MATLAB: Polar distribution, CENTERED on the center of gravity

Image Processing Toolboxpolar circlepolar distribution

hello, can someone help me to calcul number of pixels in each angle of a polar circle centred on the center of gravity of image ??

Best Answer

A=imread('31.jpg');
figure(1);imshow(A);
[gc1 gc2]=COG(A);                                               % calculating center of gravity 
gc1=round(gc1);gc2=round(gc2);                           
hold all;
figure(1);plot(gc2,gc1,'*r');
A(find(A>125))=255; A(find(A<=125))=0;               % simplify pixels range to binary
A(:,:,2)=and(A(:,:,2),A(:,:,3));A(:,:,3)=[];
A=and(A(:,:,1),A(:,:,2));
B=logical(~A);                 
figure(2);imshow(B)
figure(2);hold all;plot(gc2,gc1,'*r');
% all the points of the CSI                                       % attempts to avoid outer circle including all pixels of CSI
%  [i1,j1]=ind2sub(size(B),find(B==1))
% intersect(find(B==1),sub2ind(size(B),xc2,yc2))
t=[0:1:360];r=20;
xc=r.*cosd(t);yc=r.*sind(t);
xc=xc+gc2;yc=yc+gc1;
xc2=int64(floor(xc));yc2=int64(floor(yc));
% figure(2);hc=plot(xc2,yc2,'r','LineWidth',1.5);            % the initial circle
   L2=[]; for k=1:1:length(xc2) L2=[L2;  B(xc2(k),yc2(k))]; end   
   while sum(L2)>0    
  %     while sum(B(sub2ind(size(B),xc2(1:end),yc2(1:end))))>0   
      r=r+1;
      xc=r.*cosd(t);yc=r.*sind(t);
    xc=xc+gc2;yc=yc+gc1;
      xc2=int64(floor(xc));yc2=int64(floor(yc));
      L2=[]; 
      for k=1:1:length(xc2) 
          L2=[L2;  B(int64(round(xc2(k))),int64(round(yc2(k))))]; 
      end
  %     figure(2);hc=plot(xc2,yc2,'r','LineWidth',1.5);  
  end
figure(2);hc=plot(xc2,yc2,'r','LineWidth',1.5);
B(sub2ind(size(B),xc2(1:end),yc2(1:end))) % the values of the image on the circle points B(hc.YData,hc.XData)
prompt1={'key in degrees sector: '};                                               % input degrees per sector
dlg_title='sector degrees ';n_lines=1;
default={'30'};sctr=inputdlg(prompt1,dlg_title,n_lines,default);
sector_angle= str2num(cell2mat(sctr));
p_center=[gc2 gc1];
k=[0:sector_angle:360];                                                                  % split 360 angle into equal sector angles
rc1=r*exp(j*pi*k./180); 
x_arc1=real(rc1)+p_center(1); y_arc1=imag(rc1)+p_center(2);           % get points sectors around outer PPI circle
hold all;
figure(2);plot(x_arc1,y_arc1,'go');
sector_scan=[p_center(1) x_arc1(1) x_arc1(2) p_center(1);p_center(2) y_arc1(1) y_arc1(2) p_center(2)];
figure(2);plot(sector_scan(1,:),sector_scan(2,:),'b','LineWidth',1.5);
D=double(B);
[yq,xq]=find(D);
figure(2);plot(xq,yq,'*r');   % initiial test, just to visualise inpolygon input points are correct
[in,on]=inpolygon(xq,yq,sector_scan(1,:)',sector_scan(2,:)'); 
figure(2);plot(xq(in),yq(in),'*g')
figure(2);plot(xq(on),yq(on),'*y')
count=zeros(1,length(k)-1);
for s=1:1:length(k)-1
    sector_scan=[p_center(1) x_arc1(s) x_arc1(s+1) p_center(1);p_center(2) y_arc1(s) y_arc1(s+1) p_center(2)];
    [in,on]=inpolygon(xq,yq,sector_scan(1,:)',sector_scan(2,:)'); 
    count(s)=numel(xq(in))
end
count'
busiest_sector_count=max(count)
emptiest_sector_count=min(count)
figure(3);bar(count);grid on
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thanks in advance
John
<mailto:jgb2012@sky.com jgb2012@sky.com>
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