Deleting numpy array elements from a 3d numpy array with given array of indices

Question

I have a numpy array:

arr = array([[[ 0,  1,  2],
    [ 3,  4,  5],
    [ 6,  7,  8]],

   [[ 9, 10, 11],
    [12, 13, 14],
    [15, 16, 17]],

   [[18, 19, 20],
    [21, 22, 23],
    [24, 25, 26]]])

and an array of indices, ind = array([0, 1, 1])

What I would like to do is for the ith row in arr, delete the ind[i]th row in arr[i] using only numpy.delete.

So in essence a more pythonic way to do this:

x, y, z = arr.shape
new_arr = np.empty((x, y - 1, z))
for i, j in enumerate(ind):
    new_arr[i] = np.delete(arr[i], j, 0)

arr = new_arr.astype(int)

So the output here would be:

array([[[ 3,  4,  5],
        [ 6,  7,  8]],

       [[ 9, 10, 11],
        [15, 16, 17]],

       [[18, 19, 20],
        [24, 25, 26]]])

You probably need to create a new, smaller array, and then copy what you need. There is no way to delete what you need in a loop because the array is not fixed-sized. — Matthieu Brucher
– Matthieu Brucher, Commented Dec 20, 2018 at 17:16
So, the code I posted is the only way to do achieve this? Ah, rip, I was hoping numpy had some way to do this which was more optimized. — Spellstaker
– Spellstaker, Commented Dec 20, 2018 at 17:17
The code you posted delete the rows of a[i]'s, not the columns. Which exactly do you want to delete, rows or columns? — Quang Hoang
– Quang Hoang, Commented Dec 20, 2018 at 17:20
@QuangHoang, the array is 3d, so the names 'rows' and 'columns' are ambiguous. Here the OP is thinking of the 0 and 1 axes, you seem think of them as 1 and 2. — hpaulj
– hpaulj, Commented Dec 20, 2018 at 17:23

Spellstaker · Accepted Answer · 2018-12-20 17:42:57Z

4

A working solution:

import numpy as np

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

                [[9, 10, 11],
                 [12, 13, 14],
                 [15, 16, 17]],

                [[18, 19, 20],
                 [21, 22, 23],
                 [24, 25, 26]]])


a0, a1, a2 = arr.shape
indices = np.array([0, 1, 1])

mask = np.ones_like(arr, dtype=bool)
mask[np.arange(a0), indices, :] = False

result = arr[mask].reshape((a0, -1, a2))
print(result)

Output

[[[ 3  4  5]
  [ 6  7  8]]

 [[ 9 10 11]
  [15 16 17]]

 [[18 19 20]
  [24 25 26]]]

edited Dec 20, 2018 at 17:42

Spellstaker

618 bronze badges

answered Dec 20, 2018 at 17:31

Dani Mesejo

62.2k6 gold badges56 silver badges86 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

hpaulj Over a year ago

There's nothing hacky about that. np.delete does that sort of boolean masking, but doesn't generalize it to the 2d case as you did. I was slowly working toward this sort of solution. The last reshape can be reshape(a0, -1, a2).

Dani Mesejo Over a year ago

@hpaulj Thanks for your comments, updated the answer!

Collectives™ on Stack Overflow

Deleting numpy array elements from a 3d numpy array with given array of indices

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related