Remove empty elements with all zeros along a numpy 4D array using mask

Question

Given a sample numpy array like so:

a = np.array([[[[0,0,0], [0,0,0], [0,0,0]],
               [[0,0,0], [0,0,0], [0,0,0]]],
              [[[0,1,2], [1,1,1], [1,1,1]],
               [[1,1,1], [1,2,2], [1,1,1]]],
              [[[0,1,2], [1,1,1], [1,1,1]],
               [[1,1,1], [1,2,2], [1,1,1]]],
              [[[0,1,2], [1,1,1], [1,1,1]],
               [[1,1,1], [1,2,2], [1,1,1]]]])
#a.shape = (4, 2, 3, 3)

How can I get it to return a numpy array with shape (3,2,3,3) considering that the first element is all zeros? My dataset is a bigger one of shape (m, x, y, z) and I'll need to return non-zero (m-n, x,y,z) arrays where n are the (x,y,z) shaped arrays with all zeros.

So far I tried this:

mask = np.equal(a, np.zeros(shape=(2,3,3)))

'''
Returns:
        [[[[ True  True  True]
   [ True  True  True]
   [ True  True  True]]

  [[ True  True  True]
   [ True  True  True]
   [ True  True  True]]]


 [[[ True False False]
   [False False False]
   [False False False]]

  [[False False False]
   [False False False]
   [False False False]]]


 [[[ True False False]
   [False False False]
   [False False False]]

  [[False False False]
   [False False False]
   [False False False]]]


 [[[ True False False]
   [False False False]
   [False False False]]

  [[False False False]
   [False False False]
   [False False False]]]]
'''

But applying a[~mask] gives me a flattened array:

[1 2 1 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 2 1
 1 1 1 1 1 1 1 1 1 2 2 1 1 1] (51,)

What I need is something like this:

np.array([[[[0,1,2], [1,1,1], [1,1,1]],
           [[1,1,1], [1,2,2], [1,1,1]]],
          [[[0,1,2], [1,1,1], [1,1,1]],
           [[1,1,1], [1,2,2], [1,1,1]]],
          [[[0,1,2], [1,1,1], [1,1,1]],
           [[1,1,1], [1,2,2], [1,1,1]]]])

Bonus: I need to apply this to a separate/mirror (m, x, y, z) shaped array so maybe I'll need a masked approach?

Well a problem is that if there are for instance two sublists where we remove only for one sublist an element, then the two sublists no longer contain the same number of elements, which is a requirement in numpy. — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Feb 13, 2018 at 22:24

akuiper · Accepted Answer · 2018-02-13 22:27:07Z

2

Use all over axises other than the first axis to create the boolean array for indexing:

a[~(a == 0).all(axis=(1,2,3))]

#array([[[[0, 1, 2],
#         [1, 1, 1],
#         [1, 1, 1]],

#        [[1, 1, 1],
#         [1, 2, 2],
#         [1, 1, 1]]],


#       [[[0, 1, 2],
#         [1, 1, 1],
#         [1, 1, 1]],

#        [[1, 1, 1],
#         [1, 2, 2],
#         [1, 1, 1]]],


#       [[[0, 1, 2],
#         [1, 1, 1],
#         [1, 1, 1]],

#        [[1, 1, 1],
#         [1, 2, 2],
#         [1, 1, 1]]]])

answered Feb 13, 2018 at 22:27

akuiper

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2 Comments

weiji14 Over a year ago

Ah great, this does what I want! But I also need to generate a mask for applying it to another separate array of the same dimension (see updated question).

weiji14 Over a year ago

Nevermind, found out I can use mask = (a == 0).all(axis=(1,2,3)). Marking answer as accepted!

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