Image processing calculations edge

I have a script that performs calculations on a image. I want to have the pixels on the edge to have different weighing in this calculation.
I believe the eta20 being produced normally (without the *2) is correct. I'd like to be able to manipulate its value by changing the weight assigned to the boundary pixels
a = ['1.jpg'];
b = imread(a);
a = double(a);
[M, N] = size(a);
[x, y] = meshgrid(1:N, 1:M);
[p, q] = size(b); % Turning image into matrix or p by q dimensions
% Turn x, y, and a into column vectors.
x = x(:);
y = y(:);
a = a(:);
BW = edge(b,'sobel'); %inbuilt matlab function that gives '1' for boundary pixels
imshow(BW)
% Loop through the matrix which is assigning different moments if the pixel
% is on the boundary or not.
for r = 1:p
for c = 1:q
if BW(r,c)==1 %If value is 1 pixel is on the boundary
m.m00 = sum(a);
m.m10 = sum(x .* a);
m.m01 = sum(y .* a);
m.m20 = sum(x.^2 .* a)*2; % *2 has no on effect on eta20
else
m.m00 = sum(a);
m.m10 = sum(x .* a);
m.m01 = sum(y .* a);
m.m20 = sum(x.^2 .* a)*2; %*2 has effect on eta20
end
end
end
xbar = m.m10 / m.m00;
ybar = m.m01 / m.m00;
e.eta20 = (m.m20 - xbar*m.m10) / m.m00^2
If someone could explain what I did wrong? I know its probably more math than matlab.
Note: The final purpose is to calculate Hu's moments which consider the pixels at the edge of the image to be more important in the final calculations but that's exactly what isn't happening

 採用された回答

Image Analyst
Image Analyst 2013 年 12 月 22 日

0 投票

BW equals 1 not just for the boundary but everywhere it's true. If you want to find pixels on the boundary, use bwperim:
boundaryPixels = bwperim(BW);
[boundaryRows, boundaryCols] = find(boundaryPixels);

3 件のコメント

Minhal
Minhal 2013 年 12 月 23 日
Okay now I replaced BW = edge(b,'sobel') by BW = bwperim(b) but now its just the opposite now and the part in the else has no effect now.
I see ([boundaryRows, boundaryCols] = find(boundaryPixels);) gives the location of all pixels on the boundary. Is there a way to change my if statement so if the pixel is on the boundary it does the first calculation else it does the other? I tried several ways but it didn't work.
Image Analyst
Image Analyst 2013 年 12 月 23 日
It will not be the opposite. Edges will still be indicated by white lines.
There is nothing to change. If you do
if BW(r,c)==1
then it will do what follows ("the first calculation") if the pixel is on the boundary, just like you asked.
Minhal
Minhal 2013 年 12 月 23 日
Thanks a lot. So the the calculation now considers the boundary pixels. Perfect I really appreciate your help!

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