Why eigenvector & eigenvalue in LDA become zero?
Asked Answered
C

1

27

I'd like to implement fast PLDA (Probabilistic Linear Discriminant Analysis) in OpenCV. in this, LINK fast PLDA have been implemented in Matlab and Python. One of the parts of PLDA is LDA. I've written following code for implementing LDA in OpenCV:

int LDA_dim = 120;

// Load data

FileStorage fs("newStorageFile.yml", FileStorage::READ);

// Read data

Mat train_data, train_labels;

fs["train_data"] >> train_data;
fs["train_labels"] >> train_labels;

// LDA

if (LDA_dim > 0)
{
    LDA lda(LDA_dim);
    lda.compute(train_data, train_labels);          // compute eigenvectors

    Mat eigenvectors = lda.eigenvectors();
}

I've converted database that was introduced in above link from .mat to .yml. The result is newStorageFile.yml that I've uploaded here. train_data have 650 rows and 600 cols and train_labels have 650 rows and 1 cols. I don't know why eigenvectors and eigenvalue become zero!!? PLZ help me to fix this code.

It's better to bring the code that convert data from .mat to .yml :

function matlab2opencv( variable, fileName, flag)

[rows cols] = size(variable);

% Beware of Matlab's linear indexing
variable = variable';

% Write mode as default
if ( ~exist('flag','var') )
    flag = 'w'; 
end

if ( ~exist(fileName,'file') || flag == 'w' )
    % New file or write mode specified 
    file = fopen( fileName, 'w');
    fprintf( file, '%%YAML:1.0\n');
else
    % Append mode
    file = fopen( fileName, 'a');
end

% Write variable header
fprintf( file, '    %s: !!opencv-matrix\n', inputname(1));
fprintf( file, '        rows: %d\n', rows);
fprintf( file, '        cols: %d\n', cols);
fprintf( file, '        dt: f\n');
fprintf( file, '        data: [ ');

% Write variable data
for i=1:rows*cols
    fprintf( file, '%.6f', variable(i));
    if (i == rows*cols), break, end
    fprintf( file, ', ');
    if mod(i+1,4) == 0
        fprintf( file, '\n            ');
    end
end

fprintf( file, ']\n');

fclose(file);

Edit 1 ) I've tried LDA with some sample that myself generate:

Mat train_data = (Mat_<double>(3, 3) << 25, 45, 44, 403, 607, 494, 2900, 5900, 2200);
    Mat train_labels = (Mat_<int>(3, 1) << 1, 2, 3 );

    LDA lda(LDA_dim);

    lda.compute(train_data, train_labels);          // compute eigenvectors
    Mat_<double> eigenvectors = lda.eigenvectors();
    Mat_<double> eigenvalues = lda.eigenvalues();
    cout << eigenvectors << endl << eigenvalues;

but I've to got same result: eigenvalue and eigenvector become zero: eigenvector and eigenvalue

Contemplate answered 30/10, 2017 at 6:55 Comment(4)
@bytefish Since you've developed LDA in OpenCV, I think you can help me...Contemplate
did you find any solution for this problem? I'm facing the same issue here, I save the lda.yml, but when I open the file all vectors are 0.0Hesperides
@Amir, No, I've changed my algorithm!!Contemplate
First, mathematically eigenvector can't be zero (eigenvalues can be). Second, eigenvector and eigenvalues are defined for a square matrix. For non-square matrix are defined singular values. You are displaying three vectors (0,0), (1,0) and (0,1); and two eigenvalues 0 and 0.Haemal
R
1

It is because of the floating point imprecision that the eigen values get close to zero.

Rhianna answered 10/1, 2020 at 13:56 Comment(0)

© 2022 - 2024 — McMap. All rights reserved.