Timeout on a function call

Question

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

Answer 1

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.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

Answer 2

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()

Answer 3

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"

Answer 4

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):
    ...

Answer 5

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

Answer 6

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

class TimeoutError(Exception):

    def __init__(self, func, timeout):
        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[object], kwargs: Dict[str, object], 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) -> object:
        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.