i am also solving a similar problem in which i need to draw a circle and make every pixel outside that circle to a same value (o or 1 doesn't matter). Any suggestions on ? Thanks in Advance

Respuestas (2)

Geoff Hayes
Geoff Hayes el 6 de Sept. de 2015

0 votos

gill11a - if your image is given by myImage then you could do something like the following to colour every pixel outside of the circle to be the same colour. Suppose that (a,b) is the centre of the circle with radius r. Then
function plotcircle(a,b,r)
close all;
% create a simple 100x200 image of random pixels
nrows = 100;
ncols = 200;
myImg = randi(255,nrows,ncols,3);
% iterate over each pixel and determine if it is outside the circle or not
for x=1:ncols
for y=1:nrows
if (x-a)^2 + (y-b)^2 > r^2
% pixel is outside of the circle, so colour it blue
myImg(y,x,:) = [0 0 255];
end
end
end
% display the image
image(myImg/255);
axis equal
Then you can call the above function as
plotcircle(50,100,25)
plotcircle(100,50,40)
Try the above and see what happens!

1 comentario

bhagyashri kapre
bhagyashri kapre el 19 de En. de 2017
thank you Geoff Hayes,This code is use full in my project

Iniciar sesión para comentar.

Image Analyst
Image Analyst el 6 de Sept. de 2015

0 votos

Also see my attached demos on masking with a circle.

2 comentarios

gill11a
gill11a el 6 de Sept. de 2015
Editada: Image Analyst el 6 de Sept. de 2015
Thanks for the prompt response. I think i forgot to mention one very important aspect of the problem: I do not know the radius of the circle i wish to crop. For eg. in the picture i attached, i want the program to identify the circles themselves, and then follow the procedure you described in your link.
Image Analyst
Image Analyst el 6 de Sept. de 2015
  1. Take any of the color channels.
  2. Take the middle row of the image.
  3. Call diff() to see where the value changes.
  4. Knowing the middle is at (number of columns)/2, compute radius.

Iniciar sesión para comentar.

Categorías

Más información sobre 2-D and 3-D Plots en Centro de ayuda y File Exchange.

Preguntada:

el 6 de Sept. de 2015

Editada:

el 19 de En. de 2017

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by