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

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)]

Answer 1

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)]

Answer 2

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 3

Sorting a tuple is quite simple:

tuple(sorted(t))

Answer 4

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

or:

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