===========
Aggregation
===========

.. currentmodule:: django.db.models

The topic guide on :doc:`Django's database-abstraction API </topics/db/queries>`
described the way that you can use Django queries that create,
retrieve, update and delete individual objects. However, sometimes you will
need to retrieve values that are derived by summarizing or *aggregating* a
collection of objects. This topic guide describes the ways that aggregate values
can be generated and returned using Django queries.

Throughout this guide, we'll refer to the following models. These models are
used to track the inventory for a series of online bookstores:

.. _queryset-model-example:

.. code-block:: python

    from django.db import models


    class Author(models.Model):
        name = models.CharField(max_length=100)
        age = models.IntegerField()


    class Publisher(models.Model):
        name = models.CharField(max_length=300)


    class Book(models.Model):
        name = models.CharField(max_length=300)
        pages = models.IntegerField()
        price = models.DecimalField(max_digits=10, decimal_places=2)
        rating = models.FloatField()
        authors = models.ManyToManyField(Author)
        publisher = models.ForeignKey(Publisher, on_delete=models.CASCADE)
        pubdate = models.DateField()


    class Store(models.Model):
        name = models.CharField(max_length=300)
        books = models.ManyToManyField(Book)

Cheat sheet
===========

In a hurry? Here's how to do common aggregate queries, assuming the models
above:

.. code-block:: pycon

    # Total number of books.
    >>> Book.objects.count()
    2452

    # Total number of books with publisher=BaloneyPress
    >>> Book.objects.filter(publisher__name="BaloneyPress").count()
    73

    # Average price across all books, provide default to be returned instead
    # of None if no books exist.
    >>> from django.db.models import Avg
    >>> Book.objects.aggregate(Avg("price", default=0))
    {'price__avg': 34.35}

    # Max price across all books, provide default to be returned instead of
    # None if no books exist.
    >>> from django.db.models import Max
    >>> Book.objects.aggregate(Max("price", default=0))
    {'price__max': Decimal('81.20')}

    # Difference between the highest priced book and the average price of all books.
    >>> from django.db.models import FloatField
    >>> Book.objects.aggregate(
    ...     price_diff=Max("price", output_field=FloatField()) - Avg("price")
    ... )
    {'price_diff': 46.85}

    # All the following queries involve traversing the Book<->Publisher
    # foreign key relationship backwards.

    # Each publisher, each with a count of books as a "num_books" attribute.
    >>> from django.db.models import Count
    >>> pubs = Publisher.objects.annotate(num_books=Count("book"))
    >>> pubs
    <QuerySet [<Publisher: BaloneyPress>, <Publisher: SalamiPress>, ...]>
    >>> pubs[0].num_books
    73

    # Each publisher, with a separate count of books with a rating above and below 5
    >>> from django.db.models import Q
    >>> above_5 = Count("book", filter=Q(book__rating__gt=5))
    >>> below_5 = Count("book", filter=Q(book__rating__lte=5))
    >>> pubs = Publisher.objects.annotate(below_5=below_5).annotate(above_5=above_5)
    >>> pubs[0].above_5
    23
    >>> pubs[0].below_5
    12

    # The top 5 publishers, in order by number of books.
    >>> pubs = Publisher.objects.annotate(num_books=Count("book")).order_by("-num_books")[:5]
    >>> pubs[0].num_books
    1323

Generating aggregates over a ``QuerySet``
=========================================

Django provides two ways to generate aggregates. The first way is to generate
summary values over an entire ``QuerySet``. For example, say you wanted to
calculate the average price of all books available for sale. Django's query
syntax provides a means for describing the set of all books:

.. code-block:: pycon

    >>> Book.objects.all()

What we need is a way to calculate summary values over the objects that
belong to this ``QuerySet``. This is done by appending an ``aggregate()``
clause onto the ``QuerySet``:

.. code-block:: pycon

    >>> from django.db.models import Avg
    >>> Book.objects.all().aggregate(Avg("price"))
    {'price__avg': 34.35}

The ``all()`` is redundant in this example, so this could be simplified to:

.. code-block:: pycon

    >>> Book.objects.aggregate(Avg("price"))
    {'price__avg': 34.35}

The argument to the ``aggregate()`` clause describes the aggregate value that
we want to compute - in this case, the average of the ``price`` field on the
``Book`` model. A list of the aggregate functions that are available can be
found in the :ref:`QuerySet reference <aggregation-functions>`.

``aggregate()`` is a terminal clause for a ``QuerySet`` that, when invoked,
returns a dictionary of name-value pairs. The name is an identifier for the
aggregate value; the value is the computed aggregate. The name is
automatically generated from the name of the field and the aggregate function.
If you want to manually specify a name for the aggregate value, you can do so
by providing that name when you specify the aggregate clause:

.. code-block:: pycon

    >>> Book.objects.aggregate(average_price=Avg("price"))
    {'average_price': 34.35}

If you want to generate more than one aggregate, you add another argument to
the ``aggregate()`` clause. So, if we also wanted to know the maximum and
minimum price of all books, we would issue the query:

.. code-block:: pycon

    >>> from django.db.models import Avg, Max, Min
    >>> Book.objects.aggregate(Avg("price"), Max("price"), Min("price"))
    {'price__avg': 34.35, 'price__max': Decimal('81.20'), 'price__min': Decimal('12.99')}

Generating aggregates for each item in a ``QuerySet``
=====================================================

The second way to generate summary values is to generate an independent
summary for each object in a :class:`.QuerySet`. For example, if you are
retrieving a list of books, you may want to know how many authors contributed
to each book. Each Book has a many-to-many relationship with the Author; we
want to summarize this relationship for each book in the ``QuerySet``.

Per-object summaries can be generated using the
:meth:`~.QuerySet.annotate` clause. When an ``annotate()`` clause is
specified, each object in the ``QuerySet`` will be annotated with the
specified values.

The syntax for these annotations is identical to that used for the
:meth:`~.QuerySet.aggregate` clause. Each argument to ``annotate()`` describes
an aggregate that is to be calculated. For example, to annotate books with the
number of authors:

.. code-block:: pycon

    # Build an annotated queryset
    >>> from django.db.models import Count
    >>> q = Book.objects.annotate(Count("authors"))
    # Interrogate the first object in the queryset
    >>> q[0]
    <Book: The Definitive Guide to Django>
    >>> q[0].authors__count
    2
    # Interrogate the second object in the queryset
    >>> q[1]
    <Book: Practical Django Projects>
    >>> q[1].authors__count
    1

As with ``aggregate()``, the name for the annotation is automatically derived
from the name of the aggregate function and the name of the field being
aggregated. You can override this default name by providing an alias when you
specify the annotation:

.. code-block:: pycon

    >>> q = Book.objects.annotate(num_authors=Count("authors"))
    >>> q[0].num_authors
    2
    >>> q[1].num_authors
    1

Unlike ``aggregate()``, ``annotate()`` is *not* a terminal clause. The output
of the ``annotate()`` clause is a ``QuerySet``; this ``QuerySet`` can be
modified using any other ``QuerySet`` operation, including ``filter()``,
``order_by()``, or even additional calls to ``annotate()``.

.. _combining-multiple-aggregations:

Combining multiple aggregations
-------------------------------

Combining multiple aggregations with ``annotate()`` will :ticket:`yield the
wrong results <10060>` because joins are used instead of subqueries:

    >>> book = Book.objects.first()
    >>> book.authors.count()
    2
    >>> book.store_set.count()
    3
    >>> q = Book.objects.annotate(Count('authors'), Count('store'))
    >>> q[0].authors__count
    6
    >>> q[0].store__count
    6

For most aggregates, there is no way to avoid this problem, however, the
:class:`~django.db.models.Count` aggregate has a ``distinct`` parameter that
may help:

    >>> q = Book.objects.annotate(Count('authors', distinct=True), Count('store', distinct=True))
    >>> q[0].authors__count
    2
    >>> q[0].store__count
    3

.. admonition:: If in doubt, inspect the SQL query!

    In order to understand what happens in your query, consider inspecting the
    ``query`` property of your ``QuerySet``.

Joins and aggregates
====================

So far, we have dealt with aggregates over fields that belong to the
model being queried. However, sometimes the value you want to aggregate
will belong to a model that is related to the model you are querying.

When specifying the field to be aggregated in an aggregate function, Django
will allow you to use the same :ref:`double underscore notation
<field-lookups-intro>` that is used when referring to related fields in
filters. Django will then handle any table joins that are required to retrieve
and aggregate the related value.

For example, to find the price range of books offered in each store,
you could use the annotation:

.. code-block:: pycon

    >>> from django.db.models import Max, Min
    >>> Store.objects.annotate(min_price=Min("books__price"), max_price=Max("books__price"))

This tells Django to retrieve the ``Store`` model, join (through the
many-to-many relationship) with the ``Book`` model, and aggregate on the
price field of the book model to produce a minimum and maximum value.

The same rules apply to the ``aggregate()`` clause. If you wanted to
know the lowest and highest price of any book that is available for sale
in any of the stores, you could use the aggregate:

.. code-block:: pycon

    >>> Store.objects.aggregate(min_price=Min("books__price"), max_price=Max("books__price"))

Join chains can be as deep as you require. For example, to extract the
age of the youngest author of any book available for sale, you could
issue the query:

.. code-block:: pycon

    >>> Store.objects.aggregate(youngest_age=Min("books__authors__age"))

Following relationships backwards
---------------------------------

In a way similar to :ref:`lookups-that-span-relationships`, aggregations and
annotations on fields of models or models that are related to the one you are
querying can include traversing "reverse" relationships. The lowercase name
of related models and double-underscores are used here too.

For example, we can ask for all publishers, annotated with their respective
total book stock counters (note how we use ``'book'`` to specify the
``Publisher`` -> ``Book`` reverse foreign key hop):

.. code-block:: pycon

    >>> from django.db.models import Avg, Count, Min, Sum
    >>> Publisher.objects.annotate(Count("book"))

(Every ``Publisher`` in the resulting ``QuerySet`` will have an extra attribute
called ``book__count``.)

We can also ask for the oldest book of any of those managed by every publisher:

.. code-block:: pycon

    >>> Publisher.objects.aggregate(oldest_pubdate=Min("book__pubdate"))

(The resulting dictionary will have a key called ``'oldest_pubdate'``. If no
such alias were specified, it would be the rather long ``'book__pubdate__min'``.)

This doesn't apply just to foreign keys. It also works with many-to-many
relations. For example, we can ask for every author, annotated with the total
number of pages considering all the books the author has (co-)authored (note how we
use ``'book'`` to specify the ``Author`` -> ``Book`` reverse many-to-many hop):

.. code-block:: pycon

    >>> Author.objects.annotate(total_pages=Sum("book__pages"))

(Every ``Author`` in the resulting ``QuerySet`` will have an extra attribute
called ``total_pages``. If no such alias were specified, it would be the rather
long ``book__pages__sum``.)

Or ask for the average rating of all the books written by author(s) we have on
file:

.. code-block:: pycon

    >>> Author.objects.aggregate(average_rating=Avg("book__rating"))

(The resulting dictionary will have a key called ``'average_rating'``. If no
such alias were specified, it would be the rather long ``'book__rating__avg'``.)

Aggregations and other ``QuerySet`` clauses
===========================================

``filter()`` and ``exclude()``
------------------------------

Aggregates can also participate in filters. Any ``filter()`` (or
``exclude()``) applied to normal model fields will have the effect of
constraining the objects that are considered for aggregation.

When used with an ``annotate()`` clause, a filter has the effect of
constraining the objects for which an annotation is calculated. For example,
you can generate an annotated list of all books that have a title starting
with "Django" using the query:

.. code-block:: pycon

    >>> from django.db.models import Avg, Count
    >>> Book.objects.filter(name__startswith="Django").annotate(num_authors=Count("authors"))

When used with an ``aggregate()`` clause, a filter has the effect of
constraining the objects over which the aggregate is calculated.
For example, you can generate the average price of all books with a
title that starts with "Django" using the query:

.. code-block:: pycon

    >>> Book.objects.filter(name__startswith="Django").aggregate(Avg("price"))

.. _filtering-on-annotations:

Filtering on annotations
~~~~~~~~~~~~~~~~~~~~~~~~

Annotated values can also be filtered. The alias for the annotation can be
used in ``filter()`` and ``exclude()`` clauses in the same way as any other
model field.

For example, to generate a list of books that have more than one author,
you can issue the query:

.. code-block:: pycon

    >>> Book.objects.annotate(num_authors=Count("authors")).filter(num_authors__gt=1)

This query generates an annotated result set, and then generates a filter
based upon that annotation.

If you need two annotations with two separate filters you can use the
``filter`` argument with any aggregate. For example, to generate a list of
authors with a count of highly rated books:

.. code-block:: pycon

    >>> highly_rated = Count("book", filter=Q(book__rating__gte=7))
    >>> Author.objects.annotate(num_books=Count("book"), highly_rated_books=highly_rated)

Each ``Author`` in the result set will have the ``num_books`` and
``highly_rated_books`` attributes. See also :ref:`conditional-aggregation`.

.. admonition:: Choosing between ``filter`` and ``QuerySet.filter()``

    Avoid using the ``filter`` argument with a single annotation or
    aggregation. It's more efficient to use ``QuerySet.filter()`` to exclude
    rows. The aggregation ``filter`` argument is only useful when using two or
    more aggregations over the same relations with different conditionals.

Order of ``annotate()`` and ``filter()`` clauses
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

When developing a complex query that involves both ``annotate()`` and
``filter()`` clauses, pay particular attention to the order in which the
clauses are applied to the ``QuerySet``.

When an ``annotate()`` clause is applied to a query, the annotation is computed
over the state of the query up to the point where the annotation is requested.
The practical implication of this is that ``filter()`` and ``annotate()`` are
not commutative operations.

Given:

* Publisher A has two books with ratings 4 and 5.
* Publisher B has two books with ratings 1 and 4.
* Publisher C has one book with rating 1.

Here's an example with the ``Count`` aggregate:

.. code-block:: pycon

    >>> a, b = Publisher.objects.annotate(num_books=Count("book", distinct=True)).filter(
    ...     book__rating__gt=3.0
    ... )
    >>> a, a.num_books
    (<Publisher: A>, 2)
    >>> b, b.num_books
    (<Publisher: B>, 2)

    >>> a, b = Publisher.objects.filter(book__rating__gt=3.0).annotate(num_books=Count("book"))
    >>> a, a.num_books
    (<Publisher: A>, 2)
    >>> b, b.num_books
    (<Publisher: B>, 1)

Both queries return a list of publishers that have at least one book with a
rating exceeding 3.0, hence publisher C is excluded.

In the first query, the annotation precedes the filter, so the filter has no
effect on the annotation. ``distinct=True`` is required to avoid a :ref:`query
bug <combining-multiple-aggregations>`.

The second query counts the number of books that have a rating exceeding 3.0
for each publisher. The filter precedes the annotation, so the filter
constrains the objects considered when calculating the annotation.

Here's another example with the ``Avg`` aggregate:

.. code-block:: pycon

    >>> a, b = Publisher.objects.annotate(avg_rating=Avg("book__rating")).filter(
    ...     book__rating__gt=3.0
    ... )
    >>> a, a.avg_rating
    (<Publisher: A>, 4.5)  # (5+4)/2
    >>> b, b.avg_rating
    (<Publisher: B>, 2.5)  # (1+4)/2

    >>> a, b = Publisher.objects.filter(book__rating__gt=3.0).annotate(
    ...     avg_rating=Avg("book__rating")
    ... )
    >>> a, a.avg_rating
    (<Publisher: A>, 4.5)  # (5+4)/2
    >>> b, b.avg_rating
    (<Publisher: B>, 4.0)  # 4/1 (book with rating 1 excluded)

The first query asks for the average rating of all a publisher's books for
publisher's that have at least one book with a rating exceeding 3.0. The second
query asks for the average of a publisher's book's ratings for only those
ratings exceeding 3.0.

It's difficult to intuit how the ORM will translate complex querysets into SQL
queries so when in doubt, inspect the SQL with ``str(queryset.query)`` and
write plenty of tests.

``order_by()``
--------------

Annotations can be used as a basis for ordering. When you
define an ``order_by()`` clause, the aggregates you provide can reference
any alias defined as part of an ``annotate()`` clause in the query.

For example, to order a ``QuerySet`` of books by the number of authors
that have contributed to the book, you could use the following query:

.. code-block:: pycon

    >>> Book.objects.annotate(num_authors=Count("authors")).order_by("num_authors")

``values()``
------------

Ordinarily, annotations are generated on a per-object basis - an annotated
``QuerySet`` will return one result for each object in the original
``QuerySet``. However, when a ``values()`` clause is used to constrain the
columns that are returned in the result set, the method for evaluating
annotations is slightly different. Instead of returning an annotated result
for each result in the original ``QuerySet``, the original results are
grouped according to the unique combinations of the fields specified in the
``values()`` clause. An annotation is then provided for each unique group;
the annotation is computed over all members of the group.

For example, consider an author query that attempts to find out the average
rating of books written by each author:

    >>> Author.objects.annotate(average_rating=Avg('book__rating'))

This will return one result for each author in the database, annotated with
their average book rating.

However, the result will be slightly different if you use a ``values()`` clause:

.. code-block:: pycon

    >>> Author.objects.values("name").annotate(average_rating=Avg("book__rating"))

In this example, the authors will be grouped by name, so you will only get
an annotated result for each *unique* author name. This means if you have
two authors with the same name, their results will be merged into a single
result in the output of the query; the average will be computed as the
average over the books written by both authors.

Order of ``annotate()`` and ``values()`` clauses
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As with the ``filter()`` clause, the order in which ``annotate()`` and
``values()`` clauses are applied to a query is significant. If the
``values()`` clause precedes the ``annotate()``, the annotation will be
computed using the grouping described by the ``values()`` clause.

However, if the ``annotate()`` clause precedes the ``values()`` clause,
the annotations will be generated over the entire query set. In this case,
the ``values()`` clause only constrains the fields that are generated on
output.

For example, if we reverse the order of the ``values()`` and ``annotate()``
clause from our previous example:

.. code-block:: pycon

    >>> Author.objects.annotate(average_rating=Avg("book__rating")).values(
    ...     "name", "average_rating"
    ... )

This will now yield one unique result for each author; however, only
the author's name and the ``average_rating`` annotation will be returned
in the output data.

You should also note that ``average_rating`` has been explicitly included
in the list of values to be returned. This is required because of the
ordering of the ``values()`` and ``annotate()`` clause.

If the ``values()`` clause precedes the ``annotate()`` clause, any annotations
will be automatically added to the result set. However, if the ``values()``
clause is applied after the ``annotate()`` clause, you need to explicitly
include the aggregate column.

.. _aggregation-ordering-interaction:

Interaction with ``order_by()``
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Fields that are mentioned in the ``order_by()`` part of a queryset are used
when selecting the output data, even if they are not otherwise specified in the
``values()`` call. These extra fields are used to group "like" results together
and they can make otherwise identical result rows appear to be separate. This
shows up, particularly, when counting things.

By way of example, suppose you have a model like this::

    from django.db import models


    class Item(models.Model):
        name = models.CharField(max_length=10)
        data = models.IntegerField()

If you want to count how many times each distinct ``data`` value appears in an
ordered queryset, you might try this::

    items = Item.objects.order_by("name")
    # Warning: not quite correct!
    items.values("data").annotate(Count("id"))

...which will group the ``Item`` objects by their common ``data`` values and
then count the number of ``id`` values in each group. Except that it won't
quite work. The ordering by ``name`` will also play a part in the grouping, so
this query will group by distinct ``(data, name)`` pairs, which isn't what you
want. Instead, you should construct this queryset::

    items.values("data").annotate(Count("id")).order_by()

...clearing any ordering in the query. You could also order by, say, ``data``
without any harmful effects, since that is already playing a role in the
query.

This behavior is the same as that noted in the queryset documentation for
:meth:`~django.db.models.query.QuerySet.distinct` and the general rule is the
same: normally you won't want extra columns playing a part in the result, so
clear out the ordering, or at least make sure it's restricted only to those
fields you also select in a ``values()`` call.

.. note::
    You might reasonably ask why Django doesn't remove the extraneous columns
    for you. The main reason is consistency with ``distinct()`` and other
    places: Django **never** removes ordering constraints that you have
    specified (and we can't change those other methods' behavior, as that
    would violate our :doc:`/misc/api-stability` policy).

Aggregating annotations
-----------------------

You can also generate an aggregate on the result of an annotation. When you
define an ``aggregate()`` clause, the aggregates you provide can reference
any alias defined as part of an ``annotate()`` clause in the query.

For example, if you wanted to calculate the average number of authors per
book you first annotate the set of books with the author count, then
aggregate that author count, referencing the annotation field:

.. code-block:: pycon

    >>> from django.db.models import Avg, Count
    >>> Book.objects.annotate(num_authors=Count("authors")).aggregate(Avg("num_authors"))
    {'num_authors__avg': 1.66}

Aggregating on empty querysets or groups
----------------------------------------

When an aggregation is applied to an empty queryset or grouping, the result
defaults to its :ref:`default <aggregate-default>` parameter, typically
``None``. This behavior occurs because aggregate functions return ``NULL`` when
the executed query returns no rows.

You can specify a return value by providing the :ref:`default
<aggregate-default>` argument for most aggregations. However, since
:class:`~django.db.models.Count` does not support the :ref:`default
<aggregate-default>` argument, it will always return ``0`` for empty querysets
or groups.

For example, assuming that no book contains *web* in its name, calculating the
total price for this book set would return ``None`` since there are no matching
rows to compute the :class:`~django.db.models.Sum` aggregation on:

.. code-block:: pycon

    >>> from django.db.models import Sum
    >>> Book.objects.filter(name__contains="web").aggregate(Sum("price"))
    {"price__sum": None}

However, the :ref:`default <aggregate-default>` argument can be set when
calling :class:`~django.db.models.Sum` to return a different default value if
no books can be found:

.. code-block:: pycon

    >>> Book.objects.filter(name__contains="web").aggregate(Sum("price", default=0))
    {"price__sum": Decimal("0")}

Under the hood, the :ref:`default <aggregate-default>` argument is implemented
by wrapping the aggregate function with
:class:`~django.db.models.functions.Coalesce`.
