Python Numpy stack 2d arrays in vector

Question

So, I would like to stack couple 2d arrays to vector so it would look like this:

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

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

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

I can make smth like this:

import numpy as np
a = np.zeros((5, 5), dtype=int)
b = np.zeros((5, 5), dtype=int)
c = np.stack((a, b), 0)
print(c)

To get this:

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

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

But I cant figure out how to add third 2d array to such vector or how to create such vector of 2d arrays iteratively in a loop. Append, stack, concat just dont keep the needed shape

So, any suggestions? Thank you!

Conclusion: Thanks to Tom and Mozway we've got two answers

Tom's:

data_x_train = x_train[np.where((y_train==0) | (y_train==1))

Mozway's:

out = np.empty((0,5,5))

while condition:
    # get new array
    a = XXX
    out = np.r_[out, a[None]]
out

I didn't. This vector of len 3 is desirable result that I can't get — Gleb S
– Gleb S, Commented Sep 29, 2021 at 9:59
But, what is the logic to get it from two 2-dimensional arrays? — Dani Mesejo
– Dani Mesejo, Commented Sep 29, 2021 at 10:00
Your question is a bit ambiguous due to the zero-only arrays, if you have n arrays of same shape and want to stack them, do it as a single pass: np.stack((a1, a2, a3, ..., an), 0) — mozway
– mozway, Commented Sep 29, 2021 at 10:04
I'm trying to get specific 28x28 matrices from fashin_mnist dataset of 60000 pictures. Dataset is in shape of (60000, 28, 28) — Gleb S
– Gleb S, Commented Sep 29, 2021 at 10:04

mozway · Accepted Answer · 2021-09-29 10:20:20Z

1

Assuming the following arrays:

a = np.ones((5, 5), dtype=int)
b = np.ones((5, 5), dtype=int)*2
c = np.ones((5, 5), dtype=int)*3

You can stack all at once using:

np.stack((a, b, c), 0)

If you really need to add the arrays iteratively, you can use np.r_:

out = a[None]

for i in (b,c):
    out = np.r_[out, i[None]]

output:

array([[[1, 1, 1, 1, 1],
        [1, 1, 1, 1, 1],
        [1, 1, 1, 1, 1],
        [1, 1, 1, 1, 1],
        [1, 1, 1, 1, 1]],

       [[2, 2, 2, 2, 2],
        [2, 2, 2, 2, 2],
        [2, 2, 2, 2, 2],
        [2, 2, 2, 2, 2],
        [2, 2, 2, 2, 2]],

       [[3, 3, 3, 3, 3],
        [3, 3, 3, 3, 3],
        [3, 3, 3, 3, 3],
        [3, 3, 3, 3, 3],
        [3, 3, 3, 3, 3]]])

edit: if you do not know the arrays in advance

out = np.empty((0,5,5))

while condition:
    # get new array
    a = XXX
    out = np.r_[out, a[None]]
out

edited Sep 29, 2021 at 10:20

answered Sep 29, 2021 at 10:09

mozway

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4 Comments

Gleb S Over a year ago

What if I don't know my a, b, c's yet and I'm finding them in a loop by some condition. Like desire = np.array([]); for i in range(1000): if arr1[i] == 1: desire = np.append(arr2[i])

mozway Over a year ago

@GlebS I added an alternative

hpaulj Over a year ago

Collect the a,b,c iteratively in a list, and do one stack at the end. This is more efficient, and easier to get right.

mozway Over a year ago

@hpaulj I fully agree (see my first option), this was however OP's request to to this iteratively (maybe they need the intermediate result)

Tom McLean · Accepted Answer · 2021-10-02 08:02:27Z

1

Do you mean something like:

np.tile(a, (3, 1, 1))

array([[[0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0]],

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

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

Edit: Do you mean something like:

test = np.tile(a, (3000, 1, 1))
filtered_subset = test[[1, 10, 100], :, :]

array([[[0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0]],

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

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

edited Oct 2, 2021 at 8:02

answered Sep 29, 2021 at 10:01

Tom McLean

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11 Comments

Gleb S Over a year ago

Yeah, but I want to iterate through big set of (60000, 28, 28) shape and add specific (28, 28) shape matrixes to a new vector. So it would be like (12000, 28, 28)

Tom McLean Over a year ago

You are going to have to provide a better example, i'm not sure what you mean

Tom McLean Over a year ago

I've added what you might mean?

Gleb S Over a year ago

colab.research.google.com/drive/…

Gleb S Over a year ago

Thank you so much! I'll put your and Mozway's answers to post!

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