Obtain subscript values of common diagonal rectangle in binary matrix?
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I am trying to find where in the MM_bin binary matrix (see attached MM_bin.mat file) contains the check_mat matrix (see code below), and then extract the subscript coordinates in the MM_bin matrix of the 1s from the check_mat matrix. So far I am able to use a 2D convolution to extract the centroids of the spots where the check_mat matrix is found in the MM_bin matrix, but I am unsure how to extract the subscript coordinates in the MM_bin matrix for each centroid.
load('MM_bin.mat')
check_mat = [0 0 0 0 0 1 1 1 1;
0 0 0 0 1 1 1 1 0;
0 0 0 1 1 1 1 0 0;
0 0 1 1 1 1 0 0 0;
0 1 1 1 1 0 0 0 0;
1 1 1 1 0 0 0 0 0];
minLi = 6; % length of the 1s diagonal
minLj = 4; % width of the 1s diagonal
[i_cent, j_cent] = find(conv2(MM_bin(:,:,1), check_mat, 'same') == minLi*minLj)
5 个评论
Bruno Luong
2018-10-12
编辑：Bruno Luong
2018-10-12
Don't understand your question:
- "So far I am able to use a 2D convolution to extract the centroids of the spots where the check_mat matrix is found in the MM_bin matrix"
In what form you have the extracted centroid? It must be some sort of coordinates.
- "but I am unsure how to extract the subscript coordinates in the MM_bin matrix for each centroid."
What is the difference between those two notions????
Andrew Poissant
2018-10-12
Surely, if you have the centroid, it's trivial to find the corners, so I'm unsure what you're asking.
However, unless I misunderstood completely your convolution code does not work. It finds the pattern of 1 but ignores the pattern of zeros. e.g, if you try it on:
MM_bin = ones(7, 9);
i_cent, j_cent] = find(conv2(MM_bin(:,:,1), check_mat, 'same') == minLi*minLj)
The ouput is
i_cent =
3
4
j_cent =
5
5
when it should be empty?
You cannot use convolution for your search
Andrew Poissant
2018-10-12
Guillaume
2018-10-12
If you're just looking for the pattern of ones and you don't care if the 0s in your check_mat match a 0 or a 1, then your code is correct and I misunderstood. If the 0s must match a 0, then it cannot work with a convolution.
Assuming your code is correct, I still don't understand what final out you want. Perhaps, provide an example with smaller matrices.
E.G., with
MM_bin = [0 1 0 1 0
0 1 1 1 1
1 0 1 1 1]
check_mat = [0 1
1 0]
what final output do you want?
采纳的回答
Bruno Luong
2018-10-12
编辑：Bruno Luong
2018-10-12
% Fake data
MM_bin=rand(100)>0.2;
check_mat = [0 0 0 0 0 1 1 1 1;
0 0 0 0 1 1 1 1 0;
0 0 0 1 1 1 1 0 0;
0 0 1 1 1 1 0 0 0;
0 1 1 1 1 0 0 0 0;
1 1 1 1 0 0 0 0 0];
s = size(check_mat);
% Careful if your pattern is not symmetric you must flip it
% for each dimension when using with CONV
[i_cent, j_cent] = find(conv2(MM_bin(:,:,1), ...
fliplr(flipud(check_mat)), 'same') == sum(check_mat(:)));
[ip,jp] = find(check_mat);
i = floor(i_cent(:)-s(1)/2)+ip(:).';
j = floor(j_cent(:)-s(2)/2)+jp(:).';
% remove redundancy
ij = unique([i(:) j(:)],'rows');
% graphical check
close all
imagesc(MM_bin);
hold on
plot(ij(:,2),ij(:,1),'.r') % NOTE: x-axis is second dimension
axis equal
更多回答（3 个）
Image Analyst
2018-10-12
To find the centroid of the 1's, if you have the Image Processing Toolbox, you can use regionprops
check_mat = [0 0 0 0 0 1 1 1 1;
0 0 0 0 1 1 1 1 0;
0 0 0 1 1 1 1 0 0;
0 0 1 1 1 1 0 0 0;
0 1 1 1 1 0 0 0 0;
1 1 1 1 0 0 0 0 0];
props = regionprops(check_mat, 'Centroid');
xCentroid = props.Centroid(1)
yCentroid = props.Centroid(2)
9 个评论
Andrew Poissant
2018-10-12
Andrew Poissant
2018-10-12
Image Analyst
2018-10-12
编辑：Image Analyst
2018-10-12
No, you can't use convolution, in general. Think about it and you'll know why. You'd have to check every single element, which is what isequal does(). You simply use isequal() like I said in my answer below. Basically use a for loop like
for col = 1 : columns
for row = 1 : rows
thisWindow = binaryImage(row1:row2, col1:col2)
if isequal(thisWindow, mask)
% It's found!
and so on. Let me know if you can't finish it.
Bruno Luong
2018-10-12
The convolution is perfectly a fine method if he looks for matching the pattern of 1s and doesn't care about the surrounding 0s.
I think that what he wants to achieve.
Image Analyst
2018-10-12
In addition, you'd have to check for the image to be equal to the exact number of 1's in the mask/kernel/window, i.e. 24.
output = conv2(MM_bin, check_mat, 'same');
foundLocations = MM_bin == sum(check_mat(:));
With isequal(), you don't have to do that. But isequal() searches for exact matches while conv2() searches for a match within the 1 area and ignores any 1's in the 0 region of the mask Bruno said. Depends on what he means by "centroids of the spots where the check_mat matrix is found in the MM_bin matrix". I think he means the entire array of both 0 and 1 are found, not just where the 1's are found. If you had an MM_bin image of all solid 1's then the conv2 would be 24 everywhere and then you'd be finding the centroid everywhere which doesn't make much sense to me.
Bruno Luong
2018-10-12
编辑：Bruno Luong
2018-10-12
Why not? May be the image is the gradient of something or a skeleton, so OP just want to find a contour that matches a pattern that going in some direction. Such image doesn't have 1s in a very big area.
Matt J
2018-10-12
Bruno Luong
2018-10-12
编辑：Bruno Luong
2018-10-12
Or simply change the single conv2 check to
conv2(Im,p,'same')==sum(p(:)) & conv2(Im,1-p,'same')==sum(1-p(:))
since they are both binary images.
Image Analyst
2018-10-12
It's not clear to me what "(minus the 0s)" means. Maybe "ignoring any zeros"??? In other words, as long as that parallelogram is there, it doesn't matter if, outside the parallelogram, it's all ones, all zeros, or some random pattern or ones and zeros.
Image Analyst
2018-10-12
0 个投票
If you want to find where check_mat occurs in a much larger matrix, simple scan and use isequal().
[I0,J0]=find(check_mat);
I = I0 + (i_cent.'-3); %I subscripts
J = J0 + (j_cent.'-5); %J subscripts
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