How to sort (list/tuple) of lists/tuples?

Question

I have some data either in a list of lists or a list of tuples, like this:

data = [[1,2,3], [4,5,6], [7,8,9]]
data = [(1,2,3), (4,5,6), (7,8,9)]

And I want to sort by the 2nd element in the subset. Meaning, sorting by 2,5,8 where 2 is from (1,2,3), 5 is from (4,5,6). What is the common way to do this? Should I store tuples or lists in my list?

Answer 1

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

Answer 2

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

or:

data.sort(key=lambda tup: tup[1])  # sorts in place

Answer 3

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.

Answer 4

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

Answer 5

Without lambda:

def sec_elem(s):
    return s[1] 

sorted(data, key=sec_elem) 

Answer 6

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)