The Python Oracle

How to filter objects for count annotation in Django?

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Chapters
00:00 How To Filter Objects For Count Annotation In Django?
01:10 Answer 1 Score 52
01:47 Answer 2 Score 97
02:03 Answer 3 Score 6
03:45 Accepted Answer Score 228
04:55 Thank you

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Full question
https://stackoverflow.com/questions/3075...

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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...

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Tags
#python #django #djangomodels #djangoaggregation

#avk47



ACCEPTED ANSWER

Score 228


Conditional aggregation in Django 2.0+ allows you to further reduce the amount of faff this has been in the past. This will also use Postgres' filter logic, which is somewhat faster than a sum-case (I've seen numbers like 20-30% bandied around).

Anyway, in your case, we're looking at something as simple as:

from django.db.models import Q, Count
events = Event.objects.annotate(
    paid_participants=Count('participant', filter=Q(participant__is_paid=True))
)

There's a separate section in the docs about filtering on annotations. It's the same stuff as conditional aggregation but more like my example above. Either which way, this is a lot healthier than the gnarly subqueries I was doing before.

For more complex annotation filters, it may be more understandable to structure the filter as a separate queryset that is passed to an __in expression:

from django.db.models import Q, Count
paid_participants = Participant.objects.filter(is_paid=True)
events = Event.objects.annotate(
    paid_participants=Count(
        'participant', 
        filter=Q(participant__in=paid_participants)
    )
)

In this context, the Participant queryset does not need to explicitly filter on the outer Event ID like a subquery because Count (and other aggregation functions) implicitly filter for reverse foreign key relationships of the current row. In other words, with this format specify your filters globally for the Model that is being counted.




ANSWER 2

Score 97


Just discovered that Django 1.8 has new conditional expressions feature, so now we can do like this:

events = Event.objects.all().annotate(paid_participants=models.Sum(
    models.Case(
        models.When(participant__is_paid=True, then=1),
        default=0, output_field=models.IntegerField()
    )))



ANSWER 3

Score 52


UPDATE

The sub-query approach which I mention is now supported in Django 1.11 via subquery-expressions.

Event.objects.annotate(
    num_paid_participants=Subquery(
        Participant.objects.filter(
            is_paid=True,
            event=OuterRef('pk')
        ).values('event')
        .annotate(cnt=Count('pk'))
        .values('cnt'),
        output_field=models.IntegerField()
    )
)

I prefer this over aggregation (sum+case), because it should be faster and easier to be optimized (with proper indexing).

For older version, the same can be achieved using .extra

Event.objects.extra(select={'num_paid_participants': "\
    SELECT COUNT(*) \
    FROM `myapp_participant` \
    WHERE `myapp_participant`.`is_paid` = 1 AND \
            `myapp_participant`.`event_id` = `myapp_event`.`id`"
})



ANSWER 4

Score 6


I would suggest to use the .values method of your Participant queryset instead.

For short, what you want to do is given by:

Participant.objects\
    .filter(is_paid=True)\
    .values('event')\
    .distinct()\
    .annotate(models.Count('id'))

A complete example is as follow:

  1. Create 2 Events:

    event1 = Event.objects.create(title='event1')
    event2 = Event.objects.create(title='event2')
    
  2. Add Participants to them:

    part1l = [Participant.objects.create(event=event1, is_paid=((_%2) == 0))\
              for _ in range(10)]
    part2l = [Participant.objects.create(event=event2, is_paid=((_%2) == 0))\
              for _ in range(50)]
    
  3. Group all Participants by their event field:

    Participant.objects.values('event')
    > <QuerySet [{'event': 1}, {'event': 1}, {'event': 1}, {'event': 1}, {'event': 1}, {'event': 1}, {'event': 1}, {'event': 1}, {'event': 1}, {'event': 1}, {'event': 2}, {'event': 2}, {'event': 2}, {'event': 2}, {'event': 2}, {'event': 2}, {'event': 2}, {'event': 2}, {'event': 2}, {'event': 2}, '...(remaining elements truncated)...']>
    

    Here distinct is needed:

    Participant.objects.values('event').distinct()
    > <QuerySet [{'event': 1}, {'event': 2}]>
    

    What .values and .distinct are doing here is that they are creating two buckets of Participants grouped by their element event. Note that those buckets contain Participant.

  4. You can then annotate those buckets as they contain the set of original Participant. Here we want to count the number of Participant, this is simply done by counting the ids of the elements in those buckets (since those are Participant):

    Participant.objects\
        .values('event')\
        .distinct()\
        .annotate(models.Count('id'))
    > <QuerySet [{'event': 1, 'id__count': 10}, {'event': 2, 'id__count': 50}]>
    
  5. Finally you want only Participant with a is_paid being True, you may just add a filter in front of the previous expression, and this yield the expression shown above:

    Participant.objects\
        .filter(is_paid=True)\
        .values('event')\
        .distinct()\
        .annotate(models.Count('id'))
    > <QuerySet [{'event': 1, 'id__count': 5}, {'event': 2, 'id__count': 25}]>
    

The only drawback is that you have to retrieve the Event afterwards as you only have the id from the method above.