regarding the kernel weight output of libsvm operator
I applied libsvm operator for several data sets, and found that the kernel weight values of the built model tend to be always positive. For instance, I can have
However, according to SVM theory, the weight vector should satisfy equation of
| i 1.9108829072778841 cp 1.762460806463015 medimmune 1.630318586802012 |
wx+b =0The x is the points located on the decision hyperplane. The entries in the weight vector cannot always be larger than zero. Does the weight vector output by Rapidminer has a different physical meaning than the SVM theory?
Find more posts tagged with
Sort by:
1 - 3 of
31
Marius, thanks.
With respect to the data set, the feature vectors are constructed using binary occurrence for text files, which cannot have any negative x values. However, all of the weight values are still bigger than zero. I tried several data sets and observed the same scenario. Thanks.
With respect to the data set, the feature vectors are constructed using binary occurrence for text files, which cannot have any negative x values. However, all of the weight values are still bigger than zero. I tried several data sets and observed the same scenario. Thanks.
Marius wrote:
Well, if you have negative values in x, then w does not necessarily need negative values.
E.g. w=(1,1,1), b = 0 specifies a plane where x1 + x2 + x3 = 0, which is perfectly realizable.
Best regards,
Marius
Well, good point. Without any further investigation I can only guess: maybe the libSVM operator only outputs the absolute values of the weights.
But despite of the weight representation, the libSVM obviously does quite a good job. However, if you need signed weights, you should try the Support Vector Machine operator (without any additions to the name).
Best regards,
Marius
But despite of the weight representation, the libSVM obviously does quite a good job. However, if you need signed weights, you should try the Support Vector Machine operator (without any additions to the name).
Best regards,
Marius
E.g. w=(1,1,1), b = 0 specifies a plane where x1 + x2 + x3 = 0, which is perfectly realizable.
Best regards,
Marius