Cumulative count of duplicate values by certain axis in Numpy array

Question

Let's say I have this numpy array:

array([[4, 5, 6, 8, 5, 6],
       [5, 1, 1, 9, 0, 5],
       [7, 0, 5, 8, 0, 5],
       [9, 2, 3, 8, 2, 3],
       [1, 2, 2, 9, 2, 8]])

And going row by row, I would like to see, by column, the cumulative count of the number that appears. So for this array, the result would be:

array([[0, 0, 0, 0, 0, 0],  # (*0) 
       [0, 0, 0, 0, 0, 0],  # (*1) 
       [0, 0, 0, 1, 1, 1],  # (*2)
       [0, 0, 0, 2, 0, 0],  # (*3)
       [0, 1, 0, 1, 1, 0]]  # (*4)

(*0): first time each value appears
(*1): all values are different from the previous one (in the column)
(*2): For the last 3 columns, a 1 appears because there is already 1 value repetition.
(*3): For the 4th column, a 2 appears because it's the 3rd time that a 8 appears.
(*4): In the 4th column, a 1 appears because it's the 2nd time that a 9 appears in that column. Similarly, for the second and second to last column.

Any idea how to perform this?

Thanks!

Matias Eiletz · Accepted Answer · 2021-02-04 07:59:44Z

1

Maybe there is a faster way using numpy ufuncs, however here is a solution using standard python:

from collections import defaultdict
import numpy as np

a = np.array([[4, 5, 6, 8, 5, 6],
              [5, 1, 1, 9, 0, 5],
              [7, 0, 5, 8, 0, 5],
              [9, 2, 3, 8, 2, 3],
              [1, 2, 2, 9, 2, 8]])

# define function
def get_count(array):
    count = []
    for row in array.T:
        occurences = defaultdict(int)
        rowcount = []
        for n in row:
            occurences[n] += 1
            rowcount.append(occurences[n] - 1)
        count.append(rowcount)
    return np.array(count).T

Output:

>>> get_count(a)
array([[0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0],
       [0, 0, 0, 1, 1, 1],
       [0, 0, 0, 2, 0, 0],
       [0, 1, 0, 1, 1, 0]])

edited Feb 4, 2021 at 7:59

Matias Eiletz

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answered Feb 3, 2021 at 13:13

sunnytown

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