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
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)
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):
...
#!/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)
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
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}")