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?
# 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.
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.
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')
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
from operator import attrgetter ut.sort(key = attrgetter('count'), reverse = True)
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.
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
__lt__ operations for this to work. Then just use
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]
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.