Eccentricity of a 2D convex hull in Python

Question

How can I calculate the eccentricity of a 2D convex hull in Python?

Eccentricity: a parameter of an ellipse (or a closed shape) indicating its deviation from the circularity whose value ranging from 0 (circle) to 1 (line).

Answer 1

Well, for those who still would like to see an answer to this question: if we assume the closed shape is an ellipse or similar to an ellipse, the eccentricity is defined as sqrt(square(major_axis_length/2)-square(minor_axis_length/2)) where the major and minor axis's are shown in the figure.

The end points of the axis's are taken as min and max valued points along each axis. With 4 sample geo-points (these also could be Cartesian coordinates), we can write something like this:

import numpy as np
from scipy.spatial.distance import euclidean

points = np.array([[50.6636778,5.0939791], [50.7674881,5.4663611], [50.94594, 5.48977], [51.0380754,5.4012648]])

small_latwise = np.min(points[points[:, 0] == np.min(points[:, 0])], 0)
small_lonwise = np.min(points[points[:, 1] == np.min(points[:, 1])], 0)
big_latwise = np.max(points[points[:, 0] == np.max(points[:, 0])], 0)
big_lonwise = np.max(points[points[:, 1] == np.max(points[:, 1])], 0)
distance_lat = euclidean(big_latwise, small_latwise)
distance_lon = euclidean(big_lonwise, small_lonwise)
if distance_lat >= distance_lon:
    major_axis_length = distance_lat
    minor_axis_length = distance_lon
else:
    major_axis_length = distance_lon
    minor_axis_length = distance_lat
a = major_axis_length/2
b = minor_axis_length/2
ecc = np.sqrt(np.square(a)-np.square(b))/a
print(ecc)

Which yields 0.0854194406287