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

# 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)``