How can I loop over a binary image to get 4 equal quadrants always?
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I am following this steps to get 4 equal quadrants from a binary image:
r1=q1(1:size(q1,1)/2,1:size(q1,2)/2,:);
r2=q1(size(q1,1)/2+1:size(q1,1),1:size(q1,2)/2,:);
r3=q1(1:size(q1,1)/2,size(q1,2)/2+1:size(q1,2),:);
r4=q1(size(q1,1)/2+1:size(q1,1),size(q1,2)/2+1:size(q1,2),:);
After this, I want to get 4 equal quadrants from r1,r2,r3,r4 indivisually. This process will continue as long as we can get 4 equal quadrants . How can I simply do this with the help of any loop.
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Walter Roberson
2019년 4월 17일
To confirm: if the original image happened to be a power of 2 on each side, such as 512 x 512, then you would want the division into equal quadrants to continue right down to the point where the quadrant sizes were 1 x 1 ? But for a 513 x 513 matrix, then no sub-division would be done, since you cannot make equal quardrants ?
답변 (2개)
Jan
2019년 4월 17일
편집: Jan
2019년 4월 17일
Using 4 distinct variables is less convenient than reshaping the array:
s = size(q1);
q2 = reshape(q1, s(1)/2, 2, s(2)/2, 2);
Now you have e.g. your r4 stored in q2(:, 2, :, 2). In general:
s = size(q1); % Assuming that q1 is a binary image ==> 2D!
% How often can the size be divided by 2:
n = min(sum(factor(s(1)) == 2), sum(factor(s(2)) == 2));
twos = repmat(2, 1, n);
div = 2^n;
qq = reshape(q1, [s(1) / div, twos, s(2) / div, twos]);
Perhaps you want to permute the array (you did not mention, what you need as output).
qq = permute(qq, [1, n+2, 2:n+1, n+3:2*n+2]);
% And maybe:
qq = reshape(qq, s(1)/div, s(2)/div, []);
Now qq(:, :, i) contains the i.th quadrant.
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Jan
2019년 4월 17일
What is "no of white pixels count" and what is the relation to the original question?
Walter Roberson
2019년 4월 17일
function splitted = rsplit4(img)
[r, c, p] = size(img);
if mod(r,2) || mod(c,2)
splitted = img;
else
splitted = {rsplit4(img(1:end/2,1:end/2, :)), rsplit4(img(1:end/2,end/2+1:end, :));
rsplit4(img(end/2+1:end,1:end/2,:)), rsplit4(img(end/2+1:end,end/2+1:end,:))};
end
end
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Walter Roberson
2019년 4월 17일
So you have, in this example, an 80 x 160 image, and you want to sub-divide down to 10 x 10 ? It is not clear why you did not continue on to 5 x 5.
The centroid of the 80 x 160 would be at (40, 80), and the maximum horizontal or vertical distance from the centroid to the edge would be 80. Your previous requirements say that we must take an 80 x 80 square on each side of the centroid. However, with the centroid being at (40,80), we cannot take an 80x80 square to either side of it.
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