python - Counting occurrences without using collections.Counter

Question

I am trying to retrieve the most frequent and less frequent elements in a list.

frequency([13,12,11,13,14,13,7,11,13,14,12,14,14])

My output is:

([7], [13, 14])

I tried it with:

import collections
s = [13,12,11,13,14,13,7,11,13,14,12,14,14]
count = collections.Counter(s)
mins = [a for a, b in count.items() if b == min(count.values())]
maxes = [a for a, b in count.items() if b == max(count.values())]
final_vals = [mins, maxes]

But I don't want to use the collections module and try a more logic oriented solution.
Can you please help me to do it without collections?

Answer 1

You could use a try and except approach with a dict to emulate a Counter.

def counter(it):
    counts = {}
    for item in it:
        try:
            counts[item] += 1
        except KeyError:
            counts[item] = 1
    return counts

or alternativly you can use dict.get with a default of 0:

def counter(it):
    counts = {}
    for item in it:
        counts[item] = counts.get(item, 0) + 1
    return counts

And you should do the min() and max() outside the comprehensions to avoid repeatedly calculating that quantity (the function is now O(n) instead of O(n^2):

def minimum_and_maximum_frequency(cnts):
    min_ = min(cnts.values())
    max_ = max(cnts.values())
    min_items = [k for k, cnt in cnts.items() if cnt == min_]
    max_items = [k for k, cnt in cnts.items() if cnt == max_]
    return min_items, max_items

This would work as expected then:

>>> minimum_and_maximum_frequency(counter([13,12,11,13,14,13,7,11,13,14,12,14,14]))
([7], [13, 14])