Timeline for Why does k-means clustering algorithm use only Euclidean distance metric?
Current License: CC BY-SA 3.0
7 events
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| Jan 7, 2014 at 21:51 | comment | added | Has QUIT--Anony-Mousse | I only sketched the proof. Look up a complete proof of convergence for k-means. Then verify it works for other distance functions. | |
| Jan 7, 2014 at 15:38 | comment | added | curious | I still can't see why the mean minimizes distances in terms of euclidean distances and in terms of cosine it doesn't as part of the proof | |
| Jan 7, 2014 at 14:51 | comment | added | Verena Praher | very good explanation. I never gave the euclidean distance a second thought and didn't realize that it was actually minimizing the withing cluster sum of squares. | |
| Jan 7, 2014 at 14:46 | comment | added | ttnphns |
I agree with you answer. Note that your operational account k-means may stop converging with other distance functions is homologous to my theoretical Non-euclidean distances will generally not span euclidean space.
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| Jan 7, 2014 at 14:40 | comment | added | ttnphns |
@AnonyMousse @ttnphns answer refers to pairwise Euclidean distances! In my answer, 1st paragraph, I clearly refer both to "SS error" (direct) and "pairwise d^2" (implicit) interpretations.
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| Jan 7, 2014 at 14:22 | comment | added | curious | But at the first step of k-means each point is put in the cluster with the closest euclidean distance with the centroid of the cluster...So there is a distance metric | |
| Jan 7, 2014 at 13:57 | history | answered | Has QUIT--Anony-Mousse | CC BY-SA 3.0 |