问题描述
如何在图像上应用这些Gabor滤波器小波？
How do i apply these Gabor filter wavelets on an image ?
close all;
clear all;
clc;
% Parameter Setting
R = 128;
C = 128;
Kmax = pi / 2;
f = sqrt( 2 );
Delt = 2 * pi;
Delt2 = Delt * Delt;
% Show the Gabor Wavelets
for v = 0 : 4
for u = 1 : 8
GW = GaborWavelet ( R, C, Kmax, f, u, v, Delt2 ); % Create the Gabor wavelets
figure( 2 );
subplot( 5, 8, v * 8 + u ),imshow ( real( GW ) ,[]); % Show the real part of Gabor wavelets
end
figure ( 3 );
subplot( 1, 5, v + 1 ),imshow ( abs( GW ),[]); % Show the magnitude of Gabor wavelets
end
function GW = GaborWavelet (R, C, Kmax, f, u, v, Delt2)
k = ( Kmax / ( f ^ v ) ) * exp( 1i * u * pi / 8 );% Wave Vector
kn2 = ( abs( k ) ) ^ 2;
GW = zeros ( R , C );
for m = -R/2 + 1 : R/2
for n = -C/2 + 1 : C/2
GW(m+R/2,n+C/2) = ( kn2 / Delt2 ) * exp( -0.5 * kn2 * ( m ^ 2 + n ^ 2 ) / Delt2) * ( exp( 1i * ( real( k ) * m + imag ( k ) * n ) ) - exp ( -0.5 * Delt2 ) );
end
end
修改：这是我图片的尺寸
推荐答案
Gabor过滤器的典型用法是计算多个方向中的每个方向的过滤器响应，例如用于边缘检测。
A typical use of Gabor filters is to calculate filter responses at each of several orientations, e.g. for edge detection.
您可以使用，通过对图像和滤波器的傅里叶变换的逐元乘积进行逆傅立叶变换。以下是基本公式：
You can convolve a filter with an image using the Convolution Theorem, by taking the inverse Fourier transform of the element-wise product of the Fourier transforms of the image and the filter. Here is the basic formula:
%# Our image needs to be 2D (grayscale)
if ndims(img) > 2;
img = rgb2gray(img);
end
%# It is also best if the image has double precision
img = im2double(img);
[m,n] = size(img);
[mf,nf] = size(GW);
GW = padarray(GW,[n-nf m-mf]/2);
GW = ifftshift(GW);
imgf = ifft2( fft2(img) .* GW );
通常，FFT卷积对于大小> 20的内核来说是优越的。有关详细信息，我建议 C 中的数字配方，它具有良好的，语言无关的方法描述及其警告。
Typically, the FFT convolution is superior for kernels of size > 20. For details, I recommend Numerical Recipes in C, which has a good, language agnostic description of the method and its caveats.
你的内核已经很大了，但有了FFT方法它们可以和图像一样大，因为无论如何它们都被填充到那个大小。由于FFT的周期性，该方法执行循环卷积。这意味着滤镜将环绕图像边框，因此我们必须填充图像本身以消除此边缘效果。最后，由于我们需要对所有过滤器的总响应（至少在典型的实现中），我们需要依次将每个过滤器应用于图像，并对响应求和。通常只使用3到6个方向，但在多个比例（不同的内核大小）上进行过滤也很常见，因此在这种情况下使用了大量的过滤器。
Your kernels are already large, but with the FFT method they can be as large as the image, since they are padded to that size regardless. Due to the periodicity of the FFT, the method performs circular convolution. That means that the filter will wrap around the image borders, so we have to pad the image itself as well to eliminate this edge effect. Finally, since we want the total response to all the filters (at least in a typical implementation), we need to apply each to the image in turn, and sum the responses. Usually one uses just 3 to 6 orientations, but it is also common to do the filtering at several scales (different kernel sizes) as well, so in that context a larger number of filters is used.
您可以使用以下代码完成所有操作：
You can do the whole thing with code like this:
img = im2double(rgb2gray(img)); %#
[m,n] = size(img); %# Store the original size.
%# It is best if the filter size is odd, so it has a discrete center.
R = 127; C = 127;
%# The minimum amount of padding is just "one side" of the filter.
%# We add 1 if the image size is odd.
%# This assumes the filter size is odd.
pR = (R-1)/2;
pC = (C-1)/2;
if rem(m,2) ~= 0; pR = pR + 1; end;
if rem(n,2) ~= 0; pC = pC + 1; end;
img = padarray(img,[pR pC],'pre'); %# Pad image to handle circular convolution.
GW = {}; %# First, construct the filter bank.
for v = 0 : 4
for u = 1 : 8
GW = [GW {GaborWavelet(R, C, Kmax, f, u, v, Delt2)}];
end
end
%# Pad all the filters to size of padded image.
%# We made sure padsize will only be even, so we can divide by 2.
padsize = size(img) - [R C];
GW = cellfun( ...
@(x) padarray(x,padsize/2), ...
GW, ...
'UniformOutput',false);
imgFFT = fft2(img); %# Pre-calculate image FFT.
for i=1:length(GW)
filter = fft2( ifftshift( GW{i} ) ); %# See Numerical Recipes.
imgfilt{i} = ifft2( imgFFT .* filter ); %# Apply Convolution Theorem.
end
%# Sum the responses to each filter. Do it in the above loop to save some space.
imgS = zeros(m,n);
for i=1:length(imgfilt)
imgS = imgS + imgfilt{i}(pR+1:end,pC+1:end); %# Just use the valid part.
end
%# Look at the result.
imagesc(abs(imgS));
请记住，这基本上是最低限度的实现。您可以选择使用边框的复制而不是零来填充，对图像应用窗口函数或使垫尺寸更大以获得频率分辨率。这些都是对我上面概述的技术的标准增强，并且通过和。另请注意，我没有添加任何基本的MATLAB优化，例如预分配等。
Keep in mind that this is essentially the bare minimum implementation. You can choose to pad with replicates of the border instead of zeros, apply a windowing function to the image or make the pad size larger in order to gain frequency resolution. Each of these is a standard augmentation to the technique I've outlined above and should be trivial to research via Google and Wikipedia. Also note that I haven't added any basic MATLAB optimizations such as pre-allocation, etc.
作为最后一点，您可能想要跳过图像填充（即如果您的过滤器总是小于图像，请使用第一个代码示例。这是因为，向图像添加零会创建填充开始的人工边缘特征。如果过滤器很小，循环卷积的环绕不会引起问题，因为只会涉及过滤器的填充中的零。但是一旦过滤器足够大，环绕效应就会变得很严重。如果必须使用大型过滤器，则可能必须使用更复杂的填充方案，或裁剪图像的边缘。
As a final note, you may want to skip the image padding (i.e. use first code example) if your filters are always much smaller than the image. This is because, adding zeros to an image creates an artificial edge feature where the padding begins. If the filter is small, the wraparound from circular convolution doesn't cause problems, because only the zeros in the filter's padding will be involved. But as soon as the filter is large enough, the wraparound effect will become severe. If you have to use large filters, you may have to use a more complicated padding scheme, or crop the edge of the image.
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