According to sklearn.metrics.v_measure_score, it says

This score is identical to normalized_mutual_info_score with the 'arithmetic' option for averaging.

In the user guide, it refers to the Appendix of this paper, which concludes enter image description here.

But isn't it missing a factor of 1/2 comparing to normalized_mutual_info_score defined by enter image description here in Page 83 of the paper itself, or am I missing anything in the derivation?

  • $\begingroup$ There are many different variants of normalized mutual information, not just one definition, so your question is pointless. You will have to modify any premade package function yourself with a simple post-transformation if you disagree with its calculation $\endgroup$ – develarist Oct 7 '20 at 8:42
  • $\begingroup$ @develarist I assume when you said different variants, you meant different ways of normalization, i.e. geometric, arithmetic, min or max. In this question, the statement made by the paper is specific to the arithmetic one. What I am asking is the correctness of this statement, so I don't think the question is pointless. $\endgroup$ – Chen Chen Oct 7 '20 at 16:45
  • $\begingroup$ The reason why the paper uses the arithmetic normalization is purely based on the opinion of the authors for their specific application. It doesn't mean they were right, or that other variants were not right. The choice of normalization technique is simply just that, a choice. The original question being described as pointless is not meant to ridicule, but is merely due to the fact that the arbitrary nature of the answer to the question makes it pointless to read too much into $\endgroup$ – develarist Oct 7 '20 at 16:52
  • $\begingroup$ @develarist My question is not about which one is right, but whether it is equivalent to V-measure when it uses arithmetic normalization. $\endgroup$ – Chen Chen Oct 7 '20 at 16:55
  • $\begingroup$ Calculate them and see if they equal each other. To be honest though, your question really is asking which is right: "But isn't it missing a factor of 1/2". It's pretty obvious the authors made a misleading statement, or mistake, since $V \neq$ normalized_mutual_info_score $\endgroup$ – develarist Oct 7 '20 at 17:03

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