Extracting a 2D numpy array from a 2D numpy array

Question

I'm working with a problem using numpy arrays and I hit a roadblock, basically I have two arrays, one with a 2D numpy array and the other is a 1D numpy which represents some index of the 2D array, what I need is to use pairs of this indexes to extract a 2D numpy array from the original 2D array, I did something, but I'm sure it can be better, so I'm asking for advice. Here is my code:

import numpy as np
import itertools

x = np.arange(25).reshape(5, 5) #Original Array

#x = [[ 0  1  2  3  4]
#     [ 5  6  7  8  9]
#     [10 11 12 13 14]
#     [15 16 17 18 19]
#     [20 21 22 23 24]]

y = np.array([0, 2, 4]) #Indexes

idx = list(itertools.product(y, repeat = 2)) #This create a combination of the indexes to act as my coordinates from the array

#idx = [(0, 0), (0, 2), (0, 4), (2, 0), (2, 2), (2, 4), (4, 0), (4, 2), (4, 4)]

newarray = np.array([x[i] for i in idx]).reshape(3, 3) #This uses the tuples from before to extract the values of the original array

#newarray = [[ 0  2  4]
#            [10 12 14]             #The extracted array
#            [20 22 24]]

So it works, but I think there is a lot to improve, for example, in the final step I use a list comprehesion and then a numpy array, and then a reshape, also I'm not sure if it's okay to create all the combinations of the index array maybe there is a easier way, so any advice will be appreciated, thank you!

hpaulj · Accepted Answer · 2021-04-28 04:05:36Z

1

 x[::2, ::2]

will select every other row and column

For a less regular pattern try

x[y[:,None], y]

which uses advanced indexing

answered Apr 28, 2021 at 4:05

hpaulj

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2 Comments

Carlos Eduardo Corpus Over a year ago

I gotta check advaced indexing, not sure what the None thing is doing here, anyway, thank you so much!

hpaulj Over a year ago

Here None is np.newaxis, making a (n,1) array. In Advanced Indexing, the indexing arrays are broadcast against each other. So a (n,1) with (m,) indexes a (n,m) space. A (m,) with (m,) indexes a (m,).

Tim Roberts · Accepted Answer · 2021-04-28 03:51:13Z

1

Numpy has some sophisticated indexing options. Also remember that reshape is free; never be afraid to reshape.

import numpy as np
import itertools

x = np.arange(25).reshape(5, 5) #Original Array
y = [0, 2, 4]
idx = list(itertools.product(y, repeat = 2)) #This create a combination of the indexes to act as my coordinates from the array
idx0 = [k[0] for k in idx]
idx1 = [k[1] for k in idx]
print(idx)
newarray = x[idx0,idx1].reshape((3,3))
print(newarray)

Output:

[(0, 0), (0, 2), (0, 4), (2, 0), (2, 2), (2, 4), (4, 0), (4, 2), (4, 4)]
[[ 0  2  4]
 [10 12 14]
 [20 22 24]]

answered Apr 28, 2021 at 3:51

Tim Roberts

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Comments

de Broglie · Accepted Answer · 2021-04-28 04:15:09Z

1

I think what you hate is the for keyword (like me). And in fact you don't need itertools.
So my answer would be:

import numpy as np

x = np.arange(25).reshape(5, 5)
y = np.array([0, 2, 4])

ny = y.size
i = y.reshape(ny, 1)
j = y.repeat(ny).reshape(ny, ny).T

print(x[i, j])

Output:

[[ 0  2  4]
 [10 12 14]
 [20 22 24]]

answered Apr 28, 2021 at 4:15

de Broglie

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