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$\begingroup$ But don't we usually normalize the vectors before we use similarity measure? In that case, your user 2 in the scenario is more similar to user 1 no matter what, isn't it? $\endgroup$

IgNite
– IgNite

2019-12-01 14:38:18 +00:00
Commented Dec 1, 2019 at 14:38

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