我最近研究了 CCA 的概念，并想在 MATLAB 中实现它。但是有一个现有的 matlab 命令佳能 http://in.mathworks.com/help/stats/canoncorr.html展示。我想编写自己的代码。我对其进行了广泛的研究，发现了三种方法：
1：哈同：该方法使用拉格朗日乘子将问题分解为广义特征值问题。代码可以在这里找到：cca_hardoon http://www.davidroihardoon.com/Professional/Code_files/cca.m为了理智起见，我还在这里给出了代码：数据必须事先居中。
function [Wx, Wy, r] = cca(X,Y)
% CCA calculate canonical correlations
%
% [Wx Wy r] = cca(X,Y) where Wx and Wy contains the canonical correlation
% vectors as columns and r is a vector with corresponding canonical
% correlations.
%
% Update 31/01/05 added bug handling.
if (nargin ~= 2)
disp('Inocorrect number of inputs');
help cca;
Wx = 0; Wy = 0; r = 0;
return;
end
% calculating the covariance matrices
z = [X; Y];
C = cov(z.');
sx = size(X,1);
sy = size(Y,1);
Cxx = C(1:sx, 1:sx) + 10^(-8)*eye(sx);
Cxy = C(1:sx, sx+1:sx+sy);
Cyx = Cxy';
Cyy = C(sx+1:sx+sy,sx+1:sx+sy) + 10^(-8)*eye(sy);
%calculating the Wx cca matrix
Rx = chol(Cxx);
invRx = inv(Rx);
Z = invRx'*Cxy*(Cyy\Cyx)*invRx;
Z = 0.5*(Z' + Z); % making sure that Z is a symmetric matrix
[Wx,r] = eig(Z); % basis in h (X)
r = sqrt(real(r)); % as the original r we get is lamda^2
Wx = invRx * Wx; % actual Wx values
% calculating Wy
Wy = (Cyy\Cyx) * Wx;
% by dividing it by lamda
Wy = Wy./repmat(diag(r)',sy,1);
2.MATLAB方法请注意，数据的居中是在代码本身内完成的。
3. CCA by Normal SVD only : This approach does not require the qr decomposition and utilizes the svd decomposition only. I have referred top this article here : cca by svd http://www.cs.columbia.edu/~stratos/research/pca_cca.pdf. Please refer to the text articles below which are taken from the referred article.
我尝试自己编写这个程序，但没有成功。
function [A,B,r,U,V] = cca_by_svd(x,y)
% computing the means
N = size(x,1); mu_x = mean(x,1); mu_y = mean(y,1);
% substracting the means
x = x - repmat(mu_x,N,1); y = y - repmat(mu_y,N,1);
x = x.'; y = y.';
% computing the covariance matrices
Cxx = (1/N)*x*(x.'); Cyy = (1/N)*y*(y.'); Cxy = (1/N)*x*(y.');
%dimension
m = min(rank(x),rank(y));
%m = min(size(x,1),size(y,1));
% computing the quare root inverse of the matrix
[V,D]=eig(Cxx); d = diag(D);
% Making all the eigen values positive
d = (d+abs(d))/2; d2 = 1./sqrt(d); d2(d==0)=0; Cxx_iv=V*diag(d2)*inv(V);
% computing the quare root inverse of the matrix
[V,D]=eig(Cyy); d = diag(D);
% Making all the eigen values positive
d = (d+abs(d))/2; d2 = 1./sqrt(d); d2(d==0)=0; Cyy_iv=V*diag(d2)*inv(V);
Omega = Cxx_iv*Cxy*Cyy_iv;
[C,Sigma,D] = svd(Omega);
A = Cxx_iv*C; A = A(:,1:m);
B = Cyy_iv*D.'; B = B(:,1:m);
A = real(A); B = real(B);
U = A.'*x; V = B.'*y;
r = Sigma(1:m,1:m);
我正在运行这个代码片段：
clc;clear all;close all;
load carbig;
X = [Displacement Horsepower Weight Acceleration MPG];
nans = sum(isnan(X),2) > 0;
x = X(~nans,1:3);
y = X(~nans,4:5);
[A1, B1, r1, U1, V1] = canoncorr(x, y);
[A2, B2, r2, U2, V2] = cca_by_svd(x, y);
[A3, B3, r3] = cca(x.',y.',1);
投影向量如下：
>> A1
A1 =
0.0025 0.0048
0.0202 0.0409
-0.0000 -0.0027
>> A2
A2 =
0.0025 0.0048
0.0202 0.0410
-0.0000 -0.0027
>> A3
A3 =
-0.0302 -0.0050 -0.0022
0.0385 -0.0420 -0.0176
0.0020 0.0027 -0.0001
>> B1
B1 =
-0.1666 -0.3637
-0.0916 0.1078
>> B2
B2 =
-0.1668 -0.3642
-0.0917 0.1079
>> B3
B3 =
0.0000 + 0.0000i 0.3460 + 0.0000i 0.1336 + 0.0000i
0.0000 + 0.0000i -0.0967 + 0.0000i 0.0989 + 0.0000i
问题：有人可以告诉我哪里出了问题吗？我提到的三种方法都解决相同的问题，理想情况下它们的解决方案应该收敛。我承认我的代码'cca_by_svd'可能是错误的，但hardoon的代码和matlab的输出应该是相同的。请指出我哪里出错了。edit我已经重新检查并更正了我的代码。现在对于这个数据集，方法 2 和 3 收敛。
有几件事是cca(X,Y)不这样做canoncorr does:
一是对数据进行标准化。如果你添加X = normc(X')'（也为Y）到你的cca(X,Y)函数，输出r将匹配canoncorr。如果你调查canoncorr的代码，你会看到它从 QR 分解开始X and Y.
另一个区别是eig按升序对特征值进行排序，因此cca(X,Y)应该翻转输出eig(Z).
注意：尽管纠正了这些差异，我无法完全恢复 Wx 和 Wy 以匹配的输出canoncorr。理想情况下，Wx'*Wx 之间应该完全相同cca and canoncorr.