Remove 2d slice from 3d numpy array

Question

I need to remove the last arrays from a 3D numpy cube. I have:

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

[[9,8,7],
 [6,5,4],
 [3,2,1]],

[[0,0,0],
 [0,0,0],
 [0,0,0]],

[[0,0,0],
 [0,0,0],
 [0,0,0]]])

How do I remove the arrays with zero sub-arrays like at the bottom side of the cube, using np.delete?

(I cannot simply remove all zero values, because there will be zeros in the data on the top side)

Did either of the posted solutions work for you? Any feedback on the solutions? — Divakar, Oct 08 '19 at 20:12

rafaelc · Answer 1 · 2019-10-07T19:55:54.830

4

For a 3D cube, you might check all against the last two axes

a = np.asarray(a)
a[~(a==0).all((2,1))]

array([[[1, 2, 3],
        [4, 5, 6],
        [7, 8, 9]],

       [[9, 8, 7],
        [6, 5, 4],
        [3, 2, 1]]])

edited Oct 07 '19 at 19:55

answered Oct 07 '19 at 19:16

rafaelc

57,686
15
58
82

1

If one is specifically looking for zeros, one could even do `a[a.any((2, 1))]`, since `0` is considered equivalent to `False`. Makes the code a bit less readable for a negligible speedup though. – Daniel F Oct 08 '19 at 06:08

Matt Hall · Answer 2 · 2019-10-07T19:28:01.617

2

If you know where they are already, the easiest thing to do is slice them off:

a[:-2]

Results in:

array([[[1, 2, 3],
        [4, 5, 6],
        [7, 8, 9]],

       [[9, 8, 7],
        [6, 5, 4],
        [3, 2, 1]]])

edited Oct 07 '19 at 19:28

answered Oct 07 '19 at 19:21

Matt Hall

7,614
1
23
36

score 2 · Answer 3 · answered Oct 07 '19 at 19:21

Here's one way to remove trailing all zeros slices, as mentioned in the question that we want to keep the all zeros slices in the data on the top side -

a[:-(a==0).all((1,2))[::-1].argmin()]

Sample run -

In [80]: a
Out[80]: 
array([[[0, 0, 0],
        [0, 0, 0],
        [0, 0, 0]],

       [[9, 8, 7],
        [6, 5, 4],
        [3, 2, 1]],

       [[0, 0, 0],
        [0, 0, 0],
        [0, 0, 0]],

       [[0, 0, 0],
        [0, 0, 0],
        [0, 0, 0]]])

In [81]: a[:-(a==0).all((1,2))[::-1].argmin()]
Out[81]: 
array([[[0, 0, 0],
        [0, 0, 0],
        [0, 0, 0]],

       [[9, 8, 7],
        [6, 5, 4],
        [3, 2, 1]]])

score 0 · Answer 4 · answered Oct 07 '19 at 19:16

Hope this helps,

a_new=[] #Create a empty list
for item in a:
        if not (np.count_nonzero(item) == 0): #check if inner matrix is empty or not
            a_new.append(item) #appending to inner matrix to the list

a_new=np.array(a_new) #creating numpy matrix with removed zero elements

Output:

array([[[1, 2, 3],
        [4, 5, 6],
        [7, 8, 9]],

       [[9, 8, 7],
        [6, 5, 4],
        [3, 2, 1]]])

score 0 · Answer 5 · answered Oct 07 '19 at 19:24

0

Use any and select :)

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

[[9,8,7],
 [6,5,4],
 [3,2,1]],

[[0,0,0],
 [0,0,0],
 [0,0,0]],

[[0,0,0],
 [0,0,0],
 [0,0,0]]])
a[a.any(axis=2).any(axis=1)]

answered Oct 07 '19 at 19:24

Sebastian Wozny

16,943
7
52
69

Remove 2d slice from 3d numpy array

5 Answers5