How to timeout function in python, timeout less than a second

Question

Specification of the problem:

I’m searching through really great amount of lines of a log file and I’m distributing those lines to groups in order to regular expressions(RegExses) I have stored using the re.match() function. Unfortunately some of my RegExses are too complicated and Python sometimes gets himself to backtracking hell. Due to this I need to protect it with some kind of timeout.

Problems:

re.match, I’m using, is Python’s function and as I found out somewhere here on StackOverflow (I’m really sorry, I can not find the link now 🙁 ). It is very difficult to interrupt thread with running Python’s library. For this reason threads are out of the game.
Because evaluating of re.match function takes relatively short time and I want to analyse with this function great amount of lines, I need some timeout function that wont’t take too long to execute (this makes threads even less suitable, it takes really long time to initialise new thread) and can be set to less than one second.
For those reasons, answers here – Timeout on a function call
and here – Timeout function if it takes too long to finish with decorator (alarm – 1sec and more) are off the table.

I’ve spent this morning searching for solution to this question but I did not find any satisfactory answer.

Asked By: Jendas

||

Source

Answer 1

Solution:

I’ve just modified a script posted here: Timeout function if it takes too long to finish.

And here is the code:

from functools import wraps
import errno
import os
import signal

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)

        def wrapper(*args, **kwargs):
            signal.signal(signal.SIGALRM, _handle_timeout)
            signal.setitimer(signal.ITIMER_REAL,seconds) #used timer instead of alarm
            try:
                result = func(*args, **kwargs)
            finally:
                signal.alarm(0)
            return result
        return wraps(func)(wrapper)
    return decorator

And then you can use it like this:

from timeout import timeout 
from time import time

@timeout(0.01)
def loop():
    while True:
       pass
try:
    begin = time.time()
    loop()
except TimeoutError, e:
    print "Time elapsed: {:.3f}s".format(time.time() - begin)

Which prints

Time elapsed: 0.010s

Answered By: Jendas

Answer 2

This is how we can define a timeout slow_function function but it doesn’t stop the slow_function even after raising exception:

import threading, time

class TimeoutError(Exception):
    pass

def slow_function():
    time.sleep(1000)
    return "Done"

def run_with_timeout(func, timeout):
    def target():
        nonlocal result
        result = func()

    result = None
    thread = threading.Thread(target=target)
    thread.start()
    thread.join(timeout)
    if thread.is_alive():
        raise TimeoutError("Function timed out")
    return result

try:
    result = run_with_timeout(slow_function, 3)
except TimeoutError:
    print("Function timed out")
else:
    print("Function returned:", result)

Answered By: Mohammadreza

How to timeout function in python, timeout less than a second

Question:

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