On Thu, Feb 28, 2013 at 11:12 AM, Nathaniel Smith <n...@pobox.com> wrote: > On Thu, Feb 28, 2013 at 5:50 PM, Robert Bradshaw <rober...@gmail.com> wrote: >> On Thu, Feb 28, 2013 at 7:13 AM, Sebastian Berg >> <sebast...@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. > (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? - Robert _______________________________________________ cython-devel mailing list cython-devel@python.org http://mail.python.org/mailman/listinfo/cython-devel
