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have been straggling with the concept of feature normalization for the hyperspectral images. Respect to my problem I have attached a picture of my problem which clearly stands the issue which I have. enter image description here

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There are different methods for the normalization. If it is for visual graphs, you can divide it into bands which can be defined in realtime, so you'd write a program which presents visual graphs of frequency A->B where A and B can be changed by the user. That helps you see display it graphically. using X and Y axis of the mouse to define A->B bandwidth. Then you can scan backwards and forwards through all the frequencies interactively while also normalizing them, using an amplitude meter beside the graph, to say the overall max amplitude of that bandwidth.

If the normalizing process is for high fidelity, balanced post processing and statistics, then it's best to find the highest value at any frequency and divide all the pixel amplitudes by that high value so that it equals 1.

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  • $\begingroup$ Thanks for your remark. Actually, I want to compute the PCA feature space. Therefore, I would like to first scale the input data and then apply PCA on them since the magnitude of pixel values in different bands are different from each other. So the only thing that I want to know is that, should I compute the mean for each pixel in 200 dimensions or should I compute the mean for all pixels in the individual bands? Note that each band is a variable which might be significant in prediction for a classification algorithm. $\endgroup$ – morteza Jan 29 at 19:54
  • $\begingroup$ I'd be pretty sure that it's normalized for the entire array, rather than for individual bands, because the PCA will be comparing across amplitudes to find the loudest bands, and otherwise you would hugely amplifly some bands that are near zero and reduce the loudest ones to the same level. there is a lot about normalize PCA online... i.e. groups.google.com/forum/#!topic/kallisto-sleuth-users/… $\endgroup$ – com.prehensible Jan 30 at 10:09
  • $\begingroup$ Thank you so much. I have got the idea. :) $\endgroup$ – morteza Jan 30 at 15:56

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