"Meta Cost with SVM"
bobdobbs
New Altair Community Member
Hi,
After some work with the unbalanced training data, I had the idea of using a meta-cost operator.
Is it possible to use the libSVM operator as the inner function for the meta-cost operator???
If so, what exactly will the meta-cost operator very as it searches for the best result? (Is it just a search of class weights, or is it doing something different?)
Thanks!!
-B
After some work with the unbalanced training data, I had the idea of using a meta-cost operator.
Is it possible to use the libSVM operator as the inner function for the meta-cost operator???
If so, what exactly will the meta-cost operator very as it searches for the best result? (Is it just a search of class weights, or is it doing something different?)
Thanks!!
-B
0
Answers
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Hi,
you can do that, as you can do with al learning algorithms supporting the needed capabilities. Take a look at the operator info of the MetaCost operator in order to get this informations.
I think the meta cost operator performs a reweighting of the example weights but I'm not quite sure. You could take a look in the literature, this operator is based on.
Greetings,
Sebastian0 -
... I'm at the same point. Considering using the Find Threshold (from Classifier / Metamodelling) vs. MetaCost vs. simple Threshold Finder / Theshold Applier (from Model Application / Thresholds)
It is my understanding - but please correct me if wrong! - that Threshold Finder is not really a learner, whereas the other two do some internal X-Validation. Also, looking at the default values of 10 iterations for MetaCost vs. 200 for Find Threshold, there must be an assumption that one converges faster than the other (??)
My inner learner is an SVM classifier. Thankful for any advise!
greetings - Stefan
0 -
Hi,
as on July I don't have a clue how this method works internally. I'm not familiar with the paper of Pedro Domingos, but you might take a look into it.
And I don't think the parameter settings for the number of iterations is based upon a grounded assumption of convergence. I would rather think it was very practically chosen to avoid long calculation times.
Greetings,
Sebastian0