I am trying to implement a gabor filter for use in textured image segmentation. I am doing this in MATLAB and consulting the concepts from paper - A level set and Gabor-based Active Contour Algorithm for Segmenting Textured Images
I am highlighing the relevant parts below: The 2D gabor function is given as
where
The frequency of the span-limited sinusoidal grating is given by F and its orientation is specified by Theta. Sigma is the scale parameter. This filter is to be used on an image and as the gabor filter consists of an imaginary component, the Gabor transform is obtained as shown below.
where, GR and GI are the real and imaginary parts obtained by convoluting it with an image u0. I need to code this part in MATLAB and generate the Gabor transformed image for the different values of theta, F and sigma
My code
clc;clear all;close all;
sigma=.0075;
m_size=7;
theta=pi/4;
F=60;
[real_g,im_g]=gabor(m_size,sigma,F,theta);
//My Gabor function
function [real_g,im_g] = gabor(m_size,sigma,F,theta)
[x,y]=meshgrid(1:m_size,1:m_size);
real_g = zeros(m_size);
im_g = zeros(m_size);
g_sigma = zeros(m_size);
for i=1:size(x,1)
for j=1:size(y,1)
g_sigma(i,j) = (1./(2*pi*sigma^2)).*exp(((-1).*(i^2+j^2))./(2*sigma^2));
real_g(i,j) = g_sigma(i,j).*cos((2*pi*F).*(i.*cos(theta)+j.*sin(theta)));
im_g(i,j) = g_sigma(i,j).*sin((2*pi*F).*(i.*cos(theta)+j.*sin(theta)));
end
end
My output
>> real_g
real_g =
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
>> im_g
im_g =
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
My gabor filter is completely wrong. Please could you guys help me to construct the correct gabor filter? Please note that the data for the parameters and the formula is taken from the paper I already referred.
Any help will be greatly appreciated. PS If anybody needs the paper I can mail it too. Thanks.
Hopefully the below codes would be of some use to what you are working on.
It demonstrate how to transform an image using your Gabor filter with varying thetas (as shown in the images). Cheers.
% get image
u0=double(imread('cameraman.tif'));
% initialize parameters
sigma = 3;
m_size = 7;
F = 1;
m_size_halfed = round((m_size-1)/2);
% make up some thetas
thetas=0:pi/5:pi;
% loop through all thetas
for i = 1:numel(thetas)
theta = thetas(i);
% setup the "gabor transform"
[x,y]=meshgrid(-m_size_halfed:m_size_halfed,-m_size_halfed:m_size_halfed);
g_sigma = (1./(2*pi*sigma^2)).*exp(((-1).*(x.^2+y.^2))./(2*sigma.^2));
real_g = g_sigma.*cos((2*pi*F).*(x.*cos(theta)+y.*sin(theta)));
im_g = g_sigma.*sin((2*pi*F).*(x.*cos(theta)+y.*sin(theta)));
% perform Gabor transform
u_0sft=sqrt(conv2(u0,real_g,'same').^2+conv2(u0,im_g,'same').^2);
subplot(1,numel(thetas)+1,i+1)
imagesc(u_0sft);
colormap('gray'); axis image; axis off;
title(sprintf('theta:%1.1f',theta));
end
% visualize image
subplot(1,numel(thetas)+1,1)
imagesc(u0);
colormap('gray'); axis image; axis off;
title('original');
Related
I'm new in MATLAB and using it for some medical analysis. I have a matrix which contains a circular shape in it. Here is a sample:
0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 1 0 1 1 0 0 0
0 0 1 0 0 0 0 1 0 0
0 1 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 0 0
0 0 1 1 0 1 1 0 0 0
0 0 0 0 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 0
This matrix has been computed by this line:
egslbr= edge(slbr,'log');
Where slbr is my image. By merging the egslbr with my slbr I get the image bellow.
I want to cancel all the color pixel outside the greenish circle. Is there any way to do this?
You should be able to create a mask from your circle (assuming it's a complete circle -- the sample matrix you gave has a gap in it which makes this much more difficult, but I'm going to assume that was a mistake). Here's a simple way of doing it:
mask = ~cumsum(egslbr) | ~cumsum(egslbr,'reverse');
slbr(mask) = 0;
This should set every pixel outside the circle (not including the edge of the circle though) to zero.
I am trying to find islands of numbers in a matrix.
By an island, I mean a rectangular area where ones are connected with each other either horizontally, vertically or diagonally including the boundary layer of zeros
Suppose I have this matrix:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1
0 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0
0 0 0 0 0 0 1 0 1 0 0 0 0 1 1 1 1
0 0 0 1 0 1 0 1 1 0 0 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0
0 0 1 0 1 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 0
By boundary layer, I mean row 2 and 7, and column 3 and 10 for island#1.
This is shown below:
I want the row and column indices of the islands. So for the above matrix, the desired output is:
isl{1}= {[2 3 4 5 6 7]; % row indices of island#1
[3 4 5 6 7 8 9 10]} % column indices of island#1
isl{2}= {[2 3 4 5 6 7]; % row indices of island#2
[12 13 14 15 16 17]}; % column indices of island#2
isl{3} ={[9 10 11 12]; % row indices of island#3
[2 3 4 5 6 7 8 9 10 11];} % column indices of island#3
It doesn't matter which island is detected first.
While I know that the [r,c] = find(matrix) function can give the row and column indices of ones but I have no clues on how to detect the connected ones since they can be connected in horizontal, vertical and diagonal order.
Any ideas on how to deal with this problem?
You should look at the BoundingBox and ConvexHull stats returned by regionprops:
a = imread('circlesBrightDark.png');
bw = a < 100;
s = regionprops('table',bw,'BoundingBox','ConvexHull')
https://www.mathworks.com/help/images/ref/regionprops.html
Finding the connected components and their bounding boxes is the easy part. The more difficult part is merging the bounding boxes into islands.
Bounding Boxes
First the easy part.
function bBoxes = getIslandBoxes(lMap)
% find bounding box of each candidate island
% lMap is a logical matrix containing zero or more connected components
bw = bwlabel(lMap); % label connected components in logical matrix
bBoxes = struct2cell(regionprops(bw, 'BoundingBox')); % get bounding boxes
bBoxes = cellfun(#round, bBoxes, 'UniformOutput', false); % round values
end
The values are rounded because the bounding boxes returned by regionprops lies outside its respective component on the grid lines rather than the cell center, and we need integer values to use as subscripts into the matrix. For example, a component that looks like this:
0 0 0
0 1 0
0 0 0
will have a bounding box of
[ 1.5000 1.5000 1.0000 1.0000 ]
which we round to
[ 2 2 1 1]
Merging
Now the hard part. First, the merge condition:
We merge bounding box b2 into bounding box b1 if b2 and the island of b1 (including the boundary layer) have a non-null intersection.
This condition ensures that bounding boxes are merged when one component is wholly or partially inside the bounding box of another, but it also catches the edge cases when a bounding box is within the zero boundary of another. Once all of the bounding boxes are merged, they are guaranteed to have a boundary of all zeros (or border the edge of the matrix), otherwise the nonzero value in its boundary would have been merged.
Since merging involves deleting the merged bounding box, the loops are done backwards so that we don't end up indexing non-existent array elements.
Unfortunately, making one pass through the array comparing each element to all the others is insufficient to catch all cases. To signal that all of the possible bounding boxes have been merged into islands, we use a flag called anyMerged and loop until we get through one complete iteration without merging anything.
function mBoxes = mergeBoxes(bBoxes)
% find bounding boxes that intersect, and merge them
mBoxes = bBoxes;
% merge bounding boxes that overlap
anyMerged = true; % flag to show when we've finished
while (anyMerged)
anyMerged = false; % no boxes merged on this iteration so far...
for box1 = numel(mBoxes):-1:2
for box2 = box1-1:-1:1
% if intersection between bounding boxes is > 0, merge
% the size of box1 is increased b y 1 on all sides...
% this is so that components that lie within the borders
% of another component, but not inside the bounding box,
% are merged
if (rectint(mBoxes{box1} + [-1 -1 2 2], mBoxes{box2}) > 0)
coords1 = rect2corners(mBoxes{box1});
coords2 = rect2corners(mBoxes{box2});
minX = min(coords1(1), coords2(1));
minY = min(coords1(2), coords2(2));
maxX = max(coords1(3), coords2(3));
maxY = max(coords1(4), coords2(4));
mBoxes{box2} = [minX, minY, maxX-minX+1, maxY-minY+1]; % merge
mBoxes(box1) = []; % delete redundant bounding box
anyMerged = true; % bounding boxes merged: loop again
break;
end
end
end
end
end
The merge function uses a small utility function that converts rectangles with the format [x y width height] to a vector of subscripts for the top-left, bottom-right corners [x1 y1 x2 y2]. (This was actually used in another function to check that an island had a zero border, but as discussed above, this check is unnecessary.)
function corners = rect2corners(rect)
% change from rect = x, y, width, height
% to corners = x1, y1, x2, y2
corners = [rect(1), ...
rect(2), ...
rect(1) + rect(3) - 1, ...
rect(2) + rect(4) - 1];
end
Output Formatting and Driver Function
The return value from mergeBoxes is a cell array of rectangle objects. If you find this format useful, you can stop here, but it's easy to get to the format requested with ranges of rows and columns for each island:
function rRanges = rect2range(bBoxes, mSize)
% convert rect = x, y, width, height to
% range = y:y+height-1; x:x+width-1
% and expand range by 1 in all 4 directions to include zero border,
% making sure to stay within borders of original matrix
rangeFun = #(rect) {max(rect(2)-1,1):min(rect(2)+rect(4),mSize(1));...
max(rect(1)-1,1):min(rect(1)+rect(3),mSize(2))};
rRanges = cellfun(rangeFun, bBoxes, 'UniformOutput', false);
end
All that's left is a main function to tie all of the others together and we're done.
function theIslands = getIslandRects(m)
% get rectangle around each component in map
lMap = logical(m);
% get the bounding boxes of candidate islands
bBoxes = getIslandBoxes(lMap);
% merge bounding boxes that overlap
bBoxes = mergeBoxes(bBoxes);
% convert bounding boxes to row/column ranges
theIslands = rect2range(bBoxes, size(lMap));
end
Here's a run using the sample matrix given in the question:
M =
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1
0 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0
0 0 0 0 0 0 1 0 1 0 0 0 0 1 1 1 1
0 0 0 1 0 1 0 1 1 0 0 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0
0 0 1 0 1 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 0
>> getIslandRects(M)
ans =
{
[1,1] =
{
[1,1] =
9 10 11 12
[2,1] =
2 3 4 5 6 7 8 9 10 11
}
[1,2] =
{
[1,1] =
2 3 4 5 6 7
[2,1] =
3 4 5 6 7 8 9 10
}
[1,3] =
{
[1,1] =
2 3 4 5 6 7
[2,1] =
12 13 14 15 16 17
}
}
Quite easy!
Just use bwboundaries to get the boundaries of each of the blobs. you can then just get the min and max in each x and y direction of each boundary to build your box.
Use image dilation and regionprops
mat = [...
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1;
0 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0;
0 0 0 0 0 0 1 0 1 0 0 0 0 1 1 1 1;
0 0 0 1 0 1 0 1 1 0 0 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0;
0 0 1 0 1 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 0];
mat=logical(mat);
dil_mat=imdilate(mat,true(2,2)); %here we make bridges to 1 px away ones
l_mat=bwlabel(dil_mat,8);
bb = regionprops(l_mat,'BoundingBox');
bb = struct2cell(bb); bb = cellfun(#(x) fix(x), bb, 'un',0);
isl = cellfun(#(x) {max(1,x(2)):min(x(2)+x(4),size(mat,1)),...
max(1,x(1)):min(x(1)+x(3),size(mat,2))},bb,'un',0);
Let Y be a vector of length N, containing numbers from 1 to 10. As example code you can use:
Y = vec(1:10);
I am writing the code which must create an N x 10 matrix, each row consisting of all zeros except for a 1 only in the position which corresponds to the number in vector Y. Thus, 1 in Y becomes 10000000000, 3 becomes 0010000000, and so on.
This approach works:
cell2mat(arrayfun(#(x)eye(10)(x,:), Y, 'UniformOutput', false))
My next idea was to "optimize", so eye(10) is not generated N times, and I wrote this:
theEye = eye(10);
cell2mat(arrayfun(#(x)theEye(x,:), Y, 'UniformOutput', false))
However, now Octave is giving me error:
error: can't perform indexing operations for diagonal matrix type
error: evaluating argument list element number 1
Why do I get this error? What is wrong?
Bonus questions — do you see a better way to do what I am doing? Is my attempt to optimize making things easier for Octave?
I ran this code in Octave and eye creates a matrix of a class (or whatever this is) known as a Diagonal Matrix:
octave:3> theEye = eye(10);
octave:4> theEye
theEye =
Diagonal Matrix
1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1
In fact, the documentation for Octave says that if the matrix is diagonal, a special object is created to handle the diagonal matrices instead of a standard matrix: https://www.gnu.org/software/octave/doc/interpreter/Creating-Diagonal-Matrices.html
What's interesting is that we can slice into this matrix outside of the arrayfun call, regardless of it being in a separate class.
octave:1> theEye = eye(10);
octave:2> theEye(1,:)
ans =
Diagonal Matrix
1 0 0 0 0 0 0 0 0 0
However, as soon as we put this into an arrayfun call, it decides to crap out:
octave:5> arrayfun(#(x)theEye(x,:), 1:3, 'uni', 0)
error: can't perform indexing operations for diagonal matrix type
This to me doesn't make any sense, especially since we can slice into it outside of arrayfun. One may suspect that it has something to do with arrayfun and since you are specifying UniformOutput to be false, a cell array of elements is returned per element in Y and perhaps something is going wrong when storing these slices into each cell array element.
However, this doesn't seem to be the culprit either. I took the first three rows of theEye, placed them into a cell array and merged them together using cell2mat:
octave:6> cell2mat({theEye(1,:); theEye(2,:); theEye(3,:)})
ans =
1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
As such, I suspect that it may be some sort of internal bug (if you could call it that...). Thanks to user carandraug (see comment above), this is indeed a bug and it has been reported: https://savannah.gnu.org/bugs/?47510. What may also provide insight is that this code runs as expected in MATLAB.
In any case, one thing you can take away from this is that I would seriously refrain from using cell2mat. Just use straight up indexing:
Y = vec(1:10);
theEye = eye(10);
out = theEye(Y,:);
This would index into theEye and extract out the relevant rows stored in Y and create a matrix where each row is zero except for the corresponding value seen in each element Y.
Also, have a look at this post for a similar example: Replace specific columns in a matrix with a constant column vector
However, it is defined over the columns instead of the rows, but it's very similar to what you want to achieve.
Another approach; We start with the data:
>> len = 10; % max number
>> vec = randi(len, [1 7]) % vector of numbers
vec =
1 10 9 5 7 3 6
Now we build the indicator matrix:
>> I = full(sparse(1:numel(vec), vec, 1, numel(vec), len))
I =
1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 1 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
I have a matrix with only zero's and ones:
acc062_1 acc062_2 acc003_1 acc003_2 acc039_1 acc039_2
SL2.40ct15849 0 1 0 0 1 0
SL2.40ct15848 0 0 0 0 0 0
SL2.40ct15847 0 0 0 0 0 0
SL2.40ct15846 0 0 0 0 0 0
SL2.40ct15845 0 0 0 0 0 0
SL2.40ct15844 1 1 1 1 1 1
SL2.40ct11061 0 0 0 0 0 0
SL2.40ct11060 0 0 0 0 0 0
SL2.40sc04607 1 1 1 1 1 1
SL2.40ct11212 0 0 0 0 0 0
SL2.40ch12 1 1 1 1 1 1
With this matrix I can create an image with the image function of the graphics package. using this code:
image(x)
This gives me an image exactly as expected with the colors red for 0 and white for 1. But the labels on the x-axis and y-axis are not the rownames and column names. This is a range between zero and one, how do I change these to my column names?
When using the heatmap function:
heatmap(x)
The labels are the column names and rownames for the x-axis and y-axis. But now all rows with only zero's or only one's are blanco. Only the read with variation is drawn as expected. (Also a clustering is done, but I'm able to turn this feature off)
Does anyone know how to get the image as created with image(x) and get the labels as created with heatmap(x)? prefer of using the image function because the number of rows will be very high.
I think you should run image() without the axes (parameters xaxt and yaxt) and then add the axes with specified labels:
par( mar = par( "mar" ) + c( 2, 4, 0, 0 ) )
image( x, xaxt= "n", yaxt= "n" )
axis( 1, at=seq(0,1,length.out=ncol( x ) ), labels= colnames( x ), las= 2 )
axis( 2, at=seq(0,1,length.out=nrow( x ) ), labels= rownames( x ), las= 2)
Can someone explain what's going on here?
octave:1> t = eye(3)
t =
Diagonal Matrix
1 0 0
0 1 0
0 0 1
octave:2> diag(t(3,:))
ans =
Diagonal Matrix
0 0 0
0 0 0
0 0 1
octave:3> diag(t(2,:))
ans =
Diagonal Matrix
0 0 0
0 1 0
0 0 0
octave:4> diag(t(1,:))
ans = 1
Why do the first two give back 3x3 matrices but the last one is just a number?
The problem arises because of the way t(1,:) was created, from eye(3).
If you output the rows of t individually the results are:
octave.28> t(1,:)
ans =
**Diagonal Matrix**
1 0 0
octave.29> t(2,:)
ans =
0 1 0
octave.30> t(3,:)
ans =
0 0 1
For some reason (I can't explain) t(1,:) is still recognized as a diagonal matrix, while t(2,:) and t(3,:) are vectors. When you call diag(t(:,1)) it is not receiving a vector argument, but rather a matrix. If you convert t(:,1) to vector before evaluation you get the expected result.
octave.31> diag(vec(t(1,:)))
ans =
**Diagonal Matrix**
1 0 0
0 0 0
0 0 0