I'm trying to blind unblur an image using a gauss filter using the following code
but I know I've a problem whenever the filter contains zero, so I wonder if there is any other method of deconvolution but using the FFT
function [ out ] = imblur( file)
img = im2double(imread(file));
h = fspecial('gaussian', [15 15], 3);
img_red = img(:,:,1);
img_blue = img(:,:,2);
img_green = img(:,:,3);
[m,n] = size(img_red);
[mb,nb] = size(h);
% output size
mm = m + mb - 1;
nn = n + nb - 1;
x1(:,:,1) = (ifft2(fft2(img_red,mm,nn)./ fft2(h,mm,nn)));
x2(:,:,2) = (ifft2(fft2(img_blue,mm,nn)./ fft2(h,mm,nn)));
x3(:,:,3) = (ifft2(fft2(img_green,mm,nn)./ fft2(h,mm,nn)));
out = cat(3, x1(:,:,1), x2(:,:,2), x3(:,:,3));
imshow(out);
replacing the zeros with epsilon worked just fine.
Related
i'm making image segmentation with self organizing map. the image segement by 3 cluster. Sample image is :
and i have type the matlab code like this bellow :
clear;
clc;
i=imread('DataSet/3.jpg');
I = imresize(i,0.5);
cform = makecform('srgb2lab');
lab_I = applycform(I,cform);
ab = double(lab_I(:,:,2:3));
nrows = size(ab,1);
ncols = size(ab,2);
ab = reshape(ab,nrows*ncols,2);
a = ab(:,1);
b = ab(:,2);
normA = (a-min(a(:))) ./ (max(a(:))-min(a(:)));
normB = (b-min(b(:))) ./ (max(b(:))-min(b(:)));
ab = [normA normB];
newnRows = size(ab,1);
newnCols = size(ab,2);
cluster = 3;
% Max number of iteration
N = 90;
% initial learning rate
eta = 0.3;
% exponential decay rate of the learning rate
etadecay = 0.2;
%random weight
w = rand(2,cluster);
%initial D
D = zeros(1,cluster);
% initial cluster index
clusterindex = zeros(newnRows,1);
% start
for t = 1:N
for data = 1 : newnRows
for c = 1 : cluster
D(c) = sqrt(((w(1,c)-ab(data,1))^2) + ((w(2,c)-ab(data,2))^2));
end
%find best macthing unit
[~, bmuindex] = min(D);
clusterindex(data)=bmuindex;
%update weight
oldW = w(:,bmuindex);
new = oldW + eta * (reshape(ab(data,:),2,1)-oldW);
w(:,bmuindex) = new;
end
% update learning rate
eta= etadecay * eta;
end
%Label Every Pixel in the Image Using the Results from KMEANS
pixel_labels = reshape(clusterindex,nrows,ncols);
%Create Images that Segment the I Image by Color.
segmented_images = cell(1,3);
rgb_label = repmat(pixel_labels,[1 1 3]);
for k = 1:cluster
color = I;
color(rgb_label ~= k) = 0;
segmented_images{k} = color;
end
figure,imshow(segmented_images{1}), title('objects in cluster 1');
figure,imshow(segmented_images{2}), title('objects in cluster 2');
figure,imshow(segmented_images{3}), title('objects in cluster 3');
and after runing the matlab code, there is no image segmentation result. Matlab show 3 figure, Figure 1 show the full image, figure 2 blank, figure 3 blank .
please anyone help me to revise my matlab code, is any wrong code or something?
new = oldW + eta * (reshape(ab(data,:),2,1)-oldW);
This line looks suspicious to me, why you are subtracting old weights here, i dont think this makes any sense there, just remove oldW from there and check your results again.
Thank You
I want to use the line new command of PDE toolbox as Matlab R2015 to restore a noisy image with gaussian noise.
The PDE is:
∇.(( ∇u)/(√(1+|∇u|2))) +(f2)/(u2) = 1 in Ω (∂u)/(∂n)=0 in ∂Ω
Where f is the noisy image and u the restored image.
I tried the following code:
clear
close all
clc
img = 'AA.jpg';
mInputImage = double(imread(img));
mInputImage = rgb2gray(mInputImage);
[numRows, numCols] = size(mInputImage);
Var = 0.04;
Mean = 0;
mInputImageNoisy = imnoise((mInputImage(:,:,1)),'gaussian',Mean, Var);
% reshape the input and noisy images to vectors
mInputImageVector = reshape(mInputImage,numRows*numCols,1);
mInputImageNoisyVector = reshape(mInputImageNoisy,numRows*numCols,1);
Residu1 = norm(mInputImageVector-mInputImageNoisyVector)/norm(mInputImageVector)
RegularisationCoefficient = 0.7*ones((numRows-1)*(numCols-1),1);
mOutputImageVector = mInputImageNoisyVector;
%a = (mInputImageNoisyVector.^2) ./ mOutputImageVector.^3;
f = 1;
rtol = 1e-1;
c = '1./sqrt(1+ux.^2+uy.^2)';
% Create a PDE Model with a single dependent variable
numberOfPDE = 1;
pdem = createpde(numberOfPDE);
g = #squareg;
geometryFromEdges(pdem,g);
% Plot the geometry and display the edge labels for use in the boundary
% condition definition.
figure;
pdegplot(pdem, 'edgeLabels', 'on');
%axis([0 numRows 0 numCols]);
axis([-2 2 -2 2]);
title 'Geometry With Edge Labels Displayed'
b2 = applyBoundaryCondition(pdem,'Edge',[1 2 3 4], 'u', 0);
[p,e,t] = poimesh(g,numRows, numCols);
numCols
pdemesh(p,e,t);
axis equal
for iter = 1: numRows*numRows,
mOutputImageVector(iter) = pdenonlin(pdem,c,...
(mInputImageNoisyVector(iter).^2) ./ mOutputImageVector(iter).^3,...
f,'tol',rtol);
SaveImageVector(iter) = mOutputImageVector;
end
mOutputImage = reshape(SaveImageVector,numRows,numRows);
mOutputImage = uint8(mOutputImage);
figure()
imshow(mOutputImage)
A filter g is called separable if it can be expressed as the multiplication of two vectors grow and gcol . Employing one dimensional filters decreases the two dimensional filter's computational complexity from O(M^2 N^2) to O(2M N^2) where M and N are the width (and height) of the filter mask and the image respectively.
In this stackoverflow link, I wrote the equation of a Gabor filter in the spatial domain, then I wrote a matlab code which serves to create 64 gabor features.
According to the definition of separable filters, the Gabor filters are parallel to the image axes - theta = k*pi/2 where k=0,1,2,etc.. So if theta=pi/2 ==> the equation in this stackoverflow link can be rewritten as:
The equation above is extracted from this article.
Note: theta can be extented to be equal k*pi/4. By comparing to the equation in this stackoverflow link, we can consider that f= 1 / lambda.
By changing my previous code in this stackoverflow link, I wrote a matlab code to make the Gabor filters separable by using the equation above, but I am sure that my code below is not correct especially when I initialized the gbp and glp equations. That is why I need your help. I will appreciate your help very much.
Let's show now my code:
function [fSiz,filters1,filters2,c1OL,numSimpleFilters] = init_gabor(rot, RF_siz)
image=imread('xxx.jpg');
image_gray=rgb2gray(image);
image_gray=imresize(image_gray, [100 100]);
image_double=double(image_gray);
rot = [0 45 90 135]; % we have four orientations
RF_siz = [7:2:37]; %we get 16 scales (7x7 to 37x37 in steps of two pixels)
minFS = 7; % the minimum receptive field
maxFS = 37; % the maximum receptive field
sigma = 0.0036*RF_siz.^2 + 0.35*RF_siz + 0.18; %define the equation of effective width
lambda = sigma/0.8; % it the equation of wavelength (lambda)
G = 0.3; % spatial aspect ratio: 0.23 < gamma < 0.92
numFilterSizes = length(RF_siz); % we get 16
numSimpleFilters = length(rot); % we get 4
numFilters = numFilterSizes*numSimpleFilters; % we get 16x4 = 64 filters
fSiz = zeros(numFilters,1); % It is a vector of size numFilters where each cell contains the size of the filter (7,7,7,7,9,9,9,9,11,11,11,11,......,37,37,37,37)
filters1 = zeros(max(RF_siz),numFilters);
filters2 = zeros(numFilters,max(RF_siz));
for k = 1:numFilterSizes
for r = 1:numSimpleFilters
theta = rot(r)*pi/180;
filtSize = RF_siz(k);
center = ceil(filtSize/2);
filtSizeL = center-1;
filtSizeR = filtSize-filtSizeL-1;
sigmaq = sigma(k)^2;
for x = -filtSizeL:filtSizeR
fx = exp(-(x^2)/(2*sigmaq))*cos(2*pi*x/lambda(k));
f1(x+center,1) = fx;
end
for y = -filtSizeL:filtSizeR
gy = exp(-(y^2)/(2*sigmaq));
f2(1,y+center) = gy;
end
f1 = f1 - mean(mean(f1));
f1 = f1 ./ sqrt(sum(sum(f1.^2)));
f2 = f2 - mean(mean(f2));
f2 = f2 ./ sqrt(sum(sum(f2.^2)));
p = numSimpleFilters*(k-1) + r;
filters1(1:filtSize,p)=f1;
filters2(p,1:filtSize)=f2;
convv1=imfilter(image_double, filters1(1:filtSize,p),'conv');
convv2=imfilter(double(convv1), filters2(p,1:filtSize),'conv');
figure
imagesc(convv2);
colormap(gray);
end
end
I think the code is correct provided your previous version of Gabor filter code is correct too. The only thing is that if theta = k * pi/4;, your formula here should be separated to:
fx = exp(-(x^2)/(2*sigmaq))*cos(2*pi*x/lambda(k));
gy = exp(-(G^2 * y^2)/(2*sigmaq));
To be consistent, you may use
f1(1,x+center) = fx;
f2(y+center,1) = gy;
or keep f1 and f2 as it is but transpose your filters1 and filters2 thereafter.
Everything else looks good to me.
EDIT
My answer above works for theta = k * pi/4;, with other angles, based on your paper,
x = i*cos(theta) - j*sin(theta);
y = i*sin(theta) + j*cos(theta);
fx = exp(-(x^2)/(2*sigmaq))*exp(sqrt(-1)*x*cos(theta));
gy = exp(-(G^2 * y^2)/(2*sigmaq))*exp(sqrt(-1)*y*sin(theta));
The final code will be:
function [fSiz,filters1,filters2,c1OL,numSimpleFilters] = init_gabor(rot, RF_siz)
image=imread('xxx.jpg');
image_gray=rgb2gray(image);
image_gray=imresize(image_gray, [100 100]);
image_double=double(image_gray);
rot = [0 45 90 135];
RF_siz = [7:2:37];
minFS = 7;
maxFS = 37;
sigma = 0.0036*RF_siz.^2 + 0.35*RF_siz + 0.18;
lambda = sigma/0.8;
G = 0.3;
numFilterSizes = length(RF_siz);
numSimpleFilters = length(rot);
numFilters = numFilterSizes*numSimpleFilters;
fSiz = zeros(numFilters,1);
filters1 = zeros(max(RF_siz),numFilters);
filters2 = zeros(numFilters,max(RF_siz));
for k = 1:numFilterSizes
for r = 1:numSimpleFilters
theta = rot(r)*pi/180;
filtSize = RF_siz(k);
center = ceil(filtSize/2);
filtSizeL = center-1;
filtSizeR = filtSize-filtSizeL-1;
sigmaq = sigma(k)^2;
for x = -filtSizeL:filtSizeR
fx = exp(-(x^2)/(2*sigmaq))*exp(sqrt(-1)*x*cos(theta));
f1(1, x+center) = fx;
end
for y = -filtSizeL:filtSizeR
gy=exp(-(y^2)/(2*sigmaq))*exp(sqrt(-1)*y*sin(theta));
f2(y+center,1) = gy;
end
f1 = f1 - mean(mean(f1));
f1 = f1 ./ sqrt(sum(sum(f1.^2)));
f2 = f2 - mean(mean(f2));
f2 = f2 ./ sqrt(sum(sum(f2.^2)));
p = numSimpleFilters*(k-1) + r;
filters1(1:filtSize,p)=f1;
filters2(p,1:filtSize)=f2;
convv1=imfilter(image_double, filters1(1:filtSize,p),'conv');
convv2=imfilter(double(convv1), filters2(p,1:filtSize),'conv');
figure
imagesc(imag(convv2));
colormap(gray);
end
end
Normally, a Gabor filter, as its name suggests, is used to filter an image and extract everything that it is oriented in the same direction of the filtering.
In this question, you can see more efficient code than written in this Link
Assume 16 scales of Filters at 4 orientations, so we get 64 gabor filters.
scales=[7:2:37], 7x7 to 37x37 in steps of two pixels, so we have 7x7, 9x9, 11x11, 13x13, 15x15, 17x17, 19x19, 21x21, 23x23, 25x25, 27x27, 29x29, 31x31, 33x33, 35x35 and 37x37.
directions=[0, pi/4, pi/2, 3pi/4].
The equation of Gabor filter in the spatial domain is:
The equation of Gabor filter in the frequency domain is:
In the spatial domain:
function [fSiz,filters,c1OL,numSimpleFilters] = init_gabor(rot, RF_siz)
image=imread('xxx.jpg');
image_gray=rgb2gray(image);
image_gray=imresize(image_gray, [100 100]);
image_double=double(image_gray);
rot = [0 45 90 135]; % we have four orientations
RF_siz = [7:2:37]; %we get 16 scales (7x7 to 37x37 in steps of two pixels)
minFS = 7; % the minimum receptive field
maxFS = 37; % the maximum receptive field
sigma = 0.0036*RF_siz.^2 + 0.35*RF_siz + 0.18; %define the equation of effective width
lambda = sigma/0.8; % it the equation of wavelength (lambda)
G = 0.3; % spatial aspect ratio: 0.23 < gamma < 0.92
numFilterSizes = length(RF_siz); % we get 16
numSimpleFilters = length(rot); % we get 4
numFilters = numFilterSizes*numSimpleFilters; % we get 16x4 = 64 filters
fSiz = zeros(numFilters,1); % It is a vector of size numFilters where each cell contains the size of the filter (7,7,7,7,9,9,9,9,11,11,11,11,......,37,37,37,37)
filters = zeros(max(RF_siz)^2,numFilters); % Matrix of Gabor filters of size %max_fSiz x num_filters, where max_fSiz is the length of the largest filter and num_filters the total number of filters. Column j of filters matrix contains a n_jxn_j filter (reshaped as a column vector and padded with zeros).
for k = 1:numFilterSizes
for r = 1:numSimpleFilters
theta = rot(r)*pi/180; % so we get 0, pi/4, pi/2, 3pi/4
filtSize = RF_siz(k);
center = ceil(filtSize/2);
filtSizeL = center-1;
filtSizeR = filtSize-filtSizeL-1;
sigmaq = sigma(k)^2;
for i = -filtSizeL:filtSizeR
for j = -filtSizeL:filtSizeR
if ( sqrt(i^2+j^2)>filtSize/2 )
E = 0;
else
x = i*cos(theta) - j*sin(theta);
y = i*sin(theta) + j*cos(theta);
E = exp(-(x^2+G^2*y^2)/(2*sigmaq))*cos(2*pi*x/lambda(k));
end
f(j+center,i+center) = E;
end
end
f = f - mean(mean(f));
f = f ./ sqrt(sum(sum(f.^2)));
p = numSimpleFilters*(k-1) + r;
filters(1:filtSize^2,p)=reshape(f,filtSize^2,1);
fSiz(p)=filtSize;
end
end
% Rebuild all filters (of all sizes)
nFilts = length(fSiz);
for i = 1:nFilts
sqfilter{i} = reshape(filters(1:(fSiz(i)^2),i),fSiz(i),fSiz(i));
%if you will use conv2 to convolve an image with this gabor, so you should also add the equation below. But if you will use imfilter instead of conv2, so do not add the equation below.
sqfilter{i} = sqfilter{i}(end:-1:1,end:-1:1); %flip in order to use conv2 instead of imfilter (%bug_fix 6/28/2007);
convv=imfilter(image_double, sqfilter{i}, 'same', 'conv');
figure;
imagesc(convv);
colormap(gray);
end
phi = ij*pi/4; % ij = 0, 1, 2, 3
theta = 3;
sigma = 0.65*theta;
filterSize = 7; % 7:2:37
G = zeros(filterSize);
for i=(0:filterSize-1)/filterSize
for j=(0:filterSize-1)/filterSize
xprime= j*cos(phi);
yprime= i*sin(phi);
K = exp(2*pi*theta*sqrt(-1)*(xprime+ yprime));
G(round((i+1)*filterSize),round((j+1)*filterSize)) =...
exp(-(i^2+j^2)/(sigma^2))*K;
end
end
As of R2015b release of the Image Processing Toolbox, you can create a Gabor filter bank using the gabor function in the image processing toolbox, and you can apply it to an image using imgaborfilt.
In the frequency domain:
sigma_u=1/2*pi*sigmaq;
sigma_v=1/2*pi*sigmaq;
u0=2*pi*cos(theta)*lambda(k);
v0=2*pi*sin(theta)*lambda(k);
for u = -filtSizeL:filtSizeR
for v = -filtSizeL:filtSizeR
if ( sqrt(u^2+v^2)>filtSize/2 )
E = 0;
else
v_theta = u*cos(theta) - v*sin(theta);
u_theta = u*sin(theta) + v*cos(theta);
E=(1/2*pi*sigma_u*sigma_v)*((exp((-1/2)*(((u_theta-u0)^2/sigma_u^2))+((v_theta-v0)^2/sigma_v^2))) + (exp((-1/2)*(((u_theta+u0)^2/sigma_u^2))+((v_theta+v0)^2/sigma_v^2))));
end
f(v+center,u+center) = E;
end
end
I implemented a method for removing shadows based on invariant color features found in the paper Entropy Minimization for Shadow Removal. My implementation seems to be yielding similar computational results sometimes, but they are always off, and my grayscale image is blocky, maybe as a result of incorrectly taking the geometric mean.
Here is an example plot of the information potential from the horse image in the paper as well as my invariant image. Multiply the x-axis by 3 to get theta(which goes from 0 to 180):
And here is the grayscale Image my code outputs for the correct maximum theta (mine is off by 10):
You can see the blockiness that their image doesn't have:
Here is their information potential:
When dividing by the geometric mean, I have tried using NaN and tresholding the image so the smallest possible value is .01, but it doesn't seem to change my output.
Here is my code:
I = im2double(imread(strname));
[m,n,d] = size(I);
I = max(I, .01);
chrom = zeros(m, n, 3, 'double');
for i = 1:m
for j = 1:n
% if ((I(i,j,1)*I(i,j,2)*I(i,j,3))~= 0)
chrom(i,j, 1) = I(i,j,1)/((I(i,j,1)*I(i,j,2)*I(i,j, 3))^(1/3));
chrom(i,j, 2) = I(i,j,2)/((I(i,j,1)*I(i,j,2)*I(i,j, 3))^(1/3));
chrom(i,j, 3) = I(i,j,3)/((I(i,j,1)*I(i,j,2)*I(i,j, 3))^(1/3));
% else
% chrom(i,j, 1) = 1;
% chrom(i,j, 2) = 1;
% chrom(i,j, 3) = 1;
% end
end
end
p1 = mat2gray(log(chrom(:,:,1)));
p2 = mat2gray(log(chrom(:,:,2)));
p3 = mat2gray(log(chrom(:,:,3)));
X1 = mat2gray(p1*1/(sqrt(2)) - p2*1/(sqrt(2)));
X2 = mat2gray(p1*1/(sqrt(6)) + p2*1/(sqrt(6)) - p3*2/(sqrt(6)));
maxinf = 0;
maxtheta = 0;
data2 = zeros(1, 61);
for theta = 0:3:180
M = X1*cos(theta*pi/180) - X2*sin(theta*pi/180);
s = sqrt(std2(X1)^(2)*cos(theta*pi/180) + std2(X2)^(2)*sin(theta*pi/180));
s = abs(1.06*s*((m*n)^(-1/5)));
[m, n] = size(M);
length = m*n;
sources = zeros(1, length, 'double');
count = 1;
for x=1:m
for y = 1:n
sources(1, count) = M(x , y);
count = count + 1;
end
end
weights = ones(1, length);
sigma = 2*s;
[xc , Ak] = fgt_model(sources , weights , sigma , 10, sqrt(length) , 6 );
sum1 = sum(fgt_predict(sources , xc , Ak , sigma , 10 ));
sum1 = sum1/sqrt(2*pi*2*s*s);
data2(theta/3 + 1) = sum1;
if (sum1 > maxinf)
maxinf = sum1;
maxtheta = theta;
end
end
InvariantImage2 = cos(maxtheta*pi/180)*X1 + sin(maxtheta*pi/180)*X2;
Assume the Fast Gauss Transform is correct.
I don't know whether this makes any difference as it is more than a month now, but the blockiness and different information potential plot is simply caused by compression of the used image. You can't expect to be getting same results using this image as they had, because they have used raw, high resolution uncompressed version of it. I have to say I am fairly impressed with your results, especially with implementing the information potential. That thing went over my head a little.
John.