How can I save each part (sector) of a circle in an individual image?

I create a circle in my image. Then I subdivide the circle into 6 parts with this code.

x0=centroidsX;
y0=centroidsY;
r=10;
teta=-pi:0.01:pi;
x=r*cos(teta)+x0
y=r*sin(teta)+y0
plot(x,y)
hold on
scatter(x0,y0,'or')
axis square 
%----------------------------------------
% divide your circle to n sectors
n=6
tet=linspace(-pi,pi,n+1)
xi=r*cos(tet)+x0
yi=r*sin(tet)+y0
for k=1:numel(xi)
plot([x0 xi(k)],[y0 yi(k)],'black')
hold on
end

My need is to save each sector of the circle into an individual image.

Answer 1

[Gustavo Liñán Cembrano]

Hi, assuming your image variable is Im; I've created this small code, which provides a ThisSector image (B&W) and ThisSectorIDs (indexes to sector positions) which you can use to select part of your image to crop (or just multiply the image by the sector) to obtain black everywhere instead where the sector is one. Hope it helps. The imshow() etc. is just for you to see the produced sectors. The code allows you to interactively draw the circle around the area of interest. If you already have the circle defined, just comment the section where the circle is created interactively

[Nrows,Ncols,Ncolors]=size(Im);
hFig=imshow(Im);
%% Comment this section if x0,y0,r are already selected
hCircle=drawcircle;
x0=hCircle.Center(1);
y0=hCircle.Center(2);
r=hCircle.Radius;

%% if you already have x0,y0,r, then use
hCircle=drawcircle('Center',[x0,y0],'Radius',r);

%%


CircleMask=createMask(hCircle);
n=6;
tet=linspace(-pi,pi,n+1);
xi=r*cos(tet)+x0;
yi=r*sin(tet)+y0;

for k=1:numel(xi)-1

  if (k<0.5*(numel(xi)))
    CornerX1=xi(k);
    CornerY1=1;
    CornerX2=xi(k+1);
    CornerY2=1;
  else
    CornerX1=xi(k);
    CornerY1=Nrows;
    CornerX2=xi(k+1);
    CornerY2=Nrows;
  end
  
  
  
  hPoly=drawpolygon('Position',[x0 y0; xi(k) yi(k); CornerX1 CornerY1; CornerX2 CornerY2; xi(k+1) yi(k+1);x0 y0]);
  ThisMask=createMask(hPoly);
  ThisSector=(ThisMask & CircleMask);
  ThisSectorIDs=find(ThisSector);
  figure(k);
  imshow(ThisSector);
  
  
end

[dziri halima]

Thank you sir for your reply regarding my problem, your code is very useful, but my need is to obtain each sector with original value not binary results. I try to change your code but I failed

[Image Analyst]

Assuming your image is grayscale:

  ThisSector=(ThisMask & CircleMask);
  graySector = Im .* ThisSector;
  figure(k);
  imshow(graySector, []);
  axes('on', 'image');

[Gustavo Liñán Cembrano]

Add the following to mu previous Code Inside the for loop

ThisSectorIm=Im.*double(ThisSector); SectorName=strcat('Sector_',num2str(k),'.jpg'); imwrite(ThisSectorIm,SectorName);