Hi, On Tue, Jun 26, 2012 at 4:39 PM, Benjamin Root <ben.r...@ou.edu> wrote: > > > On Tuesday, June 26, 2012, Charles R Harris wrote: >> >> >> >> On Tue, Jun 26, 2012 at 3:42 PM, Matthew Brett <matthew.br...@gmail.com> >> wrote: >> >> Hi, >> >> On Mon, Jun 18, 2012 at 3:50 PM, Matthew Brett <matthew.br...@gmail.com> >> wrote: >> > Hi, >> > >> > On Sun, Jun 17, 2012 at 7:22 PM, Charles R Harris >> > <charlesr.har...@gmail.com> wrote: >> >> >> >> >> >> On Sat, Jun 16, 2012 at 2:33 PM, Matthew Brett >> >> <matthew.br...@gmail.com> >> >> wrote: >> >>> >> >>> Hi, >> >>> >> >>> On Sat, Jun 16, 2012 at 8:03 PM, Matthew Brett >> >>> <matthew.br...@gmail.com> >> >>> wrote: >> >>> > Hi, >> >>> > >> >>> > On Sat, Jun 16, 2012 at 10:40 AM, Nathaniel Smith <n...@pobox.com> >> >>> > wrote: >> >>> >> On Fri, Jun 15, 2012 at 4:10 AM, Charles R Harris >> >>> >> <charlesr.har...@gmail.com> wrote: >> >>> >>> >> >>> >>> >> >>> >>> On Thu, Jun 14, 2012 at 8:06 PM, Matthew Brett One potential problem is that it implies that it will always be the same as any version of matlab's tolerance. What if they change it in a future release? How likely are we to even notice?
>> >>> >>> <matthew.br...@gmail.com> >> >>> >>> wrote: >> >>> >>>> >> >>> >>>> Hi, >> >>> >>>> >> >>> >>>> I noticed that numpy.linalg.matrix_rank sometimes gives full rank >> >>> >>>> for >> >>> >>>> matrices that are numerically rank deficient: >> >>> >>>> >> >>> >>>> If I repeatedly make random matrices, then set the first column >> >>> >>>> to be >> >>> >>>> equal to the sum of the second and third columns: >> >>> >>>> >> >>> >>>> def make_deficient(): >> >>> >>>> X = np.random.normal(size=(40, 10)) >> >>> >>>> deficient_X = X.copy() >> >>> >>>> deficient_X[:, 0] = deficient_X[:, 1] + deficient_X[:, 2] >> >>> >>>> return deficient_X >> >>> >>>> >> >>> >>>> then the current numpy.linalg.matrix_rank algorithm returns full >> >>> >>>> rank >> >>> >>>> (10) in about 8 percent of cases (see appended script). >> >>> >>>> >> >>> >>>> I think this is a tolerance problem. The ``matrix_rank`` >> >>> >>>> algorithm >> >>> >>>> does this by default: >> >>> >>>> >> >>> >>>> S = spl.svd(M, compute_uv=False) >> >>> >>>> tol = S.max() * np.finfo(S.dtype).eps >> >>> >>>> return np.sum(S > tol) >> >>> >>>> >> >>> >>>> I guess we'd we want the lowest tolerance that nearly always or >> >>> >>>> always >> >>> >>>> identifies numerically rank deficient matrices. I suppose one >> >>> >>>> way of >> >>> >>>> looking at whether the tolerance is in the right range is to >> >>> >>>> compare >> >>> >>>> the calculated tolerance (``tol``) to the minimum singular value >> >>> >>>> (``S.min()``) because S.min() in our case should be very small >> >>> >>>> and >> >>> >>>> indicate the rank deficiency. The mean value of tol / S.min() for >> >>> >>>> the >> >>> >>>> current algorithm, across many iterations, is about 2.8. We >> >>> >>>> might >> >>> >>>> hope this value would be higher than 1, but not much higher, >> >>> >>>> otherwise >> >>> >>>> we might be rejecting too many columns. >> >>> >>>> >> >>> >>>> Our current algorithm for tolerance is the same as the 2-norm of >> >>> >>>> M * >> >>> >>>> eps. We're citing Golub and Van Loan for this, but now I look at >> >>> >>>> our >> >>> >>>> copy (p 261, last para) - they seem to be suggesting using u * >> >>> >>>> |M| >> >>> >>>> where u = (p 61, section 2.4.2) eps / 2. (see [1]). I think the >> >>> >>>> Golub >> >> >> I'm fine with that, and agree that it is likely to lead to fewer folks >> wondering why Matlab and numpy are different. A good explanation in the >> function documentation would be useful. >> >> Chuck >> > > One potential problem is that it implies that it will always be the same as > any version of matlab's tolerance. What if they change it in a future > release? How likely are we to even notice? I guess that matlab is unlikely to change for the same reason that we would be reluctant to change, once we've found an acceptable value. I was thinking that we would say something like: """ The default tolerance is : tol = S.max() * np.finfo(M.dtype).eps * max((m, n)) This corresponds to the tolerance suggested in NR page X, and to the tolerance used by MATLAB at the time of writing (June 2012; see http://www.mathworks.com/help/techdoc/ref/rank.html). """ I don't know whether we would want to track changes made by matlab - maybe we could have that discussion if they do change? Best, Matthew _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
