Django aggregation tutorial

By : Shabda Raaj

One of the new and most awaited features with Django 1.1 was aggregation. As usual, Django comes with a very comprehensive documentation for this. Here, I have tried to put this in how-to form.

Jump to howtos or Get source on Github.

Essentially, aggregations are nothing but a way to perform an operation on group of rows. In databases, they are represented by operators as sum, avg etc.

To do these operations Django added two new methods to querysets.

  1. aggregate
  2. annotate

When you are have a queryset you can do two operations on it,

  1. Operate over the rowset to get a single value from it. (Such as sum of all salaries in the rowset)
  2. Operate over the rowset to get a value for each row in the rowset via some related table.

The thing to notice is that option 1, will create one row from rowset, while option 2 will not change the number of rows in the rowset. If you are into analogies, you can think that option 1 is like a reduce and option 2 is like a map.

In sql terms, aggregate is a operation(SUM, AVG, MIN, MAX), without a group by, while annotate is a operation with a group by on (Unless explicitly overriden).

Ok enough talk, on to some actual work. Here is a fictional representing a HRMS application. We will use this to see how to use aggreagtion to solve some common problems.

from django.db import models

class Department(models.Model):
    dept_name = models.CharField(max_length = 100)
    established_on = models.DateField()

    def __unicode__(self):
        return self.dept_name

class Level(models.Model):
    level_name = models.CharField(max_length = 100)
    pay_min = models.PositiveIntegerField()
    pay_max = models.PositiveIntegerField()

    def __unicode__(self):
        return self.level_name

class Employee(models.Model):
    emp_name = models.CharField(max_length = 100)
    department = models.ForeignKey(Department)
    level = models.ForeignKey(Level)
    reports_to = models.ForeignKey('self', null=True, blank=True)

    pay = models.PositiveIntegerField()
    joined_on = models.DateField()

class Leave(models.Model):
    employee = models.ForeignKey(Employee)
    leave_day = models.DateField()

#Populate DB, so we can do some meaningful queries.
#Create Dept, Levels manually.
#Get the names file from
#Or the whole sqlite database from
import random
from datetime import timedelta, date
import pickle
names = pickle.load(file('/home/shabda/names.pickle'))
for i in range(1000):
    emp = Employee() = random.choice(names)
    emp.department = random.choice(list(Department.objects.all()))
    emp.level = random.choice(Level.objects.all())
    try: emp.reports_to = random.choice(list(Employee.objects.filter(department=emp.department)))
    except:pass = random.randint(emp.level.pay_min, emp.level.pay_max)
    emp.joined_on = emp.department.established_on + timedelta(days = random.randint(0, 200))

employees = list(Employees.objects.all())
for i in range(100):
    employee = random.choice(employees)
    leave = Leave(employee = employee)
    leave.leave_day = - timedelta(days = random.randint(0, 365))


Find the total number of employees.

In sql you might want to do something like,

select count(id) from hrms_employee

Which becomes,


If fact doing a connection.queries.pop() shows the exact query.

SELECT COUNT("hrms_employee"."id") AS "total" FROM "hrms_employee"

But wait, we have a builtin method already for that, Employee.objects.all().count(), so lets try something else.

Find the total pay of employees.

The CEO wants to find out what is the total salary expediture, this also converts the queryset to a single value, so we want to .aggregate here.


Gives you the total amount you are paying to your employees.

Find the total number of employees, per department.

Here we want a value per row in queryset, so we need to use aggregate here. Also, there would be one aggregated value per dpeartment, so we need to annotate Department queryset.


If you are only interested in name of department and employee count for it, you can do, Department.objects.values('dept_name').annotate(Count('employee'))

The sql is

SELECT "hrms_department"."dept_name", COUNT("hrms_employee"."id") AS "employee__count" FROM "hrms_department" LEFT OUTER JOIN "hrms_employee" ON ("hrms_department"."id" = "hrms_employee"."department_id") GROUP BY "hrms_department"."dept_name"

Find the total number of employees, for a specific department.

Here you could use either of .annotate or .aggregate,


If you see the SQLs, you will see that .annotate did a group by, while the .aggregate did not, but as there was only one row, group by had no effect.

Find the total number of employees, per department, per level

This time, we can not annotate either Department model, or the Level model, as we need to group by both department and level. So we will annotate on Employee

Employee.objects.values('department__dept_name', 'level__level_name').annotate(Count('id'))

This leads to the sql,

SELECT "hrms_department"."dept_name", "hrms_level"."level_name", COUNT("hrms_employee"."id") AS "id__count" FROM "hrms_employee" INNER JOIN "hrms_department" ON ("hrms_employee"."department_id" = "hrms_department"."id") INNER JOIN "hrms_level" ON ("hrms_employee"."level_id" = "hrms_level"."id") GROUP BY "hrms_department"."dept_name", "hrms_level"."level_name

Which combination of Employee and Deparments employes the most people

We can order on the annotated fields, so the last query is modified to,

Employee.objects.values('department__dept_name', 'level__level_name').annotate(employee_count = Count('id')).order_by('-employee_count')[:1]

Which employee name is the most common.

We can want to group by emp_name, so emp_name is added to values. After that we order on the annotated field and get the first element, to get the most common name.


This was a overview of how django annotations work. These remove a whole class of queries for which you had to use custom sql queries in the past.


  1. Source on Github
  2. sqlite file for this model to test
  3. Aggregation on Django docs

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Brandon Konkle

Thank you very much for the detailed overview! I've been looking forward to aggregation because of some challenges in a project towards the beginning of this year. This is definitely a great new feature for Django.

Vitaly Babiy

Love your posts, Another great post.


@walter: Thanks very much. I would love to see more translations. All the content here is CC-BY-SA licensed. I should put up a notice for that somewhere.

Jeff 18th Aug., 2009

Thanks for the tutorial.

Steve Roger

Hi..Thanks for an article on Django.Presently I am working on a couple of Database projexts and really it was a tough task for me to perform aggregation.Now i found a solution for it..

Walter Cruz

Great, I was in doubt about the license ;)

Walter Cruz

IN your links, bitbucket is .org, don't .com ;)


@Walter: Thanks so much. Changed it now.

spooky 20th Aug., 2009

I have a question, what if I want to get an aggregate that is a result of an arithmetic operation on two columns?
eg. sql would be

select foo*bar from buz

do I have to have a custom sql or is there some way to do this using the api? (some sort of Buz.objects.aggregate('foo*bar') )

spooky 20th Aug., 2009

select sum(foo*bar) from buz
was the thing I wanted to write ;)


Great work - thanks for the post!

Danilo Cabello

In the last query, I think the Count should be with emp_name, not with id.

Great article!

martin 13th Nov., 2009

Hi Shabda, thanks for the great tutorial! I have one scenario that i can't seem to solve with one of the examples presented, i posted it on stackoverflow here. Any chance i could persuade you to have a look at it?

Django developer

Nice tutorial. I just bookmarked in my browser for my development.

joujou 12th Sept., 2013

thank you for your post!

I have the same model:

class Employee(models.Model):
emp_name = models.CharField(max_length = 100)
department = models.ForeignKey(Department)
level = models.ForeignKey(Level)
reports_to = models.ForeignKey('self', null=True, blank=True)

the prob is: in model form admin, the dropdown list of reports_to don't display list of employee name but "employee object, employee object"!!
how can i fix that?

Akshar Raaj 12th Sept., 2013

Guess you are missing __unicode__ on Employee.

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@apgwoz Employee.objects.values(’name’).annotate(Count(’id’)) -> select count(id) from emp group by name

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