Speeding Up Celery Backends, Part 3

Posted by Alexander Todorov on Tue 11 November 2014

In the second part of this article we've seen how slow Celery actually is. Now let's explore what happens inside and see if we can't speed things up.

I've used pycallgraph to create call graph visualizations of my application. It has the nice feature to also show execution time and use different colors for fast and slow operations.

Full command line is:

pycallgraph -v --stdlib --include ... graphviz -o calls.png -- ./manage.py celery_load_test

where the --include is used to limit the graph to a particular Python module(s).

General findings

call graph

  • The first four calls is where most of the time is spent as seen on the picture.
  • As it seems most of the slow down comes from Celery itself, not the underlying messaging transport Kombu (not shown on picture)
  • celery.app.amqp.TaskProducer.publish_task takes half of the execution time of celery.app.base.Celery.send_task
  • celery.app.task.Task.delay directly executes .apply_async and can be skipped if one rewrites the code.

More findings

In celery.app.base.Celery.send_task there is this block of code:

349         with self.producer_or_acquire(producer) as P:
350             self.backend.on_task_call(P, task_id)
351             task_id = P.publish_task(
352                 name, args, kwargs, countdown=countdown, eta=eta,
353                 task_id=task_id, expires=expires,
354                 callbacks=maybe_list(link), errbacks=maybe_list(link_error),
355                 reply_to=reply_to or self.oid, **options
356             )

producer is always None because delay() doesn't pass it as argument. I've tried passing it explicitly to apply_async() as so:

from djapp.celery import *

# app = debug_task._get_app() # if not defined in djapp.celery
producer = app.amqp.producer_pool.acquire(block=True)

However this doesn't speedup anything. If we replace the above code block like this:

349         with producer as P:

it blows up on the second iteration because producer and its channel is already None !?!

If you are unfamiliar with the with statement in Python please read this article. In short the with statement is a compact way of writing try/finally. The underlying kombu.messaging.Producer class does a self.release() on exit of the with statement.

I also tried killing the with statement and using producer directly but with limited success. While it was not released(was non None) on subsequent iterations the memory usage grew much more and there wasn't any performance boost.


The with statement is used throughout both Celery and Kombu and I'm not at all sure if there's a mechanism for keep-alive connections. My time constraints are limited and I'll probably not spend anymore time on this problem soon.

Since my use case involves task producer and consumers on localhost I'll try to workaround the current limitations by using Kombu directly (see this gist) with a transport that uses either a UNIX domain socket or a name pipe (FIFO) file.

tags: Python, Django, QA

Comments !