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

Answer 1

As a python neophyte, I just wanted to mention that if the data did actually look like this:

data = [('abc', 121),('abc', 231),('abc', 148), ('abc',221)]

then sorted() would automatically sort by the second element in the tuple, as the first elements are all identical.

Answer 2

>>> 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 is also 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

Answer 3

For a lambda-avoiding method, first define your own function:

def MyFn(a):
    return a[1]

then:

sorted([('abc', 121),('abc', 231),('abc', 148), ('abc',221)], key=MyFn)