2
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • $\begingroup$ 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

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.