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I have two classes (A,B) that I would like to classify using a SVM. Say that I have a class C and a function f. Can I do this:

A' =  f(A,C) = |A-C|
B' =  f(B,C) = |B-C|

and then perform the classification on A' and B' instead? In the context of my problem A and B are classes where elements are vectors. The f function measures the Mahalanobis distance of each vector with respect to the distribution imposed by C.

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  • $\begingroup$ SVMs support binary classification by nature. Your classes can't be vectors. You're converting vectors to a scalar norm of some vector difference. These still aren't binary classes. I don't think SVM is the tool you want; maybe you can clarify the problem? $\endgroup$
    – Sean Owen
    Nov 15, 2014 at 14:22
  • $\begingroup$ Each element in classes A and B are vectors of a given dimension d. In other words you have d features to consider to classify a vector as A or B. $\endgroup$
    – Diego
    Nov 15, 2014 at 21:40
  • $\begingroup$ A class is like a label that can apply to a feature vector. It makes sense to say a vector can have label A but then what does the notation |A-C| mean? the norm of the difference of vectors that are in class A / C, is a scalar, and you say the scalar belongs to class A'? $\endgroup$
    – Sean Owen
    Nov 15, 2014 at 22:00
  • $\begingroup$ What I meant by |A-C| is actually the Mahalanobis transformation where I calculate the covariance and centroid of C and then I use those two parameters to transform my classes A and B into A' and B' which are the respective multivariate distances. Effectively now every element of A' and B' are scalars. $\endgroup$
    – Diego
    Nov 16, 2014 at 0:46
  • $\begingroup$ Basically I want to know if this is a valid transformation in the sense that I am not informing the algorithm (double dipping?) because I know that B is closer to C than A. $\endgroup$
    – Diego
    Nov 16, 2014 at 0:48

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