如何在Django中过滤用于计数注释的对象？

Consider simple Django models Event and Participant:

class Event(models.Model):
    title = models.CharField(max_length=100)

class Participant(models.Model):
    event = models.ForeignKey(Event, db_index=True)
    is_paid = models.BooleanField(default=False, db_index=True)

It’s easy to annotate events query with total number of participants:

events = Event.objects.all().annotate(participants=models.Count('participant'))

How to annotate with count of participants filtered by is_paid=True?

I need to query all events regardless of number of participants, e.g. I don’t need to filter by annotated result. If there are 0 participants, that’s ok, I just need 0 in annotated value.

The example from documentation doesn’t work here, because it excludes objects from query instead of annotating them with 0.

Update. 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()
    )))

Update 2. Django 2.0 has new Conditional aggregation feature, see the accepted answer below.

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('participants', filter=Q(participants__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.

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()
    )))

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`"
})

I would suggest to use the 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.

What result I am looking for:

• People (assignee) who have tasks added to a report. – Total Unique count of People
• People who have tasks added to a report but, for task whose billability is more than 0 only.

In general, I would have to use two different queries:

Task.objects.filter(billable_efforts__gt=0)
Task.objects.all()

But I want both in one query. Hence:

Task.objects.values('report__title').annotate(withMoreThanZero=Count('assignee', distinct=True, filter=Q(billable_efforts__gt=0))).annotate(totalUniqueAssignee=Count('assignee', distinct=True))

Result:

<QuerySet [{'report__title': 'TestReport', 'withMoreThanZero': 37, 'totalUniqueAssignee': 50}, {'report__title': 'Utilization_Report_April_2019', 'withMoreThanZero': 37, 'totalUniqueAssignee': 50}]>