Fuzzy C-Means and Principle Component Analysis Approaching to Images (Numbers 0 ans 6) Recognition

由于这学期上了Fuzzy Logic Control这门课程，所以希望借着博客的东风来记录自己的学习历程，并且加深自己的理解。入门新手，如有错误，请您指正。

This is a image classification problem to which I ask you to use fuzzy clustering function fuzzy c-mean.The attached data file contains an input array X2 with 1200  16x16 digitized images of handwritten 0 and 6. The true labels of those images are given
in the array y2. The convention used is -1 for digit 0, and 1 for digit 6. Your task is to investigate the use of fuzzy clustering to classify the input images as 0 or 6. You should aim at minimizing the classification error. The classification error is the
difference between your fuzzy classification and the true class given in the y2 array.

 分析：每张图片的分辨率是16*16灰度图像，所以每张图像的特征有256个，所以此时对于1200个样本来说，256个特征确实是太多了。但是出于测试的目标，我测试了一下大概正确区分的几率为78.2%左右。在第二个程序中我加入了Principle Component Analysis法去提取出most significant features，然后以此为数据代替原来的每张图用256个特征表示的数据，利用fcm去识别图像数字内容。

Solution 1：

%FCM applications in images clustering \

close all;

clear all;

clc;

load('digits_0_6');

%-- plot 20 images in X2

figure(1);

for i=1:20

    subplot(5,4,i); imshow(X2(:,:,i));

end

for i=1:size(X2,3)

    X2_Reshape(i,:)=reshape(X2(:,:,i).',1,256);

end

%Specify fuzzy partition matrix exponents.

M = [1.1 2.0 3.0 4.0];

for i = 1:4

    % Cluster the data.

    options = [M(i) NaN NaN 0];

    clu_n = 2;

    [centers,U] = fcm(X2_Reshape,clu_n,options);

    % Classify the data points.

    maxU = max(U);

    index1 = find(U(1,:) == maxU);

    index2 = find(U(2,:) == maxU);

    % Find data points with lower maximum membership values.

    %index3 = find(maxU < 0.6);

    % Calculate the average maximum membership value.

    averageMax = mean(maxU);

    

    len_1 = length(index1);

    len_2 = length(index2);

    

    if find(index2(:,:)==4)

        for j =1:len_1

            estimated_y2(index1(j)) =-1;

        end

        

        for j =1:len_2

            estimated_y2(index2(j)) =1;

        end

    elseif find(index1(:,:)==4)

        for j =1:len_1

            estimated_y2(index1(j)) =1;

        end

        

        for j =1:len_2

        estimated_y2(index2(j)) = -1;

        end

    end

   

    %-- calculate the clustering accuracy

    right = 0;

    

    for k=1:length(y2)

        if estimated_y2(1,k) == y2(1,k)

            right = right +1;

        end

    end

    

    %-- maximum accuracy is about 95.6% just by fcm approaching to this problem

    %-- maybe because too many features 256 for just 1200 samples,so I try

    %-- to reduce image features by pca in another solution, the accuracy is up to 96% 

    fprintf('When M = %d, Accuracy = %.3f%%, averageMax is %.3f;\n', i,right/12,averageMax); 

end

在此程序中显示了前二十个图像，输出结果是：

When M = 1, Accuracy = 76.333%, averageMax is 0.965;

When M = 2, Accuracy = 77.500%, averageMax is 0.666;

When M = 3, Accuracy = 78.167%, averageMax is 0.500;

When M = 4, Accuracy = 78.167%, averageMax is 0.500;

Solution 2：

%-- pca and fcm method to indentify number 0 and 6

%-- pca : extract the project of the images

%-- fcm : clusterinf the principle features of image 

close all;

clear all;

clc;

load('digits_0_6');

%-- plot 20 images in X2

figure(1);

for i=1:20

    subplot(5,4,i); imshow(X2(:,:,i));

end

%adopting pca method to extract principal features in images

for i=1:size(X2,3)

    %the number of the most significant principal components that are taken into account.

    m =5;   

    image = X2(:,:,i);

    [l,N]=size(image);

    % Subtracting the mean

    mean_vec=mean(image')';

    X_zero=image-mean_vec*ones(1,N);

    % Computing the covariance matrix and its eigenvalues/eigenvectors

    R=cov(X_zero');

    [V,D]=eig(R);

    eigenval=diag(D);

    [eigenval,ind]=sort(eigenval,1,'descend');

    eigenvec=V(:,ind);

    explain=eigenval/sum(eigenval);

    % Keeping the first m eigenvaules/eigenvectors

    eigenval=eigenval(1:m);

    eigenvec=eigenvec(:,1:m);

    % Computing the transformation matrix

    A=eigenvec(:,1:m)';

    % Computing the transformed data set

    Y=A*image;

    

    %-- images projection for 5 significant principal components

    X2_Projection(i,:) = reshape(Y',1,80); 

end

%Specify fuzzy partition matrix exponents.

M = [1.1 2.0 3.0 4.0];

for i = 1:4

    % Cluster the data.

    options = [M(i) NaN NaN 0];

    clu_n = 2;

    [centers,U] = fcm(X2_Projection,clu_n,options);

    % Classify the data points.

    maxU = max(U);

    index1 = find(U(1,:) == maxU);

    index2 = find(U(2,:) == maxU);

    % Find data points with lower maximum membership values.

    index3 = find(maxU < 0.6);

    % Calculate the average maximum membership value.

    averageMax = mean(maxU);

    

    len_1 = length(index1);

    len_2 = length(index2);

    

    if find(index2(:,:)==4)

        for j =1:len_1

            estimated_y2(index1(j)) =-1;

        end

        

        for j =1:len_2

            estimated_y2(index2(j)) =1;

        end

    elseif find(index1(:,:)==4)

        for j =1:len_1

            estimated_y2(index1(j)) =1;

        end

        

        for j =1:len_2

        estimated_y2(index2(j)) = -1;

        end

    end

   

    %-- calculate the clustering accuracy

    right = 0;

    

    for k=1:length(y2)

        if estimated_y2(1,k) == y2(1,k)

            right = right +1;

        end

    end

    

    %-- maximum accuracy is about 95.6% by pca and fcm approaching to this problem 

    fprintf('When M = %d, Accuracy = %.3f%%, averageMax is %.3f;\n', i,right/12,averageMax); 

end

输出结果是：

When M = 1, Accuracy = 95.667%, averageMax is 0.978;

When M = 2, Accuracy = 92.167%, averageMax is 0.571;

When M = 3, Accuracy = 91.833%, averageMax is 0.500;

When M = 4, Accuracy = 91.833%, averageMax is 0.500;

可以看出，实验结果有明显的提升，但并不一定是最优答案，希望有新的朋友继续探索。

我想把数据传上去呢，可是没发现怎么upload啊。