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
#!/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)
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"
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
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()}")
#!/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)
以防对任何人都有帮助,在@piro的回答的基础上,我做了一个函数装饰器:
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
在一个有20秒超时的函数上使用包装器看起来像这样:
@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}")