Feeding HOG into SVM: the HOG has 9 bins, but the SVM takes in a 1D matrix

In OpenCV, there is a CvSVM class which takes in a matrix of samples to train the SVM. The matrix is 2D, with the samples in the rows.

I created my own method to generate a histogram of oriented gra开发者_Go百科dients (HOG) off of a video feed. To do this, I created a 9 channeled matrix to store the HOG, where each channel corresponds to an orientation bin. So in the end I have a 40x30 matrix of type CV_32FC(9).

Also made a visualisation for the HOG and it's working.

I don't see how I'm supposed to feed this matrix into the OpenCV SVM, because if I flatten it, I don't see how the SVM is supposed to learn a 9D hyperplane from 1D input data.

The SVM always takes in a single row of data per feature vector. The dimensionality of the feature vector is thus the length of the row. If you're dealing with 2D data, then there are 2 items per feature vector. Example of 2D data is on this webpage:

http://www.csie.ntu.edu.tw/~cjlin/libsvm/

code of an equivalent demo in OpenCV http://sites.google.com/site/btabibian/labbook/svmusingopencv

The point is that even though you're thinking of the histogram as 2D with 9-bin cells, the feature vector is in fact the flattened version of this. So it's correct to flatten it out into a long feature vector. The result for me was a feature vector of length 2304 (16x16x9) and I get 100% prediction accuracy on a small test set (i.e. it's probably slightly less than 100% but it's working exceptionally well).

The reason this works is that the SVM is working on a system of weights per item of the feature vector. So it doesn't have anything to do with the problem's dimension, the hyperplane is always in the same dimension as the feature vector. Another way of looking at it is to forget about the hyperplane and just view it as a bunch of weights for each item in the feature vector. In this case, it needs one weighting for every item, then it multiplies each item by its weighting and outputs the result.