Heroku上的Django Celery任务导致高内存使用率

Question

I have a celery task on Heroku that connects to an external API and retrieves some data, stores in the database and repeats several hundred times. Very quickly (after ~10 loops) Heroku starts warning about high memory usage. Any ideas?

tasks.py

@app.task
def retrieve_details():
    for p in PObj.objects.filter(some_condition=True):
        p.fetch()

models.py

def fetch(self):
    v_data = self.service.getV(**dict(
        Number=self.v.number
    ))
    response = self.map_response(v_data)

    for key in ["some_key","some_other_key",]:
        setattr(self.v, key, response.get(key))

    self.v.save()

Heroky logs

2017-01-01 10:26:25.634
132 <45>1 2017-01-01T10:26:25.457411+00:00 heroku run.5891 - - Error R14 (Memory quota exceeded)

Go to the log: https://api.heroku.com/myapps/xxx@heroku.com/addons/logentries

You are receiving this email because your Logentries alarm "Memory quota exceeded"
has been triggered.

In context:
2017-01-01 10:26:25.568 131 <45>1 2017-01-01T10:26:25.457354+00:00 heroku run.5891 - - Process running mem=595M(116.2%)
2017-01-01 10:26:25.634 132 <45>1 2017-01-01T10:26:25.457411+00:00 heroku run.5891 - - Error R14 (Memory quota exceeded)

Answer 1

You're basically loading a bunch of data into a Python dictionary in memory. This will cause a lot of memory overhead, especially if you are grabbing a lot of objects from the local database.

Do you really need to store all of these objects in a dictionary?

What most people do for things like this is:

• Retrieve one object at a time from the database.
• Process that item (perform whatever logic you need).
• Repeat.

This way, you only end up storing a single object in memory at any given time, thereby greatly reducing your memory footprint.

If I were you, I'd look for ways to move my logic into the database query, or simply process each item individually.

Answer 2

To expand on the veritable rdegges thoughts, here are two strategies I have used in the past when working with celery/python to help reduce the memory footprint: (1) kick off subtasks that each process exactly one object and/or (2) use generators.

1. kick off subtasks that each process exactly one object:

    @app.task def retrieve_details(): qs = PObj.objects.filter(some_condition=True) for p in qs.values_list('id', flat=True): do_fetch.delay(p) @app.task def do_fetch(n_id): p = PObj.objects.get(id=n_id) p.fetch() 

   Now you can tune celery so that it kills of processes after processing N number of PObj (tasks) to keep memory footprint low using --max-tasks-per-child .

2. Using generators: you can also try this using generators so that you can (theoretically) throw away the PObj after you call fetch

    def ps_of_interest(chunk=10): n = chunk start = 0 while n == chunk: some_ps = list(PObj.objects.filter(some_condition=True)[start:start + n]) n = len(some_ps) start += chunk for p in some_ps: yield p @app.task def retrieve_details(): for p in ps_of_interest(): p.fetch() 

For my money, I'd go with option #1.