Hi, On Tue, Jun 26, 2012 at 5:04 PM, Charles R Harris <charlesr.har...@gmail.com> wrote: > > > On Tue, Jun 26, 2012 at 5:46 PM, Matthew Brett <matthew.br...@gmail.com> > wrote: >> >> 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? > > > I wouldn't bother tracking Matlab, but I think the alternative threshold > could be mentioned in the notes. Something like > > A less conservative threshold is ... > > Maybe mention that because of numerical uncertainty there will always be a > chance that the computed rank could be wrong, but that with the conservative > threshold the rank is very unlikely to be less than the computed rank.
Sounds good to me. Would anyone object to a pull request with these changes (matlab tolerance default, description in docstring)? Cheers, Matthew _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
