Creating 3d Tensor Array from 2d Array (Python)

Question

I have two numpy arrays (4x4 each). I would like to concatenate them to a tensor of (4x4x2) in which the first 'sheet' is the first array, second 'sheet' is the second array, etc. However, when I try np.stack the output of d[1] is not showing the correct values of the first matrix.

import numpy as np
x = array([[ 3.38286851e-02, -6.11905173e-05, -9.08147798e-03,
        -2.46860166e-02],
       [-6.11905173e-05,  1.74237508e-03, -4.52140165e-04,
        -1.22904439e-03],
       [-9.08147798e-03, -4.52140165e-04,  1.91939979e-01,
        -1.82406361e-01],
       [-2.46860166e-02, -1.22904439e-03, -1.82406361e-01,
         2.08321422e-01]])
print(np.shape(x)) # 4 x 4

y = array([[ 6.76573701e-02, -1.22381035e-04, -1.81629560e-02,
        -4.93720331e-02],
       [-1.22381035e-04,  3.48475015e-03, -9.04280330e-04,
        -2.45808879e-03],
       [-1.81629560e-02, -9.04280330e-04,  3.83879959e-01,
        -3.64812722e-01],
       [-4.93720331e-02, -2.45808879e-03, -3.64812722e-01,
         4.16642844e-01]])
print(np.shape(y)) # 4 x 4

d = np.dstack((x,y))
np.shape(d) # indeed it is 4,4,2... but if I do d[1] then it is not the first x matrix.
d[1] # should be y

You expect d[1] to be y, no? d[0] -> x?

Mad Physicist
– Mad Physicist

2022-02-02 01:31:47 +00:00
Commented Feb 2, 2022 at 1:31 — Mad Physicist
– Mad Physicist, Commented Feb 2, 2022 at 1:31
you are right! making correction to comment now. thanks!

maximus
– maximus

2022-02-02 01:32:33 +00:00
Commented Feb 2, 2022 at 1:32 — maximus
– maximus, Commented Feb 2, 2022 at 1:32

Mad Physicist · Accepted Answer · 2022-02-02 01:28:13Z

1

If you do np.dstack((x, y)), which is the same as the more explicit np.stack((x, y), axis=-1), you are concatenating along the last, not the first axis (i.e., the one with size 2):

(x == d[..., 0]).all()
(y == d[..., 1]).all()

Ellipsis (...) is a python object that means ": as many times as necessary" when used in an index. For a 3D array, you can equivalently access the leaves as

d[:, :, 0]
d[:, :, 1]

If you want to access the leaves along the first axis, your array must be (2, 4, 4):

d = np.stack((x, y), axis=0)
(x == d[0]).all()
(y == d[1]).all()

answered Feb 2, 2022 at 1:28

Mad Physicist

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6 Comments

maximus Over a year ago

Awesome answer! What if I don't have an initial array and want to stack iteratively?

Mad Physicist Over a year ago

@sarah. That's a frequent question. Either estimate how much you'll need and pre-allocate, or append to a list until you're done. Reallocating the entire buffer over and over is going to be a major bottleneck.

maximus Over a year ago

That is true, how would we allocate a 4x4 initial array and then stack the subsequent on top?

Mad Physicist Over a year ago

@sarah. I highly recommend you don't do that. But you can do np.zeros((0, 4, 4)) or (4, 4, 0). But again, use a list or pre-allocate the maximum size instead.

Mad Physicist Over a year ago

@sarah. The first by slicing: d = d[1:]. Interior ones, by np.delete. Again, use a list until the size stops changing. Deleting a 4-byte reference from a list is much cheaper than moving a multidimensional array.

|

user17242583 · Accepted Answer · 2022-02-02 01:28:58Z

1

Use np.stack instead of np.dstack:

>>> d = np.stack([y, x])

>>> np.all(d[0] == y)
True

>>> np.all(d[1] == x)
True

>>> d.shape
(2, 4, 4)

answered Feb 2, 2022 at 1:28

user17242583

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