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I'm studying the knn classification algorithm. Why can the euclidean distance be considered a nice measure of affinity between examples ?

In one dimension (1 attribute) this seems correct, but if I add dimensions, can the euclidean distance still be considerd a good measure of affinity? Why?

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    $\begingroup$ Your premise is incorrect. Euclidean distance is not the only distance function used for knn or k-means or etc. These models can work with any distance function. $\endgroup$ – Ricardo Cruz May 17 '18 at 21:40
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You can think of examples as vectors in $\mathbb{R}^p$, where $p$ is the number of features. Two examples will be very similar if the distance between them is close to $0$ (in the extreme case, if two examples are equal their euclidean distance is $0$). One way to measure the distance is using euclidean distance, but other distances can be used, as cosine distance or $L^p$ metrics. In fact, if $p$ is very high, then Euclidean distance is not a good measure, as it tends to make the distances too uniform (see this paper).

Edit: When $p$ is very high:

See this magnificient answer to the issues that very high $p$ may have.

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  • $\begingroup$ Growing the dimensions the distances get bigger. But maybe it's not a problem and maybe the euclidean distance still gives a good measure of affinity. Why not ? $\endgroup$ – Qwerto May 17 '18 at 8:45
  • $\begingroup$ Do you mean when $p$ is very high? $\endgroup$ – David Masip May 17 '18 at 8:46
  • $\begingroup$ When p is greater than 2 $\endgroup$ – Qwerto May 17 '18 at 8:49
  • $\begingroup$ When $p$ is not very big, the euclidean distance works. What's the problem with that? $\endgroup$ – David Masip May 17 '18 at 8:50
  • $\begingroup$ In practice, growing the dimensions and considering euclidean distance, you are giving more "degrees of freedom" to the examples to be considered affine to the query example. Am i correct ? $\endgroup$ – Qwerto May 17 '18 at 8:51

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