MATLAB: How can I calculate number of pixel in each angle of a circle?

circlepolar histogram

hello, Can someone help me please I need to calculate number of pixel in each angle of a circle arround image like this

Best Answer

Dear Zaafouri

the previous script works, the result when applied to the second image you have supplied does not have identical bars, have a look:

Explanation:

the centring and framing of the image are both of capital importance.

These 2 parameters are as important as the image itself.

When I commented to work without the polar grid, I left the outer circle, precisely because if you do not define one CENTRE and one outer circle (let me call it the SCOPE WINDOW) then one image has many different counts as combinations of centres and scope windows are guessed.

So, depending upon where you place the centre and how big you define the scope window you most certainly will have different counts, so the measurement is only valid of you define 3 things:

1.- IMAGE

2.- CENTRE

3.- SCOPE WINDOW

Now let's have a look to the second image you have supplied in your comment:

1.- the 1st cell of the previous script inverts the black background 2nd image you have supplied.

C=imread('round2_start_image3.jpg');imshow(C);
C(find(C>125))=255; C(find(C<=125))=0;                                      % simplify pixels range to binary

C(:,:,2)=and(C(:,:,2),C(:,:,3));C(:,:,3)=[];
C=and(C(:,:,1),C(:,:,2));
C=logical(~C);                 
f1=figure(1);imshow(C)
f1.Position=[100 100 1300 1300];
Let's start with same type of image as in the question, with white background.
C=imread('12.jpg');imshow(C);
C(find(C>125))=255; C(find(C<=125))=0;                                      % simplify pixels range to binary
C(:,:,2)=and(C(:,:,2),C(:,:,3));C(:,:,3)=[];
C=and(C(:,:,1),C(:,:,2));
C=logical(~C);                 
f1=figure(1);imshow(C)
f1.Position=[100 100 1300 1300];
2.- take reference points and see whether it works on 1st sector:
p_center=ginput(1);                                                                     % manual input centre PPI
p_edge=ginput(1);                                                                       % manual input any point of PPI edge
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));
% make sure chosen sector is multiple of 360
% if rem(360,sector)>0 break, or ask to input again
r=abs((p_center(1)-p_edge(1))^2-(p_center(2)-p_edge(2))^2)^.5;  % calculate PPI radius in pixels
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;
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)];
plot(sector_scan(1,:),sector_scan(2,:),'b','LineWidth',1.5);
D=double(C);
[yq,xq]=find(D);
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,:)'); 
plot(xq(in),yq(in),'*g')
plot(xq(on),yq(on),'*y')
3.- It seems it works, all we have to do now it to count the green dots for each sector, repeating the for loop of the
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

4.- voila,

    uint64(count')
    ans =
                     5864
                     3869
                    21177
                    21238
                     5239
                     5071
                     3723
                     5127
                     5637
                     5021
                     5573
                     5646
    max(count)
    min(count)
    =
          21238.00
    =
           3723.00
5.- and the bars graph:
figure(2);bar(count);grid on

Summary:

So, with clearly defined IMAGE, CENTER and SCOPE WINDOW, you get a measurement. Without aiming, it doesn't work.

so dear Zaafouri,

would you please be so kind to mark my answer as ACCEPTED ANSWER, thanks in advance.

Feel free to contact me by email for additional improvements.

Awaiting your answer

Regards

John Bofarull Guix

<mailto:jgb2012@sky.com jgb2012@sky.com>

Related Solutions

MATLAB: Polar distribution, CENTERED on the center of gravity

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

If you find this answer of any help solving your question,

please mark it as ACCEPTED ANSWER.

To any other reader, if you find my answer of any help, please click on the thumbs-up vote link,

thanks in advance

John

<mailto:jgb2012@sky.com jgb2012@sky.com>

MATLAB: Finding average radius of a binarized image of a circle

If you just want the equivalent circular diameter, you can ask regionprops().

props = regionprops(mask, 'EquivDiameter');
ECD = props.EquivDiameter; % Equivalent Circular Diameter - diameter of a circle with the same number of pixels as your blob.

If you want radius as a function of angle, then that's more complicated and involves using regionprops to find the centroid, bwboundaries() to find the outline, atan2d() to find the angle, and sqrt() to find the distance (radius) at that angle.

Best Answer

Related Solutions

MATLAB: Polar distribution, CENTERED on the center of gravity

MATLAB: Finding average radius of a binarized image of a circle

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