Download timeout function in a loop

Download timeout function in a loop - python

I am trying to download some data from the web (web scraping); I have a list of URLs, a few of the URLs take too much time, and thus the loop gets stuck there; I am implementing a function that timeout after a certain period of threshold value and loop should be continued.
For example, if the downloading_source looks like this:
import time
import numpy as np
def downloading_source(x):
wt = np.random.randint(1,50)
print("waiting time", wt)
time.sleep(wt)
return x**2
for the demo, I am taking random values as time. sleep, and the timeout function looks like this
import error
import os
import signal
import functools
class TimeoutError(Exception):
pass
def timeout(seconds=10, error_message=os.strerror(errno.ETIME)):
def decorator(func):
def _handle_timeout(signum, frame):
raise TimeoutError(error_message)
#functools.wraps(func)
def wrapper(*args, **kwargs):
signal.signal(signal.SIGALRM, _handle_timeout)
signal.alarm(seconds)
try:
result = func(*args, **kwargs)
finally:
signal.alarm(0)
return result
return wrapper
return decorator
The download loop:
# Timeout after 5 seconds
#timeout(5)
def long_running_function2(x):
return downloading_source(x)
all_urls = list(range(1,100))
downloaded_data = []
for url in all_urls:
try:
print(url)
down_data = long_running_function2(url)
except Exception as e:
pass
downloaded_data.append(down_data)
It's working; I was wondering, is there any better way to do this?

Related

How to execute another task if current task takes too much time in python [duplicate]

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?

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

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()

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

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

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):
...

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

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.

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.

Building on and and enhancing the answer by #piro , you can build a contextmanager. This allows for very readable code which will disable the alaram signal after a successful run (sets signal.alarm(0))
from contextlib import contextmanager
import signal
import time
#contextmanager
def timeout(duration):
def timeout_handler(signum, frame):
raise TimeoutError(f'block timedout after {duration} seconds')
signal.signal(signal.SIGALRM, timeout_handler)
signal.alarm(duration)
try:
yield
finally:
signal.alarm(0)
def sleeper(duration):
time.sleep(duration)
print('finished')
Example usage:
In [19]: with timeout(2):
...: sleeper(1)
...:
finished
In [20]: with timeout(2):
...: sleeper(3)
...:
---------------------------------------------------------------------------
Exception Traceback (most recent call last)
<ipython-input-20-66c78858116f> in <module>()
1 with timeout(2):
----> 2 sleeper(3)
3
<ipython-input-7-a75b966bf7ac> in sleeper(t)
1 def sleeper(t):
----> 2 time.sleep(t)
3 print('finished')
4
<ipython-input-18-533b9e684466> in timeout_handler(signum, frame)
2 def timeout(duration):
3 def timeout_handler(signum, frame):
----> 4 raise Exception(f'block timedout after {duration} seconds')
5 signal.signal(signal.SIGALRM, timeout_handler)
6 signal.alarm(duration)
Exception: block timedout after 2 seconds

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()

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

Highlights
Raises TimeoutError uses exceptions to alert on timeout - can easily be modified
Cross Platform: Windows & Mac OS X
Compatibility: Python 3.6+ (I also tested on python 2.7 and it works with small syntax adjustments)
For full explanation and extension to parallel maps, see here https://flipdazed.github.io/blog/quant%20dev/parallel-functions-with-timeouts
Minimal Example
>>> #killer_call(timeout=4)
... def bar(x):
... import time
... time.sleep(x)
... return x
>>> bar(10)
Traceback (most recent call last):
...
__main__.TimeoutError: function 'bar' timed out after 4s
and as expected
>>> bar(2)
2
Full code
import multiprocessing as mp
import multiprocessing.queues as mpq
import functools
import dill
from typing import Tuple, Callable, Dict, Optional, Iterable, List, Any
class TimeoutError(Exception):
def __init__(self, func: Callable, timeout: int):
self.t = timeout
self.fname = func.__name__
def __str__(self):
return f"function '{self.fname}' timed out after {self.t}s"
def _lemmiwinks(func: Callable, args: Tuple, kwargs: Dict[str, Any], q: mp.Queue):
"""lemmiwinks crawls into the unknown"""
q.put(dill.loads(func)(*args, **kwargs))
def killer_call(func: Callable = None, timeout: int = 10) -> Callable:
"""
Single function call with a timeout
Args:
func: the function
timeout: The timeout in seconds
"""
if not isinstance(timeout, int):
raise ValueError(f'timeout needs to be an int. Got: {timeout}')
if func is None:
return functools.partial(killer_call, timeout=timeout)
#functools.wraps(killer_call)
def _inners(*args, **kwargs) -> Any:
q_worker = mp.Queue()
proc = mp.Process(target=_lemmiwinks, args=(dill.dumps(func), args, kwargs, q_worker))
proc.start()
try:
return q_worker.get(timeout=timeout)
except mpq.Empty:
raise TimeoutError(func, timeout)
finally:
try:
proc.terminate()
except:
pass
return _inners
if __name__ == '__main__':
#killer_call(timeout=4)
def bar(x):
import time
time.sleep(x)
return x
print(bar(2))
bar(10)
Notes
You will need to import inside the function because of the way dill works.
This will also mean these functions may not be not compatible with doctest if there are imports inside your target functions. You will get an issue with __import__ not found.

Just in case it is helpful for anyone, building on the answer by #piro, I've made a function decorator:
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
Using the wrapper on a function with a 20 seconds timeout would look something like:
#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}")

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"

Another solution with asyncio :
If you want to cancel the background task and not just timeout on the running main code, then you need an explicit communication from main thread to ask the code of the task to cancel , like a 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)

#!/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)

The func_timeout package by Tim Savannah has worked well for me.
Installation:
pip install func_timeout
Usage:
import time
from func_timeout import func_timeout, FunctionTimedOut
def my_func(n):
time.sleep(n)
time_to_sleep = 10
# time out after 2 seconds using kwargs
func_timeout(2, my_func, kwargs={'n' : time_to_sleep})
# time out after 2 seconds using args
func_timeout(2, my_func, args=(time_to_sleep,))

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.'

I have face the same problem but my situation is need work on sub thread, signal didn't work for me, so I wrote a python package: timeout-timer to solve this problem, support for use as context or decorator, use signal or sub thread module to trigger a timeout interrupt:
from timeout_timer import timeout, TimeoutInterrupt
class TimeoutInterruptNested(TimeoutInterrupt):
pass
def test_timeout_nested_loop_both_timeout(timer="thread"):
cnt = 0
try:
with timeout(5, timer=timer):
try:
with timeout(2, timer=timer, exception=TimeoutInterruptNested):
sleep(2)
except TimeoutInterruptNested:
cnt += 1
time.sleep(10)
except TimeoutInterrupt:
cnt += 1
assert cnt == 2
see more: https://github.com/dozysun/timeout-timer

Here is a simple example running one method with timeout and also retriev its value if successfull.
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()}")

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)

Here is a POSIX version that combines many of the previous answers to deliver following features:
Subprocesses blocking the execution.
Usage of the timeout function on class member functions.
Strict requirement on time-to-terminate.
Here is the code and some test cases:
import threading
import signal
import os
import time
class TerminateExecution(Exception):
"""
Exception to indicate that execution has exceeded the preset running time.
"""
def quit_function(pid):
# Killing all subprocesses
os.setpgrp()
os.killpg(0, signal.SIGTERM)
# Killing the main thread
os.kill(pid, signal.SIGTERM)
def handle_term(signum, frame):
raise TerminateExecution()
def invoke_with_timeout(timeout, fn, *args, **kwargs):
# Setting a sigterm handler and initiating a timer
old_handler = signal.signal(signal.SIGTERM, handle_term)
timer = threading.Timer(timeout, quit_function, args=[os.getpid()])
terminate = False
# Executing the function
timer.start()
try:
result = fn(*args, **kwargs)
except TerminateExecution:
terminate = True
finally:
# Restoring original handler and cancel timer
signal.signal(signal.SIGTERM, old_handler)
timer.cancel()
if terminate:
raise BaseException("xxx")
return result
### Test cases
def countdown(n):
print('countdown started', flush=True)
for i in range(n, -1, -1):
print(i, end=', ', flush=True)
time.sleep(1)
print('countdown finished')
return 1337
def really_long_function():
time.sleep(10)
def really_long_function2():
os.system("sleep 787")
# Checking that we can run a function as expected.
assert invoke_with_timeout(3, countdown, 1) == 1337
# Testing various scenarios
t1 = time.time()
try:
print(invoke_with_timeout(1, countdown, 3))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
t1 = time.time()
try:
print(invoke_with_timeout(1, really_long_function2))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
t1 = time.time()
try:
print(invoke_with_timeout(1, really_long_function))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
# Checking that classes are referenced and not
# copied (as would be the case with multiprocessing)
class X:
def __init__(self):
self.value = 0
def set(self, v):
self.value = v
x = X()
invoke_with_timeout(2, x.set, 9)
assert x.value == 9

I intend to kill the process if job not done , using thread and process both to achieve this.
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
Actual class as a contextmanager
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()
How to use it
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)

How to put a "session" time out for a function in Python? [duplicate]

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?

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

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()

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

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

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):
...

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

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.

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.

Building on and and enhancing the answer by #piro , you can build a contextmanager. This allows for very readable code which will disable the alaram signal after a successful run (sets signal.alarm(0))
from contextlib import contextmanager
import signal
import time
#contextmanager
def timeout(duration):
def timeout_handler(signum, frame):
raise TimeoutError(f'block timedout after {duration} seconds')
signal.signal(signal.SIGALRM, timeout_handler)
signal.alarm(duration)
try:
yield
finally:
signal.alarm(0)
def sleeper(duration):
time.sleep(duration)
print('finished')
Example usage:
In [19]: with timeout(2):
...: sleeper(1)
...:
finished
In [20]: with timeout(2):
...: sleeper(3)
...:
---------------------------------------------------------------------------
Exception Traceback (most recent call last)
<ipython-input-20-66c78858116f> in <module>()
1 with timeout(2):
----> 2 sleeper(3)
3
<ipython-input-7-a75b966bf7ac> in sleeper(t)
1 def sleeper(t):
----> 2 time.sleep(t)
3 print('finished')
4
<ipython-input-18-533b9e684466> in timeout_handler(signum, frame)
2 def timeout(duration):
3 def timeout_handler(signum, frame):
----> 4 raise Exception(f'block timedout after {duration} seconds')
5 signal.signal(signal.SIGALRM, timeout_handler)
6 signal.alarm(duration)
Exception: block timedout after 2 seconds

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()

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

Highlights
Raises TimeoutError uses exceptions to alert on timeout - can easily be modified
Cross Platform: Windows & Mac OS X
Compatibility: Python 3.6+ (I also tested on python 2.7 and it works with small syntax adjustments)
For full explanation and extension to parallel maps, see here https://flipdazed.github.io/blog/quant%20dev/parallel-functions-with-timeouts
Minimal Example
>>> #killer_call(timeout=4)
... def bar(x):
... import time
... time.sleep(x)
... return x
>>> bar(10)
Traceback (most recent call last):
...
__main__.TimeoutError: function 'bar' timed out after 4s
and as expected
>>> bar(2)
2
Full code
import multiprocessing as mp
import multiprocessing.queues as mpq
import functools
import dill
from typing import Tuple, Callable, Dict, Optional, Iterable, List, Any
class TimeoutError(Exception):
def __init__(self, func: Callable, timeout: int):
self.t = timeout
self.fname = func.__name__
def __str__(self):
return f"function '{self.fname}' timed out after {self.t}s"
def _lemmiwinks(func: Callable, args: Tuple, kwargs: Dict[str, Any], q: mp.Queue):
"""lemmiwinks crawls into the unknown"""
q.put(dill.loads(func)(*args, **kwargs))
def killer_call(func: Callable = None, timeout: int = 10) -> Callable:
"""
Single function call with a timeout
Args:
func: the function
timeout: The timeout in seconds
"""
if not isinstance(timeout, int):
raise ValueError(f'timeout needs to be an int. Got: {timeout}')
if func is None:
return functools.partial(killer_call, timeout=timeout)
#functools.wraps(killer_call)
def _inners(*args, **kwargs) -> Any:
q_worker = mp.Queue()
proc = mp.Process(target=_lemmiwinks, args=(dill.dumps(func), args, kwargs, q_worker))
proc.start()
try:
return q_worker.get(timeout=timeout)
except mpq.Empty:
raise TimeoutError(func, timeout)
finally:
try:
proc.terminate()
except:
pass
return _inners
if __name__ == '__main__':
#killer_call(timeout=4)
def bar(x):
import time
time.sleep(x)
return x
print(bar(2))
bar(10)
Notes
You will need to import inside the function because of the way dill works.
This will also mean these functions may not be not compatible with doctest if there are imports inside your target functions. You will get an issue with __import__ not found.

Just in case it is helpful for anyone, building on the answer by #piro, I've made a function decorator:
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
Using the wrapper on a function with a 20 seconds timeout would look something like:
#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}")

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"

Another solution with asyncio :
If you want to cancel the background task and not just timeout on the running main code, then you need an explicit communication from main thread to ask the code of the task to cancel , like a 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)

#!/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)

The func_timeout package by Tim Savannah has worked well for me.
Installation:
pip install func_timeout
Usage:
import time
from func_timeout import func_timeout, FunctionTimedOut
def my_func(n):
time.sleep(n)
time_to_sleep = 10
# time out after 2 seconds using kwargs
func_timeout(2, my_func, kwargs={'n' : time_to_sleep})
# time out after 2 seconds using args
func_timeout(2, my_func, args=(time_to_sleep,))

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.'

I have face the same problem but my situation is need work on sub thread, signal didn't work for me, so I wrote a python package: timeout-timer to solve this problem, support for use as context or decorator, use signal or sub thread module to trigger a timeout interrupt:
from timeout_timer import timeout, TimeoutInterrupt
class TimeoutInterruptNested(TimeoutInterrupt):
pass
def test_timeout_nested_loop_both_timeout(timer="thread"):
cnt = 0
try:
with timeout(5, timer=timer):
try:
with timeout(2, timer=timer, exception=TimeoutInterruptNested):
sleep(2)
except TimeoutInterruptNested:
cnt += 1
time.sleep(10)
except TimeoutInterrupt:
cnt += 1
assert cnt == 2
see more: https://github.com/dozysun/timeout-timer

Here is a simple example running one method with timeout and also retriev its value if successfull.
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()}")

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)

Here is a POSIX version that combines many of the previous answers to deliver following features:
Subprocesses blocking the execution.
Usage of the timeout function on class member functions.
Strict requirement on time-to-terminate.
Here is the code and some test cases:
import threading
import signal
import os
import time
class TerminateExecution(Exception):
"""
Exception to indicate that execution has exceeded the preset running time.
"""
def quit_function(pid):
# Killing all subprocesses
os.setpgrp()
os.killpg(0, signal.SIGTERM)
# Killing the main thread
os.kill(pid, signal.SIGTERM)
def handle_term(signum, frame):
raise TerminateExecution()
def invoke_with_timeout(timeout, fn, *args, **kwargs):
# Setting a sigterm handler and initiating a timer
old_handler = signal.signal(signal.SIGTERM, handle_term)
timer = threading.Timer(timeout, quit_function, args=[os.getpid()])
terminate = False
# Executing the function
timer.start()
try:
result = fn(*args, **kwargs)
except TerminateExecution:
terminate = True
finally:
# Restoring original handler and cancel timer
signal.signal(signal.SIGTERM, old_handler)
timer.cancel()
if terminate:
raise BaseException("xxx")
return result
### Test cases
def countdown(n):
print('countdown started', flush=True)
for i in range(n, -1, -1):
print(i, end=', ', flush=True)
time.sleep(1)
print('countdown finished')
return 1337
def really_long_function():
time.sleep(10)
def really_long_function2():
os.system("sleep 787")
# Checking that we can run a function as expected.
assert invoke_with_timeout(3, countdown, 1) == 1337
# Testing various scenarios
t1 = time.time()
try:
print(invoke_with_timeout(1, countdown, 3))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
t1 = time.time()
try:
print(invoke_with_timeout(1, really_long_function2))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
t1 = time.time()
try:
print(invoke_with_timeout(1, really_long_function))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
# Checking that classes are referenced and not
# copied (as would be the case with multiprocessing)
class X:
def __init__(self):
self.value = 0
def set(self, v):
self.value = v
x = X()
invoke_with_timeout(2, x.set, 9)
assert x.value == 9

I intend to kill the process if job not done , using thread and process both to achieve this.
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
Actual class as a contextmanager
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()
How to use it
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)

Checking a lot of URLs to see if they return 200. What's the cleverest way?

I need to check a lot (~10 million) of URLs to see if they exist (return 200). I've written the following code to do this per-URL, but to do all of the URLs will take approximately forever.
def is_200(url):
try:
parsed = urlparse(url)
conn = httplib.HTTPConnection(parsed.netloc)
conn.request("HEAD", parsed.path)
res = conn.getresponse()
return res.status == 200
except KeyboardInterrupt, e:
raise e
except:
return False
The URLs are spread across about a dozen hosts, so it seems like I should be able to take advantage of this to pipeline my requests and reduce connection overhead. How would you build this? I'm open to any programming/scripting language.

Have a look at urllib3. It supports per-host connection re-using.
Additionally using multiple processes/threads or async I/O would be a good idea.

All of this is in Python, version 3.x.
I would create worker threads that check for 200. I'll give an example. The threadpool (put in threadpool.py):
# http://code.activestate.com/recipes/577187-python-thread-pool/
from queue import Queue
from threading import Thread
class Worker(Thread):
def __init__(self, tasks):
Thread.__init__(self)
self.tasks = tasks
self.daemon = True
self.start()
def run(self):
while True:
func, args, kargs = self.tasks.get()
try: func(*args, **kargs)
except Exception as exception: print(exception)
self.tasks.task_done()
class ThreadPool:
def __init__(self, num_threads):
self.tasks = Queue(num_threads)
for _ in range(num_threads): Worker(self.tasks)
def add_task(self, func, *args, **kargs):
self.tasks.put((func, args, kargs))
def wait_completion(self):
self.tasks.join()
Now, if urllist contains your urls then your main file should be along the lines of this:
numconns = 40
workers = threadpool.ThreadPool(numconns)
results = [None] * len(urllist)
def check200(url, index):
results[index] = is_200(url)
for index, url in enumerate(urllist):
try:
workers.add_task(check200, url, index)
except KeyboardInterrupt:
print("Shutting down application, hang on...")
workers.wait_completion()
break
Note that this program scales with the other suggestions posted here, this is only dependent on is_200().

Running "unique" tasks with celery

I use celery to update RSS feeds in my news aggregation site. I use one #task for each feed, and things seem to work nicely.
There's a detail that I'm not sure to handle well though: all feeds are updated once every minute with a #periodic_task, but what if a feed is still updating from the last periodic task when a new one is started ? (for example if the feed is really slow, or offline and the task is held in a retry loop)
Currently I store tasks results and check their status like this:
import socket
from datetime import timedelta
from celery.decorators import task, periodic_task
from aggregator.models import Feed
_results = {}
#periodic_task(run_every=timedelta(minutes=1))
def fetch_articles():
for feed in Feed.objects.all():
if feed.pk in _results:
if not _results[feed.pk].ready():
# The task is not finished yet
continue
_results[feed.pk] = update_feed.delay(feed)
#task()
def update_feed(feed):
try:
feed.fetch_articles()
except socket.error, exc:
update_feed.retry(args=[feed], exc=exc)
Maybe there is a more sophisticated/robust way of achieving the same result using some celery mechanism that I missed ?

Based on MattH's answer, you could use a decorator like this:
from django.core.cache import cache
import functools
def single_instance_task(timeout):
def task_exc(func):
#functools.wraps(func)
def wrapper(*args, **kwargs):
lock_id = "celery-single-instance-" + func.__name__
acquire_lock = lambda: cache.add(lock_id, "true", timeout)
release_lock = lambda: cache.delete(lock_id)
if acquire_lock():
try:
func(*args, **kwargs)
finally:
release_lock()
return wrapper
return task_exc
then, use it like so...
#periodic_task(run_every=timedelta(minutes=1))
#single_instance_task(60*10)
def fetch_articles()
yada yada...

From the official documentation: Ensuring a task is only executed one at a time.

Using https://pypi.python.org/pypi/celery_once seems to do the job really nice, including reporting errors and testing against some parameters for uniqueness.
You can do things like:
from celery_once import QueueOnce
from myapp.celery import app
from time import sleep
#app.task(base=QueueOnce, once=dict(keys=('customer_id',)))
def start_billing(customer_id, year, month):
sleep(30)
return "Done!"
which just needs the following settings in your project:
ONCE_REDIS_URL = 'redis://localhost:6379/0'
ONCE_DEFAULT_TIMEOUT = 60 * 60 # remove lock after 1 hour in case it was stale

If you're looking for an example that doesn't use Django, then try this example (caveat: uses Redis instead, which I was already using).
The decorator code is as follows (full credit to the author of the article, go read it)
import redis
REDIS_CLIENT = redis.Redis()
def only_one(function=None, key="", timeout=None):
"""Enforce only one celery task at a time."""
def _dec(run_func):
"""Decorator."""
def _caller(*args, **kwargs):
"""Caller."""
ret_value = None
have_lock = False
lock = REDIS_CLIENT.lock(key, timeout=timeout)
try:
have_lock = lock.acquire(blocking=False)
if have_lock:
ret_value = run_func(*args, **kwargs)
finally:
if have_lock:
lock.release()
return ret_value
return _caller
return _dec(function) if function is not None else _dec

I was wondering why nobody mentioned using celery.app.control.inspect().active() to get the list of the currently running tasks. Is it not real time? Because otherwise it would be very easy to implement, for instance:
def unique_task(callback, *decorator_args, **decorator_kwargs):
"""
Decorator to ensure only one instance of the task is running at once.
"""
#wraps(callback)
def _wrapper(celery_task, *args, **kwargs):
active_queues = task.app.control.inspect().active()
if active_queues:
for queue in active_queues:
for running_task in active_queues[queue]:
# Discard the currently running task from the list.
if task.name == running_task['name'] and task.request.id != running_task['id']:
return f'Task "{callback.__name__}()" cancelled! already running...'
return callback(celery_task, *args, **kwargs)
return _wrapper
And then just applying the decorator to the corresponding tasks:
#celery.task(bind=True)
#unique_task
def my_task(self):
# task executed once at a time.
pass

This solution for celery working at single host with concurency greater 1. Other kinds (without dependencies like redis) of locks difference file-based don't work with concurrency greater 1.
class Lock(object):
def __init__(self, filename):
self.f = open(filename, 'w')
def __enter__(self):
try:
flock(self.f.fileno(), LOCK_EX | LOCK_NB)
return True
except IOError:
pass
return False
def __exit__(self, *args):
self.f.close()
class SinglePeriodicTask(PeriodicTask):
abstract = True
run_every = timedelta(seconds=1)
def __call__(self, *args, **kwargs):
lock_filename = join('/tmp',
md5(self.name).hexdigest())
with Lock(lock_filename) as is_locked:
if is_locked:
super(SinglePeriodicTask, self).__call__(*args, **kwargs)
else:
print 'already working'
class SearchTask(SinglePeriodicTask):
restart_delay = timedelta(seconds=60)
def run(self, *args, **kwargs):
print self.name, 'start', datetime.now()
sleep(5)
print self.name, 'end', datetime.now()

Timeout on a function call

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?

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

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()

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

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

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):
...

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

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.

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.

Building on and and enhancing the answer by #piro , you can build a contextmanager. This allows for very readable code which will disable the alaram signal after a successful run (sets signal.alarm(0))
from contextlib import contextmanager
import signal
import time
#contextmanager
def timeout(duration):
def timeout_handler(signum, frame):
raise TimeoutError(f'block timedout after {duration} seconds')
signal.signal(signal.SIGALRM, timeout_handler)
signal.alarm(duration)
try:
yield
finally:
signal.alarm(0)
def sleeper(duration):
time.sleep(duration)
print('finished')
Example usage:
In [19]: with timeout(2):
...: sleeper(1)
...:
finished
In [20]: with timeout(2):
...: sleeper(3)
...:
---------------------------------------------------------------------------
Exception Traceback (most recent call last)
<ipython-input-20-66c78858116f> in <module>()
1 with timeout(2):
----> 2 sleeper(3)
3
<ipython-input-7-a75b966bf7ac> in sleeper(t)
1 def sleeper(t):
----> 2 time.sleep(t)
3 print('finished')
4
<ipython-input-18-533b9e684466> in timeout_handler(signum, frame)
2 def timeout(duration):
3 def timeout_handler(signum, frame):
----> 4 raise Exception(f'block timedout after {duration} seconds')
5 signal.signal(signal.SIGALRM, timeout_handler)
6 signal.alarm(duration)
Exception: block timedout after 2 seconds

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()

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

Highlights
Raises TimeoutError uses exceptions to alert on timeout - can easily be modified
Cross Platform: Windows & Mac OS X
Compatibility: Python 3.6+ (I also tested on python 2.7 and it works with small syntax adjustments)
For full explanation and extension to parallel maps, see here https://flipdazed.github.io/blog/quant%20dev/parallel-functions-with-timeouts
Minimal Example
>>> #killer_call(timeout=4)
... def bar(x):
... import time
... time.sleep(x)
... return x
>>> bar(10)
Traceback (most recent call last):
...
__main__.TimeoutError: function 'bar' timed out after 4s
and as expected
>>> bar(2)
2
Full code
import multiprocessing as mp
import multiprocessing.queues as mpq
import functools
import dill
from typing import Tuple, Callable, Dict, Optional, Iterable, List, Any
class TimeoutError(Exception):
def __init__(self, func: Callable, timeout: int):
self.t = timeout
self.fname = func.__name__
def __str__(self):
return f"function '{self.fname}' timed out after {self.t}s"
def _lemmiwinks(func: Callable, args: Tuple, kwargs: Dict[str, Any], q: mp.Queue):
"""lemmiwinks crawls into the unknown"""
q.put(dill.loads(func)(*args, **kwargs))
def killer_call(func: Callable = None, timeout: int = 10) -> Callable:
"""
Single function call with a timeout
Args:
func: the function
timeout: The timeout in seconds
"""
if not isinstance(timeout, int):
raise ValueError(f'timeout needs to be an int. Got: {timeout}')
if func is None:
return functools.partial(killer_call, timeout=timeout)
#functools.wraps(killer_call)
def _inners(*args, **kwargs) -> Any:
q_worker = mp.Queue()
proc = mp.Process(target=_lemmiwinks, args=(dill.dumps(func), args, kwargs, q_worker))
proc.start()
try:
return q_worker.get(timeout=timeout)
except mpq.Empty:
raise TimeoutError(func, timeout)
finally:
try:
proc.terminate()
except:
pass
return _inners
if __name__ == '__main__':
#killer_call(timeout=4)
def bar(x):
import time
time.sleep(x)
return x
print(bar(2))
bar(10)
Notes
You will need to import inside the function because of the way dill works.
This will also mean these functions may not be not compatible with doctest if there are imports inside your target functions. You will get an issue with __import__ not found.

Just in case it is helpful for anyone, building on the answer by #piro, I've made a function decorator:
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
Using the wrapper on a function with a 20 seconds timeout would look something like:
#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}")

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"

Another solution with asyncio :
If you want to cancel the background task and not just timeout on the running main code, then you need an explicit communication from main thread to ask the code of the task to cancel , like a 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)

#!/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)

The func_timeout package by Tim Savannah has worked well for me.
Installation:
pip install func_timeout
Usage:
import time
from func_timeout import func_timeout, FunctionTimedOut
def my_func(n):
time.sleep(n)
time_to_sleep = 10
# time out after 2 seconds using kwargs
func_timeout(2, my_func, kwargs={'n' : time_to_sleep})
# time out after 2 seconds using args
func_timeout(2, my_func, args=(time_to_sleep,))

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.'

I have face the same problem but my situation is need work on sub thread, signal didn't work for me, so I wrote a python package: timeout-timer to solve this problem, support for use as context or decorator, use signal or sub thread module to trigger a timeout interrupt:
from timeout_timer import timeout, TimeoutInterrupt
class TimeoutInterruptNested(TimeoutInterrupt):
pass
def test_timeout_nested_loop_both_timeout(timer="thread"):
cnt = 0
try:
with timeout(5, timer=timer):
try:
with timeout(2, timer=timer, exception=TimeoutInterruptNested):
sleep(2)
except TimeoutInterruptNested:
cnt += 1
time.sleep(10)
except TimeoutInterrupt:
cnt += 1
assert cnt == 2
see more: https://github.com/dozysun/timeout-timer

Here is a simple example running one method with timeout and also retriev its value if successfull.
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()}")

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)

Here is a POSIX version that combines many of the previous answers to deliver following features:
Subprocesses blocking the execution.
Usage of the timeout function on class member functions.
Strict requirement on time-to-terminate.
Here is the code and some test cases:
import threading
import signal
import os
import time
class TerminateExecution(Exception):
"""
Exception to indicate that execution has exceeded the preset running time.
"""
def quit_function(pid):
# Killing all subprocesses
os.setpgrp()
os.killpg(0, signal.SIGTERM)
# Killing the main thread
os.kill(pid, signal.SIGTERM)
def handle_term(signum, frame):
raise TerminateExecution()
def invoke_with_timeout(timeout, fn, *args, **kwargs):
# Setting a sigterm handler and initiating a timer
old_handler = signal.signal(signal.SIGTERM, handle_term)
timer = threading.Timer(timeout, quit_function, args=[os.getpid()])
terminate = False
# Executing the function
timer.start()
try:
result = fn(*args, **kwargs)
except TerminateExecution:
terminate = True
finally:
# Restoring original handler and cancel timer
signal.signal(signal.SIGTERM, old_handler)
timer.cancel()
if terminate:
raise BaseException("xxx")
return result
### Test cases
def countdown(n):
print('countdown started', flush=True)
for i in range(n, -1, -1):
print(i, end=', ', flush=True)
time.sleep(1)
print('countdown finished')
return 1337
def really_long_function():
time.sleep(10)
def really_long_function2():
os.system("sleep 787")
# Checking that we can run a function as expected.
assert invoke_with_timeout(3, countdown, 1) == 1337
# Testing various scenarios
t1 = time.time()
try:
print(invoke_with_timeout(1, countdown, 3))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
t1 = time.time()
try:
print(invoke_with_timeout(1, really_long_function2))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
t1 = time.time()
try:
print(invoke_with_timeout(1, really_long_function))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
# Checking that classes are referenced and not
# copied (as would be the case with multiprocessing)
class X:
def __init__(self):
self.value = 0
def set(self, v):
self.value = v
x = X()
invoke_with_timeout(2, x.set, 9)
assert x.value == 9

I intend to kill the process if job not done , using thread and process both to achieve this.
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
Actual class as a contextmanager
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()
How to use it
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)

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