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
伟大的,易于使用和可靠的PyPi项目超时装饰器(https://pypi.org/project/timeout-decorator/)
安装:
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()
其他回答
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"
你可以使用多处理。过程来做到这一点。
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()
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)
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,它向被装饰的函数添加了一个超时参数,该参数执行您所期望的操作,它将停止该函数。
在pypi上查看:https://pypi.python.org/pypi/stopit