Recipe 5.10. Selecting the nth Smallest Element of a Sequence

Solution
Perhaps you can do better, if the sequence is big, has been shuffled enough, and comparisons between its items are costly. Sort is very fast, but in the end (when applied to a thoroughly shuffled sequence of length n) it always takes O(n log n) time, while there exist algorithms that can be used to get the nth smallest element in time O(n). Here is a function with a solid implementation of such an algorithm:

import random
def select(data, n):
    " Find the nth rank ordered element (the least value has rank 0). "
    # make a new list, deal with <0 indices, check for valid index
    data = list(data)
    if n<0:
        n += len(data)
    if not 0 <= n < len(data):
        raise ValueError, "can't get rank %d out of %d" % (n, len(data))
    # main loop, quicksort-like but with no need for recursion
    while True:
        pivot = random.choice(data)
        pcount = 0
        under, over = [  ], [  ]
        uappend, oappend = under.append, over.append
        for elem in data:
            if elem < pivot:
                uappend(elem)
            elif elem > pivot:
                oappend(elem)
            else:
                pcount += 1
        numunder = len(under)
        if n < numunder:
            data = under
        elif n < numunder + pcount:
            return pivot
        else:
            data = over
            n -= numunder + pcount