Plotting multiple objects on same axis
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I am new to MATLAB but I am trying to plot multiple circles on the same axis. I am using rand function to generate the centers because I want some of them to overlap, but it ends up plotting just one circle. Below is my code;
r = 0.005; x = rand(100,1); y = rand(100,1); th = 0:pi/100:2*pi; for i = 1:length(x) & 1:length(y) a = r*cos(th) + x(i); b = r*sin(th) + y(i); axis tight; hold on plot(a,b) end
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Kevin Chng
2018-10-26
编辑：Kevin Chng
2018-10-26
r = 0.005;
x = rand(100,1);
y = rand(100,1);
th = 0:pi/100:2*pi;
for i = 1:1:length(x)
a = r*cos(th) + x(i);
b = r*sin(th) + y(i);
figure(1)
axis tight;
hold on
plot(a,b)
pause(0.1)
end
Slight change your code, it is very interesting, if you add pause(), then you see the animation. However, you delete the pause() if you don't want it.
更多回答（3 个）
madhan ravi
2018-10-26
编辑：madhan ravi
2018-10-26
r = 0.005;
x = rand(100,1);
y = rand(100,1);
th = 0:pi/100:2*pi;
a=zeros(1,numel(x)) % preallocation for speed and efficiency
b= zeros(1,numel(y))
for i = 1:numel(x)
a(i) = r*cos(th(i)) + x(i);
b(i) = r*sin(th(i)) + y(i);
end
axis tight;
hold on
plot(a,b,'ob')
7 个评论
Jide Williams
2018-10-26
madhan ravi
2018-10-26
编辑：madhan ravi
2018-10-26
Anytime:)
madhan ravi
2018-10-26
its always faster to use numel() for vector than length()
Just a note, as far as I can see you are not plotting the actual radius here. The radius of the circles is defined by the markersize, and not by the variable r. Therefore, if you zoom in the radius changes. In the other answer, the radius is constant, and defined by r. Perhaps I misunderstood the question.
madhan ravi
2018-10-26
the actual question was to plot a point of circles centre point and not the radius of the circle as far as I understood
jonas
2018-10-26
Yep, you are entirely correct in that it's a poorly written question. One of the things you learn on this forum is filling in blanks and interpreting ambiguities, which is actually quite useful when teaching :)
With that said, the notations r and the values of th (0 to 2pi) and the fact that OP accepted @Keving Chng's answer would imply that OP intended the radius to go as input.
madhan ravi
2018-10-26
Wow thanks Jonas completely agree with you , one thing experience is gained day by day :)
jonas
2018-10-26
If you have the image processing toolbox available, then there is a one-line solution to your problem.
viscircles([x,y],ones(size(x)).*r)
6 个评论
Jide Williams
2018-10-26
jonas
2018-10-26
Well, the full code would read
r = 0.005;
x = rand(100,1);
y = rand(100,1);
viscircles([x,y],ones(size(x)).*r)
Jide Williams
2018-10-26
jonas
2018-10-26
r is a scalar, but the input to the function must be a vector of the same size as x and y. That's why you simply multiply it by a number of ones, like:
1 0.005
1 *. 0.005 = 0.005
1 0.005
... ...
jonas
2018-10-26
Viscircles basically does the same thing as your code, but probably more effeciently. I have not actually timed and compared.
Jide Williams
2018-10-26
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