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I am reading notes on using weights for KNN and I came across an example that I don't really understand.

Suppose we have K = 7 and we obtain the following:

Decision set = {A, A, A, A, B, B, B}

If this was the standard KNN algorithm we would pick A, however the notes give an example of using weights:

By class distribution (weight inversely proportional to class frequency)

class A: 95 %, class B 5 %

This results in a class of B.

I can't seem to figure out the math that was left out to obtain B as the answer.

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We can view nearest neighbor as a voting process where we consult our $k$ nearest neighbor.

We give the $i$-th data point a voting weight $w_i$.

In your example, each data point in class $A$ has weight $\frac1{0.95}$ and each data point in class $B$ has weight $\frac1{0.05}$. There are $4$ votes from class $A$ and $3$ votes from class $B$. We give class $A$ a score of $\frac{4}{0.95}\approx 4.21$ and class $B$ a score of $\frac{3}{0.05}=60$. Class $B$ has a higher score, hence we assign it to class $B$.

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  • $\begingroup$ This makes much more sense now, the percentage 95% and 5% is the class frequency, I thought it was the weights. $\endgroup$ – Belphegor Dec 10 '18 at 2:59

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