I'm trying to create an adaptive elliptical structuring element for an image to dilate or erode it. I write this code but unfortunately all of the structuring elements are ones(2*M+1).
I = input('Enter the input image: ');
M = input('Enter the maximum allowed semi-major axes length: ');
% determining ellipse parameteres from eigen value decomposition of LST
row = size(I,1);
col = size(I,2);
SE = cell(row,col);
padI = padarray(I,[M M],'replicate','both');
padrow = size(padI,1);
padcol = size(padI,2);
for m = M+1:padrow-M
for n = M+1:padcol-M
a = (l2(m-M,n-M)+eps/l1(m-M,n-M)+l2(m-M,n-M)+2*eps)*M;
b = (l1(m-M,n-M)+eps/l1(m-M,n-M)+l2(m-M,n-M)+2*eps)*M;
if e1(m-M,n-M,1)==0
phi = pi/2;
else
phi = atan(e1(m-M,n-M,2)/e1(m-M,n-M,1));
end
% defining structuring element for each pixel of image
x0 = m;
y0 = n;
se = zeros(2*M+1);
row_se = 0;
for i = x0-M:x0+M
row_se = row_se+1;
col_se = 0;
for j = y0-M:y0+M
col_se = col_se+1;
x = j-y0;
y = x0-i;
if ((x*cos(phi)+y*sin(phi))^2)/a^2+((x*sin(phi)-y*cos(phi))^2)/b^2 <= 1
se(row_se,col_se) = 1;
end
end
end
SE{m-M,n-M} = se;
end
end
a, b and phi are semi-major and semi-minor axes length and phi is angle between a and x axis.
I used 2 MATLAB functions to compute the Local Structure Tensor of the image, and then its eigenvalues and eigenvectors for each pixel. These are the matrices l1, l2, e1 and e2.
This is the bit of your code I didn't understand:
a = (l2(m-M,n-M)+eps/l1(m-M,n-M)+l2(m-M,n-M)+2*eps)*M;
b = (l1(m-M,n-M)+eps/l1(m-M,n-M)+l2(m-M,n-M)+2*eps)*M;
I simplified the expression for b to (just removing the indexing):
b = (l1+eps/l1+l2+2*eps)*M;
For l1 and l2 in the normal range we get:
b =(approx)= (l1+0/l1+l2+2*0)*M = (l1+l2)*M;
Thus, b can easily be larger than M, which I don't think is your intention. The eps in this case also doesn't protect against division by zero, which is typically the purpose of adding eps: if l1 is zero, eps/l1 is Inf.
Looking at this expression, it seems to me that you intended this instead:
b = (l1+eps)/(l1+l2+2*eps)*M;
Here, you're adding eps to each of the eigenvalues, making them guaranteed non-zero (the structure tensor is symmetric, positive semi-definite). Then you're dividing l1 by the sum of eigenvalues, and multiplying by M, which leads to a value between 0 and M for each of the axes.
So, this seems to be a case of misplaced parenthesis.
Just for the record, this is what you need in your code:
a = (l2(m-M,n-M)+eps ) / ( l1(m-M,n-M)+l2(m-M,n-M)+2*eps)*M;
b = (l1(m-M,n-M)+eps ) / ( l1(m-M,n-M)+l2(m-M,n-M)+2*eps)*M;
^ ^
added parentheses
Note that you can simplify your code by defining, outside of the loops:
[se_x,se_y] = meshgrid(-M:M,-M:M);
The inner two loops, over i and j, to construct se can then be written simply as:
se = ((se_x.*cos(phi)+se_y.*sin(phi)).^2)./a.^2 + ...
((se_x.*sin(phi)-se_y.*cos(phi)).^2)./b.^2 <= 1;
(Note the .* and .^ operators, these do element-wise multiplication and power.)
A further slight improvement comes from realizing that phi is first computed from e1(m,n,1) and e1(m,n,2), and then used in calls to cos and sin. If we assume that the eigenvector is properly normalized, then
cos(phi) == e1(m,n,1)
sin(phi) == e1(m,n,2)
But you can always make sure they are normalized:
cos_phi = e1(m-M,n-M,1);
sin_phi = e1(m-M,n-M,2);
len = hypot(cos_phi,sin_phi);
cos_phi = cos_phi / len;
sin_phi = sin_phi / len;
se = ((se_x.*cos_phi+se_y.*sin_phi).^2)./a.^2 + ...
((se_x.*sin_phi-se_y.*cos_phi).^2)./b.^2 <= 1;
Considering trigonometric operations are fairly expensive, this should speed up your code a bit.
Related
Is there a function in MATLAB that generates the following matrix for a given scalar r:
1 r r^2 r^3 ... r^n
0 1 r r^2 ... r^(n-1)
0 0 1 r ... r^(n-2)
...
0 0 0 0 ... 1
where each row behaves somewhat like a power analog of the CUMSUM function?
You can compute each term directly using implicit expansion and element-wise power, and then apply triu:
n = 5; % size
r = 2; % base
result = triu(r.^max((1:n)-(1:n).',0));
Or, maybe a little faster because it doesn't compute unwanted powers:
n = 5; % size
r = 2; % base
t = (1:n)-(1:n).';
u = find(t>=0);
t = t(u);
result = zeros(n);
result(u) = r.^t;
Using cumprod and triu:
% parameters
n = 5;
r = 2;
% Create a square matrix filled with 1:
A = ones(n);
% Assign the upper triangular part shifted by one with r
A(triu(A,1)==1)=r;
% cumprod along the second dimension and get only the upper triangular part
A = triu(cumprod(A,2))
Well, cumsum accumulates the sum of a vector but you are asking for a specially design matrix, so the comparison is a bit problematic....
Anyway, it might be that there is a function for this if this is a common special case triangular matrix (my mathematical knowledge is limited here, sorry), but we can also build it quite easily (and efficiently=) ):
N = 10;
r = 2;
% allocate arry
ary = ones(1,N);
% initialize array
ary(2) = r;
for i = 3:N
ary(i) = ary(i-1)*r;
end
% build matrix i.e. copy the array
M = eye(N);
for i = 1:N
M(i,i:end) = ary(1:end-i+1);
end
This assumes that you want to have a matrix of size NxN and r is the value that you want calculate the power of.
FIX: a previous version stated in line 13 M(i,i:end) = ary(i:end);, but the assignment needs to start always at the first position of the ary
Assume that I have two discrete random variables X and Y.
X = {1,3,3,5,7,7,7,9,9,9,9,9}
and
Y = {5,5,9,9,10,12,13}
Where their empirical CDFs are given as:
F_x(1) = 0.0833, F_x(3) = 0.25, F_x(5) = 0.33, F_x(7) = 0.5833 and F_x(9) = 1
and
F_y(5) = 0.2857, F_y(9) = 0.5714, F_y(10) = 0.7143, F_y(12) = 0.8571 and F_y(13) = 1
Assuming their joint distribution is
H(x,y) = F_x(x) * F_y(y)
which is actually the "assumption" of X and Y are independent.
How can i calculate the Z = X + Y and F(z) in MATLAB ?
Note: I gave the H(x,y) as a simple product function for the simplicity, but it can be anything in reality which actually models the dependency between X and Y.
Given continuous probability density functions FX and FY, with a joint probability density function FX,Y, we can compute FX+Y as the integral of FX,Y over the line z=x+y. If the probability density functions are discrete, the integral above should be written as the derivative of the integral over the part of the plane given by z<=x+y.
This is fairly simple to do in MATLAB. Let's start with OP's data:
F_x = [0.0833,0.0833,0.25,0.25,0.33,0.33,0.5833,0.5833,1]; % CDF
F_x = diff([0,F_x]); % PDF
F_y = [0,0,0,0,0.2857,0.2857,0.2857,0.2857,0.5714,0.7143,0.7143,0.8571,1]; % CDF
F_y = diff([0,F_y]); % PDF
H = F_x.' .* F_y; % example joint PDF
Now we sum F_cum(z) = sum(H(x,y)) for all values z<=x+y, and then take the derivative F = diff([0,F_cum]):
[m,n] = size(H);
F_cum = zeros(1,m+n-1);
for z = 1:numel(F_cum)
s = 0;
for x = 1:numel(F_x)
y = z-x+1;
y = max(min(y,n),1); % avoid out of bounds indexing
s = s + sum(H(x,1:y));
end
F_cum(z) = s;
end
F = diff([0,F_cum]);
Note that we defined y=z-x+1, meaning z=y+x-1. Thus F(1) corresponds to z=2. This is the lowest possible value that can come out of the sum of the two distributions, which we defined to start at 1.
The above can be simplified by padding H with zeros and shifting each row by one additional element. This lines up the line z=x+y on a column of the matrix, allowing us to use a trivial sum projection:
H = [H,zeros(m)];
for ii=2:m
H(ii,:) = circshift(H(ii,:),ii-1);
end
F_cum = cumsum(sum(H,1));
F_cum = F_cum(1:end-1); % last element we don't need
F2 = diff([0,F_cum]);
But because diff([0,cumsum(F)]) == F (up to numerical precision), we can skip those two operations:
F3 = sum(H,1);
F3 = F3(1:end-1); % last element we don't need
(all(abs(F-F2)<1e-15) and all(abs(F-F3)<1e-16))
Is there any function in Matlab which calculates the correlation ratio?
Here is an implementation I tried to do, but the results are not right.
function cr = correlation_ratio(X, Y, L)
ni = zeros(1, L);
sigmai = ni;
for i = 0:(L-1)
Yn = Y(X == i);
ni(1, i+1) = numel(Yn);
m = (1/ni(1, i+1))*sum(Yn);
sigmai(1, i+1) = (1/ni(1, i+1))*sum((Yn - m).^2);
end
n = sum(ni);
prod = ni.*sigmai;
cr = (1-(1/n)*sum(prod))^0.5;
This is the equation on the Wikipedia page:
where:
η is the correlation ratio,
yx,i are the sample values (x is the class label, i the sample index),
yx (with the bar on top) is the mean of sample values for class x,
y (with the bar on top) is the mean for all samples across all classes, and
nx is the number of samples in class x.
This is how I interpreted it into code:
function eta = correlation_ratio(X, Y)
X = X(:); % make sure we've got column vectors, simplifies things below a bit
Y = Y(:);
L = max(X);
mYx = zeros(1, L+1); % we'll write mean per class here
nx = zeros(1, L+1); % we'll write number of samples per class here
for i = unique(X).'
Yn = Y(X == i);
if numel(Yn)>1
mYx(i+1) = mean(Yn);
nx(i+1) = numel(Yn);
end
end
mY = mean(Y); % mean across all samples
eta = sqrt(sum(nx .* (mYx - mY).^2) / sum((Y-mY).^2));
The loop could be replaced with accumarray.
I want to verify the convolution theorem in matlab.
Firstly, I do a 2D discrete convolution of a 2D Gaussian with
an image graymap(x, y).
Secondly, I compute the Fourier Transform of
the same 2D Gaussian and of the original image. Then perform a scalar multiplication
of these two Fourier Transforms, followed by an inverse Fourier Transform of the result.
Finally, I will calculate the MSE between the two results. However, I found the err is 800+.
This is my code:
[row, col] = size(graymap);
[row_2, col_2] = size(z);
result = zeros(row, col);
for i = 1: col
for j = 1:row
accumulation_value = 0;
for k = -4:4
for h = -4:4
if ((i+k > 0 && i+k < col + 1) && (j+h > 0 && j+h < row + 1))
value_image = double(graymap(i+k, j+h));
else
value_image = 0;
end
accumulation_value = accumulation_value + value_image * double(z(5 + k, 5 + h));
weighted_sum = weighted_sum + z(5 + k, 5 + h);
end
end
result(i,j) = (accumulation_value);
end
result_blur_1 = uint8(255*mat2gray(result));
M = size(graymap,1);
N = size(graymap,2);
resIFFT = ifft2(fft2(double(graymap), M, N) .* fft2(double(z), M, N));
result_blur_2 = uint8(255*mat2gray(resIFFT));
err = immse(result_blur_1, result_blur_2);
z is the 9*9 gaussian kernel. I don't flip it because it is symmetric.
I think my implementation of convolution is correct because the result is same as conv2(graymap, z, 'same').
Therefore, I believe there are something wrong with the second part. In fact, I am confused on how padding works. May it is the cause of the big MSE.
There are indeed problems with your implementation of the second part. The most important rule to remember when implementing convolution via fft is that you are actually calculating a circular convolution, not a linear convolution. Fortunately, there is a condition under which the two become equivalent. This condition is that the two arrays should be zero-padded to have a size equal to the sum of the sizes of each minus 1 (in all dimensions). So if you are working with an image X of size MxN, and a mask Z of size PxQ, then you should pad the two arrays with zeros to so they have at least dimensions M+P-1xN+Q-1. Any additional zeros won't hurt, so it's convenient to match a 'fft-friendly' size if possible (using nextpow2 for example). You just have to take the first M+P-1xN+Q-1 values.
Now, that would work straight forward if you just wanted the full result of the convolution. But because you want the central part of the convolution (the option 'same'), you need to select the correct indexes. The first index will be ceil(([P Q] - 1)/2) + 1, and then you take as many consecutive indexes as the image size.
Here is an example putting all together:
M = randperm(1024,1);
N = randperm(1024,1);
X = rand(M,N);
P = randperm(64,1);
Q = randperm(64,1);
Z = rand(P,Q);
% 'standard' convolution with option 'same'
C1 = conv2(X,Z,'same');
R = 2^nextpow2(M+P-1);
S = 2^nextpow2(N+Q-1);
% convolution with fft. Notice the zero-padding to R,S
C2 = real(ifft2(fft2(X,R,S) .* fft2(Z,R,S)));
n = ceil(([P Q] - 1)/2);
ind{1} = n(1) + (1:M);
ind{2} = n(2) + (1:N);
C2 = C2(ind{:});
err = immse(C1,C2)
I get errors of the order of 1e-26
I try to implement the GLCM method with the formula from wikipedia, but I have problems to fill my GLCM due to indices problems with matlab.
I have also used NitdepthQuantisation to reduce the number of Gray Levels, but for now I use the full 8 bit.
function [C] = GLCM(img, level, theta, delta)
% Quantisation of the input Image to desired value
imgQ = ImageQuantisation(img, level);
[m n] = size(imgQ);
% Get the number of gray levels
maxGV = max(img(:));
% Create GLCM initial Matrix
C = zeros(maxGV, maxGV);
% Positions
delta_x = ceil(delta*cos(theta));
delta_y = ceil(delta*sin(theta));
%% Find Occurences
for i = delta_x+1:m-delta_x
for j = delta_y+1:n-delta_y
if(imgQ(i, j) == imgQ(i+delta_x, j+delta_y))
C(, ) = C(, ) + 1;
end
end
end
end
The answer can be found by ensuring the inner nested double for loops have the correct indices to access the image. They were using the outer most pair of for loops for indices rather than the inner ones. The OP has commented that this has slight differences between what MATLAB gives to calculate the GLCM but it is good enough for the OP to overlook:
for o = 1:maxGV
for p = 1:maxGV
if(imgQ(i, j) == o & imgQ(i+delta_x, j+delta_y) == p)
C(o, p) = C(o, p) + 1;
end
end
end