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.
Essentially, aggregations are nothing but a way to perform an operation on group of rows. In databases,
they are represented by operators as
To do these operations Django added two new methods to querysets.
When you are have a queryset you can do two operations on it,
- Operate over the rowset to get a single value from it. (Such as sum of all salaries in the rowset)
- 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 rowset_table.id. (Unless explicitly overriden).
Ok enough talk, on to some actual work. Here is a fictional models.py 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 http://dl.getdropbox.com/u/271935/djaggregations/names.pickle #Or the whole sqlite database from http://dl.getdropbox.com/u/271935/djaggregations/bata.db import random from datetime import timedelta, date import pickle names = pickle.load(file('/home/shabda/names.pickle')) for i in range(1000): emp = Employee() emp.name = 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 emp.pay = random.randint(emp.level.pay_min, emp.level.pay_max) emp.joined_on = emp.department.established_on + timedelta(days = random.randint(0, 200)) emp.save() """ """ employees = list(Employees.objects.all()) for i in range(100): employee = random.choice(employees) leave = Leave(employee = employee) leave.leave_day = date.today() - timedelta(days = random.randint(0, 365)) leave.save() """
Find the total number of employees.
In sql you might want to do something like,
select count(id) from hrms_employee
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
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,
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
If you see the SQLs, you will see that
.annotate did a
group by, while the
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
group by both department and level. So we will annotate on Employee
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.
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Thank you for reading the Agiliq blog. This article was written by shabda on Aug 18, 2009 in aggreagtion .
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