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

Question

I have a list of Python objects that I want to sort by a specific attribute of each object:

>>> ut
[Tag(name="toe", count=10), Tag(name="leg", count=2), ...]

How do I sort the list by .count in descending order?

Answer 1

# 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 .

Answer 2

A way that can be fastest, especially if your list has a lot of records, is to use operator.attrgetter("count") . However, this might run on an pre-operator version of Python, so it would be nice to have a fallback mechanism. You might want to do the following, then:

try: import operator
except ImportError: keyfun= lambda x: x.count # use a lambda if no operator module
else: keyfun= operator.attrgetter("count") # use operator since it's faster than lambda

ut.sort(key=keyfun, reverse=True) # sort in-place

Answer 3

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 4

Object-oriented approach

It's good practice to make object sorting logic, if applicable, a property of the class rather than incorporated in each instance the ordering is required.

This ensures consistency and removes the need for boilerplate code.

At a minimum, you should specify __eq__ and __lt__ operations for this to work. Then just use sorted(list_of_objects) .

class Card(object):

    def __init__(self, rank, suit):
        self.rank = rank
        self.suit = suit

    def __eq__(self, other):
        return self.rank == other.rank and self.suit == other.suit

    def __lt__(self, other):
        return self.rank < other.rank

hand = [Card(10, 'H'), Card(2, 'h'), Card(12, 'h'), Card(13, 'h'), Card(14, 'h')]
hand_order = [c.rank for c in hand]  # [10, 2, 12, 13, 14]

hand_sorted = sorted(hand)
hand_sorted_order = [c.rank for c in hand_sorted]  # [2, 10, 12, 13, 14]

Answer 5

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

Answer 6

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 7

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

---

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

Answer 8

If the attribute you want to sort by is a property , then you can avoid importing operator.attrgetter and use the property'sfget 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 9

Also if someone wants to sort list that contains strings and numbers for eg

 eglist=[
     "some0thing3",
     "some0thing2",
     "some1thing2",
     "some1thing0",
     "some3thing10",
     "some3thing2",
     "some1thing1",
     "some0thing1"]

Then here is the code for that:

import re

def atoi(text):
    return int(text) if text.isdigit() else text

def natural_keys(text):
    return [ atoi(c) for c in re.split(r'(\d+)', text) ]

eglist=[
         "some0thing3",
         "some0thing2",
         "some1thing2",
         "some1thing0",
         "some3thing10",
         "some3thing2",
         "some1thing1",
         "some0thing1"
]

eglist.sort(key=natural_keys)
print(eglist)