fit curve with parameter

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Stefan Lang 2020-10-19

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评论： J. Alex Lee 2020-10-19

I have some points, annotated from an xray, along the spine. I need to fit a curve through these points. After that, i want to parametrize my fitted curve from 0 to 1, so i can move along the curve in equal intervals. If i just project my points onto the x axis, my line will be distorted.

So i want to choose point 0, which should give me the x/y coordinates of the fitted curve and the slope of the curve at this point. Point 0.5 should give me the point in the middle of the curve and its x/y values and the slope there, and point 1 the last point of the curve.

Any ideas how to do this? I thought about spline/cscvn, but i don't really get it...

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Stefan Lang 2020-10-19

Which two curves? I'm just fitting one curve through some points (x,y). E.g. through 8 points, but then I want to have a function that can give me all the x and y values along the fitted curve. And this should be with a parameter t between 0 and 1, so when i set t = 0.5, i get the points (x,y) that are on the curve in the middle of the curve length.

Rik 2020-10-19

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E.g. this:

a=linspace(0,pi,100);x=5*sin(a)+rand(size(a));y=6*cos(a)+rand(size(a));%replace with your data
subplot(1,2,1)
plot(x,y)
title('actual data')
subplot(1,2,2)
t=linspace(0,1,numel(x));
plot(t,x),hold on,plot(t,y),hold off
title('as a function of t')

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Answer 1

J. Alex Lee 2020-10-19

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If i understand Rik correctly, I think you'd first have to estimate the t data by computing the cumulative arc length between your x and y data points. But for that to work, your annotated points need to be ordered (from your 0 to 1). If they are ordered, your cummulative arc length can be estimated by something like

function s = cummArcLength(x,y)
    s = zeros(size(x));
    
    for i = 2:length(x)
        s(i) = s(i-1) + sqrt( ...
          (x(i)-x(i-1))^2 ...
        + (y(i)-y(i-1))^2 ...
        );
    end
end

And then just divide s by s(end) to force it to [0,1].

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Rik 2020-10-19

This would be more precise than what I was thinking of: if you have two arrays (x and y), you can use the index as your parameter (and divide by numel(x) instead of s(end)).

J. Alex Lee 2020-10-19

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[Approximate] arc length would be safer if points were very unevenly distributed.

And for completeness, in either case you'd want to do the interpolation one way or another on both x and y parametrically on t

tData = cummArcLength(xData,yData)
xFit = interp1(sData,xData,tFit) % or spline or whatever
yFit = interp1(sData,yData,tFit) % or spline or whatever

If you have a predetermined set of tFit you are interested in. Maybe using something like splines makes it easier to create a function that can be evaluated at arbitrary t

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