How to fit the data points into a closed curve?

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Wesley 2020-11-26

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https://ww2.mathworks.cn/matlabcentral/answers/662758-how-to-fit-the-data-points-into-a-closed-curve

回答： Wesley 2020-11-27

Hello everyone, I’d like to ask you a question. Now I have a series of data points and want to fit these points into a closed curve. But the effect is not good, I would be very grateful if anyone can help me. The data required for closed curve fitting is shown below.

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John D'Errico 2020-11-26

Please don't use an answer just to make a comment. I've moved your answer into this comment.

John D'Errico 2020-11-26

What about that data do you want to be a closed curve? No, you cannot use interpolation using tools like interp1 to fit the black curve.

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Answer 1

Stephan 2020-11-26

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https://ww2.mathworks.cn/matlabcentral/answers/662758-how-to-fit-the-data-points-into-a-closed-curve#answer_556873

在 MATLAB Online 中打开

DataFit.xlsx

This will not solve your problem finally, but it shows a possible way to tackle this. The idea here is to use a cardiod curve and find parameters that try to minimize the difference between the single data points:

data = readmatrix('DataFit.xlsx');
a = [72.0839, 1.3643, 338.485, 289.4648, -21];
[x,y] = calcCurve(a);
figure
scatter(data(:,1),data(:,2))
hold on
scatter(x,y)
hold off
function [x,y] = calcCurve(a)
    theta = linspace(-pi,pi,1187);
    r = @(theta) a(1).*(a(2)+cos(theta));
    rho = r(theta);
    [x,y] = pol2cart(theta,rho);
    x = -x';
    y = -y';
    xy = sortrows([x,y],1);
    x = xy(:,1) + a(3);
    y = xy(:,2) + a(4);
end

The parameters for a were found by optimization techniques - maybe this is a local optimum and we could do better. But the point is, that you have to think about your mathematical problem and analyze what you have there. After that it makes sense to start coding:

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Answer 2

Wesley 2020-11-27

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https://ww2.mathworks.cn/matlabcentral/answers/662758-how-to-fit-the-data-points-into-a-closed-curve#answer_557248

Data_fitting_by_G1_rational_cubic_Bezier_curves_us.pdf

The curve you fit out is a bit larger than the original image. I am thinking of a fitting method that is more appropriate to the original curve. The method described in this article feels very feasible. But I don't know how to implement it with programs. The article is as follows: