import cv2
import poly_point_isect
import numpy as np

points = [{'x': 622, 'y': 339}, {'x': 677, 'y': 337}, {'x': 792, 'y': 331}, {'x': 873, 'y': 317}, {'x': 900, 'y': 309}, {'x': 918, 'y': 300}, {'x': 926, 'y': 289}, {'x': 926, 'y': 283}, {'x': 902, 'y': 270}]
img = np.zeros((800, 1500, 3), np.uint8)
b, g, r = 0x3E, 0x88, 0xE5  # orange
img[:, :, 0] = b
img[:, :, 1] = g
img[:, :, 2] = r
draw_connected_points(img, points)
points = np.array([[p['x'], p['y']] for p in points])
reg = LsqEllipse().fit(points)
center, width, height, phi = reg.as_parameters()
print(center)
cv2.imshow("test", img)
cv2.waitKey(0)
cv2.destroyAllWindows()

The fitted center is (-78086.55916753855, -4782.9191357491145) which is strange for the input points:

Thanks for your code, it is very helpful.
I am trying to find the distance between data points and the ellipse, so that I can estimate the error. Is there an easy way to do that? Thanks in advance!

The error occurs when I call fit_obj.as_parameters if the given data failed to fit an ellipse. Since len(fit_ellipse_P.coefficients) = 0 in this situation. As a user, I could only add a condition outside, But obviously, you guys could add this condition into the built-in code.

AttributeError: module 'ellipse' has no attribute 'make_test_ellipse'

Great piece of code! Is it by any chance possible to fit using a fixed center for the ellipse?

I have some points and want to fit an ellipse on them. But the fitted ellipse is big. is there any way to limit its size?

Thank you for making this code available open source. It is very useful for my work.

Would you be willing to upload it to PyPI? This would be helpful for me because my package currently contains a copy of ellipses.py and I would rather import it.

I can help create the setup.py and MANIFEST.in files if you'd like.

Hello,
Many thanks for this great software that has been very useful to me.

Upgrading from 2.0.1 to 2.2.1 I noticed that somehow the axis in which the parameters are returned are inverted.
Digging a bit in the code I noticed that now width and height are inverted with respect to the definition of https://mathworld.wolfram.com/Ellipse.html, resulting in an ellipse with major axis aligned with the x-axis to have phi=90deg

Any reason to do that?

I would have thought the one below to be the the implementation complying with the Mathworld documentation.

Many thanks

class myLsqEllipse(LsqEllipse):
def as_parameters(self):
"""Returns the definition of the fitted ellipse as localized parameters

    Returns
    _______
    center : tuple
        (x0, y0)
    width : float
        Total length (diameter) of horizontal axis.
    height : float
        Total length (diameter) of vertical axis.
    phi : float
        The counterclockwise angle [radians] of rotation from the x-axis to the semimajor axis
    """

    # Eigenvectors are the coefficients of an ellipse in general form
    # the division by 2 is required to account for a slight difference in
    # the equations between (*) and (**)
    # a*x^2 +   b*x*y + c*y^2 +   d*x +   e*y + f = 0  (*)  Eqn 1
    # a*x^2 + 2*b*x*y + c*y^2 + 2*d*x + 2*f*y + g = 0  (**) Eqn 15
    # We'll use (**) to follow their documentation
    a = self.coefficients[0]
    b = self.coefficients[1] / 2.
    c = self.coefficients[2]
    d = self.coefficients[3] / 2.
    f = self.coefficients[4] / 2.
    g = self.coefficients[5]

    # Finding center of ellipse [eqn.19 and 20] from (**)
    x0 = (c*d - b*f) / (b**2 - a*c)
    y0 = (a*f - b*d) / (b**2 - a*c)
    center = (x0, y0)

    # Find the semi-axes lengths [eqn. 21 and 22] from (**)
    numerator = 2 * (a*f**2 + c*d**2 + g*b**2 - 2*b*d*f - a*c*g)
    denominator1 = (b**2 - a*c) * ( np.sqrt((a-c)**2+4*b**2) - (c+a))  # noqa: E201
    denominator2 = (b**2 - a*c) * (-np.sqrt((a-c)**2+4*b**2) - (c+a))
    width = np.sqrt(numerator / denominator1)
    height = np.sqrt(numerator / denominator2)

    # Angle of counterclockwise rotation of major-axis of ellipse to x-axis
    # [eqn. 23] from (**)
    # w/ trig identity eqn 9 form (***)
    if b == 0 and a < c:
        phi = 0.0
    elif b == 0 and a > c:
        phi = np.pi/2
    elif b != 0 and a < c:
        phi = 0.5 * np.arctan(2*b/(a-c))
    elif b != 0 and a > c:
        phi = 0.5 * (np.pi + np.arctan(2*b/(a-c)))
    elif a == c:
        phi = 0.0
    else:
        raise RuntimeError("Unreachable")

    return center, width, height, phi

Hello,

It seems that the angle calculation is not always correct.
I have this case and angle is -0.13946737246282517
This seems the angle from Y axis instead of the X axis

EllipseIn.txt

I have cloned the repository and downloaded the pip package specified in the README, but I still end up getting this error.

Hi,

Nice code, very helpful, thanks for making it public!

I was a little confused by one point in the code, but so far as I could spot, it's only a docstring (minor) issue:

• In ref (*) (paper) the ellipse is defined as a x^2 + b x y + ..., which are the coefficients returned in the corresponding attribute ellipse.LsqEllipse.coefficients.
• However, the docstring in that attribute ellipse.LsqEllipse.coefficients points to ref (**) (MathWorld), which defines the ellipse with additional factors: a x^2 + 2 b x y + ..., and is the source of my confusion.

In the method as_parameters, those factors are implemented where needed, but wouldn't it be clearer to:

• either modify the docstring where it needs to be modified,
• or incorporate the incriminated factors directly in the fitting routine (and harmonize the notation throughout the code)?

Happy holidays!

there is the possibility to derive uncertainties of this fitting ?
Thanks

Hello and thank you for your nice package.

Currently I use your package for fitting ellipses and it works perfect but I would like to know if there is way to output the uncertainties of the computed ellipse parameters ? If it's not implemented can you guide me to a paper explaining how to compute them for an ellipse fit.

Thank you in advance

as shown in the picture in your repository, the ellipse fits the center of the data.
https://github.com/bdhammel/least-squares-ellipse-fitting/blob/master/media/ellipse_fit.png
I would like to know which condition should I change, to fit the whole data inside the ellipse?

instead of multiplying (width, height) by 4

could be very useful to add weights to this, so some points could be regarded with less significance than others