python object list, sort in descending order based on object attribute

Question

I have a list of objects, which I need to sort by one of the objects attributes.

I can sort in ascending order with the following code

list1 = sorted(lis         


        

Answer 1

Specify reverse=True argument:

list1 = sorted(list1, key=lambda object1: object1.fitness, reverse=True)

Demo (simple list of integers):

>>> l = [6, 0, 2, 3, 1, 5, 4]
>>> sorted(l)
[0, 1, 2, 3, 4, 5, 6]
>>> sorted(l, reverse=True)
[6, 5, 4, 3, 2, 1, 0]

Demo (datetime.dates, using operator.attrgetter instead of lambda as @SethMMorton suggested):

>>> from datetime import date
>>> from operator import attrgetter
>>> l = [date(2014, 4, 11), date(2014, 4, 2), date(2014, 4, 3), date(2014, 4, 8)]

>>> sorted(l, key=attrgetter('day'))
[datetime.date(2014, 4, 2), 
 datetime.date(2014, 4, 3), 
 datetime.date(2014, 4, 8), 
 datetime.date(2014, 4, 11)]
>>> sorted(l, key=attrgetter('day'), reverse=True)
[datetime.date(2014, 4, 11), 
 datetime.date(2014, 4, 8), 
 datetime.date(2014, 4, 3), 
 datetime.date(2014, 4, 2)]

Answer 2

See this link to craete your own comparator. You can even compare by multiple attributes or complex logic.

It works as this: cmp_to_key expects a function giving key = cmp_to_key(obj) where obj is an object to be sorted. Finally an object implementing comparator operators is returned, and can be used so you can do any custom comparisson logic.