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

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

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?

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 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
``````

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)
``````

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.

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
``````

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)]
``````

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)
``````

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)