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
[,
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
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]