Count of Each Element in 2d NumPy Array [duplicate]

Question

Imagine you have a 2D-array (as a NumPy int array) like:

[[2,2,3,3],
 [2,3,3,3],
 [3,3,4,4]]

Now you want to get an array of the same shape, but instead of the original values, you want to replace the number by its occurrences. Which means, the number 2 changes to 3, since it occurred 3 times, the 3s become 7s and the 4s become 2s.

So the output would be:

[[3,3,7,7],
 [3,7,7,7],
 [7,7,2,2]]

My solution was first to create a dictionary, which saves all original values as keys and as values the number of occurrences. But for arrays of shape 2000x2000, this seemed to be quite slow.

How could I achieve this more efficiently?

Thanks!

Answer 1

I believe you should be able to stay in NumPy here by using return_inverse within np.unique():

If True, also return the indices of the unique array (for the specified axis, if provided) that can be used to reconstruct ar.

>>> import numpy as np

>>> a = np.array([[2,2,3,3],
...               [2,3,3,3],
...               [3,3,4,4]])

>>> _, inv, cts = np.unique(a, return_inverse=True, return_counts=True)
>>> cts[inv].reshape(a.shape)

array([[3, 3, 7, 7],
       [3, 7, 7, 7],
       [7, 7, 2, 2]])

This will also work for the case where the flattened array is not sorted, such as b = np.array([[1, 2, 4], [4, 4, 1]]).