First, the Lanczos decomposition in Mahout is pretty new. It has passed some tests, but has a very reasonable probability of significant bugs, especially if you are giving it a matrix that does not have a strongly decreasing sequence of singular values.
Remember also that the singular values are not necessarily returned in the order you might expect. Can you say a bit more about your data? How large? How sparse? This may also be of interest to you: https://issues.apache.org/jira/browse/MAHOUT-376 I see this stochastic decomposition as the way forward to larger decomposition than can be done with the Lanczos solver. 2010/11/19 Fernando Fernández <[email protected]> > Hi Pedro, > > Which version of Mahout are you working with? I think > DistributedLanczosSolver is available from from version 0.4. I have more or > less the same doubts about the results as you have, so i'm willing to read > the answers to these questions :) > > > Best. > Fernando. > > 2010/11/19 PEDRO MANUEL JIMENEZ RODRIGUEZ <[email protected]> > > > > > Dear Mahout developers, > > > > I'm a Computer Science student from the National University of Distance > > Education in Spain. I'm currently developing my final year project which > is > > about Diffusion Maps. > > > > This method is used for dimensionality reduction and it uses the Lanczos > > algorithm during its operations. The method is already implemented in the > > last release version > > of Mahout in the LanczosSolver class but we foresee the need to use the > > algorithm with distributed calculations. This implementation of Diffusion > > Maps has to deal > > with extremely large matrices and the distributed calculation is critical > > for me. > > > > I have noticed that there is a DistributedLanzcosSolver class implemented > > in the Mahout library but I can’t have access to the source code because > it > > isn't in the > > last release version of Mahout. > > > > > > Could you please let me know if I could have access to the source code of > > this class?Also I would like to ask you about how the LanczosSolver > > implementation works. I have made some test between this class and other > > program which has been implemented in R. This program is using a library > > called Arpack, which also uses the Lanczos algorithm. When I calculate > the > > eigenvalues and the eigenvectors of a symmetric matrix. I haven’t the > same > > results. For example: > > For this matrix: > > > > > > 4.42282138 1.51744077 0.07690571 0.93650042 2.19609401 > > 1.51744077 1.73849477 -0.11856149 0.76555191 1.3673608 > > 0.07690571 -0.11856149 0.55065932 1.72163263 -0.2283693 > > 0.93650042 0.76555191 1.72163263 0.09470345 -1.16626194 > > 2.19609401 1.3673608 -0.2283693 -1.16626194 -0.37321311 > > Results for R: > > > > > > Eigenvalues > > > > -0.6442398 1.1084103 2.3946915 6.2018925 > > > > Eigenvectors [,1] [,2] [,3] [,4] > > > > [1,] -0.17050824 0.46631043 -0.010360993 0.83660453 > > [2,] -0.06455473 -0.87762807 -0.008814402 0.40939079 > > [3,] 0.68602882 0.04706265 -0.666429293 0.02602181 > > [4,] -0.39567054 -0.07491643 -0.670834157 0.12161492 > > [5,] 0.58272541 -0.06705358 0.325066897 0.34208875 > > > > > > Results for Java: > > > > > > Eigenvalues > > > > 0.0 0.007869004183962289 0.023293016691817894 0.10872358093523908 > > 0.13087002850143611 > > I never get the same eigenvalues, I think this is because the > documentation > > of the class says: > > To avoid floating point overflow problems which arise in power-methods > like > > Lanczos, an initial pass is made through the input matrix to generate a > good > > starting seed vector by summing all the rows of the input matrix, and > > compute the trace(inputMatrixt*matrix) > > This latter value, being the sum of all of the singular values, is used > to > > rescale the entire matrix, effectively forcing the largest singular value > to > > be strictly > > less than one, and transforming floating point overflow problems into > > floating point underflow (ie, very small singular values will become > > invisible, as they will > > appear to be zero and the algorithm will terminate). > > Is it possible to return the eigenvalues to theirs original value? > > Eigenvectors > > > > -0.83660453 0.23122937 0.010360993 0.46631043 -0.17050824 > > -0.40939079 0.24067227 0.008814402 -0.87762807 -0.06455473 > > -0.02602181 0.28695718 0.666429293 0.04706265 0.68602882 > > -0.12161492 -0.61075665 0.670834157 -0.07491643 -0.39567054 > > -0.34208875 -0.65821099 -0.325066897 -0.06705358 0.58272541 > > > > > > > > Always happens the same. I have to force the calculation of N vectors > (with > > an N x N matrix) to obtain the same values for the eigenvectors, > > except in the sign of some of the values, which is acceptable. I thought > > this implementation of the algorithm should return the eigenvectors > sorted > > but all the time I’m obtaining a vector which I don’t want to calculate > > between them. > > In the example above it’s the second one starting from the left.Why is > this > > happen? > > > > Thanks in advance. > > > > K.r.Pedro >
