algorithm - MATLAB中的图像处理算法

Question

我正在尝试实现本文中描述的算法:

Decomposition of biospeckle images in temporary spectral bands

算法解释如下:

We recorded a sequence of N successive speckle images with a sampling frequency fs. In this way it was possible to observe how a pixel evolves through the N images. That evolution can be treated as a time series and can be processed in the following way: Each signal corresponding to the evolution of every pixel was used as input to a bank of filters. The intensity values were previously divided by their temporal mean value to minimize local differences in reflectivity or illumination of the object. The maximum frequency that can be adequately analyzed is determined by the sampling theorem and s half of sampling frequency fs. The latter is set by the CCD camera, the size of the image, and the frame grabber. The bank of filters is outlined in Fig. 1.

In our case, ten 5° order Butterworth filters were used, but this number can be varied according to the required discrimination. The bank was implemented in a computer using MATLAB software. We chose the Butter-worth filter because, in addition to its simplicity, it is maximally flat. Other filters, an infinite impulse response, or a finite impulse response could be used.

By means of this bank of filters, ten corresponding signals of each filter of each temporary pixel evolution were obtained as output. Average energy Eb in each signal was then calculated:

where pb(n) is the intensity of the filtered pixel in the nth image for filter b divided by its mean value and N is the total number of images. In this way, En values of energy for each pixel were obtained, each of hem belonging to one of the frequency bands in Fig. 1.

With these values it is possible to build ten images of the active object, each one of which shows how much energy of time-varying speckle there is in a certain frequency band. False color assignment to the gray levels in the results would help in discrimination.

这是我基于此的 MATLAB 代码:

for i=1:520
    for j=1:368
        ts = [];
        for k=1:600
            ts = [ts D{k}(i,j)]; %%% kth image pixel i,j --- ts is time series
        end
        ts = double(ts);
          temp = mean(ts);        
           if (temp==0)
                for l=1:10
                    filtImag1{l}(i,j)=0;
                end
                continue;
           end

         ts = ts-temp;          
         ts = ts/temp;    
        N = 5; % filter order
        W = [0.0 0.10;0.10 0.20;0.20 0.30;0.30 0.40;0.40 0.50;0.50 0.60 ;0.60 0.70;0.70 0.80 ;0.80 0.90;0.90 1.0];      
        [B,A]=butter(N,0.10,'low');
        ts_f(1,:) = filter(B,A,ts);         
        N1 = 5;                        
        for ind = 2:9           
            Wn = W(ind,:);
            [B,A] = butter(N1,Wn);            
            ts_f(ind,:) = filter(B,A,ts);            
        end        
        [B,A]=butter(N,0.90,'high');
        ts_f(10,:) = filter(B,A,ts); 

        for ind=1:10
          %Following Paper Suggestion          
           filtImag1{ind}(i,j) =sum(ts_f(ind,:).^2);
        end                 
    end
end

for i=1:10
  figure,imshow(filtImag1{i});  
  colorbar
end

pre_max = max(filtImag1{1}(:));
for i=1:10
   new_max = max(filtImag1{i}(:));
    if (pre_max<new_max)
        pre_max=max(filtImag1{i}(:));
    end
end
new_max = pre_max;

pre_min = min(filtImag1{1}(:));
for i=1:10
   new_min = min(filtImag1{i}(:));
    if (pre_min>new_min)
        pre_min = min(filtImag1{i}(:));
    end
end

new_min = pre_min;

%normalize
 for i=1:10
 temp_imag = filtImag1{i}(:,:);
 x=isnan(temp_imag);
 temp_imag(x)=0;
 t_max = max(max(temp_imag));
 t_min = min(min(temp_imag));
 temp_imag = (double(temp_imag-t_min)).*((double(new_max)-double(new_min))/double(t_max-t_min))+(double(new_min));

 %median filter
 %temp_imag = medfilt2(temp_imag);
 imag_test2{i}(:,:) = temp_imag;
 end

for i=1:10
  figure,imshow(imag_test2{i});
  colorbar
 end

for i=1:10
    A=imag_test2{i}(:,:);
    B=A/max(max(A));
    B=histeq(A);
 figure,imshow(B); 
 colorbar
 imag_test2{i}(:,:)=B;
end

但我没有得到与论文相同的结果。有没有人知道为什么？或者我哪里出错了？

编辑 通过从@Amro 获得帮助并使用他的代码，我得到了以下图像: 这是我来自 72 小时发芽扁 bean 的原始图像(400 张图像，每秒 5 帧):

这是 10 个不同波段的结果图片:

Answer 1

我能发现几个问题:

• 当您将信号除以其均值时，您需要检查它是否不为零。否则结果将为 NaN。

• 作者(我正在关注 this article)使用了一组滤波器，其频带覆盖整个范围，直至奈奎斯特频率。你正在做一半。您传递给 butter 的归一化频率应该一直上升到 1(对应于 fs/2)

• 在计算每个过滤信号的能量时，我认为你不应该除以它的平均值(你之前已经考虑过了)。而是简单地为每个过滤后的信号做:E = sum(sig.^2);

• 在最后的后处理步骤中，您应该归一化到范围 [0,1]，然后应用中值过滤算法 medfilt2。计算看起来不对，应该是这样的:

  img = ( img - min(img(:)) ) ./ ( max(img(:)) - min(img(:)) );
  

---

编辑:

考虑到以上几点，我尝试用向量化的方式重写代码。由于您没有发布样本输入图像，我无法测试结果是否符合预期...另外我不确定如何解释最终图像:)

%# read biospeckle images
fnames = dir( fullfile('folder','myimages*.jpg') );
fnames = {fnames.name};
N = numel(fnames);                    %# number of images
Fs = 1;                               %# sampling frequency in Hz
sz = [209 278];                       %# image sizes
T = zeros([sz N],'uint8');            %# store all images
for i=1:N
    T(:,:,i) = imread( fullfile('folder',fnames{i}) );
end

%# timeseries corresponding to every pixel
T = reshape(T, [prod(sz) N])';        %# columns are the signals
T = double(T);                        %# work with double class

%# normalize signals before filtering (avoid division by zero)
mn = mean(T,1);
T = bsxfun(@rdivide, T, mn+(mn==0));  %# divide by temporal mean

%# bank of filters
numBanks = 10;
order = 5;                                       % butterworth filter order
fCutoff = linspace(0, Fs/2, numBanks+1)';        % lower/upper cutoff freqs
W = [fCutoff(1:end-1) fCutoff(2:end)] ./ (Fs/2); % normalized frequency bands
W(1,1) = W(1,1) + 1e-5;                          % adjust first freq
W(end,end) = W(end,end) - 1e-5;                  % adjust last freq

%# filter signals using the bank of filters
Tf = cell(numBanks,1);                %# filtered signals using each filter
for i=1:numBanks
    [b,a] = butter(order, W(i,:));    %# bandpass filter
    Tf{i} = filter(b,a,T);            %# apply filter to all signals
end
clear T                               %# cleanup unnecessary stuff

%# compute average energy in each signal across frequency bands
Tf = cellfun(@(x)sum(x.^2,1), Tf, 'Uniform',false);

%# normalize each to [0,1], and build corresponding images
Tf = cellfun(@(x)reshape((x-min(x))./range(x),sz), Tf, 'Uniform',false);

%# show images
for i=1:numBanks
    subplot(4,3,i), imshow(Tf{i})
    title( sprintf('%g - %g Hz',W(i,:).*Fs/2) )
end
colormap(gray)

(我使用了来自 here 的图像作为上述结果)

编辑#2

对上面的代码做了一些修改和简化。这将减少内存占用。例如，我使用元胞数组而不是单个多维矩阵来存储结果。这样我们就不会分配一大块连续的内存。我还重复使用了相同的变量，而不是在每个中间步骤引入新变量...