在Matlab中使用可分离Gabor滤波器 [英] Working on Separable Gabor filters in matlab

问题描述

如果可以将过滤器g表示为两个向量 grow 和 gcol 的乘法，则过滤器g称为可分离。使用一维滤波器将二维滤波器的计算复杂度从 O（M ^ 2 N ^ 2）减少到 O（2M N ^ 2） code>其中M和N分别是过滤器掩码和图片的宽度（和高度）。

在。

注意：θ可以扩展到等于 k * pi / 4。通过与这个stackoverflow链接，我们可以认为 f = 1 / lambda 。

通过更改我之前的代码，我写了一个matlab代码，使Gabor过滤器可分离通过使用上面的等式，但我是确保我的代码下面是不正确的，特别是当我初始化 gbp 和 glp 方程。这就是为什么我需要你的帮助。

现在让我们来看看我的代码：

  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]; ％我们有四个方向
 RF_siz = [7：2：37]; ％我们得到16个刻度（7x7到37x37，以两个像素为单位）
 minFS = 7; ％最小接受字段
 maxFS = 37; ％最大接受场
 sigma = 0.0036 * RF_siz。^ 2 + 0.35 * RF_siz + 0.18; ％定义有效宽度方程
 lambda = sigma / 0.8; ％it波长方程（λ）
 G = 0.3; ％空间纵横比：0.23 < γ， 0.92 
 
 
 numFilterSizes = length（RF_siz）; ％我们得到16 
 
 numSimpleFilters = length（rot）; ％我们得到4 
 
 numFilters = numFilterSizes * numSimpleFilters; ％我们得到16x4 = 64个过滤器
 
 fSiz = zeros（numFilters，1）; ％它是一个大小为numFilters的向量，其中每个单元格包含过滤器的大小（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 
  

解决方案

我认为代码是正确的，只要您的以前版本的Gabor过滤器代码也是正确的。唯一的是如果 theta = k * pi / 4; ，你的公式这里应该分隔：

  fx = exp（ - （x ^ 2）/（2 * sigmaq））* cos（2 * pi * x / lambda（k） 
 gy = exp（ - （G ^ 2 * y ^ 2）/（2 * sigmaq））; 
  

为了保持一致，您可以使用

  f1（1，x + center）= fx; 
 f2（y + center，1）= gy; 
  

或保留 f1 和 f2 ，但它之后转换你的 filters1 和 filters2
其他一切对我看起来不错。

EDIT

回答上面的工作为 theta = k * pi / 4; ，其他角度，根据你的论文，

  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 
 gy = exp（ - （G ^ 2 * y ^ 2）/（2 * sigmaq））* exp（sqrt（-1）* y * sin 
  

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

解决方案

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));

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