How to Read or access diagonal pixels of an Image from (R1,C1) to (R2,C2)?

Question

AMAR P 2018-9-19

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https://ww2.mathworks.cn/matlabcentral/answers/419791-how-to-read-or-access-diagonal-pixels-of-an-image-from-r1-c1-to-r2-c2

评论： Saurabh Patel 2018-9-21

Following is the image am working on. * it's more of splitting an image in sort-of-45degree.* I have taken a random samples along ROW and COLUMNS.

SelectRowSample = 1   112   223   334   445   555
SelectColSample = 1   141   281   421   561   700

1st Blue line start from (141,1) and should end on (1,112).
2nd from (281,1) and should end on (1,223).and onwards.

ImageSize = 555x700

what I have tried?

for ColNum = 1:AllSampleCols
    RowElem = 1;
    for ColElem = SelectColSample(ColNum):-1:1     
        text(ColElem,RowElem,'.','Color','b');
        if (RowElem < SelectRowSample(ColNum))
            RowElem = RowElem + 1;
        end
    end
end

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AMAR P 2018-9-21

Answer/Solution:

Get Co-Ords of the Points that you wish to join.
Calculate length of a every line segment.
Calculate absolute Co-Ords of the each pixel using linspace.
Use X and y Co-ords to draw Plot.

Thanks..

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Answer 1

Saurabh Patel 2018-9-20

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/419791-how-to-read-or-access-diagonal-pixels-of-an-image-from-r1-c1-to-r2-c2#answer_337612

在 MATLAB Online 中打开

Try the following:

 % Specify the value of (r1,c1) and (r2,c2)
r1 = 1; c1 = 112;
r2 = 141; c2 = 1;
 % alpha is parameter used to reach from point 1 to point 2
alpha = (0:0.1:1)';
 % (r,c) are the points on the line
r = round(r1+alpha*(r2-r1));
c = round(c1+alpha*(c2-c1));
 % extracting the image intensity at (r,c)
index = sub2ind(size(I),r,c);
selectedI = I(index);
 % Check if this is what you desire
figure(), hold on
imagesc(I)
plot(c,r)

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AMAR P 2018-9-21

在 MATLAB Online 中打开

I think this can work. Will give it a try. What do you think about using.

linspace()

instead of

r = round(r1+alpha*(r2-r1));
c = round(c1+alpha*(c2-c1));

Saurabh Patel 2018-9-21

Yes, you can use linspace(). It essentially does the same. The reason I used parametric form is to keep it clear that we are getting the image coordinates of the pixels lying on the line (the r and c expressed here as typical parametric form of a line).

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