[Numpy-discussion] Funny business with 'is' operator?

James Bergstra bergstrj at iro.umontreal.ca
Fri Sep 6 13:50:48 EDT 2013

Thanks, this is exactly what I was looking for. I'll look into what this
Diophantine equation is. Also, relatedly, a few months ago Julian Taylor at
least wrote what was there in C, which made it faster, if not better.

- James

On Fri, Sep 6, 2013 at 1:27 PM, Robert Kern <robert.kern at gmail.com> wrote:

> On Fri, Sep 6, 2013 at 5:58 PM, James Bergstra <bergstrj at iro.umontreal.ca>
> wrote:
> >
> > I'm stumped. I can't figure out how to extract from e.g.
> >
> > view = A[:, 3]
> >
> > that the view starts at element 3 of A. I was planning to make a
> may_share_memory implementation based on the idea of swapping in a buffer
> of 0s, and using the shapes, strides, itemsize etc. to increment just the
> parts of the 0s buffer touched by the two ndarrays. If there  are any 2s in
> the incremented buffer, it's because the two views overlapped. It's not the
> best algorithm for comparing tiny views of huge arrays, I was wondering if
> in my case it would have been quicker than the built-in method (I don't
> know how it works).
> It certainly won't be faster. may_share_memory() is very simplistic. It
> just checks if the outermost bounds of the each array overlap. Thus, it can
> give false positives for arrays that are interleaved but do not actually
> intersect. This is why it is named may_share_memory() instead of
> does_share_memory().
> > I actually have another data structure around to pull out that shape and
> stride info, but it's a shame to use it, because then I can't use the
> algorithm to compare ndarrays in general (or at least ones that have been
> created by typical construction methods and slicing).
> All of the information you need is stored in the __array_interface__
> attribute of an ndarray.
> The right way to solve this is to solve a bounded, linear Diophantine
> equation. That's where you should be looking if you want to crack this
> problem.
> --
> Robert Kern
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