(4,1,2) Numpy Array Sort Clockwise

Question

I have a numpy array as the following:

my_array = np.float32([[[ 323. , 143.]], [[ 237. , 143.]], [[ 227. , 230.]], [[ 318. , 233.]]])

This 4 points represent the vertices of a rectangle that lies on a image, I need to reorder them clockwise and save it to a new np array, (top-left-> top-right -> bottom - right -> bottom - left). In my example it will be:

[237,  143] -> [323, 143] -> [318, 233] -> [227, 230]

I have read this but my skills on numpy aren't as good to implement it...

Thanks!

Have you taken a look at this? stackoverflow.com/questions/2706605/… Maybe it helps — rafaelc
– rafaelc, Commented May 6, 2015 at 22:54

Divakar · Accepted Answer · 2015-05-22 06:24:01Z

4

You could do something like this -

import numpy as np
from scipy.spatial import distance

def sortpts_clockwise(A):
    # Sort A based on Y(col-2) coordinates
    sortedAc2 = A[np.argsort(A[:,1]),:]

    # Get top two and bottom two points
    top2 = sortedAc2[0:2,:]
    bottom2 = sortedAc2[2:,:]

    # Sort top2 points to have the first row as the top-left one
    sortedtop2c1 = top2[np.argsort(top2[:,0]),:]
    top_left = sortedtop2c1[0,:]

    # Use top left point as pivot & calculate sq-euclidean dist against
    # bottom2 points & thus get bottom-right, bottom-left sequentially
    sqdists = distance.cdist(top_left[None], bottom2, 'sqeuclidean')
    rest2 = bottom2[np.argsort(np.max(sqdists,0))[::-1],:]

    # Concatenate all these points for the final output
    return np.concatenate((sortedtop2c1,rest2),axis =0)

Sample input, output -

In [85]: A
Out[85]: 
array([[ 281.,  147.],
       [ 213.,  170.],
       [ 239.,  242.],
       [ 307.,  219.]], dtype=float32)

In [86]: sortpts_clockwise(A)
Out[86]: 
array([[ 213.,  170.],
       [ 281.,  147.],
       [ 307.,  219.],
       [ 239.,  242.]], dtype=float32)

edited May 22, 2015 at 6:24

answered May 7, 2015 at 4:47

Divakar

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3 Comments

Divakar Over a year ago

Note to OP: If the input array is of shape 4 x 1 x 2, it could be squeezed to a shape of 4 x 2 at the start before proceeding with the rest of the code. So, A = np.squeeze(A) at the start could be used in such a case.

Luis Enrique Over a year ago

It works great but I need to change the tolerance for some matrices like A = np.float32([[281., 147.],[213. ,170.],[239. ,242.], [307. ,219.]]), so I made a variable tolerance, any advise how to pick this tolerance given the points?

Divakar Over a year ago

@LuisEnrique Check out the edits please. Now, it uses eucl. dist., so must be more robust.

cosmoscalibur · Accepted Answer · 2015-05-07 03:33:02Z

0

If you need such you show in your example

new_array = my_array[[1,0,3,2]]

Or exactly clockwise (and in general, not only for 4 points)

n = len(my_array)
order = [i for i in range(0, n-2)]
order.insert(0, n-1)
new_array = my_array[order]

answered May 7, 2015 at 3:33

cosmoscalibur

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