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I have been trying to find all connected components using 8 neighbors in a binary image, without using the function "bwlabel".
For example, my input matrix is:
a =
1 1 0 0 0 0 0
1 1 0 0 1 1 0
1 1 0 0 0 1 0
1 1 0 0 0 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
I would to have something like this:
a =
1 1 0 0 0 0 0
1 1 0 0 2 2 0
1 1 0 0 0 2 0
1 1 0 0 0 0 0
0 0 0 0 0 3 0
0 0 0 0 0 0 0
There are 3 connected objects in this image.
This is a common problem in image processing. There are many variations, such as flood filling a region in an image, or finding what pixels belong to the same region. One common approach is to use depth first search. The idea is that you traverse your image from left to right and top to bottom and for any pixels encountered that are equal to 1, you add them to a stack. For each pixel in your stack, you pop off the stack, then look at the neighbouring pixels that are surrounding this pixel. Any pixels that are 1 you add to the stack. You need to keep an additional variable where any pixels you have already visited, you don't add these to the stack. When the stack is empty, we have found those pixels that are an entire region, so you mark these with a unique ID. You then repeat this procedure until you run out of regions in your image.
As such, given that your matrix is stored in A, this is the basic algorithm:
Initialize an array that's the same size as A that is logical. This will record which pixels we have examined or visited. Also initialize an output array B to all zeroes that gives you all of the connected components that you are seeking. Any locations that are zero in the end don't belong to any connected components. Also initialize an ID counter that keeps track of what connected component label each of these will have.
For each location that's in our matrix:
a. If the location is 0, mark this location as visited and continue.
b. If we have already visited this location, then continue.
c. If we have not visited this location... go to Step #3.
Add this unvisited location to a stack.
a. While this stack is not empty...
b. Pop this location off of the stack
c. If we have visited this location, then continue.
d. Else, mark this location as visited and mark this location with the connected components ID.
e. Given this location, look at the 8 neighbouring pixels.
f. Remove those pixels in this list that have been visited, not equal to 1 or out of bounds of the matrix
g. Whatever locations are remaining, add these to the stack.
Once the stack is empty, increment the counter, then go back to Step #2.
Keep going until we have visited all of the locations in our array.
Without further ado, here's the code.
%// Step #1
visited = false(size(A));
[rows,cols] = size(A);
B = zeros(rows,cols);
ID_counter = 1;
%// Step 2
%// For each location in your matrix...
for row = 1 : rows
for col = 1 : cols
%// Step 2a
%// If this location is not 1, mark as visited and continue
if A(row,col) == 0
visited(row,col) = true;
%// Step 2b
%// If we have visited, then continue
elseif visited(row,col)
continue;
%// Step 2c
%// Else...
else
%// Step 3
%// Initialize your stack with this location
stack = [row col];
%// Step 3a
%// While your stack isn't empty...
while ~isempty(stack)
%// Step 3b
%// Pop off the stack
loc = stack(1,:);
stack(1,:) = [];
%// Step 3c
%// If we have visited this location, continue
if visited(loc(1),loc(2))
continue;
end
%// Step 3d
%// Mark location as true and mark this location to be
%// its unique ID
visited(loc(1),loc(2)) = true;
B(loc(1),loc(2)) = ID_counter;
%// Step 3e
%// Look at the 8 neighbouring locations
[locs_y, locs_x] = meshgrid(loc(2)-1:loc(2)+1, loc(1)-1:loc(1)+1);
locs_y = locs_y(:);
locs_x = locs_x(:);
%%%% USE BELOW IF YOU WANT 4-CONNECTEDNESS
% See bottom of answer for explanation
%// Look at the 4 neighbouring locations
% locs_y = [loc(2)-1; loc(2)+1; loc(2); loc(2)];
% locs_x = [loc(1); loc(1); loc(1)-1; loc(1)+1];
%// Get rid of those locations out of bounds
out_of_bounds = locs_x < 1 | locs_x > rows | locs_y < 1 | locs_y > cols;
locs_y(out_of_bounds) = [];
locs_x(out_of_bounds) = [];
%// Step 3f
%// Get rid of those locations already visited
is_visited = visited(sub2ind([rows cols], locs_x, locs_y));
locs_y(is_visited) = [];
locs_x(is_visited) = [];
%// Get rid of those locations that are zero.
is_1 = A(sub2ind([rows cols], locs_x, locs_y));
locs_y(~is_1) = [];
locs_x(~is_1) = [];
%// Step 3g
%// Add remaining locations to the stack
stack = [stack; [locs_x locs_y]];
end
%// Step 4
%// Increment counter once complete region has been examined
ID_counter = ID_counter + 1;
end
end %// Step 5
end
With your example matrix, this is what I get for B:
B =
1 1 0 0 0 0 0
1 1 0 0 2 2 0
1 1 0 0 0 2 0
1 1 0 0 0 0 0
0 0 0 0 0 3 0
0 0 0 0 0 0 0
To search in a 4-connected neighbourhood
To modify the code to search in a 4-connected region, that is only North, East, West and South, the section where you see %// Look at the 8 neighbouring locations, that is:
%// Look at the 8 neighbouring locations
[locs_y, locs_x] = meshgrid(loc(2)-1:loc(2)+1, loc(1)-1:loc(1)+1);
locs_y = locs_y(:);
locs_x = locs_x(:);
To search in a 4-connected fashion, you simply have to modify this code to only give those cardinal directions:
%// Look at the 4 neighbouring locations
locs_y = [loc(2)-1; loc(2)+1; loc(2); loc(2)];
locs_x = [loc(1); loc(1); loc(1)-1; loc(1)+1];
The rest of the code remains untouched.
To match MATLAB's bwlabel function
If you want to match the output of MATLAB's bwlabel function, bwlabel searches for connected components in column major or FORTRAN order. The above code searches in row major or C order. Therefore you simply have to search along columns first rather than rows as what the above code is doing and you do this by swapping the order of the two for loops.
Specifically, instead of doing:
for row = 1 : rows
for col = 1 : cols
....
....
You would do:
for col = 1 : cols
for row = 1 : rows
....
....
This should now replicate the output of bwlabel.
I would like to calculate a couple of texture features (namely: small/ large number emphasis, number non-uniformity, second moment and entropy). Those can be computed from Neighboring gray-level dependence matrix. I'm struggling with understanding/implementation of this. There is very little info on this method (publicly available).
According to this paper:
This matrix takes the form of a two-dimensional array Q, where Q(i,j) can be considered as frequency counts of grayness variation of a processed image. It has a similar meaning as histogram of an image. This array is Ng×Nr where Ng is the number of possible gray levels and Nr is the number of possible neighbours to a pixel in an image.
If the image function f(i,j) is discrete, then it is easy to computer the Q matrix (for positive integer d, a) by counting the number of times the difference between each element in f(i,j) and its neighbours is equal or less than a at a certain distance d.
Here is the example from the same paper (d = 1, a = 0):
Input (image) matrix and output matrix Q:
I've been looking at this example for hours now and still can't figure out how they got that Q matrix. Anyone?
The method was originally created by C. Sun and W. Wee and was described in a paper called: "Neighboring gray level dependence matrix for texture classification" to which I got access, but can't download (after pressing download the page reloads and that's it).
In the example that you have provided, d=1 and a=0. When d=1, we consider pixels in an 8-pixel neighbourhood. When a=0, this means that we look for pixels that have the same value as the centre of the neighbourhood.
The basic algorithm is the following:
Initialize your NGLDM matrix to all zeroes. The total number of rows corresponds to the total number of possible intensities / values in your image. The total number of columns corresponds to how many pixels are in your neighbourhood plus 1. As such for d=1, we have an 8-pixel neighbourhood and so 8 + 1 = 9. Because there are 4 possible intensities (0,1,2,3), we thus have a 4 x 9 matrix. Let's call this matrix M.
For each pixel in your matrix, take note of this pixel. This goes in the Ng row.
Write out how many valid neighbours there are that surround this pixel.
Count how many times you see the neighbouring pixels matching that pixel in Step #1. This is your Nr column.
Once you figure out the numbers in Step #1 and Step #2, increment this location by 1.
Here's a slight gotcha: They ignore the border locations. As such, you don't do this procedure for the first row, last row, first column or last column. My guess is that they want to be sure that you have an 8-pixel neighbourhood all the time. This is also dictated by the distance d=1. You must be able to grab every valid pixel given a centre location at d=1. If d=2, then you would have to make sure that every pixel in the centre of the neighbourhood has a 25 pixel neighbourhood and so on.
Let's start from the second row, second column location of this matrix. Let's go through the steps:
Ng = 1 as the location is 1.
Valid neighbours - Starting from the top left pixel in this neighbourhood, and scanning left to right and omitting the centre, we have: 1, 1, 2, 0, 1, 0, 2, 2.
How many values are equal to 1? Three times. Therefore Nr = 3
M(Ng,Nr) += 1. Access row Ng = 1, and access row Nr = 3, and increment this spot by 1.
Want to know how I figured out they don't count the borders? Let's do the bottom left pixel. That location is 0, so Ng = 0. If you repeat the algorithm that I just said, you would expect Ng = 0, Nr = 1, and so you would expect at least one entry in that location in your matrix... but you don't! If you do similar checks around the border of the image, you'll see that entries that are supposed to be there... aren't. Take a look at the third row, fifth column. You would think that Ng = 1 and Nr = 1, but we don't see that in the matrix.
One more example. Why is M(Ng,Nr) = 4, Ng = 2, Nr = 4? Well, take a look at every pixel that has a 2 in it. The only valid locations where we can capture an 8 pixel neighbourhood successfully are the row=2, col=4, row=3, col=3, row=3, col=4, row=4, col=3, and row=4, col=4. By applying the same algorithm that we have seen, you'll see that for each of those locations, Nr = 4. As such, we see this combination of Ng = 2, Nr = 4 four times, and that's why the location is set to 4. However, in row=3, col=4, this actually is Nr = 5, as there are five 2s in that neighbourhood at that centre. That's why you see Ng = 2, Nr = 5, M(Ng,Nr) = 1.
As an example, let's do one of the locations. Let's do the 2 smack dab in the middle of the matrix (row=3, col=3):
Ng = 2
What are the valid neighbouring pixels? 1, 1, 2, 0, 2, 3, 2, 2 (omit the centre)
Count how many pixels equal to 2. There are four of them, so Nr = 4
M(Ng,Nr) += 1. Take Ng = 2, Nr = 4 and increment this spot by 1.
If you do this with the other valid locations that have 2, you'll see that Nr = 4 each time with the exception of the third row and fourth column, where Nr = 5.
So how would we implement this in MATLAB? What you can do is use im2col to transform each valid neighbourhood into columns. What I'm also going to do is extract the centre of each neighbourhood. This is actually the middle row of the matrix. We will then figure out how many pixels for each neighbourhood equal the centre, sum them up, and this will determine our Nr values. The Ng values will be the middle row values themselves. Once we do this, we can compute a histogram based on these values just like how the algorithm is doing to get our matrix. In other words, try doing this:
% // Your example
A = [1 1 2 3 1; 0 1 1 2 2; 0 0 2 2 1; 3 3 2 2 1; 0 0 2 0 1];
B = im2col(A, [3 3]); %//Convert neighbourhoods to columns - 3 x 3 means d = 1
C = bsxfun(#eq, B, B(5,:)); %//Figure out a logical matrix where each column tells
%//you how many elements equals the one in each centre
D = sum(C, 1) - 1; %// Must subtract by 1 to discount centre pixel
Ng = B(5,:).' + 1; % // We must make this into a column vector, and we also must
% // offset by 1 as MATLAB starts indexing by 1.
%// Column vector is for accumarray input
Nr = D.' + 1; %// Do the same for Nr. We could have simply left out the + 1 here and
%// took out the subtraction of -1 for D, but I want to explicitly show
%// the steps
Q = accumarray([Ng Nr], 1, [4 9]); %// 4 unique intensities, 9 possible locations (0-8)
... and here is our matrix:
Q =
0 0 1 0 0 0 0 0 0
0 0 1 1 0 0 0 0 0
0 0 0 0 4 1 0 0 0
0 1 0 0 0 0 0 0 0
If you check this, you'll see this matches with Q.
Bonus
If you want to be able to accommodate for the algorithm in general, where you specify d and a, we can simply follow the guidelines of your text. For each neighbourhood, you find the difference between the centre pixel and all of the other pixels. You count how many pixels are <= a for any positive integer d. Note that this will create a 2*d + 1 x 2*d + 1 neighbourhood we need to examine. We can also make this into a function. Without further ado:
%// Set A up yourself, then use a and d as inputs
%// Precondition - a and d are both integers. a can be 0 and d is positive!
function [Q] = calculateGrayDepMatrix(A, a, d)
neigh = 2*d + 1; % //Calculate rows/columns of neighbourhood
numTotalNeigh = neigh*neigh; % //Calculate total number of pixels in neighbourhood
middleRow = ceil(numTotalNeigh / 2); %// Figure out which index the middle row is
B = im2col(A, [neigh neigh]); %// Make into columns
Cdiff = abs(bsxfun(#minus, B, B(middleRow,:))); %// For each neighbourhood, subtract with its centre
C = Cdiff <= a; %// For each neighbourhood, figure out which differences are <= a
D = sum(C, 1) - 1; % //For each neighbourhood, add them up
Ng = B(middleRow,:).' + 1; % // Determine Ng and Nr, and find Q
Nr = D.' + 1;
Q = accumarray([Ng Nr], 1, [max(Ng) numTotalNeigh]);
end
We can recreate the scenario we showed above with the example matrix by:
A = [1 1 2 3 1; 0 1 1 2 2; 0 0 2 2 1; 3 3 2 2 1; 0 0 2 0 1];
Q = calculateGrayDepMatrix(A, 0, 1);
Q is thus:
Q =
0 0 1 0 0 0 0 0 0
0 0 1 1 0 0 0 0 0
0 0 0 0 4 1 0 0 0
0 1 0 0 0 0 0 0 0
Hope this helps!
I need an algorithm in Matlab which counts how many adjacent and non-overlapping (1,1) I have in each row of a matrix A mx(n*2) without using loops. E.g.
A=[1 1 1 0 1 1 0 0 0 1; 1 0 1 1 1 1 0 0 1 1] %m=2, n=5
Then I want
B=[2;3] %mx1
Specific case
Assuming A to have ones and zeros only, this could be one way -
B = sum(reshape(sum(reshape(A',2,[]))==2,size(A,2)/2,[]))
General case
If you are looking for a general approach that must work for all integers and a case where you can specify the pattern of numbers, you may use this -
patt = [0 1] %%// pattern to be found out
B = sum(reshape(ismember(reshape(A',2,[])',patt,'rows'),[],2))
Output
With patt = [1 1], B = [2 3]
With patt = [0 1], B = [1 0]
you can use transpose then reshape so each consecutive values will now be in a row, then compare the top and bottom row (boolean compare or compare the sum of each row to 2), then sum the result of the comparison and reshape the result to your liking.
in code, it would look like:
A=[1 1 1 0 1 1 0 0 0 1; 1 0 1 1 1 1 0 0 1 1] ;
m = size(A,1) ;
n = size(A,2)/2 ;
Atemp = reshape(A.' , 2 , [] , m ) ;
B = squeeze(sum(sum(Atemp)==2))
You could pack everything in one line of code if you want, but several lines is usually easier for comprehension. For clarity, the Atemp matrix looks like that:
Atemp(:,:,1) =
1 1 1 0 0
1 0 1 0 1
Atemp(:,:,2) =
1 1 1 0 1
0 1 1 0 1
You'll notice that each row of the original A matrix has been broken down in 2 rows element-wise. The second line will simply compare the sum of each row with 2, then sum the valid result of the comparisons.
The squeeze command is only to remove the singleton dimensions not necessary anymore.
you can use imresize , for example
imresize(A,[size(A,1),size(A,2)/2])>0.8
ans =
1 0 1 0 0
0 1 1 0 1
this places 1 where you have [1 1] pairs... then you can just use sum
For any pair type [x y] you can :
x=0; y=1;
R(size(A,1),size(A,2)/2)=0; % prealocarting memory
for n=1:size(A,1)
b=[A(n,1:2:end)' A(n,2:2:end)']
try
R(n,find(b(:,1)==x & b(:,2)==y))=1;
end
end
R =
0 0 0 0 1
0 0 0 0 0
With diff (to detect start and end of each run of ones) and accumarray (to group runs of the same row; each run contributes half its length rounded down):
B = diff([zeros(1,size(A,1)); A.'; zeros(1,size(A,1))]); %'// columnwise is easier
[is js] = find(B==1); %// rows and columns of starts of runs of ones
[ie je] = find(B==-1); %// rows and columns of ends of runs of ones
result = accumarray(js, floor((ie-is)/2)); %// sum values for each row of A
I wasn't quite sure how to phrase this question. Suppose I have the following matrix:
A=[1 0 0;
0 0 1;
0 1 0;
0 1 1;
0 1 2;
3 4 4]
Given row 1, I want to find all rows where:
the elements that are unique in row 1, are unique in the same column in the other row, but don't necessarily have the same value
and if there are elements with duplicate values in row 1, there are be duplicate values in the same columns in the other row, but not necessarily the same value
For example, in matrix A, if I was given row 1 I would like to find rows 4 and 6.
Can't test this right now, but I think the following will work:
A=[1 0 0;
0 0 1;
0 1 0;
0 1 1;
0 1 2;
3 4 4];
B = zeros(size(A));
for ii = 1:size(A,1)
r = A(ii,:);
B(ii,1) = 1;
for jj = 2:size(A,2)
c = find(r(1:jj-1)==r(jj));
if numel(c) > 0
B(ii,jj) = B(ii,c);
else
B(ii,jj) = B(ii,jj-1)+1;
end
end
end
At the end of this we have an array B in which "like indices have like values" and the rows you are looking for are now identical.
Now you can do
[C, ia, ic] = unique(B,'rows','stable');
disp('The answer you want is ');
disp(ia);
And the answer you want will be in the variable ia. See http://www.mathworks.com/help/matlab/ref/unique.html#btb0_8v . I am not 100% sure that you can use the rows and stable parameters in the same call - but I think you can.
Try it and see if it works - and ask questions if you need more info.
Here is a simple method
B = NaN(size(A)); %//Preallocation
for row = 1:size(A,1)
[~,~,B(row,:)] = unique(A(row,:), 'stable');
end
find(ismember(B(2:end,:), B(1,:), 'rows')) + 1
A simple solution without loops:
row = 1; %// row used as reference
equal = bsxfun(#eq, A, permute(A, [1 3 2]));
equal = reshape(equal,size(A,1),[]); %// linearized signature of each row
result = find(ismember(equal,equal(row,:),'rows')); %// find matching rows
result = setdiff(result,row); %// remove reference row, if needed
The key is to compute a "signature" of each row, meaning the equality relationship between all combinations of its elements. This is done with bsxfun. Then, rows with the same signature can be easily found with ismember.
Thanks, Floris. The unique call didn't work correctly and I think you meant to use matrix B in it, too. Here's what I managed to do, although it's not as clean:
A=[1 0 0 1;
0 0 1 3;
0 1 0 1;
0 1 1 0;
0 1 2 2;
3 4 4 3;
5 9 9 4];
B = zeros(size(A));
for ii = 1:size(A,1)
r = A(ii,:);
B(ii,1) = 1;
for jj = 2:size(A,2)
c = find(r(1:jj-1)==r(jj));
if numel(c) > 0
B(ii,jj) = B(ii,c);
else
B(ii,jj) = max(B(ii,:))+1; % need max to generalize to more columns
end
end
end
match = zeros(size(A,1)-1,size(A,2));
for i=2:size(A,1)
for j=1:size(A,2)
if B(i,j) == B(1,j)
match(i-1,j)=1;
end
end
end
index=find(sum(match,2)==size(A,2));
In the nested loops I check if the elements in the rows below it match up in the correct column. If there is a perfect match the row should sum to the row dimension.
When I generalize this for the specific problem I'm working on the matrix fills with a certain set of base size(A,2) numbers. So for base 4 and greater, a max statement is needed in the else statement for no matches. Otherwise, for certain number combinations in a given row, a duplication of an element may occur when there is none.
A overview would be to reduce each row into a "signature" counting element repeats, i.e., your row 1 becomes 1, 2. Then check for equal signatures.
Is it possible to decompose a matrix A having n rows and n columns to sum of m [n x n] permutation matrices. where m is the number of 1's in each row and each column in matrix A?
UPDATE:
yes, this is possible. I came across such an exmaple which is shown below - but How can we generalize the answer?
What you want is called a 1-factorization. One algorithm is repeatedly to find a perfect matching and remove it; probably there are others.
For the first permutation matrix, take the first 1 in the first row. For the second row, take the first 1 that is in a column you don't already have. For the third row, take the first 1 that is in a column you don't already have. And so on. Do this for all rows.
You now have one permutation matrix.
Next subtract your first permutation matrix from the original. This new matrix now has m-1 ones in each row and column. So repeat the process m-1 more times, and you'll have your m permutation matrices.
You can skip the last step, because a matrix with one 1 in each row and column already is a permutation matrix. There's no need to do any calculations.
This is a greedy algorithm that doesn't always work. We can make it work by changing the selection rule slightly. See below:
For your example:
1 0 1 1
A = 1 1 0 1
1 1 1 0
0 1 1 1
In the first step, we pick (1,1) for the first row, (2,2) for the second row, (3,3) for the thrid row and (4,4) for the 4th row. We then have:
1 0 0 0 0 0 1 1
A = 0 1 0 0 + 1 0 0 1
0 0 1 0 1 1 0 0
0 0 0 1 0 1 1 0
The first matrix is a permutation matrix. The second matrix has exactly two 1's in each row and column. So we pick, in order: (1,3), (2,1), (3,2) and... we're in trouble: the rows that contain a 1 in column 4 have already been used.
So how do we fix this? Well, we can keep track of the number of 1's remaining in each column. Instead of picking the first column that is unused, we pick the column with the lowest number of 1's remaining. For the second matrix above:
0 0 1 1 0 0 X 0 0 0 X 0 0 0 X 0
B = 1 0 0 1 --> 1 0 0 1 --> 0 0 0 X --> 0 0 0 X
1 1 0 0 1 1 0 0 1 1 0 0 X 0 0 0
0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0
------- ------- ------- -------
2 2 2 2 2 2 X 1 1 2 X X X 1 X X
So we would pick column 4 in the second step, column 1 in the 3rd step, and column 2 in the 4th step.
There can always be only one column with one remaining 1. The other 1's must have been taken away in m-1 previous rows. If you had two such columns, one of them would have had to have been picked as the minimum column before.
This can be done easily using a recursive (backtracking OR depth-first traversal) algorithm. Here is the pseudo-code for its solution:
void printPermutationMatrices(const int OrigMat[][], int permutMat[], int curRow, const int n){
//curPermutMatrix is 1-D array where value of ith element contains the value of column where 1 is placed in ith row
if(curRow == n){//Base case
//do stuff with permutMat[]
printPermutMat(permutMat);
return;
}
for(int col=0; col<n; col++){//try to place 1 in cur_row in each col if possible and go further to next row in recursion
if(origM[cur_row][col] == 1){
permutMat[cur_row] = col;//choose this col for cur_row
if there is no conflict to place a 1 in [cur_row, col] in permutMat[]
perform(origM, curPermutMat, curRow+1, n);
}
}
}
Here is how to call from your main function:
int[] permutMat = new int[n];
printPermutationMatrices(originalMatrix, permutMat, 0, n);