使用OpenCV进行特征向量计算

我有这个矩阵A,表示图像像素强度的相似性.例如:考虑10 x 10图像.在这种情况下,矩阵A具有尺寸100 x 100,并且元素A(i,j)将具有0到1范围内的值,表示像素i与j在强度方面的相似性.

我使用OpenCV进行图像处理,开发环境是Linux上的C语言.

目标是计算矩阵A的特征向量,我使用了以下方法:

static CvMat mat, *eigenVec, *eigenVal;
static double A[100][100]={}, Ain1D[10000]={};
int cnt=0;

//Converting matrix A into a one dimensional array
//Reason: That is how cvMat requires it
for(i = 0;i < affnDim;i++){
  for(j = 0;j < affnDim;j++){
 Ain1D[cnt++] = A[i][j];
  }
}

mat = cvMat(100, 100, CV_32FC1, Ain1D); 

cvEigenVV(&mat, eigenVec, eigenVal, 1e-300);

for(i=0;i < 100;i++){
  val1 = cvmGet(eigenVal,i,0); //Fetching Eigen Value

  for(j=0;j < 100;j++){   
 matX[i][j] = cvmGet(eigenVec,i,j); //Fetching each component of Eigenvector i    
  }
}

问题:执行后,我几乎将所有特征向量的所有分量都归零.我尝试了不同的图像,并尝试使用0到1之间的随机值填充A,但结果相同.

返回的顶部特征值很少如下所示:

9805401476911479666115491135488.000000  
-9805401476911479666115491135488.000000  
-89222871725331592641813413888.000000  
89222862280598626902522986496.000000  
5255391142666987110400.000000

我现在正在思考使用cvSVD()的行,它执行实数浮点矩阵的奇异值分解,并可能产生特征向量.但在此之前,我想过在这里问一下.我目前的做法有什么荒谬之处吗？我是否使用正确的API,即cvEigenVV()用于正确的输入矩阵(我的矩阵A是浮点矩阵)？

干杯

读者注意:这篇文章最初可能与该主题无关,但请参阅上述评论中的讨论.

以下是我在MATLAB中实现应用于图像像素的光谱聚类算法的尝试.我严格遵循本文由@Andriyev提到:

Andrew Ng,Michael Jordan和Yair Weiss(2002).关于谱聚类:分析和算法.在T. Dietterich,S.Becker和Z. Ghahramani(编辑),神经信息处理系统的进展14.麻省理工学院出版社

代码:

%# parameters to tune
SIGMA = 2e-3;       %# controls Gaussian kernel width
NUM_CLUSTERS = 4;   %# specify number of clusters

%% Loading and preparing a sample image
%# read RGB image, and make it smaller for fast processing
I0 = im2double(imread('house.png'));
I0 = imresize(I0, 0.1);
[r,c,~] = size(I0);

%# reshape into one row per-pixel: r*c-by-3
%# (with pixels traversed in columwise-order)
I = reshape(I0, [r*c 3]);

%% 1) Compute affinity matrix
%# for each pair of pixels, apply a Gaussian kernel
%# to obtain a measure of similarity
A = exp(-SIGMA * squareform(pdist(I,'euclidean')).^2);

%# and we plot the matrix obtained
imagesc(A)
axis xy; colorbar; colormap(hot)

%% 2) Compute the Laplacian matrix L
D = diag( 1 ./ sqrt(sum(A,2)) );
L = D*A*D;

%% 3) perform an eigen decomposition of the laplacian marix L
[V,d] = eig(L);

%# Sort the eigenvalues and the eigenvectors in descending order.
[d,order] = sort(real(diag(d)), 'descend');
V = V(:,order);

%# kepp only the largest k eigenvectors
%# In this case 4 vectors are enough to explain 99.999% of the variance
NUM_VECTORS = sum(cumsum(d)./sum(d) < 0.99999) + 1;
V = V(:, 1:NUM_VECTORS);

%% 4) renormalize rows of V to unit length
VV = bsxfun(@rdivide, V, sqrt(sum(V.^2,2)));

%% 5) cluster rows of VV using K-Means
opts = statset('MaxIter',100, 'Display','iter');
[clustIDX,clusters] = kmeans(VV, NUM_CLUSTERS, 'options',opts, ...
    'distance','sqEuclidean', 'EmptyAction','singleton');

%% 6) assign pixels to cluster and show the results
%# assign for each pixel the color of the cluster it belongs to
clr = lines(NUM_CLUSTERS);
J = reshape(clr(clustIDX,:), [r c 3]);

%# show results
figure('Name',sprintf('Clustering into K=%d clusters',NUM_CLUSTERS))
subplot(121), imshow(I0), title('original image')
subplot(122), imshow(J), title({'clustered pixels' '(color-coded classes)'})

...并使用我在Paint中绘制的简单房屋图像,结果如下:

顺便说一下,使用的前4个特征值是:

1.0000
0.0014
0.0004
0.0002

和相应的特征向量[长度为r*c = 400的列]:

-0.0500    0.0572   -0.0112   -0.0200
-0.0500    0.0553    0.0275    0.0135
-0.0500    0.0560    0.0130    0.0009
-0.0500    0.0572   -0.0122   -0.0209
-0.0500    0.0570   -0.0101   -0.0191
-0.0500    0.0562   -0.0094   -0.0184
......

请注意,执行的步骤在您的问题中没有提及(拉普拉斯矩阵,并对其行进行规范化)