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 <
> 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
> >>>> and Van Loan suggestion corresponds to:
> >>>>
> >>>> tol = np.linalg.norm(M, np.inf) * np.finfo(M.dtype).eps / 2
> >>>>
> >>>> This tolerance gives full rank for these rank-deficient matrices in
> >>>> about 39 percent of cases (tol / S.min() ratio of 1.7)
> >>>>
> >>>> We see on p 56 (section 2.3.2) that:
> >>>>
> >>>> m, n = M.shape
> >>>> 1 / sqrt(n) . |M|_{inf} <= |M|_2
> >>>>
> >>>> So we can get an upper bound on |M|_{inf} with |M|_2 * sqrt(n).
> Setting:
> >>>>
> >>>> tol = S.max() * np.finfo(M.dtype).eps / 2 * np.sqrt(n)
> >>>>
> >>>> gives about 0.5 percent error (tol / S.min() of 4.4)
> >>>>
> >>>> Using the Mathworks threshold [2]:
> >>>>
> >>>> tol = S.max() * np.finfo(M.dtype).eps * max((m, n))
> >>>>
> >>>> There are no false negatives (0 percent rank 10), but tol / S.min() is
> >>>> around 110 - so conservative, in this case.
> >>>>
> >>>> So - summary - I'm worrying our current threshold is too small,
> >>>> letting through many rank-deficient matrices without detection. I may
> >>>> have misread Golub and Van Loan, but maybe we aren't doing what they
> >>>> suggest. Maybe what we could use is either the MATLAB threshold or
> >>>> something like:
> >>>>
> >>>> tol = S.max() * np.finfo(M.dtype).eps * np.sqrt(n)
> >>>>
> >>>> - so 2 * the upper bound for the inf norm = 2 * |M|_2 * sqrt(n) . This
> >>>> gives 0 percent misses and tol / S.min() of 8.7.
> >>>>
> >>>> What do y'all think?
> >>>>
> >>>> Best,
> >>>>
> >>>> Matthew
> >>>>
> >>>> [1]
> >>>>
> http://matthew-brett.github.com/pydagogue/floating_error.html#machine-epsilon
> >>>> [2] http://www.mathworks.com/help/techdoc/ref/rank.html
> >>>>
> >>>> Output from script:
> >>>>
> >>>> Percent undetected current: 9.8, tol / S.min(): 2.762
> >>>> Percent undetected inf norm: 39.1, tol / S.min(): 1.667
> >>>> Percent undetected upper bound inf norm: 0.5, tol / S.min(): 4.367
> >>>> Percent undetected upper bound inf norm * 2: 0.0, tol / S.min(): 8.734
> >>>> Percent undetected MATLAB: 0.0, tol / S.min(): 110.477
> >>>>
> >>>> <script>
> >>>> import numpy as np
> >>>> import scipy.linalg as npl
> >>>>
> >>>> M = 40
> >>>> N = 10
> >>>>
> >>>> def make_deficient():
> >>>> X = np.random.normal(size=(M, N))
> >>>> deficient_X = X.copy()
> >>>> if M > N: # Make a column deficient
> >>>> deficient_X[:, 0] = deficient_X[:, 1] + deficient_X[:, 2]
> >>>> else: # Make a row deficient
> >>>> deficient_X[0] = deficient_X[1] + deficient_X[2]
> >>>> return deficient_X
> >>>>
> >>>> matrices = []
> >>>> ranks = []
> >>>> ranks_inf = []
> >>>> ranks_ub_inf = []
> >>>> ranks_ub2_inf = []
> >>>> ranks_mlab = []
> >>>> tols = np.zeros((1000, 6))
> >>>> for i in range(1000):
> >>>> m = make_deficient()
> >>>> matrices.append(m)
> >>>> # The SVD tolerances
> >>>> S = npl.svd(m, compute_uv=False)
> >>>> S0 = S.max()
> >>>> # u in Golub and Van Loan == numpy eps / 2
> >>>> eps = np.finfo(m.dtype).eps
> >>>> u = eps / 2
> >>>> # Current numpy matrix_rank algorithm
> >>>> ranks.append(np.linalg.matrix_rank(m))
> >>>> # Which is the same as:
> >>>> tol_s0 = S0 * eps
> >>>> # ranks.append(np.linalg.matrix_rank(m, tol=tol_s0))
> >>>> # Golub and Van Loan suggestion
> >>>> tol_inf = npl.norm(m, np.inf) * u
> >>>> ranks_inf.append(np.linalg.matrix_rank(m, tol=tol_inf))
> >>>> # Upper bound of |X|_{inf}
> >>>> tol_ub_inf = tol_s0 * np.sqrt(N) / 2
> >>>> ranks_ub_inf.append(np.linalg.matrix_rank(m, tol=tol_ub_inf))
> >>>> # Times 2 fudge
> >>>> tol_ub2_inf = tol_s0 * np.sqrt(N)
> >>>> ranks_ub2_inf.append(np.linalg.matrix_rank(m, tol=tol_ub2_inf))
> >>>> # MATLAB algorithm
> >>>> tol_mlab = tol_s0 * max(m.shape)
> >>>> ranks_mlab.append(np.linalg.matrix_rank(m, tol=tol_mlab))
> >>>> # Collect tols
> >>>> tols[i] = tol_s0, tol_inf, tol_ub_inf, tol_ub2_inf, tol_mlab,
> S.min()
> >>>>
> >>>> rel_tols = tols / tols[:, -1][:, None]
> >>>>
> >>>> fmt = 'Percent undetected %s: %3.1f, tol / S.min(): %2.3f'
> >>>> max_rank = min(M, N)
> >>>> for name, ranks, mrt in zip(
> >>>> ('current', 'inf norm', 'upper bound inf norm',
> >>>> 'upper bound inf norm * 2', 'MATLAB'),
> >>>> (ranks, ranks_inf, ranks_ub_inf, ranks_ub2_inf, ranks_mlab),
> >>>> rel_tols.mean(axis=0)[:5]):
> >>>> pcnt = np.sum(np.array(ranks) == max_rank) / 1000. * 100
> >>>> print fmt % (name, pcnt, mrt)
> >>>> </script>
> >>>
> >>>
> >>> The polynomial fitting uses eps times the largest array dimension for
> the
> >>> relative condition number. IIRC, that choice traces back to numerical
> >>> recipes.
> >
> > Chuck - sorry - I didn't understand what you were saying, and now I
> > think you were proposing the MATLAB algorithm. I can't find that in
> > Numerical Recipes - can you? It would be helpful as a reference.
> >
> >> This is the same as Matlab, right?
> >
> > Yes, I believe so, i.e:
> >
> > tol = S.max() * np.finfo(M.dtype).eps * max((m, n))
> >
> > from my original email.
> >
> >> If the Matlab condition is the most conservative, then it seems like a
> >> reasonable choice -- conservative is good so long as your false
> >> positive rate doesn't become to high, and presumably Matlab has enough
> >> user experience to know whether the false positive rate is too high.
> >
> > Are we agreeing to go for the Matlab algorithm?
>
> As extra data, current Numerical Recipes (2007, p 67) appears to prefer:
>
> tol = S.max() * np.finfo(M.dtype).eps / 2. * np.sqrt(m + n + 1.)
>
That's interesting, as something like that with a square root was my first
choice for the least squares, but then someone mentioned the NR choice.
That was all on the mailing list way several years back when I was fixing
up the polynomial fitting routine. The NR reference is on page 517 of the
1986 edition (FORTRAN), which might be hard to come by these days ;)
> There's a discussion of algorithms in:
>
> @article{konstantinides1988statistical,
> title={Statistical analysis of effective singular values in matrix
> rank determination},
> author={Konstantinides, K. and Yao, K.},
> journal={Acoustics, Speech and Signal Processing, IEEE Transactions on},
> volume={36},
> number={5},
> pages={757--763},
> year={1988},
> publisher={IEEE}
> }
>
> Yes, restricted access:
> http://ieeexplore.ieee.org/xpls/abs_all.jsp?arnumber=1585&tag=1
>
> Cheers,
>
>
Chuck
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