = SingletonTimeOut.run(to_run, TIME_OUT)

Answer 16

Tim Savannah 的 func_timeout 包 对我来说效果很好。

安装：

pip install func_timeout

使用方法：

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,))

Answer 17

#!/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)

Answer 18

这是一个简单易用的装饰器，它返回给定的默认值如果函数的执行时间到期，灵感来自 此问题的第一个答案：

import signal
from functools import wraps
import time

def timeout(seconds, default=None):
    def decorator(func):
        @wraps(func)
        def wrapper(*args, **kwargs):
            def signal_handler(signum, frame):
                raise TimeoutError("Timed out!")
            # Set up the signal handler for timeout
            signal.signal(signal.SIGALRM, signal_handler)

            # Set the initial alarm for the integer part of seconds
            signal.setitimer(signal.ITIMER_REAL, seconds)

            
            try:
                result = func(*args, **kwargs)
            except TimeoutError:
                return default
            finally:
                signal.alarm(0)
            
            return result
        
        return wrapper
    
    return decorator

@timeout(0.2, default="Timeout!")
def long_function_call(meal):
    time.sleep(3)
    return f"I have executed fully, {meal} is ready"

@timeout(1.3, default="Timeout!")
def less_long_function_call(meal):
    time.sleep(1)
    return f"I have executed fully, {meal} is ready"

result = long_function_call("bacon")
print(result)  # Prints "Timeout!" if the function execution exceeds 0.2 seconds
result = less_long_function_call("bacon")
print(result)  # Prints "Timeout!" if the function execution exceeds 1.3 seconds

Answer 19

我需要可嵌套定时中断（SIGALARM 无法做到），并且不会被 time.sleep 阻塞（基于线程的方法无法做到）。 我最终从这里复制并稍微修改了代码： http://code.activestate.com/recipes/577600-queue-for-managing-multiple-sigalrm-alarms-concurr/

代码本身：

#!/usr/bin/python

# lightly modified version of http://code.activestate.com/recipes/577600-queue-for-managing-multiple-sigalrm-alarms-concurr/


"""alarm.py: Permits multiple SIGALRM events to be queued.

Uses a `heapq` to store the objects to be called when an alarm signal is
raised, so that the next alarm is always at the top of the heap.
"""

import heapq
import signal
from time import time

__version__ = '$Revision: 2539 
以及使用示例：
import alarm
from time import sleep

try:
    with alarm.Timeout(id_='a', seconds=5):
        try:
            with alarm.Timeout(id_='b', seconds=2):
                sleep(3)
        except alarm.TimeoutError as e:
            print 'raised', e.id_
        sleep(30)
except alarm.TimeoutError as e:
    print 'raised', e.id_
else:
    print 'nope.'

.split()[1]

alarmlist = []

__new_alarm = lambda t, f, a, k: (t + time(), f, a, k)
__next_alarm = lambda: int(round(alarmlist[0][0] - time())) if alarmlist else None
__set_alarm = lambda: signal.alarm(max(__next_alarm(), 1))


class TimeoutError(Exception):
    def __init__(self, message, id_=None):
        self.message = message
        self.id_ = id_


class Timeout:
    ''' id_ allows for nested timeouts. '''
    def __init__(self, id_=None, seconds=1, error_message='Timeout'):
        self.seconds = seconds
        self.error_message = error_message
        self.id_ = id_
    def handle_timeout(self):
        raise TimeoutError(self.error_message, self.id_)
    def __enter__(self):
        self.this_alarm = alarm(self.seconds, self.handle_timeout)
    def __exit__(self, type, value, traceback):
        try:
            cancel(self.this_alarm) 
        except ValueError:
            pass


def __clear_alarm():
    """Clear an existing alarm.

    If the alarm signal was set to a callable other than our own, queue the
    previous alarm settings.
    """
    oldsec = signal.alarm(0)
    oldfunc = signal.signal(signal.SIGALRM, __alarm_handler)
    if oldsec > 0 and oldfunc != __alarm_handler:
        heapq.heappush(alarmlist, (__new_alarm(oldsec, oldfunc, [], {})))


def __alarm_handler(*zargs):
    """Handle an alarm by calling any due heap entries and resetting the alarm.

    Note that multiple heap entries might get called, especially if calling an
    entry takes a lot of time.
    """
    try:
        nextt = __next_alarm()
        while nextt is not None and nextt <= 0:
            (tm, func, args, keys) = heapq.heappop(alarmlist)
            func(*args, **keys)
            nextt = __next_alarm()
    finally:
        if alarmlist: __set_alarm()


def alarm(sec, func, *args, **keys):
    """Set an alarm.

    When the alarm is raised in `sec` seconds, the handler will call `func`,
    passing `args` and `keys`. Return the heap entry (which is just a big
    tuple), so that it can be cancelled by calling `cancel()`.
    """
    __clear_alarm()
    try:
        newalarm = __new_alarm(sec, func, args, keys)
        heapq.heappush(alarmlist, newalarm)
        return newalarm
    finally:
        __set_alarm()


def cancel(alarm):
    """Cancel an alarm by passing the heap entry returned by `alarm()`.

    It is an error to try to cancel an alarm which has already occurred.
    """
    __clear_alarm()
    try:
        alarmlist.remove(alarm)
        heapq.heapify(alarmlist)
    finally:
        if alarmlist: __set_alarm()

以及使用示例：

Answer 20

我遇到了同样的问题，但我的情况是需要在子线程上工作，信号对我不起作用，所以我编写了一个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：https://github.com/dozysun/timeout-定时器

Answer 21

这是一个简单的示例，运行一个超时方法，如果成功则检索其值。

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()}")

Answer 22

这是对给定的基于线程的解决方案的轻微改进。

下面的代码支持异常：

def runFunctionCatchExceptions(func, *args, **kwargs):
    try:
        result = func(*args, **kwargs)
    except Exception, message:
        return ["exception", message]

    return ["RESULT", result]


def runFunctionWithTimeout(func, args=(), kwargs={}, timeout_duration=10, default=None):
    import threading
    class InterruptableThread(threading.Thread):
        def __init__(self):
            threading.Thread.__init__(self)
            self.result = default
        def run(self):
            self.result = runFunctionCatchExceptions(func, *args, **kwargs)
    it = InterruptableThread()
    it.start()
    it.join(timeout_duration)
    if it.isAlive():
        return default

    if it.result[0] == "exception":
        raise it.result[1]

    return it.result[1]

以 5 秒超时调用它：

result = timeout(remote_calculate, (myarg,), timeout_duration=5)

Answer 23

这是一个 POSIX 版本，它结合了许多以前的答案，以提供以下功能：

1. 子进程阻止执行。
2. 在类成员函数上使用超时函数。
3. 对终止时间有严格要求。

这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

Answer 24

如果工作没有完成，我打算杀死进程，使用线程和进程来实现这一点。

from concurrent.futures import ThreadPoolExecutor

from time import sleep
import multiprocessing


# test case 1
def worker_1(a,b,c):
    for _ in range(2):
        print('very time consuming sleep')
        sleep(1)

    return a+b+c

# test case 2
def worker_2(in_name):
    for _ in range(10):
        print('very time consuming sleep')
        sleep(1)

    return 'hello '+in_name

作为上下文管理器的实际类

class FuncTimer():
    def __init__(self,fn,args,runtime):
        self.fn = fn
        self.args = args
        self.queue = multiprocessing.Queue()
        self.runtime = runtime
        self.process = multiprocessing.Process(target=self.thread_caller)

    def thread_caller(self):
        with ThreadPoolExecutor() as executor:
            future = executor.submit(self.fn, *self.args)
            self.queue.put(future.result())

    def  __enter__(self):
        return self

    def start_run(self):
        self.process.start()
        self.process.join(timeout=self.runtime)
        if self.process.exitcode is None:
            self.process.kill()
        if self.process.exitcode is None:
            out_res = None
            print('killed premature')
        else:
            out_res = self.queue.get()
        return out_res


    def __exit__(self, exc_type, exc_value, exc_traceback):
        self.process.kill()

如何使用它

print('testing case 1') 
with FuncTimer(fn=worker_1,args=(1,2,3),runtime = 5) as fp: 
    res = fp.start_run()
    print(res)

print('testing case 2')
with FuncTimer(fn=worker_2,args=('ram',),runtime = 5) as fp: 
    res = fp.start_run()
    print(res)