Timeline for Perform K-means (or its close kin) clustering with only a distance matrix, not points-by-features data
Current License: CC BY-SA 3.0
9 events
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| Jul 31, 2012 at 8:17 | comment | added | ttnphns | You are right: if that doesn't matter in the context of K-means clustering then it doesn't matter, all said. I was thinking of other contexts, where D and D_new must be as close as possible. | |
| Jul 31, 2012 at 8:06 | comment | added | blubb | @ttnphns: I'm sorry, I don't follow. The paper shows that running k-means on the vectors of the constant shift embedding will yield the same cluster assignments as running pairwise clustering on the original dissimilarity matrix. Given that I don't see why it should matter that the transformed distances $D_{new}$ are far from the original distances $D$. | |
| Jul 30, 2012 at 12:10 | comment | added | ttnphns | ...but this way is very bad in the sense that the resultant euclidean data X produce euclidean distances D_new which are very far from original dissimilarities D. So, I wouldn't recommend your step 5. It seems much better simply to set negative eigenvalues to 0 and skip to step 7. Or, slightly more fine approach: set negative eigenvalues to 0, rescale positive eigenvalues so that they sum be original (=trace(S)), and then skip to step 7. That's how it appears to me. | |
| Jul 30, 2012 at 12:03 | comment | added | ttnphns | Subtracting the sum of negative eigenvalues from all the eigenvalues and then restoration S matrix is equivalent to subtracting that sum from the diagonal elements of S. This endeed makes S positive (semi)definite, but... | |
| Jul 30, 2012 at 9:36 | history | edited | blubb | CC BY-SA 3.0 |
added 46 characters in body
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| Jul 30, 2012 at 9:35 | comment | added | blubb | @ttnphns: It basically is PCA, yes, but it doesn't require the distances to be metric. The description of step 5 was unfortunate, thanks for spotting it. Is it now clear? | |
| Jul 30, 2012 at 2:26 | comment | added | ttnphns | Please, exemplify your step 5. Substracting the last (negative) eigenvalue(s) from S matrix elements seem to not help make S positive semidefinite. | |
| Jul 30, 2012 at 0:52 | comment | added | ttnphns | The steps described are nothing less than Principal Coordinates Analysis which I mention in my answer. | |
| Jul 25, 2012 at 12:15 | history | answered | blubb | CC BY-SA 3.0 |