How to extract values within a circular region from a griddata?

Question

aneps 2022-6-23

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1746515-how-to-extract-values-within-a-circular-region-from-a-griddata

评论： Bjorn Gustavsson 2022-6-23

采纳的回答： Bjorn Gustavsson

I have a 2d matrix (100 x 100 size) obtained from the piece of code is as follows:

N = 100;

xv = linspace(min(CoordsX), max(CoordsX), N);

yv = linspace(min(CoordsY), max(CoordsY), N);

[X,Y] = ndgrid(xv, yv);

Data = griddata(CoordsX, CoordsY, Field, X, Y);

figure()

surf(X, Y, Data) % This make a surface plot of Data.

'Data' outputs a 100 x 100 matrix and then I plot a surface plot as it is in the code. Here I want to select a region with in a circle centered at xc, yc and radius 'r'. Then make another surface plot. How can I extract the vlaues from 'Data' those are within a circular region or radius 'r'?

Thank you

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Answer 1

Bjorn Gustavsson 2022-6-23

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1746515-how-to-extract-values-within-a-circular-region-from-a-griddata#answer_991845

在 MATLAB Online 中打开

You "simply" only ask for the values inside that circular region. You can do it at least two different ways:

N = 100;
xv = linspace(min(CoordsX), max(CoordsX), N);
yv = linspace(min(CoordsY), max(CoordsY), N);
[X,Y] = ndgrid(xv, yv);
Data = griddata(CoordsX, CoordsY, Field, X, Y);
figure()
subplot(2,2,1)
surf(X, Y, Data) % This make a surface plot of Data. 
Ioutside = find((X(:)-xc).^2+(Y(:)-yc).^2>r);
Data(Ioutside) = nan;
surf(X, Y, Data) % This make a surface plot of Data inside the circular region. 
phi360 = (0:360)*pi/180;
r = linspace(0,r);
[Phi360,R] = meshgrid(phi360,r);
Data = griddata(CoordsX, CoordsY, Field, R.*cos(Phi360), R.*sin(Phi360));
subplot(2,2,3)
surf(R.*cos(Phi360), R.*sin(Phi360), Data) % This make the surface-plot with cylindrical coords

You could(should) also take a look at the scatteredInterpolant function since that might be a useful and more flexible multi-call replacement of the griddata-function.

HTH