How to sort a list of objects based on an attribute of the objects?

Question

I\'ve got a list of Python objects that I\'d like to sort by an attribute of the objects themselves. The list looks like:

>>> ut
[,         


        

Answer 1

Add rich comparison operators to the object class, then use sort() method of the list.
See rich comparison in python.

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Update: Although this method would work, I think solution from Triptych is better suited to your case because way simpler.

Answer 2

from operator import attrgetter
ut.sort(key = attrgetter('count'), reverse = True)

Answer 3

If the attribute you want to sort by is a property, then you can avoid importing operator.attrgetter and use the property's fget method instead.

For example, for a class Circle with a property radius we could sort a list of circles by radii as follows:

result = sorted(circles, key=Circle.radius.fget)

This is not the most well-known feature but often saves me a line with the import.

Answer 4

Readers should notice that the key= method:

ut.sort(key=lambda x: x.count, reverse=True)

is many times faster than adding rich comparison operators to the objects. I was surprised to read this (page 485 of "Python in a Nutshell"). You can confirm this by running tests on this little program:

#!/usr/bin/env python
import random

class C:
    def __init__(self,count):
        self.count = count

    def __cmp__(self,other):
        return cmp(self.count,other.count)

longList = [C(random.random()) for i in xrange(1000000)] #about 6.1 secs
longList2 = longList[:]

longList.sort() #about 52 - 6.1 = 46 secs
longList2.sort(key = lambda c: c.count) #about 9 - 6.1 = 3 secs

My, very minimal, tests show the first sort is more than 10 times slower, but the book says it is only about 5 times slower in general. The reason they say is due to the highly optimizes sort algorithm used in python (timsort).

Still, its very odd that .sort(lambda) is faster than plain old .sort(). I hope they fix that.

Answer 5

It looks much like a list of Django ORM model instances.

Why not sort them on query like this:

ut = Tag.objects.order_by('-count')

Answer 6

# To sort the list in place...
ut.sort(key=lambda x: x.count, reverse=True)

# To return a new list, use the sorted() built-in function...
newlist = sorted(ut, key=lambda x: x.count, reverse=True)

More on sorting by keys.