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- ===========
- 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:
- .. code-block:: pycon
- >>> 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:
- .. code-block:: pycon
- >>> 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:
- .. code-block:: pycon
- >>> 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`.
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