Python: Counting maximum occurences of an item in a list with ties allowed

Question

Right now I am using the classic counting of occurences in a list max(lst,key=lst.count) but if there is a tie in the maximum occurences of two different elements eg lst = [1,1,2,2] , max returns only one element. In our example it will return 1 or 2 . But I need a function which will return both 1 and 2 .

Answer 1

Build a Counter from the list and take the items that match the highest count using a list comprehension:

from collections import Counter

lst = [1,1,2,2]

c = Counter(lst)
maximums = [x for x in c if c[x] == c.most_common(1)[0][1]]
print(maximums)
# [1, 2]

The Counter approach take counts in one pass (O(n)) whereas the list.count approach has O(n^2) time complexity as it passes through the list with each call.

Answer 2

This may not be the most elegant solution but it definitely works. First you find the number of the maximum occurrence, then find the elements with that count.

lst = [1,1,2,2,3,4]
max_count = max([lst.count(i) for i in set(lst)])
max_count_values = [i for i in set(lst) if lst.count(i)==max_count]

Answer 3

Good question, as finding the single most common number in a list is trivial with scipy.stats.mode , but there is nothing quite so simple if you want to find all of the different modes.

Calling list.count for each value is needlessly expensive, as this uses O(n^2) time, but computing the counts of each element only requires O(n) time.

Here is a solution using NumPy:

>>> import numpy as np
>>> lst = [1,1,2,2,3]
>>> values, counts = np.unique(lst, return_counts=True)
>>> values[np.argwhere(counts == counts.max())].flatten()
array([1, 2])

Answer 4

A one-liner, but perhaps not the clearest.

set(x for x in lst if lst.count(x)==max(lst.count(x) for x in lst))