如何使用减少的数据 - 主成分分析的输出

Question

如何使用减少的数据 - 主成分分析的输出

SKM*_*SKM 7 matlab image-processing pca dimensionality-reduction

我发现很难将理论与实施联系起来.我很感激帮助知道我的理解错误.

符号 - 粗体大写的矩阵和粗体字母小写字母的向量

是一个数据集观察,每个变量.所以,鉴于这些观察- 维数据向量, - 维主轴是 ,为在哪里是目标维度.

该观测数据矩阵的主要组成部分是矩阵 ,矩阵和矩阵 .

列形成一个正交的基础功能和输出是最小化平方重建误差的主成分投影:

和最优重建是(谁)给的 .

数据模型是

X(i,j) = A(i,:)*S(:,j) + noise

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其中PCA应在X上完成以获得输出S. S必须等于Y.

问题1:减少的数据Y不等于模型中使用的S. 我的理解在哪里错了？

问题2:如何重建以使错误最小化？

请帮忙.谢谢.

   clear all
clc
n1 = 5; %d dimension
n2 = 500; % number of examples

ncomp = 2; % target reduced dimension
%Generating data according to the model
% X(i,j) = A(i,:)*S(:,j) + noise
Ar = orth(randn(n1,ncomp))*diag(ncomp:-1:1);
T = 1:n2;
%generating synthetic data from a dynamical model
S = [ exp(-T/150).*cos( 2*pi*T/50 )
       exp(-T/150).*sin( 2*pi*T/50 ) ];
% Normalizing to zero mean and unit variance
S = ( S - repmat( mean(S,2), 1, n2 ) );
S = S ./ repmat( sqrt( mean( Sr.^2, 2 ) ), 1, n2 );
Xr = Ar * S;
Xrnoise = Xr + 0.2 * randn(n1,n2);

h1 = tsplot(S);


    X = Xrnoise;

XX = X';
[pc, ~] = eigs(cov(XX), ncomp);
Y = XX*pc;

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更新[8月10日]

根据答案,这里是完整的代码

 clear all
clc
n1 = 5; %d dimension
n2 = 500; % number of examples

ncomp = 2; % target reduced dimension
%Generating data according to the model
% X(i,j) = A(i,:)*S(:,j) + noise
Ar = orth(randn(n1,ncomp))*diag(ncomp:-1:1);
T = 1:n2;
%generating synthetic data from a dynamical model
S = [ exp(-T/150).*cos( 2*pi*T/50 )
       exp(-T/150).*sin( 2*pi*T/50 ) ];
% Normalizing to zero mean and unit variance
S = ( S - repmat( mean(S,2), 1, n2 ) );
S = S ./ repmat( sqrt( mean( S.^2, 2 ) ), 1, n2 );
Xr = Ar * S;
Xrnoise = Xr + 0.2 * randn(n1,n2);

    X = Xrnoise;

XX = X';
[pc, ~] = eigs(cov(XX), ncomp);
Y = XX*pc; %Y are the principal components of X' 
%what you call pc is misleading, these are not the principal components
%These Y columns are orthogonal, and should span the same space 
%as S approximatively indeed (not exactly, since you introduced noise).

%If you want to reconstruct 
%the original data can be retrieved by projecting 
%the principal components back on the original space like this:
Xrnoise_reconstructed = Y*pc';

%Then, you still need to project it through 
%to the S space, if you want to reconstruct S
S_reconstruct = Ar'*Xrnoise_reconstructed';


plot(1:length(S_reconstruct),S_reconstruct,'r')
hold on
 plot(1:length(S),S)

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情节是这与答案中显示的非常不同.只有S的一个组件与S_reconstructed的组件完全匹配.不应该重建源输入S的整个原始二维空间吗？即使我切断了噪音,那么S的一个组成部分也会被精确地重建.