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. Chuck
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