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

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.