[Cython] Be more forgiving about memoryview strides

[Sebastian Berg]

On Thu, 2013-02-28 at 23:25 -0800, Robert Bradshaw wrote:
> On Thu, Feb 28, 2013 at 11:12 AM, Nathaniel Smith <njs at pobox.com> wrote:
> > On Thu, Feb 28, 2013 at 5:50 PM, Robert Bradshaw <robertwb at gmail.com> wrote:
> >> On Thu, Feb 28, 2013 at 7:13 AM, Sebastian Berg
> >> <sebastian at sipsolutions.net> wrote:
> >>> Hey,
> >>>
> >>> Maybe someone here already saw it (I don't have a track account, or I
> >>> would just create a ticket), but it would be nice if Cython was more
> >>> forgiving about contiguous requirements on strides. In the future this
> >>> would make it easier for numpy to go forward with changing the
> >>> contiguous flags to be more reasonable for its purpose, and second also
> >>> to allow old (and maybe for the moment remaining) corner cases in numpy
> >>> to slip past (as well as possibly the same for other programs...). An
> >>> example is (see also https://github.com/numpy/numpy/issues/2956 and the
> >>> PR linked there for more details):
> >>>
> >>> def add_one(array):
> >>>     cdef double[::1] a = array
> >>>     a[0] += 1.
> >>>     return array
> >>>
> >>> giving:
> >>>
> >>>>>> add_one(np.ascontiguousarray(np.arange(10.)[::100]))
> >>> ValueError: Buffer and memoryview are not contiguous in the same
> >>> dimension.
> >>>
> >>> This could easily be changed if MemoryViews check the strides as "can be
> >>> interpreted as contiguous". That means that if shape[i] == 1, then
> >>> strides[i] are arbitrary (you can just change them if you like). This is
> >>> also the case for 0-sized arrays, which are arguably always contiguous,
> >>> no matter their strides are!
> >>
> >> I was under the impression that the primary value for contiguous is
> >> that it a foo[::1] can be interpreted as a foo*. Letting strides be
> >> arbitrary completely breaks this, right?
> >
> > Nope. The natural definition of "C contiguous" is "the array entries
> > are arranged in memory in the same way they would be if they were a
> > multidimensional C array" (i.e., what you said.) But it turns out that
> > this is *not* the definition that numpy and cython use!
> >
> > The issue is that the above definition is a constraint on the actual
> > locations of items in memory, i.e., given a shape, it tells you that
> > for every index,
> >  (a)  sum(index * strides) == sum(index * cumprod(shape[::-1])[::-1] * itemsize)
> > Obviously this equality holds if
> >  (b)  strides == cumprod(shape[::-1])[::-1] * itemsize
> > (Or for F-contiguity, we have
> >  (b')  strides == cumprod(shape) * itemsize
> > )
> >
> > (a) is the natural definition of "C contiguous". (b) is the definition
> > of "C contiguous" used by numpy and cython. (b) implies (a). But (a)
> > does not imply (b), i.e., there are arrays that are C-contiguous which
> > numpy and cython think are discontiguous. (Also in numpy there are
> > some weird cases where numpy accidentally uses the correct definition,
> > I think, which is the point of Sebastian's example.)
> >
> > In particular, if shape[i] == 1, then the value of stride[i] really
> > should be irrelevant to judging contiguity, because the only thing you
> > can do with strides[i] is multiply it by index[i], and if shape[i] ==
> > 1 then index[i] is always 0. So an array of int8's with shape = (10,
> > 1), strides = (1, 73) is contiguous according to (a), but not
> > according to (b). Also if shape[i] is 0 for any i, then the entire
> > contents of the strides array becomes irrelevant to judging
> > contiguity; all zero-sized arrays are contiguous according to (a), but
> > not (b).
> 
> Thanks for clarifying.
> 
> Yes, I think it makes a lot of sense to loosen our definition for
> Cython. Internally, I think the only way we use this assumption is in
> not requiring that the first/final index be multiplied by the stride,
> which should be totally fine. But this merits closer inspection as
> there may be something else.

The only problem I saw was code that used strides[-1] instead of the
itemsize (e.g. using strides[i]/strides[-1] to then index the typed
buffer instead of using strides[i]/itemsize). But that should be easy to
check, numpy had two or so cases of that itself...

> 
> > (This is really annoying for numpy because given, say, a column vector
> > with shape (n, 1), it is impossible to be both C- and F-contiguous
> > according to the (b)-style definition. But people expect expect
> > various operations to preserve C versus F contiguity, so there are
> > heuristics in numpy that try to guess whether various result arrays
> > should pretend to be C- or F-contiguous, and we don't even have a
> > consistent idea of what it would mean for this code to be working
> > correctly, never mind test it and keep it working. OTOH if we just fix
> > numpy to use the (a) definition, then it turns out a bunch of
> > third-party code breaks, like, for example, cython.)
> 
> Can you give some examples?
> 

Not sure for what :). Maybe this is an example:

In [1]: a = np.asmatrix(np.arange(9).reshape(3,3).T)

In [2]: a.flags.f_contiguous
Out[2]: True

In [3]: a[:,0].flags
Out[3]: 
  C_CONTIGUOUS : True
  F_CONTIGUOUS : False
  ...

Where that view could just as well be F-contiguous, and the fact that
numpy, when in doubt, prefers C-contiguous might be surprising. And
since it would be less strict to begin with, numpy may safe a copy here
or there (without adding weird stride fixing code).

Examples for code breakage would be this check as well as scikit-learn
and scipy in 3 or 4 cases making the assumption above of itemsize ==
strides[-1] for c-contiguous arrays.

> - Robert
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