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

Question

I have a list of tuples that looks something like this:

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

I want to sort this list in ascending order by the integer value inside the tuples. Is it possible?

Answer 1

Try using the key keyword argument of sorted(), which sorts in increasing order by default:

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 operator.itemgetter(1), which is essentially a faster version of lambda x: x[1].

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

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 4

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 5

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 6

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 7

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)

Answer 8

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)

Answer 9

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')]