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
