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

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].

>>> from operator import itemgetter>>> data = [('abc', 121),('abc', 231),('abc', 148), ('abc',221)]>>> sorted(data,key=itemgetter(1))[('abc', 121), ('abc', 148), ('abc', 221), ('abc', 231)]

IMO using itemgetter is more readable in this case than the solution by @cheeken. It isalso faster since almost all of the computation will be done on the c side (no pun intended) rather than through the use of lambda.

>python -m timeit -s "from operator import itemgetter; data = [('abc', 121),('abc', 231),('abc', 148), ('abc',221)]" "sorted(data,key=itemgetter(1))"1000000 loops, best of 3: 1.22 usec per loop>python -m timeit -s "data = [('abc', 121),('abc', 231),('abc', 148), ('abc',221)]" "sorted(data,key=lambda x: x[1])"1000000 loops, best of 3: 1.4 usec per loop

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)