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()
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