python – Timeout on a function call

The Question :

339 people think this question is useful

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?

The Question Comments :

The Answer 1

250 people think this answer is useful

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)
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:

The Answer 2

168 people think this answer is useful

You can use multiprocessing.Process to do exactly that.


import multiprocessing
import time

# bar
def bar():
    for i in range(100):
        print "Tick"

if __name__ == '__main__':
    # Start bar as a process
    p = multiprocessing.Process(target=bar)

    # Wait for 10 seconds or until process finishes

    # 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
        # OR Kill - will work for sure, no chance for process to finish nicely however
        # p.kill()


The Answer 3

85 people think this answer is useful

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
    import thread
except ImportError:
    import _thread as thread

Use version independent code:

    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:

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__])
                result = fn(*args, **kwargs)
            return result
        return inner
    return outer


And here’s the usage that directly answers your question about exiting after 5 seconds!:

def countdown(n):
    print('countdown started', flush=True)
    for i in range(n, -1, -1):
        print(i, end=', ', flush=True)
    print('countdown finished')


>>> 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

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.:

def sleep10():
    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

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

The Answer 4

58 people think this answer is useful

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):

    def handler(signum, frame):
        raise TimeoutError()

    # set the timeout handler
    signal.signal(signal.SIGALRM, handler) 
        result = func(*args, **kwargs)
    except TimeoutError as exc:
        result = default

    return result

The Answer 5

33 people think this answer is useful

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."""
        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 Answer 6

28 people think this answer is useful

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:

The Answer 7

17 people think this answer is useful

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.

The Answer 8

13 people think this answer is useful

Great, easy to use and reliable PyPi project timeout-decorator (


pip install timeout-decorator


import time
import timeout_decorator

def mytest():
    print "Start"
    for i in range(1,10):
        print "%d seconds have passed" % i

if __name__ == '__main__':

The Answer 9

11 people think this answer is useful

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.

The Answer 10

5 people think this answer is useful

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

Code Sample:

import time
from wrapt_timeout_decorator import *

def mytest(message):
    for i in range(1,10):
        print('{} seconds have passed'.format(i))

def main():

if __name__ == '__main__':

Gives the following exception:

TimeoutError: Function mytest timed out after 5 seconds

The Answer 11

3 people think this answer is useful

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))

def timeout(duration):
    def timeout_handler(signum, frame):
        raise Exception(f'block timedout after {duration} seconds')
    signal.signal(signal.SIGALRM, timeout_handler)

def sleeper(duration):

Example usage:

In [19]: with timeout(2):
    ...:     sleeper(1)

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)

<ipython-input-7-a75b966bf7ac> in sleeper(t)
      1 def sleeper(t):
----> 2     time.sleep(t)
      3     print('finished')

<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

The Answer 12

3 people think this answer is useful

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

    def run(cls, to_run: functools.partial, timeout: float):
        pool = cls.get_pool()
        loop = cls.get_loop()
            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

    def get_pool(cls):
        if cls.pool is None:
            cls.pool = ThreadPoolExecutor(multiprocessing.cpu_count())
        return cls.pool

    def get_loop(cls):
            return asyncio.get_event_loop()
        except RuntimeError:
            # 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,

results =, TIME_OUT)

The Answer 13

2 people think this answer is useful

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

except myException:
    print "whoops"

The Answer 14

2 people think this answer is useful
import sys, subprocess, threading
proc = subprocess.Popen(sys.argv[2:])
timer = threading.Timer(float(sys.argv[1]), proc.terminate)

The Answer 15

2 people think this answer is useful


  • 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

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)

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

        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)

    def _inners(*args, **kwargs) -> Any:
        q_worker = mp.Queue()
        proc = mp.Process(target=_lemmiwinks, args=(dill.dumps(func), args, kwargs, q_worker))
            return q_worker.get(timeout=timeout)
        except mpq.Empty:
            raise TimeoutError(func, timeout)
    return _inners

if __name__ == '__main__':
    def bar(x):
        import time
        return x



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.

The Answer 16

1 people think this answer is useful

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:

The code itself:


# lightly modified version of

""" 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):
        except ValueError:

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.
        nextt = __next_alarm()
        while nextt is not None and nextt <= 0:
            (tm, func, args, keys) = heapq.heappop(alarmlist)
            func(*args, **keys)
            nextt = __next_alarm()
        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()`.
        newalarm = __new_alarm(sec, func, args, keys)
        heapq.heappush(alarmlist, newalarm)
        return newalarm

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.
        if alarmlist: __set_alarm()

and a usage example:

import alarm
from time import sleep

    with alarm.Timeout(id_='a', seconds=5):
            with alarm.Timeout(id_='b', seconds=2):
        except alarm.TimeoutError as e:
            print 'raised', e.id_
except alarm.TimeoutError as e:
    print 'raised', e.id_
    print 'nope.'

The Answer 17

0 people think this answer is useful

Here is a slight improvement to the given thread-based solution.

The code below supports exceptions:

def runFunctionCatchExceptions(func, *args, **kwargs):
        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):
            self.result = default
        def run(self):
            self.result = runFunctionCatchExceptions(func, *args, **kwargs)
    it = InterruptableThread()
    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)

The Answer 18

0 people think this answer is useful

Here is a POSIX version that combines many of the previous answers to deliver following features:

  1. Subprocesses blocking the execution.
  2. Usage of the timeout function on class member functions.
  3. 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.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
        result = fn(*args, **kwargs)
    except TerminateExecution:
        terminate = True
        # Restoring original handler and cancel timer
        signal.signal(signal.SIGTERM, old_handler)

    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)
    print('countdown finished')
    return 1337

def really_long_function():

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()
    print(invoke_with_timeout(1, countdown, 3))
except BaseException:
    assert(time.time() - t1 < 1.1)
    print("All good", time.time() - t1)

t1 = time.time()
    print(invoke_with_timeout(1, really_long_function2))
except BaseException:
    assert(time.time() - t1 < 1.1)
    print("All good", time.time() - t1)

t1 = time.time()
    print(invoke_with_timeout(1, really_long_function))
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

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