legend for circle matlab

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ha ha
ha ha 2018 年 10 月 21 日
コメント済み: ha ha 2018 年 10 月 22 日
How can i add the legend for "red circle" in the image? The circle was generated using the below syntax: (I already tried this syntax , but failed)
%%INPUT DATA
clear;clc;
R = 10; x_c = 5; y_c = 8;thetas = 0:pi/64:pi;
xs = x_c + R*cos(thetas); ys = y_c + R*sin(thetas);%create input data
% Now add some random noise to make the problem a bit more challenging:
mult = 0.5;
xs = xs+mult*randn(size(xs));
ys = ys+mult*randn(size(ys));
xfit=4.9641;yfit=8.1215;Rfit=9.8862;
%%PLOT RESULT
figure; plot(xs,ys,'b.');%plot data
rectangle('position',[xfit-Rfit,yfit-Rfit,Rfit*2,Rfit*2],'curvature',[1,1],'linestyle','-','edgecolor','r');
hold on;title(sprintf('Best fit: Radius = %0.1f; Center = (%0.3f,%0.3f)',Rfit,xfit,yfit));%create title
plot(xfit,yfit,'g.');xlim([xfit-Rfit-2,xfit+Rfit+2]);ylim([yfit-Rfit-2,yfit+Rfit+2]);axis equal;
legend('input data','center circle','fitted circle');
Hope the result as below:
THANKS,

採用された回答

madhan ravi
madhan ravi 2018 年 10 月 21 日
編集済み: madhan ravi 2018 年 10 月 21 日
R = 10; x_c = 5; y_c = 8;thetas = 0:pi/64:pi;
xs = x_c + R*cos(thetas); ys = y_c + R*sin(thetas);%create input data
% Now add some random noise to make the problem a bit more challenging:
mult = 0.5;
xs = xs+mult*randn(size(xs));
ys = ys+mult*randn(size(ys));
xfit=4.9641;yfit=8.1215;Rfit=9.8862;
figure(1);
h1=plot(xs,ys,'b.');%plot data
hold on;title(sprintf('Best fit: Radius = %0.1f; Center = (%0.3f,%0.3f)',Rfit,xfit,yfit));%create title
h2=plot(xfit,yfit,'g.');xlim([xfit-Rfit-2,xfit+Rfit+2]);ylim([yfit-Rfit-2,yfit+Rfit+2]);axis equal;
h3=rectangle('position',[xfit-Rfit,yfit-Rfit,Rfit*2,Rfit*2],'curvature',[1,1],'linestyle','-','edgecolor','r');
syms x y %symbolic toolbox
h=fimplicit((x-xfit).^2+(y-yfit).^2-Rfit.^2)
h.Color='r'
axis equal
legend([h1 h2 h],{'input data','center circle','fitted circle'})
  5 件のコメント
ha ha
ha ha 2018 年 10 月 21 日
編集済み: ha ha 2018 年 10 月 21 日
plz, check the code again. I already revised it. How can I add the legend for the "red circle line" as expectation figure?
madhan ravi
madhan ravi 2018 年 10 月 21 日
see my edited answer

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その他の回答 (1 件)

Image Analyst
Image Analyst 2018 年 10 月 21 日
Sorry, but you're not doing a fit of your data to a circle at all.
You're just drawing a circle of a predefined size.
  1 件のコメント
ha ha
ha ha 2018 年 10 月 22 日
Thanks,

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