Use numpy.delete(), which returns a new array with sub-arrays along an axis deleted.

numpy.delete(a, index)

For your specific question:

import numpy as np

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

new_a = np.delete(a, index)

print(new_a)
# Output: [1, 2, 5, 6, 8, 9]

Note that numpy.delete() returns a new array since array scalars are immutable, similar to strings in Python, so each time a change is made to it, a new object is created. I.e., to quote the delete() docs:

"A copy of arr with the elements specified by obj removed. Note that delete does not occur in-place..."

If the code I post has output, it is the result of running the code.

In this particular case you can use slicing:

after = before[:,::2]

Note, this creates a new np.ndarray object that is a view over the original buffer.

A more general approach might be something like:

after = np.delete(before, [1, 3], axis=1)

This shouldn't be a view over the original buffer.

your code works for me..but as an alternative try defining the indices as a np.array themselves:

indexes = np.array([0,1,2])
x = np.delete(descritoresFaciais, indexes, axis=0)

if for some weird reason it's still not working..take the set difference between all rows and rows to delete, and return the rest of the array:

indexes = np.array([0,1,2])
allrows = np.array(range(len(descritoresFaciais)))
x = d[np.setdiff1d(allrows,indexes)]

I get the same answer your way and these 2 ways, so my gut says something weird happened when you defined your array or something changed before you tried to modify it