I have a uint16 image of size 512*512. The problem is that the image is all black except there is a medium object of uniform intensity of value 40000. This object has a black hole and I need to extract the indexes of this hole.
So the matrix is in the form of all zeroes, and in some region of the matrix we have a submatrix filled with the value 40000, and in this submatrix we have another submatrix of zeroes. This zero submatrix is the required one to be extracted in terms of its indices. Any suggestions?
A very simple solution would be to use bwlabel. For example:
>> data = 40000.*[0 0 0 0 0; 0 1 1 1 0; 0 1 0 1 0; 0 1 1 1 0; 0 0 0 0 0]
data =
0 0 0 0 0
0 40000 40000 40000 0
0 40000 0 40000 0
0 40000 40000 40000 0
0 0 0 0 0
>> labelMatrix = bwlabel(~data)
labelMatrix =
1 1 1 1 1
1 0 0 0 1
1 0 2 0 1
1 0 0 0 1
1 1 1 1 1
>> holeIndex = find(labelMatrix == 2)
holeIndex =
13
The success of this is based on many assumptions, such as there only being one bright object and one hole in it. A more robust solution would require an example of the images you're analyzing, and might involve some preprocessing of the image and the use of functions such as regionprops to identify areas of the image.
Related
I want to visualize all of the paths on 2D square in matlab.
This code gives me the following figure which consists of a 2D square and randomly distributed of 1 and 0.
https://i.hizliresim.com/Ey4G4D.png
Each 1's must connected with lines from top to bottom.
If there is 1, then there is a way and I have to plot line. Otherwise therre is no way and stop.
Without the boundary elements, there are 3 way for each elements. Each element can go left side, right side or down side.
The top left hand corner's element can go right and down direction.
The top right hand corner's element can go left and down direction.
This is the algorithm of the modelling.
https://i.hizliresim.com/Dy0z0y.jpg
How can I write this code ?
I am waiting your advise :)
Problem analysis
To get an information on the possible paths in your matrix/image you can use the diff function. It calculates the difference between two neighbouring matrix elements along the specified dimension.
The conditions for the existence of a path are:
The difference between the element and its neighbour must be 0
The element itself must be 1
Solution
The following matlab program will create 3 matrices containing the value 1 or true for each element with a path existing to its neighbour.
matrix = logical([1 1 1 1 0; ...
1 1 0 1 1; ...
0 0 0 1 0; ...
0 0 1 1 0])
hasPathtoRight = false(size(matrix));
hasPathtoRight(:,1:end-1) = (diff(matrix,1,2)==0) & (matrix(:,1:end-1)==1)
hasPathtoLeft = false(size(matrix));
hasPathtoLeft(:,2:end) = (diff(matrix,1,2)==0) & (matrix(:,2:end)==1)
hasPathDown = false(size(matrix));
hasPathDown(1:end-1,:) = (diff(matrix,1,1)==0) & (matrix(1:end-1,:)==1)
Result
The result for the example matrix is shown here:
matrix =
1 1 1 1 0
1 1 0 1 1
0 0 0 1 0
0 0 1 1 0
hasPathtoRight =
1 1 1 0 0
1 0 0 1 0
0 0 0 0 0
0 0 1 0 0
hasPathtoLeft =
0 1 1 1 0
0 1 0 0 1
0 0 0 0 0
0 0 0 1 0
hasPathDown =
1 1 0 1 0
0 0 0 1 0
0 0 0 1 0
0 0 0 0 0
You can use these matrices to draw the paths in a graphical display.
I have got a 2D matrix. There is some region in the matrix where the elements are non-zero, in particular everywhere around the edge they are zero.
I plot the matrix using image as a colorplot and would like to add the curve that shows the boundary between non-zero values to zero values in the matrix. Is there any neat way to do this without loops?
This looks like a job for convhull :
To illustrate this code i'll take a dummy example :
A=zeros(10);
B=binornd(1,0.5,8,8);
A(2:end-1,2:end-1)=B
A =
0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 1 1 1 1 1 1 0 0
0 0 1 1 0 0 0 0 1 0
0 0 0 1 0 0 0 1 0 0
0 1 0 0 0 0 0 1 0 0
0 0 0 1 1 1 1 1 1 0
0 0 1 0 1 1 1 1 0 0
0 1 0 1 1 1 1 0 1 0
0 0 0 0 0 0 0 0 0 0
1/ Find the locations of all non zero entries :
[row,col]=find(A);
2/ Take the convex hull of these locations
k=convhull(row,col);
3/ Plot the convex hull (I plot the non zero points aswell but in your problem it will be your image points)
plot(row(k),col(k),'r-',row,col,'b*')
Result :
Another option is using the image processing toolbox and the bwperim function. This will work if you know that your area is completely closed (i.e. has no holes in the boundary)
This is an example using a black and white image, and you have 2 options: fill the inner gaps before, or not. You can see in the result the differences.
A = imread('circles.png');
Afill=imfill(A,'holes'); % optional
Abound1=bwperim(Afill);
Abound2=bwperim(A);
imshow([A,Abound, Abound2])
You can plot one on top of the other with:
[x,y]= find(Abound2);
hold on
image(A*255) %// If A is logical, else use just A (not *255)
colormap('gray')
plot(y,x,'r.')
hold off
axis tight
If you have a gray-scale image (or a matrix with a single value in each position (2D matrix), then you can binarize it first by either:
If you know everything outside your object is EXACTLY zero
A=yourA>0;
If you want to separate your object from the background, and the background is not exactly zero by A=im2bw(yourA,level), by choosing your own level, or letting Otsu do it for you with level=graythresh(yourA)
I have an array of zeros and ones and I need to know if the data is spread out across the columns or concentrated in clumps.
For example:
If I have array x and it has these values:
Column 1 values: 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1
Column 2 values: 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1
if we counted the number of ones we can know that it is the same number but the ones are more well spread out and distributed in column 2 compared with column 1.
I am trying to make a score that gives me a high value if the spreading is good and low value if the spreading is bad... any ideas??
Sample of Data:
1 0 0 0 5 0 -2 -3 0 0 1
1 0 0 0 0 0 0 0 0 0 1
2 0 0 0 0 0 0 3 -3 1 0
1 2 3 0 5 0 2 13 4 5 1
1 0 0 0 0 0 -4 34 0 0 1
I think what you're trying to measure is the variance of the distribution of the number of 0s between the 1s, i.e:
f = #(x)std(diff(find(x)))
So for you data:
a = [1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1]
b = [1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1]
f(a)
= 8.0498
f(b)
= 2.0736
But I still think you're essentially trying to measure the disorder of the system which is what I imagine entropy measures but I don't know how
Note that this gives a low value if the "spreading" is good and a high value if it is bad (i.e. the opposite of your request).
Also if you want it per column then it becomes a little more complicated:
f = #(x)arrayfun(#(y)std(diff(find(x(:,y)))), 1:size(x,2))
data = [a', b'];
f(data)
WARNING: This method pretty much does not consider trailing and leading 0s. I don't know if that's a problem or not. but basically f([0; 0; 0; 1; 1; 1; 0; 0; 0]) returns 0 where as f([1; 0; 0; 1; 0; 1; 0; 0; 0]) returns a positive indicating (incorrectly) that first case is more distributed. One possible fix might be to prepend and append a row of ones to the matrix...
I think you would need an interval to find the "spreadness" locally, otherwise the sample 1 (which is named as Column 1 in the question) would appear as spread too between the 2nd and 3rd ones.
So, following that theory and assuming input_array to be the input array, you can try this approach -
intv = 10; %// Interval
diff_loc = diff(find(input_array))
spread_factor = sum(diff_loc(diff_loc<=intv)) %// desired output/score
For sample 1, spread_factor gives 4 and for sample 2 it is 23.
Another theory that you can employ would be if you assume an interval such that distance between consecutive ones must be greater than or equal to that interval. This theory would lead us to a code like this -
intv = 3; %// Interval
diff_loc = diff(find(input_array))
spread_factor = sum(diff_loc>=intv)
With this new approach - For sample 1, spread_factor is 1 and for sample 2 it is 5.
I have a 480-by-640 matrix A. For each pixel, I want to check its neighbors. The neighbors of the pixel are determined by a value N. For example, this is a part of matrix A where all the zeros are the neighbours of pixel X when N=3:
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 X 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
As shown, because N=3, all these zeros are pixel X's neighbors. The problem is if X is located before the index N=3. Here the neighbors will be pixels with one values:
X 1 1 1 0 0 0
1 1 1 1 0 0 0
1 1 1 1 0 0 0
1 1 1 1 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
Could anyone advise on how to handle this?
The simplest way to proceed is just to pad your array with with values that do not return true for whatever you are checking (say, if you're looking for nonzeros, pad with zeros, or if you're looking for finite values, pad with NaN.) The padarray function can do this for you, but requires the Image Processing Toolbox*. Otherwise, you can pad arrays yourself. For example, an unoptimized way to proceed might be
A = rand(m,n);
Apadded = [zeros(N,2*N+n); [zeros(m,N), A, zeros(m,N)]; zeros(N,2*N+n)];
for i = N+1:N+m+1
for j = N+1:N+n+1
% Process neighborhood of A(i,j)
end
end
*Also note that these sorts of "sliding neighborhood" operations, being common in image processing, are implemented for you in the Image Processing Toolbox.
Given a matrix where 1 is the current subset
test =
0 0 0 0 0 0
0 0 0 0 0 0
0 0 1 1 0 0
0 0 1 1 0 0
0 0 0 0 0 0
0 0 0 0 0 0
Is there a function, or quick method to get change the subset to the boundary of the current subset?
Eg. Get this subset from 'test' above
test =
0 0 0 0 0 0
0 1 1 1 1 0
0 1 0 0 1 0
0 1 0 0 1 0
0 1 1 1 1 0
0 0 0 0 0 0
In the end I just want to get the minimum of the cells surrounding a subset of a matrix. Sure I could loop through and get the minimum of the boundary (cell by cell), but there must be a way to do it with the method i've shown above.
Note the subset WILL be connected, but may not be rectangular. This may be the big catch.
This is a possible subset.... (Would pad this with a NaN border)
test =
0 0 0 0 0 0
0 0 0 0 0 0
0 0 1 1 0 0
0 0 1 1 0 0
0 0 1 1 1 1
0 0 1 1 1 1
Ideas?
The basic steps I'd use are:
Perform a dilation on the shape to get a new area which is the shape plus its boundary
Subtract the original shape from the dilated shape to leave just the boundary
Use the boundary to index your data matrix, then take the minimum.
Dilation
What I want to do here is pass a 3x3 window over each cell and take the maximum value in that window:
[m, n] = size(A); % assuming A is your original shape matrix
APadded = zeros(m + 2, n + 2);
APadded(2:end-1, 2:end-1) = A; % pad A with zeroes on each side
ADilated = zeros(m + 2, n + 2); % this will hold the dilated shape.
for i = 1:m
for j = 1:n
mask = zeros(size(APadded));
mask(i:i+2, j:j+2) = 1; % this places a 3x3 square of 1's around (i, j)
ADilated(i + 1, j + 1) = max(APadded(mask));
end
end
Shape subtraction
This is basically a logical AND and a logical NOT to remove the intersection:
ABoundary = ADilated & (~APadded);
At this stage you may want to remove the border we added to do the dilation, since we don't need it any more.
ABoundary = ABoundary(2:end-1, 2:end-1);
Find the minimum data point along the boundary
We can use our logical boundary to index the original data into a vector, then just take the minimum of that vector.
dataMinimum = min(data(ABoundary));
You should look at this as morphology problem, not set theory. This can be solved pretty easily with imdilate() (requires the image package). You basically only need to subtract the image to its dilation with a 3x3 matrix of 1.
octave> test = logical ([0 0 0 0 0 0
0 0 0 0 0 0
0 0 1 1 0 0
0 0 1 1 0 0
0 0 1 1 1 1
0 0 1 1 1 1]);
octave> imdilate (test, true (3)) - test
ans =
0 0 0 0 0 0
0 1 1 1 1 0
0 1 0 0 1 0
0 1 0 0 1 1
0 1 0 0 0 0
0 1 0 0 0 0
It does not, however, pads with NaN. If you really want that, you could pad your original matrix with false, do the operation, and then check if there's any true values in the border.
Note that you don't have to use logical() in which case you'll have to use ones() instead of true(). But that takes more memory and has worse performance.
EDIT: since you are trying to do it without using any matlab toolbox, take a look at the source of imdilate() in Octave. For the case of logical matrices (which is your case) it's a simple usage of filter2() which belongs to matlab core. That said, the following one line should work fine and be much faster
octave> (filter2 (true (3), test) > 0) - test
ans =
0 0 0 0 0 0
0 1 1 1 1 0
0 1 0 0 1 0
0 1 0 0 1 1
0 1 0 0 0 0
0 1 0 0 0 0
One possible solution is to take the subset and add it to the original matrix, but ensure that each time you add it, you offset its position by +1 row, -1 row and +1 column, -1 column. The result will then be expanded by one row and column all around the original subset. You then use the original matrix to mask the original subet to zero.
Like this:
test_new = test + ...
[[test(2:end,2:end);zeros(1,size(test,1)-1)],zeros(size(test,1),1)] + ... %move subset up-left
[[zeros(1,size(test,1)-1);test(1:end-1,2:end)],zeros(size(test,1),1)] + ... %move down-left
[zeros(size(test,1),1),[test(2:end,1:end-1);zeros(1,size(test,1)-1)]] + ... %move subset up-right
[zeros(size(test,1),1),[zeros(1,size(test,1)-1);test(1:end-1,1:end-1)]]; %move subset down-right
test_masked = test_new.*~test; %mask with original matrix
result = test_masked;
result(result>1)=1; % ensure that there is only 1's, not 2, 3, etc.
The result for this on your test matrix is:
result =
0 0 0 0 0 0
0 1 1 1 1 0
0 1 0 0 1 0
0 1 0 0 1 1
0 1 0 0 0 0
0 1 0 0 0 0
Edited - it now grabs the corners as well, by moving the subset up and to the left, up and to the right, down then left and down then right.
I expect this would be a very quick way to achieve this - it doesn't have any loops, nor functions - just matrix operations.