Do not plot the shape in the middle and polar transformation

Hello Guys, i have the above seen plot and i would like not to plot the shape seen within the circle and the do a polar tranformation. So that that circle appears to be a line.

[Mathieu NOE]

hello

do you have the data and the code used to do this plot ?

[Alex Perrakis]

pic01=imread("1.tiff");
pic1=im2gray(pic01);
i=1;
boundaries=[];

pic050=im2gray( imread("50.tiff") );
%pic50=im2gray(pic050);

newpicture=imsubtract(pic01,pic050);
%imshow(newpicture)
level = graythresh(newpicture);
level = 0.1405;
newbinpic = imbinarize(newpicture,level);
%imshowpair(newpicture,newbinpic,'montage')
newbinpic2=bwpropfilt(newbinpic,'perimeter',1);
% imshowpair(newbinpic2,newpicture,'montage'); 
boundaries = bwboundaries(newbinpic2,'noholes');
boundaries=[boundaries;boundaries];
binaryImage = 0.6< newbinpic2 & newbinpic2<1;




%figure; 
%imshow(pic01);
t = 0:1:350; %setup time
clors= hsv(numel(t)); %create a color map of different color for each time
figure(1),clf,hold on
%hold on
t1=1;

for k=1:length(boundaries)
thisBoundary = boundaries{k};
x = thisBoundary(:,2);
y = thisBoundary(:,1);
[theta,rho] = cart2pol(x,y);
plot(x,y,'color',clors(t1,:),'Linewidth',2);

t1 = t1+10;
end
grid off

Οf course :D

[Alex Perrakis]

Ah scratch the cart2pol line it does not work.

[Adam Danz]

We can't run that code without having 1.tiff. You could attach that file.

[Alex Perrakis]

I tried to attach em as tiff but it does not accept em. so i attached em as png

[Alex Perrakis]

and here is the 50

Answer 1

[Adam Danz]

This demo uses a cleaned up version of your code and then does the following,

1. Computes center of noisy circle
2. Shifts noisy circle to (0,0)
3. Converts shifted image to polar coordinates
4. finds the spike in rho values (radii) based on thresholding
5. Removes the spike.

If you want a smooth circle, you could fit the final (x,y) values to a cirlce (see example) or just use the median rho values in polar coordinates.

% Your code cleaned up
pic01=imread("1.png");
pic050=im2gray( imread("50.png") );
newpicture=imsubtract(pic01,pic050);
%imshow(newpicture)
% level = graythresh(newpicture);
level = 0.1405;
newbinpic = imbinarize(newpicture,level);
%imshowpair(newpicture,newbinpic,'montage')
newbinpic2=bwpropfilt(newbinpic,'perimeter',1);
% imshowpair(newbinpic2,newpicture,'montage');
boundaries = bwboundaries(newbinpic2,'noholes');

% Plot raw (x,y) values
x = boundaries{1}(:,2);
y = boundaries{1}(:,1);
figure()
subplot(2,2,1)
plot(x,y, 'r-')
title('raw values')
axis equal tight
hold on

% Assuming (x,y) is a noisy circle, find center 
cnt = [min(x)+range(x)/2, min(y)+range(y)/2];
xline(cnt(1))
yline(cnt(2))

% Shift to (0,0) and compute polar coordinates
[theta, rho] = cart2pol(x-cnt(1), y-cnt(2));
% Plot in polar coord
subplot(2,2,2)
polarplot(theta, rho)
title('Polar coord.')

% Define a reasonable threshold to isolate the spike in RHO values
% 90% of the median looks OK
subplot(2,2,3)
plot(rho)
threshold = median(rho)*.9; 
yline(threshold, '-k', 'Threshold')
ylabel('rho'); xlabel('index')
title('rho values')

% Find start and end of peak
peakIdx = find(rho < threshold); 
peakIdx(2:end-1) = []; 
xline(peakIdx(1))  % plot start of peak
xline(peakIdx(2))  % plot end of peak
% Remove the peak
theta(peakIdx(1):peakIdx(2)) = [];  % you can do the same with x 
rho(peakIdx(1):peakIdx(2)) = [];    % you can do the same with y 

% Plot the circle without the interior object
subplot(2,2,4)
polarplot(theta,rho)
title('Inner removed')

[Alex Perrakis]

Awesome Bro, thanks!

[Alex Perrakis]

Check for incorrect argument data type or missing argument in call to function 'range'.

Error in aha2 (line 53)

cnt = [min(x)+range(x)/2, min(y)+range(y)/2];

For some reason i get this error, is there a toolbox you use?

[Adam Danz]

It looks like range is in the stats toolbox.

Here's an anonymous function you can include at the top of the code that will replace range()

range = @(x)max(x(:))-min(x(:));

[Alex Perrakis]

thanks bro you did me a solid.