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用LDA和PCA模式識別方法對人臉特征進(jìn)行提取計(jì)算的Matlab程序

代碼片段和文件信息
load(‘ORLFace.mat‘);
%?basic?information
dim=size(Iv1);
tal=size(Iv2);
class=size(Iv3);
ell=5;?%?ell?training?sample;
ellsample=5;?%?ellsample?test?sample;
t=1e7;?%?Similarity?matrix?的參數(shù)
order=1;??
dita=1e7;?
NumTotal=ell*class;
lpp=NumTotal-class;
lda=class-1;
lda=35
polynomial=1
%-------------------?step?1?????KPCA???????------------------------------%
Itr=zeros(dimellclass);??%?Training?sample?feature?vector
for?classnum=1:class
????for?e=1:ell
????????Itr(:eclassnum)=Iv(:eclassnum);??%
????end????
end
It=zeros(dimellsampleclass);%?Testing?sample?feature?vector
for?classnum=1:class
????for?e=1:ellsample
????????It(:eclassnum)=Iv(:e+ellclassnum);??%
????end????
end
%----------------------------------------------------------------%
%???倒入向量?Iv(dimell)
%???樣本類數(shù)??class
Imean=zeros(dim1);
for?classnum=1:class
????for?i=1:ell
????????Imean=Imean+Itr(:iclassnum);?
????end
end
Imean=(1/(ell*class))*Imean;???????????????????????%求平均向量
Q=zeros(dimell*class);
for?classnum=1:class
????for?num=1:ell
????????Q(:num+(classnum-1)*ell)=Itr(:numclassnum)-Imean(:1);
????end
end
R=zeros(ell*classell*class);
R=Q‘*Q;????%?R‘s?size?is?ell?*?ell
d=rank(R);
d=40
[UL]=eigs(Rd‘LM‘);??%?求出?R?的eigenvector?and?eigenvalue
Wpca=zeros(dimd);
for?p=1:d
????Wpca(:p)=(1/(sqrt(L(pp))))*Q*U(:p);
end
%%?程序至此得到了線形變換的矩陣??W??.
%----------------------------------------------------------------%
Iy=zeros(dellclass);?%?training?feature?vector??%???
for?classnum=1:class
????for?num=1:ell
????????Y=zeros(d1);
????????Iy(:numclassnum)=Wpca‘*Itr(:numclassnum);
????end
end
%?-------------------------------------------------------------------------
Tpca=zeros(dellsampleclass);
for?classnum=1:class
????for?num=1:ellsample
????????Tpca(:numclassnum)=Wpca‘*It(:numclassnum);
????end
end
%?----------------???????????????????????step?3??????LDA???????????????????????????-----------------%
Minclass=zeros(dclass);
for?classnum=1:class??????%?得到?每一個(gè)類內(nèi)的樣本的平均值
????for?i=1:ell
????????Minclass(:classnum)=Minclass(:classnum)+Iy(:iclassnum);?
????end
????Minclass(:classnum)=(1/ell)*Minclass(:classnum);??
end
Sw=zeros(dd);????%?計(jì)算wihin-class?matrix?Sw
for?classnum=1:class??????
????for?i=1:ell
????????
????????Sw=Sw+(Iy(:iclassnum)-Minclass(:classnum))*(Iy(:iclassnum)-Minclass(:classnum))‘;
????????
????end
end
Msample=zeros(d1);?%?得到?所有樣本的平均值
for?classnum=1:class
????for?i=1:ell
????????Msample=Msample+Iy(:iclassnum);?
????end
end
Msample=(1/(ell*class))*Msample;???????????????????????%求平均向量
Sb=zeros(dd);????%?計(jì)算between-class?matrix?Sb
for?classnum=1:class??????
????Sb=Sb+ell*(Minclass(:classnum)-Msample)*(Minclass(:classnum)-Msample)‘;
end
%?d=dim;?????????????????????????????????%?選擇?d?=dim?個(gè)最大的特征值
[FiCi]=eigs(Swd‘LM‘);??%?求出?R?的eigenvector?and?eigenvalue
Diag=zeros(1d)
for?p=1:d
????Diag(1p)=1/(sqrt(Ci(pp)));
end
Cmodify=diag(Dia
?屬性????????????大小?????日期????時(shí)間???名稱
-----------?---------??----------?-----??----
?????文件???????5429??2006-04-07?21:22??LDA.m
?????文件???????3822??2006-04-16?13:56??PCA.m
-----------?---------??----------?-----??----
?????????????????9469????????????????????3
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