numpy 3D array shape

Question

I created a numpy array of shape (4,3,2); I expected following code can print out a array shaped 4 X 3 or 3 X 4

import numpy as np

x = np.zeros((4,3,2), np.int32)

print(x[:][:][0])

However, I got

[[0 0]
 [0 0]
 [0 0]]

Looks like a 2 X 3? I am really confused on numpy matrix now. Shouldn't I get

  [[0 0 0]
   [0 0 0]
   [0 0 0]
   [0 0 0]]

in stead? How to do map a 3D image into a numpy 3D matrix?

For example, in MATLAB, the shape (m, n, k) means (row, col, slice) in a context of an (2D/3D) image.

Thanks a lot

In numpy the display is (panel, row, col). The first dimension is the outermost one. In MATLAB x(:)` ravels the array, making a (24,1) size. x(:,:,1) selects the first panel. Why did you try x[:][:][0]? — hpaulj
– hpaulj, Commented Mar 5, 2018 at 22:48

akuiper · Accepted Answer · 2018-03-05 22:44:08Z

2

x[:] slices all elements along the first dimension, so x[:] gives the same result as x and x[:][:][0] is thus equivalent to x[0].

To select an element on the last dimension, you can do:

x[..., 0]
#array([[0, 0, 0],
#       [0, 0, 0],
#       [0, 0, 0],
#       [0, 0, 0]], dtype=int32)

or:

x[:,:,0]
#array([[0, 0, 0],
#       [0, 0, 0],
#       [0, 0, 0],
#       [0, 0, 0]], dtype=int32)

answered Mar 5, 2018 at 22:44

akuiper

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Fred · Accepted Answer · 2018-03-05 22:46:14Z

1

You need to specify all slices at the same time in a tuple, like so:

x[:, :, 0]

If you do x[:][:][0] you are actually indexing the first dimension three times. The first two create a view for the entire array and the third creates a view for the index 0 of the first dimension

answered Mar 5, 2018 at 22:46

Fred

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