Plotting multiple objects on same axis

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

Answer 1

[Kevin Chng]

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.

Answer 2

[madhan ravi]

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')

[Jide Williams]

Absolutely! Thank you, You know i thought of this, but just wasn't aware of the numel function. Thanks again

[madhan ravi]

Anytime:)

[madhan ravi]

its always faster to use numel() for vector than length()

[jonas]

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]

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]

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]

Wow thanks Jonas completely agree with you , one thing experience is gained day by day :)

Answer 3

[jonas]

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)

[Jide Williams]

Thanks Jonas and noted....hmm so the code may look like viscircles([x(i),y(i)], ones(size(x(i))).*r? or can you help define your variables? Thanks

[jonas]

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]

And what does r in my code stands for then? A multiplyer?

[jonas]

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]

Viscircles basically does the same thing as your code, but probably more effeciently. I have not actually timed and compared.

[Jide Williams]

Thanks again

Answer 4

[Jide Williams]

Ok, Thanks