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ellipse fitting with matlab using fittype

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Johannes Tischer
Johannes Tischer 2020 年 8 月 25 日
コメント済み: Matt J 2020 年 8 月 27 日
Hello,
I want to use fitoptions and fittype for a staistical nonlinear least squares fitting approach. The problem is, I cant find the correct equation for my model (ellipse). I tried multiple like: to fittype ('sqrt(b^2*(1-(x^2/a^2)))' ). In this example, the square root has to be positive. I tried to tightening the upper and lower bounderies, but it doesnt work.
Can someone make an example of the following form:
fo = fitoptions('Method','NonLinearLeastSquares')
ft = fittype('ellipse equation')
[myfit, gof] = fit(x_data, y_data, ft, fo)

採用された回答

Matt J
Matt J 2020 年 8 月 26 日
編集済み: Matt J 2020 年 8 月 26 日
You could use fit() if you convert your data to polar cooridnates. In polar coordinates, an ellipse is given by an explicit function,
  2 件のコメント
Johannes Tischer
Johannes Tischer 2020 年 8 月 27 日
After this I converted it back to cartesian coordinates and plotted the data with my original data.
The pictures shows the original data (blue) and the ellipse fit (red).
Thank you Matt !
Matt J
Matt J 2020 年 8 月 27 日
The problem with this method, of course, is that it assumes that the ellipse is centered at the origin. In general, that may not be true or known to be true.

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その他の回答 (3 件)

Alan Stevens
Alan Stevens 2020 年 8 月 25 日
It shows you how to construct your own function correctly.
  6 件のコメント
Johannes Tischer
Johannes Tischer 2020 年 8 月 26 日
This is the basic data. I want to use fittype because it has so many different options to fit and compare.
Alan Stevens
Alan Stevens 2020 年 8 月 26 日
Hmm. I'm struggling to see your basic data as elliptical. Nevertheless, the code below tries to fit them using an ellipse (I've deleted a few obvious non-elliptical outliers below; however, you could easily retain them if you wish):
load('EllipseData.mat')
% The next three lines remove outliers that are obviously not part of an
% ellipse! Delete the three lines if you wish to retain the outliers.
ix = find(data_x<-600); data_x(ix) = [];data_y(ix) = [];
iylo = find(data_y<-220); data_x(iylo) = [];data_y(iylo) = [];
iyhi = find(data_y>220); data_x(iyhi) = [];data_y(iyhi) = [];
% Set the origin at the means of the x and y data
x = data_x - mean(data_x);
y = data_y - mean(data_y);
% "Linearise" and do a straightforward least-squares fit
N = length(x);
X = x.^2; Y = y.^2;
M = [N -sum(X); sum(X) -sum(X.^2)];
V = [sum(Y); sum(X.*Y)];
AB = M\V; % AB = [A; B]
% Extract constants
A = AB(1);
B = AB(2);
% Create ellipse semimajor axes from "linear" constants
a = sqrt(A/B);
b = sqrt(A);
% Separate upper and lower halves of "elliptical" values.
xhi = x(x>=0); xlo = x(x<0);
yhi = b*sqrt(1 - (xhi/a).^2);
ylo = -b*sqrt(1 - (xlo/a).^2);
% Put them back together and shift the origin back
xf = [xhi xlo] + mean(data_x);
yf = [yhi ylo] + mean(data_y);
% Plot and compare Data and fit.
plot(data_x,data_y,'o',x,yf,'*'), grid
xlabel('x'),ylabel('y')
legend('Data','"Elliptical" fit')
This is the result:

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Matt J
Matt J 2020 年 8 月 26 日
編集済み: Matt J 2020 年 8 月 26 日
You cannot use fit() for implicit function fitting, but you could use dedicated ellipse fitting methods from the File Exchange, e.g.,
  1 件のコメント
Johannes Tischer
Johannes Tischer 2020 年 8 月 26 日
so theres no way to use this method for ellipse fitting ?
Thank you both !!! I really appreciate your support.

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Image Analyst
Image Analyst 2020 年 8 月 26 日
For what it's worth, attached is a paper that describes the method.
Least Squares orthogonal distances fitting of circle, sphere, ellipse, hyperbola, and parabola.pdf

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