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?
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
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()}")
其他回答
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。
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"
我需要一个不会被时间阻塞的可嵌套定时中断(SIGALARM不能做到)。Sleep(基于线程的方法不能做到)。我最终复制了这里的代码并对其进行了轻微修改: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 $'.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()
还有一个用法示例:
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.'
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
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)