A fast algorithm for feature extraction of hyperspectral images using the first order statistics

Abstract

A new supervised feature extraction method appropriate for small sample size situations is proposed in this work. The proposed method is based on the first-order statistics, in which there is no need to estimate the scatter matrices. Thus, the presented method not only can avoid the singularity problem in small sample size situations but also can achieve high performance in such situations. In addition, due to the fact that the proposed algorithm only exploits the first order statistical moments, it is very fast making it suitable for real-time hyperspectral scene analysis. The proposed method makes a matrix whose columns are obtained by averaging training samples of different classes. Then, a new transform is used to map the features from the original space into a new low-dimensional space such that the new features are as different from each other as possible. Subsequently, to capture the inherent nonlinearity of the original data, the algorithm is improved using the kernel trick. In experiments, four widely-used hyperspectral datasets, namely, Indian Pines, University of Pavia, Salinas, and Botswana are classified. The experimental results show that the proposed algorithm achieves state-of-the-art results in small sample size situations.

Appendix

Firstly, the derivation of eq. (8) of the paper is described. Considering equality $2\left({\sum}_{i=1}^d\left({{\boldsymbol{P}}_i}^T{\boldsymbol{P}}_i\right)\right){\mathbf{WAA}}^T-2\lambda \mathbf{W}=0$ (see Eq. (7)), one can apply the cs operator to the both sides of the equality:

$$ 2\mathrm{cs}\left(\left({\sum}_{i=1}^d\left({{\boldsymbol{P}}_i}^T{\boldsymbol{P}}_i\right)\right){\mathbf{WAA}}^T\right)-2\lambda \mathrm{cs}\left(\mathbf{W}\right)=0 $$

(16)

Using the equality cs(abc) = (c^T ⊗ a)cs(b) [28], where ⊗ is the Kronecker product, one can rewrite Eq. (16) as:

$$ \left({\mathbf{AA}}^T\otimes \left({\sum}_{i=1}^d\left({{\boldsymbol{P}}_i}^T{\boldsymbol{P}}_i\right)\right)-\lambda {\mathbf{I}}_{ld}\right)\mathrm{cs}\mathbf{W}=0 $$

(17)

which is equal to Eq. (8).

Secondly, the reason that why csW is the eigenvector corresponding to the largest eigenvalue of ${\mathbf{AA}}^T\otimes \left({\sum}_{i=1}^d\left({{\boldsymbol{P}}_i}^T{\boldsymbol{P}}_i\right)\right)$ is explained. Eq. (5) can be rewritten as:

$$ g={\sum}_{i=1}^d tr\left({\mathbf{A}}^T{\mathbf{W}}^T{\boldsymbol{P}}_i^T{\boldsymbol{P}}_i\mathbf{WA}\right) $$

(18)

Because the trace is invariant under cyclic permutations, one may write:

$$ g={\sum}_{i=1}^d tr\left({\mathbf{W}}^T{\boldsymbol{P}}_i^T{\boldsymbol{P}}_i{\mathbf{W}\mathbf{AA}}^T\right) $$

(19)

Using the equality tr(a^Tb) = cs(a)^Tcs(b) [28], Eq. (19) can be rewritten as:

$$ g={\sum}_{i=1}^d\mathrm{cs}{\left(\mathbf{W}\right)}^T\mathrm{cs}\left({\boldsymbol{P}}_i^T{\boldsymbol{P}}_i{\mathbf{WAA}}^T\right) $$

(20)

Using the equality cs(abc) = (c ⊗ a)cs(b) [28], one can write:

$$ {\displaystyle \begin{array}{c}g={\sum}_{i=1}^d\mathrm{cs}{\left(\mathbf{W}\right)}^T\left({\mathbf{AA}}^T\otimes {\boldsymbol{P}}_i^T{\boldsymbol{P}}_i\right)\mathrm{cs}\left(\mathbf{W}\right)=\\ {}\mathrm{cs}{\left(\mathbf{W}\right)}^T\left({\mathbf{AA}}^T\otimes {\sum}_{i=1}^d\left({\boldsymbol{P}}_i^T{\boldsymbol{P}}_i\right)\right)\mathrm{cs}\left(\mathbf{W}\right)\end{array}} $$

(21)

Due to the fact that csW is the eigenvector of ${\mathbf{AA}}^T\otimes \left({\sum}_{i=1}^d\left({{\boldsymbol{P}}_i}^T{\boldsymbol{P}}_i\right)\right)$ (see Eq. (8)), one can conclude:

$$ g=\lambda \mathrm{cs}{\left(\mathbf{W}\right)}^T\mathrm{cs}\left(\mathbf{W}\right)=\lambda tr\left({\mathbf{W}}^T\mathbf{W}\right)=\lambda $$

(22)

So, the maximum of g is obtained if csW is the eigenvector corresponding to the largest eigenvalue of ${\mathbf{AA}}^T\otimes \left({\sum}_{i=1}^d\left({{\boldsymbol{P}}_i}^T{\boldsymbol{P}}_i\right)\right)$.