Numpy Remove Rows and Columns from 3d Array

Question

Lets say you have a 5x5x5 numpy array

    a = np.ones((5,5,5))
    a[:,3,:] = 0
    a[:,:,3] = 0

(I know it is ugly)

This returns

    [[[1. 1. 1. 0. 1.]
  [1. 1. 1. 0. 1.]
  [1. 1. 1. 0. 1.]
  [0. 0. 0. 0. 0.]
  [1. 1. 1. 0. 1.]]

 [[1. 1. 1. 0. 1.]
  [1. 1. 1. 0. 1.]
  [1. 1. 1. 0. 1.]
  [0. 0. 0. 0. 0.]
  [1. 1. 1. 0. 1.]]

 [[1. 1. 1. 0. 1.]
  [1. 1. 1. 0. 1.]
  [1. 1. 1. 0. 1.]
  [0. 0. 0. 0. 0.]
  [1. 1. 1. 0. 1.]]

 [[1. 1. 1. 0. 1.]
  [1. 1. 1. 0. 1.]
  [1. 1. 1. 0. 1.]
  [0. 0. 0. 0. 0.]
  [1. 1. 1. 0. 1.]]

 [[1. 1. 1. 0. 1.]
  [1. 1. 1. 0. 1.]
  [1. 1. 1. 0. 1.]
  [0. 0. 0. 0. 0.]
  [1. 1. 1. 0. 1.]]]

What i want to do is to remove all rows and columns on all axis that is only 0 returning a new 4x4x4 array with only 1s in it.

I can do this for a 2 dimensional array with

a = np.delete(a,np.where(~a.any(axis=0))[0], axis=1)
a = a[~np.all(a == 0, axis=1)]

But can't figure how to do it with 3 dimensions

Anyone have an idea how that can be done?

Answer 1

You can find the indices of rows with all zero items separately for second and third axes and then remove them using np.delete :

In [25]: mask = (a == 0)

In [26]: sec = np.where(mask.all(1))[1]

In [27]: third = np.where(mask.all(2))[1]

In [28]: new = np.delete(np.delete(a, sec[1], 1), third, 2)

Note that instead of creating a new array you can reassign the result to a if you intended to do so.