Friday 16 January 2015

Principal Component Analysis Well Explained With an Example in MATLAB


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Consider an image as shown above and the image is made as first column elements as shown below

X =    [1     2     4     3     5     9     4     2;
           5     4     7     4     3     2     1     3;
           3     2     4     5     6     2     1     2;
           4     1     3     2     2     1     3     4]

First column represents one feature vectorand has four dimensions and there are 8 feature vectors.
Here, dimensions < number of feature vectors.
[vecs,val]=eigs(X*X',2);% X*X' gives matrix of size 4x4 as 4x8*8x4 . Only first two largest eigen vectors are considered as eigs(X*X',2)
%So,  instead of storing 4 vectors, you store 2 vectors
wt1=vecs'*X(:,1);
reconstructed =vecs*wt1; % approximate of the original
wt2=vecs'*X(:,2);
reconstructed =vecs*wt2;

For example: if you have 4 feature vector and each feature has 8 dimensions as shown below

X =[     1     5     3     4;
            2     4     2     1;
            4     7     4     3;
            3     4     5     2;
            5     3     6     2;
            9     2     2     1;
            4     1     1     3;
            2     3     2     4];
[vecs,val]=eigs(X'*X,2);%This is done to simplify computation as X*X' gives matrix of size 8x8. This is the trick used by Pentland and Turk
ef=X*vecs;  % in order to get eigen vectors of X*X', we have to multiply X with eigen vectors of X'*X..
for i=1:size(ef,2)
ef(:,i)=ef(:,i)./norm(ef(:,i)); % u divide each shape by its norm
end
wt1=ef'*X(:,1); % Each shape of 8 dimensions is now represented in 4 dimensions as 2 eigen vectors considered
reconstructed =ef*wt1; % you  get first shape back
wt2=ef'*X(:,2);
reconstructed =vecs*wt2;


You  can get back the image

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