The current arrangement makes it easier to compute the singular values of new documents. It requires an extra operation either to transpose or to do a multiply to get the vectors for the documents (rows).
On Mon, Jun 6, 2011 at 10:36 AM, Danny Bickson <danny.bick...@gmail.com>wrote: > If I understand your question correctly, you need to simply transpose M > before you start the run and that way > you will get the other singular vectors. > > May I ask what is the problem you are working on and why do you need the > singular vectors? > Can you consider using another matrix decomposition technique for example > alternating least squares > which gives you two lower rank matrices which simulates the large > decomposed > matrix? > > On Mon, Jun 6, 2011 at 1:30 PM, Stefan Wienert <ste...@wienert.cc> wrote: > > > Hi Danny! > > > > I understand that for M*M' (and for M'*M) the left and right > > eigenvectors are identical. But that is not exactly what I want. The > > lanczos solver from mahout gives me the eigenvectors of M*M', which > > are the left singular vectors of M. But I need the right singular > > vectors of M (and not M*M'). How do I get them? > > > > Sorry, my matrix math is not as good as it should be, but I hope you > > can help me! > > > > Thanks, > > Stefan > > > > 2011/6/6 Danny Bickson <danny.bick...@gmail.com>: > > > Hi Stefan! > > > For a positive semidefinite matrix, the lest and right eigenvectors are > > > identical. > > > See SVD wikipeida text: When *M* is also positive > > > semi-definite<http://en.wikipedia.org/wiki/Positive-definite_matrix>, > > > the decomposition *M* = *U**D**U* * is also a singular value > > decomposition. > > > So you don't need to be worried about the other singular vectors. > > > > > > Hope this helps! > > > > > > On Mon, Jun 6, 2011 at 12:57 PM, Stefan Wienert <ste...@wienert.cc> > > wrote: > > > > > >> Hi. > > >> > > >> Thanks for the help. > > >> > > >> The important points from wikipedia are: > > >> - The left singular vectors of M are eigenvectors of M*M' . > > >> - The right singular vectors of M are eigenvectors of M'*M. > > >> > > >> as you describe, the mahout lanczos solver calculate A=M'*M (I think > > >> it does A=M*M', but it is not a problem). Therefore it does already > > >> calculate the right (or left) singular vector of M. > > >> > > >> But my question is, how can I get the other singular vector? I can > > >> transpose M, but then I have to calculated two SVDs, one for the right > > >> and one for the left singular value... I think there is a better way > > >> :) > > >> > > >> Hope you can help me with this... > > >> Thanks > > >> Stefan > > >> > > >> > > >> 2011/6/6 Danny Bickson <danny.bick...@gmail.com>: > > >> > Hi > > >> > Mahout SVD implementation computes the Lanzcos iteration: > > >> > http://en.wikipedia.org/wiki/Lanczos_algorithm > > >> > Denote the non-square input matrix as M. First a symmetric matrix A > is > > >> > computed by A=M'*M > > >> > Then an approximating tridiagonal matrix T and a vector matrix V are > > >> > computed such that A =~ V*T*V' > > >> > (this process is done in a distributed way). > > >> > > > >> > Next the matrix T is next decomposed into eigenvectors and > > eignevalues. > > >> > Which is the returned result. (This process > > >> > is serial). > > >> > > > >> > The third step makes the returned eigenvectors orthogonal to each > > other > > >> > (which is optional IMHO). > > >> > > > >> > The heart of the code is found at: > > >> > > > >> > > > ./math/src/main/java/org/apache/mahout/math/decomposer/lanczos/LanczosSolver.java > > >> > At least that is where it was in version 0.4 I am not sure if there > > are > > >> > changes in version 0.5 > > >> > > > >> > Anyway, Mahout does not compute directly SVD. If you are interested > in > > >> > learning more about the relation to SVD > > >> > look at: http://en.wikipedia.org/wiki/Singular_value_decomposition, > > >> > subsection: relation to eigenvalue decomposition. > > >> > > > >> > Hope this helps, > > >> > > > >> > Danny Bickson > > >> > > > >> > On Mon, Jun 6, 2011 at 9:35 AM, Stefan Wienert <ste...@wienert.cc> > > >> wrote: > > >> > > > >> >> After reading this thread: > > >> >> > > >> >> > > >> > > > http://mail-archives.apache.org/mod_mbox/mahout-user/201102.mbox/%3caanlktinq5k4xrm7nabwn8qobxzgvobbot2rtjzsv4...@mail.gmail.com%3E > > >> >> > > >> >> Wiki-SVD: M = U S V* (* = transposed) > > >> >> > > >> >> The output of Mahout-SVD is (U S) right? > > >> >> > > >> >> So... How do I get V from (U S) and M? > > >> >> > > >> >> Is V = M (U S)* (because this is, what the calculation in the > example > > >> is)? > > >> >> > > >> >> Thanks > > >> >> Stefan > > >> >> > > >> >> 2011/6/6 Stefan Wienert <ste...@wienert.cc>: > > >> >> > > > >> > > https://cwiki.apache.org/confluence/display/MAHOUT/Dimensional+Reduction > > >> >> > > > >> >> > What is done: > > >> >> > > > >> >> > Input: > > >> >> > tf-idf-matrix (docs x terms) 6076937 x 20444 > > >> >> > > > >> >> > "SVD" of tf-idf-matrix (rank 100) produces the eigenvector (and > > >> >> > eigenvalues) of tf-idf-matrix, called: > > >> >> > svd (concepts x terms) 87 x 20444 > > >> >> > > > >> >> > transpose tf-idf-matrix: > > >> >> > tf-idf-matrix-transpose (terms x docs) 20444 x 6076937 > > >> >> > > > >> >> > transpose svd: > > >> >> > svd-transpose (terms x concepts) 20444 x 87 > > >> >> > > > >> >> > matrix multiply: > > >> >> > tf-idf-matrix-transpose x svd-transpose = result > > >> >> > (terms x docs) x (terms x concepts) = (docs x concepts) > > >> >> > > > >> >> > so... I do understand, that the "svd" here is not SVD from > > wikipedia. > > >> >> > It only does the Lanczos algorithm and some magic which produces > > the > > >> >> >> Instead either the left or right (but usually the right) > > eigenvectors > > >> >> premultiplied by the diagonal or the square root of the > > >> >> >> diagonal element. > > >> >> > from > > >> >> > > >> > > > http://mail-archives.apache.org/mod_mbox/mahout-user/201102.mbox/%3CAANLkTi=rta7tfrm8zi60vcfya5xf+dbfrj8pcds2n...@mail.gmail.com%3E > > >> >> > > > >> >> > so my question: what is the output of the SVD in mahout. And what > > do I > > >> >> > have to calculate to get the "right singular value" from svd? > > >> >> > > > >> >> > Thanks, > > >> >> > Stefan > > >> >> > > > >> >> > 2011/6/6 Stefan Wienert <ste...@wienert.cc>: > > >> >> >> > > >> >> > > >> > > https://cwiki.apache.org/confluence/display/MAHOUT/Dimensional+Reduction > > >> >> >> > > >> >> >> the last step is the matrix multiplication: > > >> >> >> --arg --numRowsA --arg 20444 \ > > >> >> >> --arg --numColsA --arg 6076937 \ > > >> >> >> --arg --numRowsB --arg 20444 \ > > >> >> >> --arg --numColsB --arg 87 \ > > >> >> >> so the result is a 6,076,937 x 87 matrix > > >> >> >> > > >> >> >> the input has 6,076,937 (each with 20,444 terms). so the result > of > > >> >> >> matrix multiplication has to be the right singular value > regarding > > to > > >> >> >> the dimensions. > > >> >> >> > > >> >> >> so the result is the "concept-document vector matrix" (as I > think, > > >> >> >> these is also called "document vectors" ?) > > >> >> >> > > >> >> >> 2011/6/6 Ted Dunning <ted.dunn...@gmail.com>: > > >> >> >>> Yes. These are term vectors, not document vectors. > > >> >> >>> > > >> >> >>> There is an additional step that can be run to produce document > > >> >> vectors. > > >> >> >>> > > >> >> >>> On Sun, Jun 5, 2011 at 1:16 PM, Stefan Wienert > <ste...@wienert.cc > > > > > >> >> wrote: > > >> >> >>> > > >> >> >>>> compared to SVD, is the result is the "right singular value"? > > >> >> >>>> > > >> >> >>> > > >> >> >> > > >> >> >> > > >> >> >> > > >> >> >> -- > > >> >> >> Stefan Wienert > > >> >> >> > > >> >> >> http://www.wienert.cc > > >> >> >> ste...@wienert.cc > > >> >> >> > > >> >> >> Telefon: +495251-2026838 > > >> >> >> Mobil: +49176-40170270 > > >> >> >> > > >> >> > > > >> >> > > > >> >> > > > >> >> > -- > > >> >> > Stefan Wienert > > >> >> > > > >> >> > http://www.wienert.cc > > >> >> > ste...@wienert.cc > > >> >> > > > >> >> > Telefon: +495251-2026838 > > >> >> > Mobil: +49176-40170270 > > >> >> > > > >> >> > > >> >> > > >> >> > > >> >> -- > > >> >> Stefan Wienert > > >> >> > > >> >> http://www.wienert.cc > > >> >> ste...@wienert.cc > > >> >> > > >> >> Telefon: +495251-2026838 > > >> >> Mobil: +49176-40170270 > > >> >> > > >> > > > >> > > >> > > >> > > >> -- > > >> Stefan Wienert > > >> > > >> http://www.wienert.cc > > >> ste...@wienert.cc > > >> > > >> Telefon: +495251-2026838 > > >> Mobil: +49176-40170270 > > >> > > > > > > > > > > > -- > > Stefan Wienert > > > > http://www.wienert.cc > > ste...@wienert.cc > > > > Telefon: +495251-2026838 > > Mobil: +49176-40170270 > > >
