Timeline for Getting scores from PCA
Current License: CC BY-SA 4.0
8 events
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| Aug 22, 2019 at 15:36 | history | edited | Sjoerd Smit | CC BY-SA 4.0 |
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| Aug 12, 2019 at 10:19 | history | edited | Sjoerd Smit | CC BY-SA 4.0 |
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| Aug 12, 2019 at 10:18 | comment | added | Sjoerd Smit | Yes, that is another possibility I ran into. Is there a good reason to prefer SVD for this? I found that computing the eigenvectors of the covariance matrix is generally faster for large datasets than computing the (truncated) SVD. Are there other numerical reasons for preferring one over the other? In addition: the covariance matrix is often a quantity of interest anyway, so if you're going to compute that, you might as well use it for the principle components too (since it's just a small step from there). That's my reasoning, at least. | |
| Aug 12, 2019 at 9:31 | history | edited | J. M.'s missing motivation♦ | CC BY-SA 4.0 |
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| Aug 12, 2019 at 9:29 | comment | added | J. M.'s missing motivation♦ |
I am personally more inclined to use SVD rather than the eigendecomposition for this. Using the equivalences presented here, we have the identity PrincipalComponents[data] == Apply[Dot, Most[SingularValueDecomposition[Standardize[data, Mean, 1 &]]]]. Additionally, Eigenvectors[Covariance[data]] is just the same as Last[SingularValueDecomposition[Standardize[data, Mean, 1 &]]] (modulo changes in signs).
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| Aug 9, 2019 at 20:17 | vote | accept | Nico A | ||
| Aug 9, 2019 at 16:58 | history | edited | Sjoerd Smit | CC BY-SA 4.0 |
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| Aug 9, 2019 at 16:02 | history | answered | Sjoerd Smit | CC BY-SA 4.0 |