NumPy - Find most common value in array, use largest value in case of a tie

Question

There are so many questions around that deal with finding the most common value in an array, but all of them return the "first" element in case of a tie. I need the highest value from the list of tied elements, imagine something like this:

import numpy as np

my_array = [1, 1, 3, 3]
most_common = np.bincount(my_array).argmax()

This gives me 1, which is obviously not wrong, but I have a tied result here and I want to decide on my own what I want to do in that case. For my application, in case of such a tie I want the highest value from my_array, which is 3. How can I do that?

PS: need a Python 2.7 answer... sorry, but can't change that at the moment

Warren Weckesser · Accepted Answer · 2021-06-15 14:34:16Z

1

You could apply argmax to the reversed output of bincount, and then adjust to take into account the reversal:

In [73]: x
Out[73]: array([3, 0, 2, 3, 1, 0, 1, 3, 2, 1, 1, 2, 1, 3, 3, 4])

In [74]: b = np.bincount(x)

In [75]: b
Out[75]: array([2, 5, 3, 5, 1])

In [76]: most_common = len(b) - 1 - b[::-1].argmax()

In [77]: most_common
Out[77]: 3

answered Jun 15, 2021 at 14:34

Warren Weckesser

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s6hebern Over a year ago

This one works also for arrays without ties and still gives the correct result in any case. So I'll go with this one.

Will Da Silva · Accepted Answer · 2021-06-16 12:59:03Z

1

Not sure how you'd go about solving this with Numpy, as there is no way to alter the tie-breaking logic of argmax, but you can do it with collections.Counter easily:

from collections import Counter

my_array = [1, 1, 3, 3]
counter = Counter(my_array)
most_common, num_occurances = max(counter.most_common(), key=lambda x: (x[1], x[0]))

edited Jun 16, 2021 at 12:59

answered Jun 15, 2021 at 14:30

Will Da Silva

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s6hebern Over a year ago

After the initial euphoria, I just found that this only works in such a very specific case and does not really give me a correct result in case of non-ties. For example, if my_array = [1, 1, 3, 3, 3, 4], the result is 4, which is absolutely wrong.

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