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Filter by property

Posted by: admin November 30, 2017 Leave a comment


Is it possible to filter by property?

i have a method in my model:

def myproperty(self):

and now i want to filter by this property like:


is this somehow possible?


Nope. Django filters operate at the database level, generating SQL. To filter based on Python properties, you have to load the object into Python to evaluate the property–and at that point, you’ve already done all the work to load it.


I might be misunderstanding your original question, but there is a filter builtin in python.

filtered = filter(myproperty, MyModel.objects)

But it’s better to use a list comprehension:

filtered = [x for x in MyModel.objects if x.myproperty()]

or even better, a generator expression:

filtered = (x for x in MyModel.objects if x.myproperty())


Looks like using F() with annotations will be my solution to this.

It’s not going to filter by @property, since F talks to the databse before objects are brought into python. But still putting it here as an answer since my reason for wanting filter by property was really wanting to filter objects by the result of simple arithmetic on two different fields.

so, something along the lines of:

companies = Company.objects\
    .annotate(chairs_needed=F('num_employees') - F('num_chairs'))\

rather than defining the property to be:

def chairs_needed(self):
    return self.num_employees - self.num_chairs

then doing a list comprehension across all objects.


Riffing off @TheGrimmScientist’s suggested workaround, you can make these “sql properties” by defining them on the Manager or the QuerySet, and reuse/chain/compose them:

With a Manager:

class CompanyManager(models.Manager):
    def with_chairs_needed(self):
        return self.annotate(chairs_needed=F('num_employees') - F('num_chairs'))

class Company(models.Model):
    # ...
    objects = CompanyManager()


With a QuerySet:

class CompanyQuerySet(models.QuerySet):
    def many_employees(self, n=50):
        return self.filter(num_employees__gte=n)

    def needs_fewer_chairs_than(self, n=5):
        return self.with_chairs_needed().filter(chairs_needed__lt=n)

    def with_chairs_needed(self):
        return self.annotate(chairs_needed=F('num_employees') - F('num_chairs'))

class Company(models.Model):
    # ...
    objects = CompanyQuerySet.as_manager()


See https://docs.djangoproject.com/en/1.9/topics/db/managers/ for more.
Note that I am going off the documentation and have not tested the above.


PLEASE someone correct me, but I guess I have found a solution, at least for my own case.

I want to work on all those elements whose properties are exactly equal to … whatever.

But I have several models, and this routine should work for all models. And it does:

def selectByProperties(modelType, specify):
    clause = "SELECT * from %s" % modelType._meta.db_table

    if len(specify) > 0:
        clause += " WHERE "
        for field, eqvalue in specify.items():
            clause += "%s = '%s' AND " % (field, eqvalue)
        clause = clause [:-5]  # remove last AND

    print clause
    return modelType.objects.raw(clause)

With this universal subroutine, I can select all those elements which exactly equal my dictionary of ‘specify’ (propertyname,propertyvalue) combinations.

The first parameter takes a (models.Model),

the second a dictionary like:
{“property1” : “77” , “property2” : “12”}

And it creates an SQL statement like

SELECT * from appname_modelname WHERE property1 = '77' AND property2 = '12'

and returns a QuerySet on those elements.

This is a test function:

from myApp.models import myModel

def testSelectByProperties ():

    specify = {"property1" : "77" , "property2" : "12"}
    subset = selectByProperties(myModel, specify)

    nameField = "property0"
    ## checking if that is what I expected:
    for i in subset:
        print i.__dict__[nameField], 
        for j in specify.keys():
             print i.__dict__[j], 

And? What do you think?


i know it is an old question, but for the sake of those jumping here i think it is useful to read the question below and the relative answer:

How to customize admin filter in Django 1.4