Retry Celery tasks with exponential back off

PythonCeleryDjango Celery

Python Problem Overview


For a task like this:

from celery.decorators import task

@task()
def add(x, y):
    if not x or not y:
        raise Exception("test error")
    return self.wait_until_server_responds(

if it throws an exception and I want to retry it from the daemon side, how can apply an exponential back off algorithm, i.e. after 2^2, 2^3,2^4 etc seconds?

Also is the retry maintained from the server side, such that if the worker happens to get killed then next worker that spawns will take the retry task?

Python Solutions


Solution 1 - Python

The task.request.retries attribute contains the number of tries so far, so you can use this to implement exponential back-off:

from celery.task import task

@task(bind=True, max_retries=3)
def update_status(self, auth, status):
    try:
        Twitter(auth).update_status(status)
    except Twitter.WhaleFail as exc:
        raise self.retry(exc=exc, countdown=2 ** self.request.retries)

To prevent a Thundering Herd Problem, you may consider adding a random jitter to your exponential backoff:

import random
self.retry(exc=exc, countdown=int(random.uniform(2, 4) ** self.request.retries))

Solution 2 - Python

As of Celery 4.2 you can configure your tasks to use an exponential backoff automatically: http://docs.celeryproject.org/en/master/userguide/tasks.html#automatic-retry-for-known-exceptions

@app.task(autoretry_for=(Exception,), retry_backoff=2)
def add(x, y):
    ...

(This was already in the docs for Celery 4.1 but actually wasn't released then, see merge request)

Solution 3 - Python

FYI, celery has a util function to calculate exponential backoff time with jitter here, so you don't need to write your own.

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionQuintin ParView Question on Stackoverflow
Solution 1 - PythonasksolView Answer on Stackoverflow
Solution 2 - PythonRupert AngermeierView Answer on Stackoverflow
Solution 3 - PythonlgylymView Answer on Stackoverflow