How to sort a list/tuple of lists/tuples by the element at a given index?

Stephen’s answer is the one I’d use. For completeness, here’s the DSU (decorate-sort-undecorate) pattern with list comprehensions:

decorated = [(tup[1], tup) for tup in data]
decorated.sort()
undecorated = [tup for second, tup in decorated]

Or, more tersely:

[b for a,b in sorted((tup[1], tup) for tup in data)]

As noted in the Python Sorting HowTo, this has been unnecessary since Python 2.4, when key functions became available.

from operator import itemgetter
data.sort(key=itemgetter(1))

Sorting a tuple is quite simple:

tuple(sorted(t))

I just want to add to Stephen’s answer if you want to sort the array from high to low, another way other than in the comments above is just to add this to the line:

reverse = True

and the result will be as follows:

data.sort(key=lambda tup: tup[1], reverse=True)

For sorting by multiple criteria, namely for instance by the second and third elements in a tuple, let

data = [(1,2,3),(1,2,1),(1,1,4)]

and so define a lambda that returns a tuple that describes priority, for instance

sorted(data, key=lambda tup: (tup[1],tup[2]) )
[(1, 1, 4), (1, 2, 1), (1, 2, 3)]

Without lambda:

def sec_elem(s):
    return s[1]

sorted(data, key=sec_elem)

itemgetter() is somewhat faster than lambda tup: tup[1], but the increase is relatively modest (around 10 to 25 percent).

(IPython session)

>>> from operator import itemgetter
>>> from numpy.random import randint
>>> values = randint(0, 9, 30000).reshape((10000,3))
>>> tpls = [tuple(values[i,:]) for i in range(len(values))]

>>> tpls[:5]    # display sample from list
[(1, 0, 0), 
 (8, 5, 5), 
 (5, 4, 0), 
 (5, 7, 7), 
 (4, 2, 1)]

>>> sorted(tpls[:5], key=itemgetter(1))    # example sort
[(1, 0, 0), 
 (4, 2, 1), 
 (5, 4, 0), 
 (8, 5, 5), 
 (5, 7, 7)]

>>> %timeit sorted(tpls, key=itemgetter(1))
100 loops, best of 3: 4.89 ms per loop

>>> %timeit sorted(tpls, key=lambda tup: tup[1])
100 loops, best of 3: 6.39 ms per loop

>>> %timeit sorted(tpls, key=(itemgetter(1,0)))
100 loops, best of 3: 16.1 ms per loop

>>> %timeit sorted(tpls, key=lambda tup: (tup[1], tup[0]))
100 loops, best of 3: 17.1 ms per loop

In order to sort a list of tuples (<word>, <count>), for count in descending order and word in alphabetical order:

data = [
('betty', 1),
('bought', 1),
('a', 1),
('bit', 1),
('of', 1),
('butter', 2),
('but', 1),
('the', 1),
('was', 1),
('bitter', 1)]

I use this method:

sorted(data, key=lambda tup:(-tup[1], tup[0]))

and it gives me the result:

[('butter', 2),
('a', 1),
('betty', 1),
('bit', 1),
('bitter', 1),
('bought', 1),
('but', 1),
('of', 1),
('the', 1),
('was', 1)]

@Stephen ‘s answer is to the point! Here is an example for better visualization,

Shout out for the Ready Player One fans! =)

>>> gunters = [('2044-04-05', 'parzival'), ('2044-04-07', 'aech'), ('2044-04-06', 'art3mis')]
>>> gunters.sort(key=lambda tup: tup[0])
>>> print gunters
[('2044-04-05', 'parzival'), ('2044-04-06', 'art3mis'), ('2044-04-07', 'aech')]

key is a function that will be called to transform the collection’s items for comparison.. like compareTo method in Java.

The parameter passed to key must be something that is callable. Here, the use of lambda creates an anonymous function (which is a callable).
The syntax of lambda is the word lambda followed by a iterable name then a single block of code.

Below example, we are sorting a list of tuple that holds the info abt time of certain event and actor name.

We are sorting this list by time of event occurrence – which is the 0th element of a tuple.

Note – s.sort([cmp[, key[, reverse]]]) sorts the items of s in place

I use this in my code:

#To sort the list based on each element's second integer (elem[1])
sorted(d2, key=lambda elem: elem[1])

Depending on which element you want to sort it by you can put it in the

(elem[*insert the index of the element you are sorting it by*])