[Numpy-discussion] Matrix rank default tolerance - is it too low?
Nathaniel Smith
njs at pobox.com
Sat Jun 16 16:39:05 EDT 2012
On Sat, Jun 16, 2012 at 9:03 PM, Matthew Brett <matthew.brett at gmail.com> wrote:
> Hi,
>
> On Sat, Jun 16, 2012 at 10:40 AM, Nathaniel Smith <njs at pobox.com> wrote:
>> On Fri, Jun 15, 2012 at 4:10 AM, Charles R Harris
>> <charlesr.harris at gmail.com> wrote:
>>>
>>>
>>> On Thu, Jun 14, 2012 at 8:06 PM, Matthew Brett <matthew.brett at 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?
>
> If so, how should this be managed? Just changing it may change the
> output of code using numpy >= 1.5.0, but then again, the threshold is
> probably incorrect.
>
> Fix and break or FutureWarning with something like:
>
> def matrix_rank(M, tol=None):
>
> where ``tol`` can be a string like ``maxdim``?
I dunno, I don't think we should do a big deprecation dance for every
bug fix. Is this a bug fix, so numpy will simply start producing more
accurate results on a given problem? I guess there isn't really a
right answer here (though claiming that [a, b, a+b] is full-rank is
clearly broken, and the matlab algorithm seems reasonable for
answering the specific question of whether a matrix is full rank), so
we'll have to hope some users speak up...
-N
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