python: sort a list of lists by an item in the sublist [duplicate]

Answer 1

l = [["user1", 100], ["user2", 234], ["user3", 131]]
l.sort(lambda x, y: x[1])

Answer 2

>>> li = [["user1", 100], ["user2", 234], ["user3", 131]]
>>> 
>>> import operator
>>> 
>>> sorted(li, key=operator.itemgetter(1))   # Ascending order
[['user1', 100], ['user3', 131], ['user2', 234]]

>>> sorted(li, key=operator.itemgetter(1), reverse=True)  # Reverse Sort
[['user2', 234], ['user3', 131], ['user1', 100]]

Answer 3

sorted accept optional key function and reverse parameter.

>>> sorted([["user1", 100], ["user2", 234], ["user3", 131]], key=lambda x: x[1], reverse=True)
[['user2', 234], ['user3', 131], ['user1', 100]]

Answer 4

You can custom the sort behaviour by pass a key and reverse. sorted will return a new list. If in-place sort wanted, use list.sort.

>>> a = [["user1", 100], ["user2", 234], ["user3", 131]]
>>> sorted(a, key=lambda x: x[1], reverse=True)
[['user2', 234], ['user3', 131], ['user1', 100]]