问题：函数调用超时
我正在Python中调用一个函数，该函数可能会停滞并迫使我重新启动脚本。
如何调用该函数或将其包装在其中，以便如果花费的时间超过5秒，脚本将取消该函数并执行其他操作？
I’m calling a function in Python which I know may stall and force me to restart the script.
How do I call the function or what do I wrap it in so that if it takes longer than 5 seconds the script cancels it and does something else?
回答 0
如果您在UNIX上运行，则可以使用信号包：
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秒钟，将调用alarm.alarm(10)处理程序。这引发了一个异常，您可以从常规Python代码中拦截该异常。
该模块不能很好地与线程配合使用（但是，谁可以呢？）
请注意，由于发生超时时会引发异常，因此它可能最终在函数内部被捕获并被忽略，例如一个这样的函数：
def loop_forever():
while 1:
print('sec')
try:
time.sleep(10)
except:
continue
You may use the signal package if you are running on UNIX:
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
10 seconds after the call signal.alarm(10), the handler is called. This raises an exception that you can intercept from the regular Python code.
This module doesn’t play well with threads (but then, who does?)
Note that since we raise an exception when timeout happens, it may end up caught and ignored inside the function, for example of one such function:
def loop_forever():
while 1:
print('sec')
try:
time.sleep(10)
except:
continue
回答 1
您可以multiprocessing.Process用来精确地做到这一点。
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
p.terminate()
p.join()
You can use multiprocessing.Process to do exactly that.
Code
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()
回答 2
如何调用该函数或将其包装起来，以便如果花费的时间超过5秒钟，脚本将取消该函数？
我发布了要点，用装饰器和来解决此问题threading.Timer。这是一个细分。
导入和设置以实现兼容性
它已经通过Python 2和3进行了测试。它也应该在Unix / Linux和Windows下运行。
首先是进口。这些尝试使代码保持一致，而不管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
使用版本无关代码：
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，我们已经从标准库中导入了我们的功能。
exit_after 装饰工
接下来，我们需要一个函数来终止main()子线程：
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
这是直接回答您有关5秒后退出的问题的用法！：
@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结束，而是该过程应退出并回溯！
KeyboardInterrupt 并不总是停止休眠线程
请注意，在Windows上的Python 2上，睡眠不会总是被键盘中断打断，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
除非它明确检查PyErr_CheckSignals()，否则也不太可能中断在扩展程序中运行的代码，请参见 Cython，Python和KeyboardInterrupt被忽略
在任何情况下，我都避免将线程休眠超过一秒钟-这是处理器时间的永恒。
如何调用该函数或将其包装在其中，以便如果花费的时间超过5秒，脚本将取消该函数并执行其他操作？
要捕获它并执行其他操作，可以捕获KeyboardInterrupt。
>>> try:
... countdown(10)
... except KeyboardInterrupt:
... print('do something else')
...
countdown started
10, 9, 8, 7, 6, countdown took too long
do something else
How do I call the function or what do I wrap it in so that if it takes longer than 5 seconds the script cancels it?
I posted a gist that solves this question/problem with a decorator and a threading.Timer. Here it is with a breakdown.
Imports and setups for compatibility
It was tested with Python 2 and 3. It should also work under Unix/Linux and Windows.
First the imports. These attempt to keep the code consistent regardless of the Python version:
from __future__ import print_function
import sys
import threading
from time import sleep
try:
import thread
except ImportError:
import _thread as thread
Use version independent code:
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
Now we have imported our functionality from the standard library.
exit_after decorator
Next we need a function to terminate the main() from the child thread:
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
And here is the decorator itself:
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
Usage
And here’s the usage that directly answers your question about exiting after 5 seconds!:
@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')
Demo:
>>> 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
The second function call will not finish, instead the process should exit with a traceback!
KeyboardInterrupt does not always stop a sleeping thread
Note that sleep will not always be interrupted by a keyboard interrupt, on Python 2 on Windows, e.g.:
@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
nor is it likely to interrupt code running in extensions unless it explicitly checks for PyErr_CheckSignals(), see Cython, Python and KeyboardInterrupt ignored
I would avoid sleeping a thread more than a second, in any case – that’s an eon in processor time.
How do I call the function or what do I wrap it in so that if it takes longer than 5 seconds the script cancels it and does something else?
To catch it and do something else, you can catch the KeyboardInterrupt.
>>> try:
... countdown(10)
... except KeyboardInterrupt:
... print('do something else')
...
countdown started
10, 9, 8, 7, 6, countdown took too long
do something else
回答 3
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），并且似乎可以正常工作（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
I have a different proposal which is a pure function (with the same API as the threading suggestion) and seems to work fine (based on suggestions on this thread)
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
回答 4
在搜索单元测试的超时调用时，我遇到了这个线程。我没有在答案或第三方软件包中找到任何简单的东西，因此我在下面编写了装饰器，您可以直接进入代码：
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，使测试或您喜欢的任何功能超时就这么简单：
@timeout(5.0) # if execution takes longer than 5 seconds, raise a TimeoutError
def test_base_regression(self):
...
I ran across this thread when searching for a timeout call on unit tests. I didn’t find anything simple in the answers or 3rd party packages so I wrote the decorator below you can drop right into code:
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
Then it’s as simple as this to timeout a test or any function you like:
@timeout(5.0) # if execution takes longer than 5 seconds, raise a TimeoutError
def test_base_regression(self):
...
回答 5
的 stopit在pypi上找到软件包似乎可以很好地处理超时问题。
我喜欢@stopit.threading_timeoutable装饰器，它增加了一个timeout向装饰的函数参数，该参数完成了您所期望的操作，从而停止了该函数。
在pypi上查看：https ://pypi.python.org/pypi/stopit
The stopit package, found on pypi, seems to handle timeouts well.
I like the @stopit.threading_timeoutable decorator, which adds a timeout parameter to the decorated function, which does what you expect, it stops the function.
Check it out on pypi: https://pypi.python.org/pypi/stopit
回答 6
有很多建议，但没有一个建议使用current.futures，我认为这是处理此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秒钟，然后引发一个TimeoutError，您可以根据需要捕获并处理它。
原生于python 3.2+，并反向移植到2.7（点安装期货）。
线程和进程之间的切换很简单，只要更换ProcessPoolExecutor用ThreadPoolExecutor。
如果您想在超时时终止进程，建议您调查Pebble。
There are a lot of suggestions, but none using concurrent.futures, which I think is the most legible way to handle this.
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)
Super simple to read and maintain.
We make a pool, submit a single process and then wait up to 5 seconds before raising a TimeoutError that you could catch and handle however you needed.
Native to python 3.2+ and backported to 2.7 (pip install futures).
Switching between threads and processes is as simple as replacing ProcessPoolExecutor with ThreadPoolExecutor.
If you want to terminate the Process on timeout I would suggest looking into Pebble.
回答 7
出色，易于使用且可靠的PyPi项目超时装饰器（https://pypi.org/project/timeout-decorator/）
安装方式：
pip install timeout-decorator
用法：
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()
Great, easy to use and reliable PyPi project timeout-decorator (https://pypi.org/project/timeout-decorator/)
installation:
pip install timeout-decorator
Usage:
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()
回答 8
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上都无法正常工作-因为我们有fork（）和signal（）-但是在Windows上，情况看起来有些不同。当涉及到Linux上的子线程时，您将无法再使用Signals。
为了在Windows下产生一个进程，它必须是可挑选的-许多修饰函数或Class方法不是。
因此，您需要使用更好的Pickler，如莳萝和多进程（而不是Pickle和多进程）-这就是为什么您不能使用ProcessPoolExecutor（或仅在功能有限的情况下）的原因。
对于超时本身-您需要定义超时的含义-因为在Windows上将花费大量（且不确定）的时间来生成该进程。如果超时时间短，这可能会很棘手。让我们假设，生成该过程大约需要0.5秒（很容易！！！）。如果您给出0.2秒的超时时间，应该怎么办？函数是否应在0.5 + 0.2秒后超时（让方法运行0.2秒）？还是被调用的进程应该在0.2秒后超时（在这种情况下，修饰的函数将始终超时，因为在那个时间内它甚至没有生成）？
嵌套的装饰器也很讨厌，您不能在子线程中使用Signals。如果要创建真正的通用跨平台装饰器，则需要考虑所有这些因素（并进行测试）。
其他问题会将异常传递回调用者以及日志记录问题（如果在修饰的函数中使用-不支持在另一个进程中记录文件）
我试图涵盖所有的极端情况，您可能会考虑wrapt_timeout_decorator包，或者至少测试那里使用的单元测试启发的自己的解决方案。
@Alexis Eggermont-很遗憾，我没有足够的意见要发表-也许其他人可以通知您-我想我已经解决了您的导入问题。
I am the author of wrapt_timeout_decorator
Most of the solutions presented here work wunderfully under Linux on the first glance – because we have fork() and signals() – but on windows the things look a bit different.
And when it comes to subthreads on Linux, You cant use Signals anymore.
In order to spawn a process under Windows, it needs to be picklable – and many decorated functions or Class methods are not.
So You need to use a better pickler like dill and multiprocess (not pickle and multiprocessing) – thats why You cant use ProcessPoolExecutor (or only with limited functionality).
For the timeout itself – You need to define what timeout means – because on Windows it will take considerable (and not determinable) time to spawn the process. This can be tricky on short timeouts. Lets assume, spawning the process takes about 0.5 seconds (easily !!!). If You give a timeout of 0.2 seconds what should happen ?
Should the function time out after 0.5 + 0.2 seconds (so let the method run for 0.2 seconds)?
Or should the called process time out after 0.2 seconds (in that case, the decorated function will ALWAYS timeout, because in that time it is not even spawned) ?
Also nested decorators can be nasty and You cant use Signals in a subthread. If You want to create a truly universal, cross-platform decorator, all this needs to be taken into consideration (and tested).
Other issues are passing exceptions back to the caller, as well as logging issues (if used in the decorated function – logging to files in another process is NOT supported)
I tried to cover all edge cases, You might look into the package wrapt_timeout_decorator, or at least test Your own solutions inspired by the unittests used there.
@Alexis Eggermont – unfortunately I dont have enough points to comment – maybe someone else can notify You – I think I solved Your import issue.
回答 9
timeout-decorator不能在Windows系统上正常运行，因为Windows不能signal很好地支持。
如果您在Windows系统中使用timeout-decorator，则会得到以下信息
AttributeError: module 'signal' has no attribute 'SIGALRM'
有些人建议使用use_signals=False但对我没有用。
作者@bitranox创建了以下软件包：
pip install https://github.com/bitranox/wrapt-timeout-decorator/archive/master.zip
代码样例：
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
timeout-decorator don’t work on windows system as , windows didn’t support signal well.
If you use timeout-decorator in windows system you will get the following
AttributeError: module 'signal' has no attribute 'SIGALRM'
Some suggested to use use_signals=False but didn’t worked for me.
Author @bitranox created the following package:
pip install https://github.com/bitranox/wrapt-timeout-decorator/archive/master.zip
Code Sample:
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()
Gives the following exception:
TimeoutError: Function mytest timed out after 5 seconds
回答 10
我们可以使用相同的信号。我认为以下示例对您有用。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"
We can use signals for the same. I think the below example will be useful for you. It is very simple compared to threads.
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"
回答 11
#!/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)
#!/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)
回答 12
我需要不会被time.sleep（基于线程的方法无法做到）阻止的可嵌套定时中断（SIGALARM无法做到）。我最终从这里复制并修改了代码：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 $'.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()
和用法示例：
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.'
I had a need for nestable timed interrupts (which SIGALARM can’t do) that won’t get blocked by time.sleep (which the thread-based approach can’t do). I ended up copying and lightly modifying code from here: http://code.activestate.com/recipes/577600-queue-for-managing-multiple-sigalrm-alarms-concurr/
The code itself:
#!/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()
and a usage example:
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.'
回答 13
这是对给定的基于线程的解决方案的一点改进。
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]
在5秒钟的超时时间内调用它：
result = timeout(remote_calculate, (myarg,), timeout_duration=5)
Here is a slight improvement to the given thread-based solution.
The code below supports exceptions:
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]
Invoking it with a 5 second timeout:
result = timeout(remote_calculate, (myarg,), timeout_duration=5)
