python process pool with timeout on each process not all of the pool

Question

I need to run many processes, but not all together, for example 4 processes at same time. multiprocessing.Pool is exactly what I need. But the problem is that I need to terminate a process if it lasts more than a timeout (e.g. 3 seconds). Pool just supports wait for a timeout for all processes not each of them. This is what I need:

def f():
    process_but_kill_if_it_takes_more_than_3_sec()
pool.map(f, inputs)

I couldn't find a simple way to use Pool with timeouts. There is a solution from Eli Bendersky. It's a function that limits execution time of an arbitrary function via Thread.join(timeout). It works, (although it's stop method doesn't work well). But this method runs a new unnecessary thread while main thread of process is just waiting, because we need a timeout controller. It should be possible to control all timeouts from a single point, something like this:

import time
from multiprocessing import Process


def f(n):
    time.sleep(n)

timeout = 3
tasks = [1, 2, 4, 1, 8, 2]

procs = []
pool_len = 4
while len(tasks) > 0 or len(procs) > 0:
    if len(tasks) > 0 and len(procs) < pool_len:
        n = tasks.pop(0)
        p = Process(target=f, args=(n,))
        p.start()
        procs.append({'n': n, 'p': p, 't': time.time() + timeout})
    for d in procs:
        if not d['p'].is_alive():
            procs.remove(d)
            print '%s finished' % d['n']
        elif d['t'] < time.time():
            d['p'].terminate()
            procs.remove(d)
            print '%s killed' % d['n']
    time.sleep(0.05)

The output should be:

1 finished
1 finished
2 finished
4 killed
2 finished
8 killed

Question: Is there a way to use Pool to solve this?

Answer 1

You could make f(n) cooperative so that it always finishes within a timeout (like in GUI/network event handlers).

If you can't make it cooperative then the only reliable option is to kill the process that is running the function:

import multiprocessing as mp

def run_with_timeout(timeout, func, *args):
    receive_end, send_end = mp.Pipe(duplex=False)
    p = mp.Process(target=func, args=args, kwargs=dict(send_end=send_end))
    p.daemon = True
    p.start()
    send_end.close() # child must be the only one with it opened
    p.join(timeout)
    if p.is_alive():
        ####debug('%s timeout', args)
        p.terminate()
    else:
        return receive_end.recv()  # get value from the child

The disadvantage is that it requires a new process for each function call (maxtasksperchild=1 Pool's analog).

It is easy to run 4 processes at the same time using a thread pool:

#!/usr/bin/env python
import logging
import time
from functools import partial
from multiprocessing.pool import ThreadPool

debug = logging.getLogger(__name__).debug

def run_mp(n, send_end):
    start = time.time()
    debug('%d starting', n)
    try:
        time.sleep(n)
    except Exception as e:
        debug('%d error %s', n, e)
    finally:
        debug('%d done, elapsed: %.3f', n, time.time() - start)
    send_end.send({n: n*n})

if __name__=="__main__":
    tasks = [1, 2, 4, 1, 8, 2]

    logging.basicConfig(format="%(relativeCreated)04d %(message)s", level=logging.DEBUG)
    print(ThreadPool(processes=4).map(partial(run_with_timeout, 3, run_mp), tasks))

Output

0027 1 starting
0028 2 starting
0030 4 starting
0031 1 starting
1029 1 done, elapsed: 1.002
1032 1 done, elapsed: 1.002
1033 8 starting
1036 2 starting
2031 2 done, elapsed: 2.003
3029 (4,) timeout
3038 2 done, elapsed: 2.003
4035 (8,) timeout
[{1: 1}, {2: 4}, None, {1: 1}, None, {2: 4}]

Beware: there could be forking + threading issues; you could use a fork-server process to workaround them.