0
$\begingroup$

I'm attempting to generate some one-to-rest weight vectors for a multi-class SVC for feature selection purposes. However, I'm current using one of the several LibSVM ports, which inherently performs one-to-one in multi-class classifications.

The columns of the sv_conf output does not appear to me to be duals of one-to-rest classifications. But is there a way to re-construct one-to-rest feature weights from the output of one-to-one classifications, or I'd have to run several one-to-one and obtain the weight vectors separately for each class?

$\endgroup$
3
  • $\begingroup$ Too much specific jargon is being used. The question is so focused it has more to do with interpreting a specific program than interpreting statistics, methods or anything relating to this site. If the question can be made into one that asks something about interpretation of a statistical method as opposed to a specific computer software, that should be done. $\endgroup$
    – Carl
    Commented Apr 7, 2017 at 1:23
  • $\begingroup$ I have re-focused the question; please review if it's still off-topic. $\endgroup$
    – John M.
    Commented Apr 7, 2017 at 19:16
  • $\begingroup$ Better, and since you seem to have someone answering you, I have suggested reopening. $\endgroup$
    – Carl
    Commented Apr 7, 2017 at 19:41

1 Answer 1

2
$\begingroup$

If you do one-to-one classification mode, there is no single weight vector; there are $\binom{m}{2}$ if you have $m$ classes. You can reconstruct each of those vectors from the appropriate entries of sv_coef, though.

$\endgroup$
1
  • $\begingroup$ What I'm not so sure about extracting these sv_coef entries is, because every one-to-one classifier has two SVs, do I need to extract the sv_coef entries for both SVs, and if so, how do I combine the results? I saw at least one literature (e1071's internals) that would sum both SVs' * sv_coef from the same classifier, but I haven't seen this elsewhere. $\endgroup$
    – John M.
    Commented Apr 6, 2017 at 17:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.