CODEWITHSUNDEEP
pythoncelerydjango-celery
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Reputation: 6667

celery - chaining groups and subtasks. -> out of order execution

When I have something like the following

group1 = group(task1.si(), task1.si(), task1.si())
group2 = group(task2.si(), task2.si(), task2.si())

workflow = chain(group1, group2, task3.si())

The intuitive interpretation is that task3 should only execute after all tasks in group 2 have finished.

In reality, task 3 executes while group1 has started but hasn't completed yet.

What am i doing wrong?

Upvotes: 31

Views: 23661

Answers (2)

Jonathan Adami
Jonathan Adami

Reputation: 356

I have the same issue with celery, trying to have a workflow where the first step is "spawn a million tasks". Tried groups of groups, subtasks, eventually my step2 kicks off before step1 is over.

Long story short I might have found a solution with the use of chords and a dumb finisher:

@celery.task
def chordfinisher( *args, **kwargs ):
  return "OK"

Doing nothing much, but it enables me to do this:

tasks = []
for id in ids:
    tasks.append( mytask.si( id ) )
step1 = chord( group( tasks ), chordfinisher.si() )

step2 = ...

workflow = chain( step1, step2 )

Originally I wanted to have step1 in a subtask but for the same reason as suspected, the action of calling a group ends, the task is considered finished, and my workflow moves on...

If someone has something better, I'm interested!

Upvotes: 23

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Reputation: 6667

So as it turns out, in celery you cannot chain two groups together.
I suspect this is because groups chained with tasks automatically become a chord
--> Celery docs: http://docs.celeryproject.org/en/latest/userguide/canvas.html

Chaining a group together with another task will automatically upgrade it to be a chord:

Groups return a parent task. When chaining two groups together, I suspect that when the first group completes, the chord starts the callback "task". I suspect this "task" is actually the "parent task" of the second group. I further suspect that this parent task completes as soon as it finishes kicking off all the subtasks within the group and as a result the next item after the 2nd group is executed.

To demonstrate this here is some sample code. You'll need to already have a running celery instance.

# celery_experiment.py

from celery import task, group, chain, chord
from celery.signals import task_sent, task_postrun, task_prerun

import time
import logging

import random
random.seed()

logging.basicConfig(level=logging.DEBUG)

### HANDLERS ###    
@task_prerun.connect()
def task_starting_handler(sender=None, task_id=None, task=None, args=None, kwargs=None, **kwds):    
    try:
        logging.info('[%s] starting' % kwargs['id'])
    except KeyError:
        pass

@task_postrun.connect()
def task_finished_handler(sender=None, task_id=None, task=None, args=None, kwargs=None, retval=None, state=None, **kwds):
    try:    
        logging.info('[%s] finished' % kwargs['id'])
    except KeyError:
        pass


def random_sleep(id):
    slp = random.randint(1, 3)
    logging.info('[%s] sleep for %ssecs' % (id, slp))
    time.sleep(slp)

@task()
def thing(id):
    logging.info('[%s] begin' % id)
    random_sleep(id)
    logging.info('[%s] end' % id)


def exec_exp():
    st = thing.si(id='st')
    st_arr = [thing.si(id='st_arr1_a'), thing.si(id='st_arr1_b'), thing.si(id='st_arr1_c'),]
    st_arr2 = [thing.si(id='st_arr2_a'), thing.si(id='st_arr2_b'),]
    st2 = thing.si(id='st2')
    st3 = thing.si(id='st3')
    st4 = thing.si(id='st4')

    grp1 = group(st_arr)
    grp2 = group(st_arr2)

    # chn can chain two groups together because they are seperated by a single subtask
    chn = (st | grp1 | st2 | grp2 | st3 | st4)

    # in chn2 you can't chain two groups together. what will happen is st3 will start before grp2 finishes
    #chn2 = (st | st2 | grp1 | grp2 | st3 |  st4)

    r = chn()
    #r2 = chn2()

Upvotes: 27

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