I agree with Stefan that option 2 is what NumPy should go with for .copy() If you want to get an identical memory copy you should be getting the .data attribute and doing something with that buffer.
My $0.02 -Travis On Thu, Apr 11, 2019 at 6:01 PM Stefan van der Walt <stef...@berkeley.edu> wrote: > Hi Marten, > > On Thu, 11 Apr 2019 09:51:10 -0400, Marten van Kerkwijk wrote: > > From the discussion so far, it > > seems the logic has boiled down to a choice between: > > > > (1) Copy is a contract that the dtype will not vary (e.g., we also do not > > change endianness); > > > > (2) Copy is a contract that any access to the data in the array will > return > > exactly the same result, without wasting memory and possibly optimized > for > > access with different strides. E.g., `array[::10].copy() also compacts > the > > result. > > I think you'll get different answers, depending on whom you ask—those > interested in low-level memory layout, vs those who use the higher-level > API. Given that higher-level API use is much more common, I would lean > in the direction of option (2). > > From that perspective, we already don't make consistency guarantees about > memory > layout and other flags. E.g., > > In [16]: x = np.arange(12).reshape((3, 4)) > > > In [17]: x.strides > > > Out[17]: (32, 8) > > In [18]: x[::2, 1::2].strides > > Out[18]: (64, 16) > > In [19]: np.copy(x[::2, 1::2]).strides > > > Out[19]: (16, 8) > > Not to mention this odd copy contract: > > >>> x = np.array([[1,2,3],[4,5,6]], order='F') > >>> print(np.copy(x).flags['C_CONTIGUOUS']) > >>> print(x.copy().flags['C_CONTIGUOUS']) > > False > True > > > The objection about arrays that don't behave identically in [0] feels > somewhat arbitary to me. As shown above, you can always find attributes > that differ between a copied array and the original. > > The user's expectation is that they'll get an array that behaves the > same way as the original, not one that is byte-for-byte compatible. The > most common use case is to make sure that the original array doesn't get > overwritten. > > Just to play devil's advocate with myself: if you do choose option (2), > how would you go about making an identical memory copy of the original > array? > > Best regards, > Stéfan > > > [0] https://github.com/numpy/numpy/issues/13299#issuecomment-481912827 > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@python.org > https://mail.python.org/mailman/listinfo/numpy-discussion >
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