sort list of tuples by reordering tuples

Question

Given a list of tuples to sort, python will sort them according to first element in tuple, then second element and so on.

>>> A
[(3, 2, 1), (0, 3, 0), (2, 1, 0), (2, 2, 3), (0, 3, 2), (2, 1, 1), (3, 3, 2), (3, 2, 0)]
>>> sorted(A)
[(0, 3, 0), (0, 3, 2), (2, 1, 0), (2, 1, 1), (2, 2, 3), (3, 2, 0), (3, 2, 1), (3, 3, 2)]

This works great. Now I want to sort them by third element, then first element and then the second element which I can do by either providing a key function or a cmp function.

>>> A
[(3, 2, 1), (0, 3, 0), (2, 1, 0), (2, 2, 3), (0, 3, 2), (2, 1, 1), (3, 3, 2), (3, 2, 0)]
>>> sorted(A, key = lambda x: (x[2], x[0], x[1]))
[(0, 3, 0), (2, 1, 0), (3, 2, 0), (2, 1, 1), (3, 2, 1), (0, 3, 2), (3, 3, 2), (2, 2, 3)]

except i take a major performance penalty

s ="""\
from numpy.random import randint as rr
A=[tuple(rr(0,10,3)) for i in range(100)]
def tuplecmp(t1, t2):
    return t1[0] - t2[0]
"""
c1 = """\
sorted(A)
"""
c2 = """\
sorted(A, key=lambda x: (x[2], x[0], x[1]))
"""
c3 = """\
sorted(A, cmp = tuplecmp)
"""
import timeit
print timeit.timeit(c1,number=10000, setup= s)
print timeit.timeit(c2,number=10000, setup= s)
print timeit.timeit(c3,number=10000, setup= s)

giving

0.60133600235,
0.980231046677,
2.68837809563

Further, the order in which I compare the individual tuple elements need not remain same. I may need to compare by 'second, first, then third' element etc. Is there a better method to do provide arbitrary comparator functions without a major performance penalty;

Answer 1

Using operator.itemgetter for your key function may be faster; you'd have to try it.

import operator
sorted(A, key=operator.itemgetter(2, 0, 1))

Answer 2

I'm not aware of a better way to do what you're trying to do. I believe you're always going to have to supply your own comparator, and then pay the cost of calling that over and over again.