DOC Clarify what the decision function in SVM calculates #12708
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sklearn/svm/base.py
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| The function values are proportional to the distance of the samples X to the separating hyperplane. If the exact distances are required, divide the function values by the norm of the weight vector (``coef_``). See also `this question <https://stats.stackexchange.com/questions/14876/interpreting-distance-from-hyperplane-in-svm>`_ for further details. |
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Thanks for the note. Is fixed. This is a nice CI system you have here :-)
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Great clarification |
sklearn/svm/base.py
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| Notes | ||
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| The function values are proportional to the distance of the samples X |
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This is only relevant to decision_function_shape='ovo' isn't it? For ovr it does something more complicated
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Good point. As far as I understand the code, the decision values are somehow weighted up together and then scaled to a defined range. It does probably not make any sense to derive to the real distances based on the confidence values. So, yes, the note is only relevant to 'ovo' decision functions.
I would suggest changing the intro of the note to
If decision_function_shape=’ovo’, the function values are proportional...
is that ok?
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Yes, just qualifying the statement like that s fine.
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Thanks @Milania1 |
…kit-learn#12708)" This reverts commit 543d640.
…kit-learn#12708)" This reverts commit 543d640.
What does this implement/fix? Explain your changes.
Strictly speaking, the decision function of an SVM does not calculate the distance to the hyperplane. Instead the function value is only proportional to this distance. Since I stumbled over this, I made this small clarification and added a note for users who need the exact distance.
If this simplification was intentional, I would suggest to keep at least the note.