How do I 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:

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

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

Answer 1

@Jose M Vidal's answer mentions something important: using rich comparisons (__lt__, __eq__ etc. as in @jpp's answer) makes sorting much slower than passing a key function (as in the accepted answer). However, they end up showing a test using __cmp__ method which doesn't exist in Python 3 (which tbf was indeed so much slower than passing key in Python 2).

I want to point out that even in Python 3.12.0, using rich comparisons as in @jpp's answer makes sorting much slower than passing a key function.

The results of a little timeit test is shown below where sorting using rich comparisons vs a lambda key function vs operator.attrgetter as key are compared. For a list with 10k items, it took about 18.8 ms when rich comparisons were used whereas it took 2.93 ms when a lambda key function was used and 2.47 ms when operator.attrgetter was used.

So, as @tzot mentioned, operator.attrgetter is faster than a lambda; however, using a key function in the first place instead of rich comparisons makes sorting over 5 times faster.

import timeit
import random
from operator import attrgetter

class Card(object):

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

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

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

n = 100
random.seed(0)
lst = [Card(random.randrange(10000)) for _ in range(10000)]


min(timeit.repeat(lambda: sorted(lst), number=n))/n
# 0.018813106999732553

min(timeit.repeat(lambda: sorted(lst, key=lambda card: card.rank), number=n))/n
# 0.0029304770001908763

min(timeit.repeat(lambda: sorted(lst, key=attrgetter('rank')), number=n))/n
# 0.00247172600007616

Answer 2

To sort the list in place:

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

To return a new list, use sorted:

new_list = sorted(orig_list, key=lambda x: x.count, reverse=True)

Explanation:

• key=lambda x: x.count sorts by count.
• reverse=True sorts in descending order.

More on sorting by keys.

Answer 3

Also if someone wants to sort list that contains strings and numbers for e.g.

 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)

Answer 4

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 5

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 6

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 7

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 8

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

Answer 9

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 10

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.