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Looks good - thanks! |
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Thanks also @hskarpetis for its valuable contributions. |
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Thanks for the contribution! It would be nice to have a small test case to check for correctness of the factorization. For instance, we can check the factors for a tensor where we know the correct factorization via numpy.allclose() We can load matlab files via scipy.io. So checking against matlab results would work. |
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Will do - I'll try a few fabricated examples of various ranks and tensor
orders, likely before Monday. I'm also curious to see Marble (
http://ideal.ece.utexas.edu/pubs/pdf/2014/Marble-kdd14.pdf) because I'm not
grokking why it isn't the same as adding one more rank.
…On Sat, Nov 26, 2016 at 11:00 AM, Maximilian Nickel < ***@***.***> wrote:
Thanks for the contribution!
It would be nice to have a small test case to check for correctness of the
factorization. For instance, we can check the factors for a tensor where we
know the correct factorization via numpy.allclose()
We can load matlab files via scipy.io. So checking against matlab results
would work.
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sktensor/cp.py
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| for j in range(inner_iter): | ||
| phi[n] = np.dot(X.unfold(n) / np.maximum(np.dot(b, pi), e), | ||
| pi.transpose()) | ||
| if np.amax(np.abs(np.ravel(np.minimum(M.U[n], 1-phi[n])))) < t: |
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Here do we need b instead of M.U[n] to match the paper? The paper has a small floor on M.U[n] also to keep zeros from entering in.
sktensor/cp.py
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| S[(phi[n] > 1) & (M.U[n] < k_tol)] = k | ||
| b = np.dot((M.U[n] + S), np.diag(M.lmbda)) | ||
| pi = khatrirao(tuple( | ||
| [M.U[i] for i in range(n) + range(n + 1, N)])).transpose() |
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Factors need to be in reversed order I think
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CP-APR algorithm
https://arxiv.org/pdf/1112.2414.pdf
The results of this implementation were tested against of that of MATLAB tensor toolbox, using the method described here.
https://pdfs.semanticscholar.org/71b5/d7ad21cf479c0816b9df00b19df87f8787a1.pdf