Python collections.Counter: most_common complexity

问题:

I'm wondering what is the complexity of the function most_common provided by the collections.Counter object in python 2.7.

More specifically, is the Counter keeping some kind of sorted list while it is being updated, allowing to perform the most_common operation faster than O(n) when n is the number of (unique) items added to the counter? For you information, I am processing some large amount of text data trying to find the n-th most frequent tokens.

I checked the official documentation (https://docs.python.org/2/library/collections.html#collections.Counter), the CPython wiki (https://wiki.python.org/moin/TimeComplexity) but I could not find the answer. Thank you in advance!

Romain.

回答1:

From the source code of collections.py, we see that if we don't specify a number of returned elements, most_common returns a sorted list of the counts. This is an O(n log n) algorithm.

If we use most_common to return k > 1 elements, then we use nlargest method of heapq. This is an O(k) + O((n - k) log k) + O(k log k) algorithm, which is very good for a small constant k, since it's essentialy linear. The O(k) part comes from heapifying the initial k counts, the second part from n - k calls to heappushpop method and the third part from sorting the final heap of k elements. Since k <= n we can conclude that the complexity is:

O(n log k)

If k = 1 then it's easy to show that the complexity is:

O(n)

回答2:

The source show exactly what happens:

def most_common(self, n=None):
    '''List the n most common elements and their counts from the most
    common to the least.  If n is None, then list all element counts.

    >>> Counter('abracadabra').most_common(3)
    [('a', 5), ('r', 2), ('b', 2)]

    '''
    # Emulate Bag.sortedByCount from Smalltalk
    if n is None:
        return sorted(self.iteritems(), key=_itemgetter(1), reverse=True)
    return _heapq.nlargest(n, self.iteritems(), key=_itemgetter(1))