put 2d coordinate axis at the center of an array in python

Question

I am going to simulate a 2d area using a matrix in python and I should change the indexes somehow!. I need to find a relation between the original matrix and a new axis coordinate. I think it can be more clarified if I explain by a piece of my code:

originalMatrix=np.random.randint(0, 10, size=(5,5))

the output is:

[[4 8 3 2 5]
 [2 2 2 6 5]
 [2 4 7 9 9]
 [6 2 6 6 6]
 [2 8 3 8 2]]

we can access to number '7' by originalMatrix[2][2]. but in the coordinate system the index must be (0,0).Index of '4' in the left side of '7' in originalMatrix is [2][1] but in the new coordinate should be (-1,0) and so on... I hope I could explain well!

Just to clarify, you want the new coordinate system to be relative to the center of the original coordinate system, right? Or is the logic something else? — Dilith Jayakody
– Dilith Jayakody, Commented Jan 23, 2022 at 5:35
What happens if you have a n x n matrix where n is even? How do you determine the (0,0) coordinate in this case? — Richard K Yu
– Richard K Yu, Commented Jan 23, 2022 at 5:45
Actually I need to move (trace) step by step. and the center element is my starting point. so I need to start from (0,0).... — mohamad bari
– mohamad bari, Commented Jan 23, 2022 at 5:57

mozway · Accepted Answer · 2022-01-23 05:52:03Z

2

An easy way is to roll your array:

a = np.array([[4, 8, 3, 2, 5],
              [2, 2, 2, 6, 5],
              [2, 4, 7, 9, 9],
              [6, 2, 6, 6, 6],
              [2, 8, 3, 8, 2]])
center = np.array(a.shape)//2
# b will be indexed with the previous center being 0,0
b = np.roll(a, -center, axis=(0,1))

b[0, -1]
# 4

b[0, 0]
# 7

NB. You have to decide what is the center in case you have an even number as a dimension.

answered Jan 23, 2022 at 5:52

mozway

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5 Comments

mohamad bari Over a year ago

It doesn't work for b[2,-1], b[0,-2],...

mozway Over a year ago

@mohamad How does it not work? It does, work perfectly(b[2,-1] is 8 as expected), but keep in mind that numpy indexing is b[row, column], if you want the other way around use the transpose b.T

mohamad bari Over a year ago

shouldn't be b[2,-1] equal to 6? '7' is located at (0,0) so '6' should be located at (2,-1). Am I right?

mozway Over a year ago

There are many 6, so the question is ambiguous, but assuming the last 6 of the fourth row in the original data, it is one row after and two columns after the 7, so: b[1, 2] NOT b[-1, 2] that is one row before and two columns after (so a 8). But as I said, if you want a [column, row] indexing, use b = b.T

mozway Over a year ago

Also note that row indexing starts from the top, but you can use flipud to flip the array up/down

Richard K Yu · Accepted Answer · 2022-01-23 05:52:46Z

0

Here is another possible approach, assuming the matrix is nxn where n is odd:

import numpy as np


def access_coordinate(matrix, x, y):
    center = (len(matrix)//2, len(matrix)//2)
    translated_coordinates = (center[0]+y, center[1] + x)
    return matrix[translated_coordinates[0], translated_coordinates[1]]
    

originalMatrix=np.array([[4, 8, 3, 2, 5],
 [2, 2, 2 ,6, 5],
 [2, 4, 7, 9, 9],
 [6, 2, 6, 6, 6],
 [2, 8, 3, 8, 2]])


print(access_coordinate(originalMatrix, -1, 0)) #4

answered Jan 23, 2022 at 5:52

Richard K Yu

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Comments

Mateo Vial · Accepted Answer · 2022-01-23 05:53:47Z

0

There is no way to do that with standard numpy arrays, but you could define a function that does it for you

def arrayByCoordinates(array, coords):
    coords = np.array(coords)
    center = np.array(array.shape) // 2
    return array[tuple(coords + center)]

>>> x
array([[ 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]])
>>> arrayByCoordinates(x,(0,0))
12

But keep in mind that if the shape of the array is even, there is no real center, so this will convert to the lowest one

edited Jan 23, 2022 at 5:53

answered Jan 23, 2022 at 5:48

Mateo Vial

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