# Find pixel coordinates to the right and to the left of a given target pixel

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Matheus Ferreira on 8 Jun 2022
In the image below (also attached), the central line of the object is burned with the value 1. The pixel values of the object are equal to two, and the background pixels (black) have values equal to zero.
Given a pixel from the central line, I want to extract all pixels to the right and the left of it, in the same direction of the object's orientation.
Any ideias on how to do this?

Matheus Ferreira on 9 Jun 2022
I find a way to do this. I will post my answer to anyone interested.
Given the image "I.png" (attached) I fitted a linear model to find the central line and move the intercept to move the central line to the right and to the left. The pixel value of a given pixel represent the number of pixels this pixel is far from the center.
% Fit a linear model using pixel coordinates to find the center line
[y,x] = find(I) ;
p = polyfit(x,y,1);
xi = linspace(min(x),max(x),max(x)) ;
yi = polyval(p,xi) ;
frameLength=100;
Izeros=zeros(size(I));
% Left side
for i= 1:frameLength
aux_p=p;
aux_p(1,2)=p(1,2)+i; % Change the intercept
yi = polyval(aux_p,xi);
centralLinePixValue=i+1;
for k = 1 : length(xi)
row = round(yi(k));
col = round(xi(k));
if row>size(I,1)||col>size(I,2)
row=size(I,1);
col=size(I,2);
end
Izeros(row, col) = centralLinePixValue;
end
end
% Right side
frameLength=70;
for i= 1:70
aux_p=p;
aux_p(1,2)=p(1,2)-i; % Change the intercept
yi = polyval(aux_p,xi);
centralLinePixValue=i+1;
for k = 1 : length(xi)
row = abs(round(yi(k)));
col = abs(round(xi(k)));
if row==0||col==0
row=1;
col=1;
end
if row>size(I,1)||col>size(I,2)
row=size(I,1);
col=size(I,2);
end
Izeros(row, col) = centralLinePixValue;
end
end
Izeros=Izeros.*I;
figure, imshow(Izeros,[])

DGM on 8 Jun 2022
Edited: DGM on 8 Jun 2022
I'm going to assume you already have a method for finding the central line. I'm just going to extract it from the example image given.
The easy way would simply be to find the angle of the object centerline and then rotate the image so that it's grid-aligned. At that point, it's just a matter of addressing columns from the image.
A = rgb2gray(A);
% crop off border artifacts
A = A(3:end-2,3:end-2);
% find angle of object line
objline = A==128;
[H,T,R] = hough(objline);
P = houghpeaks(H,1,'threshold',ceil(0.3*max(H(:))));
lines = houghlines(objline,T,R,P,'FillGap',50,'MinLength',7);
theta = sign(lines.rho)*90 - lines.theta
theta = -23
% rotate the object image to correct for presentation angle
objpict = logical(A);
objpict = imrotate(objpict,-theta);
% close-crop image to object extents
S = regionprops(objpict,'image');
objpict = S(1).Image;
imshow(objpict)
Of course, it's usually unclear what people mean when they say they want to "extract pixels". If you were to literally extract the pixels perpendicular to the centerline through a given point, all you would have is a variable-length vector of ones. The character of ones are not unique. The only information you might get from the process is the length of said vector -- in which case, the question is "how do I calculate the width of the object".
On the other hand, if the logical image is merely a reference to another (unseen) image and you want to extract the pixels from that image, then you'd obviously have to apply the rotation to it instead.
Matheus Ferreira on 9 Jun 2022
Thanks, I found a way to do. Check out the answer above.

R2021b

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