Sort a list of tuples by 2nd item (integer value) [duplicate]

Answer 1

Try using the key keyword with sorted().

sorted([('abc', 121),('abc', 231),('abc', 148), ('abc',221)], key=lambda x: x[1])

key should be a function that identifies how to retrieve the comparable element from your data structure. In your case, it is the second element of the tuple, so we access [1].

For optimization, see jamylak's response using itemgetter(1), which is essentially a faster version of lambda x: x[1].

Answer 2

From python wiki:

>>> from operator import itemgetter, attrgetter    
>>> sorted(student_tuples, key=itemgetter(2))
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]    
>>> sorted(student_objects, key=attrgetter('age'))
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]

Answer 3

For an in-place sort, use

foo = [(list of tuples)]
foo.sort(key=lambda x:x[0]) #To sort by first element of the tuple

Answer 4

Adding to Cheeken's answer, This is how you sort a list of tuples by the 2nd item in descending order.

sorted([('abc', 121),('abc', 231),('abc', 148), ('abc',221)],key=lambda x: x[1], reverse=True)

Answer 5

The fact that the sort values in the OP are integers isn't relevant to the question per se. In other words, the accepted answer would work if the sort value was text. I bring this up to also point out that the sort can be modified during the sort (for example, to account for upper and lower case).

>>> sorted([(121, 'abc'), (231, 'def'), (148, 'ABC'), (221, 'DEF')], key=lambda x: x[1])
[(148, 'ABC'), (221, 'DEF'), (121, 'abc'), (231, 'def')]
>>> sorted([(121, 'abc'), (231, 'def'), (148, 'ABC'), (221, 'DEF')], key=lambda x: str.lower(x[1]))
[(121, 'abc'), (148, 'ABC'), (231, 'def'), (221, 'DEF')]

Answer 6

For Python 2.7+, this works which makes the accepted answer slightly more readable:

sorted([('abc', 121),('abc', 231),('abc', 148), ('abc',221)], key=lambda (k, val): val)