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


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?

Asked By: Amyth



>>> 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
Answered By: jamylak

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].

Answered By: cheeken

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)]
Answered By: Dmitry Zagorulkin

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.

Answered By: Angus

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)
Answered By: Vignesh Jayavel

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

def MyFn(a):
    return a[1]


sorted([('abc', 121),('abc', 231),('abc', 148), ('abc',221)], key=MyFn)
Answered By: paulm

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)
Answered By: Neil

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')]
Answered By: DaveL17

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
Answered By: Shivank Tibrewal
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