Numpy create four other points arrays based on one point array-CodePudding

Suppose I have an array a representing 3 center points of 3 rectangles. I want to create four other copied points based on each of points in array a by add 1 or minus 1 in x, y coordinates like shown in the picture.

a = np.arange(9).reshape(3,3)
>>>a
>>>out:[[ 0  1  2]
        [ 3  4  5]
        [ 6  7  8]]

I'm very new to numpy. What I can think of is that I can make 4 coupies of a, for the first do a[:,0] =1, then a[:,1] =1. for the second do a[:,0] =1,then a[:,1]-=1.for the third do a[:,0]-=1 then a[:,1] =1, for the fourth do a[:,0]-=1then a[:,1]-=1. But I know it's stupid. So I 'm wondering if there is a clearer way to do it in numpy?

CodePudding user response：

it seems that what you need is to loop over a Cartesian product, there are many ways of doing so, one is to use itertools, here goes:

import numpy as np
import itertools

a = np.arange(9).reshape(3,3)

list_of_arrays = []
for seq in itertools.product([1, -1], repeat=2):
    b = a.copy() 
    b[:,0] =seq[0]
    b[:,1] =seq[1]
    list_of_arrays.append(b)

list_of_arrays:

[array([[1, 2, 2], [4, 5, 5], [7, 8, 8]]), array([[1, 0, 2], [4, 3, 5], [7, 6, 8]]), array([[-1, 2, 2], [ 2, 5, 5], [ 5, 8, 8]]), array([[-1, 0, 2], [ 2, 3, 5], [ 5, 6, 8]])]

CodePudding user response：

Using numpy broadcasting:

import itertools
import numpy as np

a = np.arange(9).reshape(3, 3)

shifts = np.array([(dx, dy, 0) for dx, dy in itertools.product([1, -1], repeat=2)])
shifted_a = a   shifts[:, None]