Using Subquery to annotate a Count
Solution 1
Counting all Doctor
s per Specialization
I think you make things overly complicated, probably because you think that Count('doctor')
will count every doctor per specialization (regardless the specialization of that doctor). It does not, if you Count
such related object, Django implicitly looks for related objects. In fact you can not Count('unrelated_model')
at all, it is only through relations (reversed included) like a ForeignKey
, ManyToManyField
, etc. that you can query these, since otherwise these are not very sensical.
I would like to annotate all specializations in a queryset with the number of doctors that have this specialization.
You can do this with a simple:
# Counting all doctors per specialization (so not all doctors in general)
from django.db.models import Count
Specialization.objects.annotate(
num_doctors=Count('doctor')
)
Now every Specialization
object in this queryset will have an extra attribute num_doctors
that is an integer (the number of doctors with that specialization).
You can also filter on the Specialization
s in the same query (for example only obtain specializations that end on 'my'
). As long as you do not filter on the related doctor
s set, the Count
will work (see section below how to do this).
If you however filter on the related doctor
s, then the related counts will filter out these doctors. Furthermore if you filter on another related object, then this will result in an extra JOIN
, that will act as a multiplier for the Count
s. In that case it might be better to use num_doctors=Count('doctor', distinct=True)
instead. You can always use the distinct=True
(regardless if you do extra JOIN
s or not), but it will have a small performance impact.
The above works because Count('doctor')
does not simply adds all doctors to the query, it makes a LEFT OUTER JOIN
on the doctor
s table, and thus checks that the specialization_id
of that Doctor
is exactly the one we are looking for. So the query Django will construct looks like:
SELECT specialization.*
COUNT(doctor.id) AS num_doctors
FROM specialization
LEFT OUTER JOIN doctor ON doctor.specialization_id = specialization.id
GROUP BY specialization.id
Doing the same with a subquery will functionally get the same results, but if the Django ORM and the database management system do not find a way to optimize this, this can result in an expensive query, since for every specialization, it then can result in an extra subquery in the database.
Counting specific Doctor
s per Specialization
Say however you want to count only doctors that have a name that starts with Joe, then you can add a filter on the related doctor
, like:
# counting all Doctors with as name Joe per specialization
from django.db.models import Count
Specialization.objects.filter(
doctor__name__startswith='Joe' # sample filter
).annotate(
num_doctors=Count('doctor')
)
Solution 2
Problem
The problem is that Django adds GROUP BY
as soon as it sees using an aggregate function.
Solution
So you can just create your own aggregate function but so that Django thinks it is not aggregate. Just like this:
doctors = Doctor.objects.filter(
specialization=OuterRef("id")
).order_by().annotate(
count=Func('id', 'Count')
).values('count')
spec_qs_with_counts = Specialization.objects.annotate(
num_applicable_doctors=Subquery(doctors)
)
The more details on this method you can see in this answer: https://stackoverflow.com/a/69020732/10567223
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Ysrninja
Updated on September 15, 2022Comments
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Ysrninja over 1 year
Please help me I've been stuck on this for way too long :(
What I want to do:
I have these two models:
class Specialization(models.Model): name = models.CharField("name", max_length=64) class Doctor(models.Model): name = models.CharField("name", max_length=128) # ... specialization = models.ForeignKey(Specialization)
I would like to annotate all specializations in a queryset with the number of doctors that have this specialization.
My Solution so Far:
I went through a loop and I made a simple:
Doctor.objects.filter(specialization=spec).count()
however this proved to be too slow and inefficient. The more I've read the more I realized that it would make sense to use aSubQuery
here to filter the doctors for theOuterRef
specialization. This is what I came up with:doctors = Doctor.objects.all().filter(specialization=OuterRef("id")) \ .values("specialization_id") \ .order_by() add_doctors_count = doctors.annotate(cnt=Count("specialization_id")).values("cnt")[:1] spec_qs_with_counts = Specialization.objects.all().annotate( num_applicable_doctors=Subquery(add_doctors_count, output_field=IntegerField()) )
The output I get is just 1 for every speciality. The code just annotates every doctor object with its
specialization_id
and then annotates the count within that group, meaning it will be 1.This doesn't make complete sense to me unfortunately. In my initial attempt I used an aggregate for the count, and while it works on its own it doesn't work as a
SubQuery
, I get this error:This queryset contains a reference to an outer query and may only be used in a subquery.
I posted this question before and someone suggested doing
Specialization.objects.annotate(count=Count("doctor"))
However this doesn't work because I need to count a specific queryset of Doctors.
I've followed these links
However, I'm not getting the same result:
https://docs.djangoproject.com/en/1.11/ref/models/expressions/
https://medium.com/@hansonkd/the-dramatic-benefits-of-django-subqueries-and-annotations-4195e0dafb16
If you have any questions that would make this clearer please tell me.
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Ysrninja over 5 yearsThanks for the reply, I appreciate it. I just tried this and initially thought this wasn't going to work, but it did! You're right I was overcomplicating this, I just thought the count for the doctors is going to count ALL the doctors for that Specialization, but the filter actually takes them out of the count. Thanks for the help.
-
little_birdie over 5 yearsThe problem with
annotate
+Count
is that (as I have discovered), Django generates a left outer join and then selects aCOUNT(DISTINCT)
.. which can be VERY SLOW because so many intermediate rows are generated in the result set -
Willem Van Onsem over 5 years@little_birdie: yes, but the alternative of doing it one per element is slower, since then this yields a round trip per subquery. The idea of an annotate is that you do this if you need the count of all
Specialization
s, in that case this is, querywise more interesting than counting one perSpecialization
, since this is the famous N+1 problem. -
little_birdie over 5 yearsI'm not suggesting multiple round trips. I'm suggesting that the query which django generates when you annotate(Count()) on a reverse join can be VERY slow, especially when you are doing multiple counts.. because it does an outer join to every table and then does a COUNT(DISTINCT).. this results in a huge number of rows for the server to sort through. In my application, it took 5 seconds for postgres to execute django's query whereas my hand coded query which does
(SELECT COUNT(*) .. ) AS acount
executes in 34 milliseconds. -
Willem Van Onsem over 5 years@little_birdie: but here we do not
COUNT(DISTINCT ...)
note that there is noCount(..., distinct=True)
in the ORM query, you can obtain the query withprint(str(my_django_query.query))
, and I've tested in on both a MySQL and PostgreSQL, and both result, as expected in aCOUNT(doctor_id)
(as listed in the answer). Especially if there would be multiple aggregates involved, a JOIN will eventually outperform a subquery, since it performs multiple scans (source: itprotoday.com/sql-server/…) -
Willem Van Onsem over 5 yearsThe
DISTINCT
is key here, since for a reasonable modern database system, the query plan will "collapse" the count to an accumulator, or it can even make use of theMUL
index.