Hi All, An issue [1] about the copying of arrays with structured dtype raised a question about what the expected behaviour is: does copy always preserve the dtype as is, or should it remove padding?
Specifically, consider an array with a structure with many fields, say 'a' to 'z'. Since numpy 1.16, if one does a[['a', 'z']]`, a view will be returned. In this case, its dtype will include a large offset. Now, if we copy this view, should the result have exactly the same dtype, including the large offset (i.e., the copy takes as much memory as the original full array), or should the padding be removed? 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. An argument in favour of (2) is that, before numpy 1.16, `a[['a', 'z']].copy()` did return an array without padding. Of course, this relied on `a[['a', 'z']]` already returning a copy without padding, but still this is a regression. More generally, there should at least be a clear way to get the compact copy. Also, it would make sense for things like `np.save` to remove any padding (it currently does not). What do people think? All the best, Marten [1] https://github.com/numpy/numpy/issues/13299
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