timeoutd 0.7.0

Simple way to add a timeout to any Python code.

timeoutd

Installation

From PyPI:

pip install timeoutd

From source code:

git clone https://github.com/juhannc/timeoutd.git
pip install -e .

Usage

The timeoutd module provides a decorator that can be used to limit the execution time of a function. The decorator takes a single argument, the number of seconds or a specific date (as a datetime object) after which the function should be terminated.

import time

import timeoutd

@timeoutd.timeout(5)
def mytest():
    print("Start")
    for i in range(1, 10):
        time.sleep(1)
        print(f"{i} seconds have passed")

if __name__ == '__main__':
    mytest()

The timeout decorator allows for multiple different ways to specify the timeout, for example with a datetime object:

import time
import datetime

import timeoutd

@timeoutd.timeout(datetime.datetime.now() + datetime.timedelta(0, 5))
def mytest():
    print("Start")
    for i in range(1, 10):
        time.sleep(1)
        print(f"{i} seconds have passed")

if __name__ == '__main__':
    mytest()

It also take a timedelta object:

import time
import datetime

import timeoutd

@timeoutd.timeout(datetime.timedelta(0, 5))
def mytest():
    print("Start")
    for i in range(1, 10):
        time.sleep(1)
        print(f"{i} seconds have passed")

if __name__ == '__main__':
    mytest()

But it can also take a delta in form of hours, minutes, and seconds via the kwargs:

import time

import timeoutd

@timeoutd.timeout(hours=0, minutes=0, seconds=5)
def mytest():
    print("Start")
    for i in range(1, 10):
        time.sleep(1)
        print(f"{i} seconds have passed")

if __name__ == '__main__':
    mytest()

The timeout decorator also accepts a custom exception to raise on timeout:

import time

import timeoutd

@timeoutd.timeout(5, exception_type=StopIteration)
def mytest():
    print("Start")
    for i in range(1, 10):
        time.sleep(1)
        print(f"{i} seconds have passed")

if __name__ == '__main__':
    mytest()

You can also specify a function to be called on timeout instead of raising an exception:

import time

import timeoutd

def add_two_numbers(i: int, j: int | None = None):
    if j is None:
        j = 0
    print(f"The sum of {i = } and {j = } is {i + j}")

@timeoutd.timeout(
    5,
    on_timeout=add_two_numbers,
    on_timeout_args=(1,),
    on_timeout_kwargs={"j": 2}
)
def mytest():
    print("Start")
    for i in range(1, 10):
        time.sleep(1)
        print(f"{i} seconds have passed")

if __name__ == '__main__':
    mytest()

Multithreading

Note: This feature appears to be broken in some cases for the original timeout-decorator. Some issues might still exist in this fork.

By default, timeoutd uses signals to limit the execution time of the given function. This approach does not work if your function is executed not in a main thread (for example if it's a worker thread of the web application). There is alternative timeout strategy for this case - by using multiprocessing. To use it, just pass use_signals=False to the timeout decorator function:

import time

import timeoutd

@timeoutd.timeout(5, use_signals=False)
def mytest():
    print "Start"
    for i in range(1, 10):
        time.sleep(1)
        print("{} seconds have passed".format(i))

if __name__ == '__main__':
    mytest()

Warning: Make sure that in case of multiprocessing strategy for timeout, your function does not return objects which cannot be pickled, otherwise it will fail at marshalling it between master and child processes.

Acknowledgement

Derived from http://www.saltycrane.com/blog/2010/04/using-python-timeout-decorator-uploading-s3/, https://code.google.com/p/verse-quiz/source/browse/trunk/timeout.py, and https://github.com/pnpnpn/timeout-decorator