Exception thrown in multiprocessing Pool not detected

Question:

It seems that when an exception is raised from a multiprocessing.Pool process, there is no stack trace or any other indication that it has failed. Example:

from multiprocessing import Pool 

def go():
    print(1)
    raise Exception()
    print(2)

p = Pool()
p.apply_async(go)
p.close()
p.join()

prints 1 and stops silently. Interestingly, raising a BaseException instead works. Is there any way to make the behavior for all exceptions the same as BaseException?

Asked By: Rob Lourens

||

Answers:

I’d try using pdb:

import pdb
import sys
def handler(type, value, tb):
  pdb.pm()
sys.excepthook = handler
Answered By: Clarus

I have a reasonable solution for the problem, at least for debugging purposes. I do not currently have a solution that will raise the exception back in the main processes. My first thought was to use a decorator, but you can only pickle functions defined at the top level of a module, so that’s right out.

Instead, a simple wrapping class and a Pool subclass that uses this for apply_async (and hence apply). I’ll leave map_async as an exercise for the reader.

import traceback
from multiprocessing.pool import Pool
import multiprocessing

# Shortcut to multiprocessing's logger
def error(msg, *args):
    return multiprocessing.get_logger().error(msg, *args)

class LogExceptions(object):
    def __init__(self, callable):
        self.__callable = callable

    def __call__(self, *args, **kwargs):
        try:
            result = self.__callable(*args, **kwargs)

        except Exception as e:
            # Here we add some debugging help. If multiprocessing's
            # debugging is on, it will arrange to log the traceback
            error(traceback.format_exc())
            # Re-raise the original exception so the Pool worker can
            # clean up
            raise

        # It was fine, give a normal answer
        return result

class LoggingPool(Pool):
    def apply_async(self, func, args=(), kwds={}, callback=None):
        return Pool.apply_async(self, LogExceptions(func), args, kwds, callback)

def go():
    print(1)
    raise Exception()
    print(2)

multiprocessing.log_to_stderr()
p = LoggingPool(processes=1)

p.apply_async(go)
p.close()
p.join()

This gives me:

1
[ERROR/PoolWorker-1] Traceback (most recent call last):
  File "mpdebug.py", line 24, in __call__
    result = self.__callable(*args, **kwargs)
  File "mpdebug.py", line 44, in go
    raise Exception()
Exception
Answered By: Rupert Nash

Maybe I’m missing something, but isn’t that what the get method of the Result object returns? See Process Pools.

class multiprocessing.pool.AsyncResult

The class of the result returned by Pool.apply_async() and Pool.map_async().get([timeout])
Return the result when it arrives. If timeout is not None and the result does not arrive within
timeout seconds then multiprocessing.TimeoutError is raised. If the remote
call raised an exception then that exception will be reraised by get().

So, slightly modifying your example, one can do

from multiprocessing import Pool

def go():
    print(1)
    raise Exception("foobar")
    print(2)

p = Pool()
x = p.apply_async(go)
x.get()
p.close()
p.join()

Which gives as result

1
Traceback (most recent call last):
  File "rob.py", line 10, in <module>
    x.get()
  File "/usr/lib/python2.6/multiprocessing/pool.py", line 422, in get
    raise self._value
Exception: foobar

This is not completely satisfactory, since it does not print the traceback, but is better than nothing.

UPDATE: This bug has been fixed in Python 3.4, courtesy of Richard Oudkerk. See the issue get method of multiprocessing.pool.Async should return full traceback.

Answered By: Faheem Mitha

I created a module RemoteException.py that shows the full traceback of a exception in a process. Python2. Download it and add this to your code:

import RemoteException

@RemoteException.showError
def go():
    raise Exception('Error!')

if __name__ == '__main__':
    import multiprocessing
    p = multiprocessing.Pool(processes = 1)
    r = p.apply(go) # full traceback is shown here
Answered By: User

I’ve had success logging exceptions with this decorator:

import traceback, functools, multiprocessing

def trace_unhandled_exceptions(func):
    @functools.wraps(func)
    def wrapped_func(*args, **kwargs):
        try:
            func(*args, **kwargs)
        except:
            print 'Exception in '+func.__name__
            traceback.print_exc()
    return wrapped_func

with the code in the question, it’s

@trace_unhandled_exceptions
def go():
    print(1)
    raise Exception()
    print(2)

p = multiprocessing.Pool(1)

p.apply_async(go)
p.close()
p.join()

Simply decorate the function you pass to your process pool. The key to this working is @functools.wraps(func) otherwise multiprocessing throws a PicklingError.

code above gives

1
Exception in go
Traceback (most recent call last):
  File "<stdin>", line 5, in wrapped_func
  File "<stdin>", line 4, in go
Exception
Answered By: Mark Foreman

The solution with the most votes at the time of writing has a problem:

from multiprocessing import Pool

def go():
    print(1)
    raise Exception("foobar")
    print(2)

p = Pool()
x = p.apply_async(go)
x.get()  ## waiting here for go() to complete...
p.close()
p.join()

As @dfrankow noted, it will wait on x.get(), which ruins the point of running a task asynchronously. So, for better efficiency (in particular if your worker function go takes a long time) I would change it to:

from multiprocessing import Pool

def go(x):
    print(1)
    # task_that_takes_a_long_time()
    raise Exception("Can't go anywhere.")
    print(2)
    return x**2

p = Pool()
results = []
for x in range(1000):
    results.append( p.apply_async(go, [x]) )

p.close()

for r in results:
     r.get()

Advantages: the worker function is run asynchronously, so if for example you are running many tasks on several cores, it will be a lot more efficient than the original solution.

Disadvantages: if there is an exception in the worker function, it will only be raised after the pool has completed all the tasks. This may or may not be the desirable behaviour. EDITED according to @colinfang’s comment, which fixed this.

Answered By: gozzilli

Since you have used apply_sync, I guess the use case is want to do some synchronize tasks. Use callback for handling is another option. Please note this option is available only for python3.2 and above and not available on python2.7.

from multiprocessing import Pool

def callback(result):
    print('success', result)

def callback_error(result):
    print('error', result)

def go():
    print(1)
    raise Exception()
    print(2)

p = Pool()
p.apply_async(go, callback=callback, error_callback=callback_error)

# You can do another things

p.close()
p.join()
Answered By: Asoul
import logging
from multiprocessing import Pool

def proc_wrapper(func, *args, **kwargs):
    """Print exception because multiprocessing lib doesn't return them right."""
    try:
        return func(*args, **kwargs)
    except Exception as e:
        logging.exception(e)
        raise

def go(x):
    print x
    raise Exception("foobar")

p = Pool()
p.apply_async(proc_wrapper, (go, 5))
p.join()
p.close()
Answered By: erezarnon

Since there are already decent answers for multiprocessing.Pool available, I will provide a solution using a different approach for completeness.

For python >= 3.2 the following solution seems to be the simplest:

from concurrent.futures import ProcessPoolExecutor, wait

def go():
    print(1)
    raise Exception()
    print(2)


futures = []
with ProcessPoolExecutor() as p:
    for i in range(10):
        futures.append(p.submit(go))

results = [f.result() for f in futures]

Advantages:

  • very little code
  • raises an exception in the main process
  • provides a stack trace
  • no external dependencies

For more info about the API please check out this

Additionally, if you are submitting a large number of tasks and you would like your main process to fail as soon as one of your tasks fail, you can use the following snippet:

from concurrent.futures import ProcessPoolExecutor, wait, FIRST_EXCEPTION, as_completed
import time


def go():
    print(1)
    time.sleep(0.3)
    raise Exception()
    print(2)


futures = []
with ProcessPoolExecutor(1) as p:
    for i in range(10):
        futures.append(p.submit(go))

    for f in as_completed(futures):
        if f.exception() is not None:
            for f in futures:
                f.cancel()
            break

[f.result() for f in futures]

All of the other answers fail only once all tasks have been executed.

Answered By: Vladimirs