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

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?

Answer 1

[Matheus Ferreira]

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,[])

Answer 2

[DGM]

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 = imread('tiltrect.png');
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.

[Image Analyst]

I totally agree. If you just get the pixels on either side of the line, that's not really useful information since it's just a vector of ones (first while you're inside the object) and then zeros (once you're outside of the white object). What good is that? Why don't you explain what you really need?

[Matheus Ferreira]

Thank for the prompt reply! I will better formulate my question.

I need to compute the sensor zenith angle for each pixel of this image. The center line represents the nadir view, that is, the sensor is looking straight down and the sensor zenith angle is zero. It increases in the direction of the image boarders.

Basically, for a given a line of pixels, in the same direction of the object’s orientation, I need to know for each pixel how far it is from the pixel of the center line (value=1 in the image).

[Image Analyst]

You know how to use the Pythagorean Theorem to find the Euclidean distance, right? So if the centerline pixel is at (row, column) = (100, 200) and the other pixel is at (101, 200), then it's one pixel away. Not sure what good that does you, but there you have it.

There are also some posts on point to line distance if you want that instead.

[DGM]

I'm not really sure what that description means, or whether my answer still is appropriate to your needs.

[Matheus Ferreira]

A first start is: how to isolate a single line of pixels of the object in the same direction of the object?

For example, the regionprops function returns the “Orientation” (Angle between the x-axis and the major axis of the ellipse that has the same second-moments as the region), how to isolate a line of pixels in this orientation?

[Image Analyst]

Note that the Orientation will in general not be down the centerline of the rectangle since it's the angle of an ellipse fitted to the data. So usually it's tilted towards one of the diagonals. So you might have to use

https://www.mathworks.com/matlabcentral/fileexchange/34767-a-suite-of-minimal-bounding-objects?focused=120b85be-0407-823c-5262-52a45a39de03&tab=function

or bwferet to find the centerline. Then you can rotate it with imrotate if you want. If you just want to find the distance of all pixels from the centerline, and again I'm not sure you need to, then you can just make the center line a binary image and get the distance transform with bwdist. This will get you the distance of all pixels to the closest point in the line.

[Matheus Ferreira]

Thanks, I found a way to do. Check out the answer above.