How to transform a ellipse to circle

I have some data which looks like it can be fitted with an ellips. However, I want to transform it so it would be in a cricle.How can I do it?

Answer 1

[Matt J]

If the equation for the ellipse is

x.'*Q*x =1

then you simply take the cholesky decomposition of the 2x2 matrix Q,

 L=chol(Q)

and make the change of variables y=L*x. The equation for y is

 y.'y=1

which is the equation for a circle of radius 1.

[mina sh]

Sorry my background is not strong in that. How do I get Q, given the only thing I have is the coordinates of orange dots>

[Bruno Luong]

How do you plot the blue ellipse curve? You must have an equation of it, one form or another.

[Bruno Luong]

Those verbal description is so confuse without a picture, at least to me.

Please post the MATLAB code (and nothing else please) that draws the blue ellipseoid using whatever output parameters that are required for the plot.

[mina sh]

I found this code on the internet:

%the mean should be at zero
% X is a NX2 matrix
Mu = mean(X);
X0 = bsxfun(@minus, X, Mu);
%scale the size of the ellipse based on standard deviatioon
STD = 2;       
%covers around 95% of population
conf = 2*normcdf(STD)-1;     
%inverse chi-squared with dof=dimensions
scale = chi2inv(conf,2);     
Cov = cov(X0) * scale;
% eigen decomposition [sorted by eigen values]
[V, D] = eig( Cov );     
[D, order] = sort(diag(D), 'descend');
D = diag(D);
V = V(:, order);
t = linspace(0,2*pi,100);
% unit circle
e = [cos(t) ; sin(t)];        
% scale eigenvectors
VV = V*sqrt(D);               
% project circle back to orig space
e = bsxfun(@plus, VV*e, Mu'); 
% plot cov and major/minor axes
figure(1)
plot(X(:,1),X(:,2),'.r')
hold on;
plot(e(1,:), e(2,:), 'Color','b');

[Matt J]

The parameters in the code give you the transformation to a circle directly. The transformation is,

circle_points = W\(ellipse_points-Mu(:))

[mina sh]

by W do you mean VV? and by ellipse_points do you mean X?

I know how to make that ellipse to a circle by I don't know how to transform the orange dots so that they would be inside a circular plane instead of oval plane!

[Bruno Luong]

The transformation is

P -> Q := VV \ (P - Mu')

where P is any point in R^2; it can be orange points or ellipse boundary points.

Explanation:

Look at the line of code

% unit circle
e = [cos(t) ; sin(t)];

This is the equation of the (unit) circle then transformed later by (I change e to and combine the 2 line of codes)

P = VV*Q + Mu'

before plotted as an ellipse.

If you do the opposite (solve for Q from P) you will transform the ellipse back to the circle. If you don't care whereas the resulting circle is centered at the origin, you can ignore the mean Mu and simply apply the linear transformation

Q = VV \ P

to your (orange) points P.

Now how VV related to the parameters of your ellipse found by fitting; you are in trouble since you just plug two independent codes on the internet without trying to understand what they mean.

[mina sh]

Thanks Bruno. so I did the following : Q= VV\P and P are the coordinates for the orange dots, to get Q which would be dots in the circular plane. What I am now getting is the following: left is the points P and right is Q. Hope this is correct.

[Matt J]

That looks right to me, although you might get a better ellipse fit if you fit only the convex hull of the points

k=convhull(P(1,:),P(2,:));
P=P(:,k);

[Bruno Luong]

Why? The points are not in convex hull are still as valid as points in the convex hull. What is the logic behind of throwing them away?

[Matt J]

I don't understand the logic behind "the points are not in convex hull are still as valid as points in the convex hull". The boundary points of the cloud the OP has shown is clearly elliptical, but the interior points are a random mess.

[Bruno Luong]

From my understanding the ellipse is the blue curve of the original plot not the orange points. The blue curve derived from fitting of something else (not showed here) and certainly not the orange points (otherwise it cannot contain all the orange points). So actually I don't understand why you recommend to take the envelop, envelop of what?

If you fit a closed curve in a point cloud some of them will fall outside, some falls inside, never all of them fall inside. This is a consequence of the euler lagrange condition of the curve that trully fits.

If the only data are the orange then it is clearly not right since if all the points are used to fit an ellipse, then the transformation cannot be entirely fit within a disc of radius 1 as showed in the right picture (about 0.75 as I read the plot).

So no something is not right.

[mina sh]

The orange data is X and the blue ellipse which is a fitted ellipse is e. Both of them are shown in the code above as well:

e = [cos(t) ; sin(t)];        
% scale eigenvectors
VV = V*sqrt(D);               
% project circle back to orig space
e = bsxfun(@plus, VV*e, Mu'); 

Now to plot those two subplots I used this code:

K=VV\(X')
K2=VV\(e)
figure
subplot(1,2,1);plot(X(:,1),X(:,2),'r.');axis square
subplot(1,2,2);plot(K(1,:),K(2,:),'r.');axis square
hold on
subplot(1,2,1);plot(e(1,:),e(2,:),'b.');axis square
hold on
subplot(1,2,2);plot(K2(1,:),K2(2,:),'b.');axis square

And this is what I get:

[Bruno Luong]

Now that looks right.

One think you must pay attention is the determinant of V.

I don't know where it comes from but the determinant must be +/-1. Some algorithm return V with det=+1, other doesn't specify the sign of V.

If det(V)==-1 you could get the transformation that flips your plots (however still maps the ellipse to a circle), like looks at the mirror. If that kind of transformation is problem for you, just reverse the sign of one colume of V, then do the rest identical.

[mina sh]

Thanks Bruno. You helped a lot. I checked the deteminant of V and it is +1.

Answer 2

[Image Analyst]

Are you just worried that your x and y are distorted, like digitally it's roughly round but because of the shape of the figure window and axes inside it, the shape looks stretched? If so experiment around with

axis equal

You might also want to play with xlim() and ylim()