how can I calculate number of pixel in each angle of a circle??

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

Answer 1

[John BG]

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:[e-mail] [e-mail]>

[zaafouri rabaa]

Hi John BG, Attached is the article that I want to make the implementation of the method proposed in it. and please do not forget my question:calcul of the histogram according first to sections and secondly by sections(as presented in the article) my input picture is

Answer 2

[Image Analyst]

What is the resolution of your digital image? You have to know that first. Otherwise you could have a thousand pixels, or a million, or a billion. Who's to say until you tell us how many pixels wide and tall your image is?

Then use the FAQ http://matlab.wikia.com/wiki/FAQ#How_do_I_create_an_arc.3F to create a mask for your image, then just count the pixels:

numPixels = sum(mask(:));

[zaafouri rabaa]

thanks, I'm new in matlab I develop a application of human activity recognition and I want to calculate an histogram that shows CSI's pixel numbers in different sections of bounding circle. CSI is my input image,this image is a superposed human skeleton from video this histogram is defined as the angle and normalized Euclidean distance of each pixel from center point.

[Image Analyst]

I think you forgot or overlooked the part where I asked "tell us how many pixels wide and tall your image is?" because I didn't see an answer.

[zaafouri rabaa]

sorry, I forgot to answer your question, about the number of pixels, I don't know how many pixels in each angle of the circle arround image and about the size of my image is 144x188.

[Image Analyst]

So did you try what I said? My guess is no. Alright, here is code you can use. Modify as needed.

clc;    % Clear the command window.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;

rows = 144;
columns = 188;
myImage = ones(rows, columns);
imshow(myImage);
hold on;

% Define parameters of the arc.
xCenter = columns/2;
yCenter = rows/2;
radius = 50;
% Define the angle theta as going from 30 to 150 degrees in 100 steps.
theta = linspace(120, 150, 60);
% Define x and y using "Degrees" version of sin and cos.
x = radius * cosd(theta) + xCenter; 
y = radius * sind(theta) + yCenter; 
% Tack on center point at the beginning and end of the array
x = [xCenter, x, xCenter];
y = [yCenter, y, yCenter];
% Now plot the points.
plot(x, y, 'b-', 'LineWidth', 2); 
axis equal; 
axis on;
grid on;

% Get a mask.
mask = poly2mask(x, y, rows, columns);
% Sum up the number of pixels
numPixelsInMask = sum(mask(:))
message = sprintf('The number of pixels in the sector is %d', numPixelsInMask);
uiwait(helpdlg(message));

[zaafouri rabaa]

thanks for your help

but it's not that I want I want to calculate a histogram that shows the distrubution of pixels in each angle of a circle arround the image for example at the angle 30 : 300 pixels on the angle 60: 160 pixels .............

I tested the code but he return one value

[zaafouri rabaa]

I am really sorry. I need your help

[Image Analyst]

An angle won't have any pixels in it because it's infinitely thin. A sector will, because it goes from one angle to another. However to calculate a histogram you should have gray levels in your image. Your image is just black/0 and white/255 so your histogram for a given sector will just have two values. Is that what you want?

Or, your image seems to be a polar plot of some (r, theta) list of points. If you want the histogram of the original data, rather than the image which is what you asked for, then you can just take histogram(theta) - the histogram of the theta values that you plotted.

[zaafouri rabaa]

yes I know I have only white and black in my image. (binary image) I want to have how many white pixels in each sector.

[Image Analyst]

Then, just add up the number of white pixels in your binary image that are in the sector-shaped mask. Change the last few lines of my code like this:

% Sum up the number of white pixels in your 
% binary image that are in the sector-shaped mask
numPixelsInMask = sum(binaryImage(mask))
message = sprintf('The number of white pixels in the sector is %d', numPixelsInMask);

Answer 3

[John BG]

Hi Zaafouri

You don't really need to go down to counting the pixels of the arcs closing each sector.

1.- acquire image

    C=imread('original_without_grid.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];

on the left your input image, without the polar grid, you will see in point 3 of this answer why you need to remove the grid, and on the left, the acquired image as binary map, whites=1 blacks=0.

2.- get the centre point, and just a reference point on the right hand side of the PPI outer circle

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

note that there is no check whether the sector angle is multiple of 360º, let's leave this as possible improvement.

3.- calculating sector parameters

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')

on the left, the cluttered result when loading the input image WITH polar grid, on the right, the points of interest in green colour, when loading the input image WITHOUT the polar grid. Note there is still a bit of clutter caught on the outer corners of the sector. Don't worry, I repeat, there is no need for the pixels of the arc you have been asked for earlier on, I will show you how to draw arcs to close the sectors, it's fairly easy, but first let's get the result.

4.- so, the loop you may be most interested on

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

and the result:

     count'
     =
            124.00
            615.00
           1882.00
           2112.00
            829.00
            114.00
            111.00
            110.00
            260.00
           1186.00
            178.00
             38.00
    max(count)
     =
           2112.00

    min(count)
    =
             38.00

a bit of graphing

figure(2);bar(count);grid on

I wonder if you would not mind, Zaafori, if you find this answer useful olving your question, please mark it as ACCEPTED ANSWER.

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

thanks in advance

John BG

<mailto:[e-mail] [e-mail]>

[zaafouri rabaa]

thank you very much your answer really solves a big problem. but my input image is without grid I have a database that I test it contains several actions with this new image the code he remains the same??

[zaafouri rabaa]

I tested this code with the previous image, but the histogram result is always the same the first 8 sectors are equal but the last 4 sectors are 0 the résulat is incorrect Please you know the problem ?

[Image Analyst]

1. Why do you NOT have the original data that was used to create the polar plots in the first place? Like I said, it would make a lot more sense to do the histogram of the original data than on an image of a graph of it.
2. Secondly, the grid lines and tick marks will affect the number of white pixels in the image. Do you want that to be the case? I'd guess not. If not, then post the images without the gridlines and tick labels.

[zaafouri rabaa]

here is my input image I need to calculate the white pixels in each sector of a circle engloubante image

[Image Analyst]

OK. And for the answer to #1.....???????

[zaafouri rabaa]

I really did not understand your question. I have a picture and i want calculate histogram That shows number of white pixels in Each sector

[Image Analyst]

The image is of a graph, right? Well, how did the graph get made? Somebody must have plotted some data. Where is that data? Why do you have only the image, like a screenshot of your graph, but not the original data used to create the plot?

[zaafouri rabaa]

I have an input video first I detected the moving object in each frame and then I used a skeletonization algorithm to each frame of the video finallent I accumulated the skeleton at the center of a single image it is the image that I want to compute its histogram

[Image Analyst]

So you're aiming a video camera at a paper printout of a polar graph? If you don't like my solution, then have you tried John or Thorsten's solution yet?

[John BG]

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 coherent 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:[e-mail] [e-mail]>

Answer 4

[Thorsten]

 I = imread('../../Downloads/testimage.jpg');

 % use ginput to determine the center: click where the center should be,
 % then hit return
 % [cx cy] = ginput

 cx = 71.0823;
 cy = 74.9057;

 cx = round(cx);
 cy = round(cy);

 % idea: determine an angle for each pixel A and then select those pixels
 % AI for which the image is > 0
 x = (1:size(I, 2)) - cx;
 y = (1:size(I,1)) - cy;
 [X, Y] = meshgrid(x, y);
 A = atan2(X, Y);

 AI = A(I>0);

 hc = deg2rad((-180:30:150)+15); % center of sectors
 hist(AI(:), hc) % x runs from -pi to pi, the angles returned by atan2