我在Python中调用一个函数,我知道这个函数可能会暂停,并迫使我重新启动脚本。

我怎么调用这个函数或者我把它包装在什么里面,这样如果它花费超过5秒脚本就会取消它并做其他事情?


当前回答

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

其他回答

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

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

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)

asyncio的另一个解决方案:

如果你想取消后台任务,而不仅仅是在运行的主代码上超时,那么你需要一个来自主线程的显式通信,要求任务的代码取消,比如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)

超时装饰器不能在Windows系统上工作,因为Windows不太支持信号。

如果你在windows系统中使用超时装饰器,你会得到以下结果

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

有些人建议使用use_signals=False,但对我没用。

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()

给出以下例外:

TimeoutError: Function mytest timed out after 5 seconds

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