Dear all
I have object array of arrays, which I compare element-wise to None in various places: >>> a = numpy.array([numpy.arange(5),None,numpy.nan,numpy.arange(6),None],dtype=numpy.object) >>> a array([array([0, 1, 2, 3, 4]), None, nan, array([0, 1, 2, 3, 4, 5]), None], dtype=object) >>> numpy.equal(a,None) FutureWarning: comparison to `None` will result in an elementwise object comparison in the future. So far, I always ignored the warning, for lack of an idea how to resolve it. Now, with Numpy 1.13, I have to resolve the issue, because it fails with: ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() It seem that the numpy.equal is applied to each inner array, returning a Boolean array for each element, which cannot be coerced to a single Boolean. The expression >>> numpy.vectorize(operator.is_)(a,None) gives the desired result, but feels a bit clumsy. Is there a cleaner, efficient way to do an element-wise (but shallow) comparison? Thank you and best regards, Martin Gfeller, Swisscom _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion |
On Mon, 2017-07-17 at 09:13 +0000, [hidden email] wrote:
> Dear all > > I have object array of arrays, which I compare element-wise to None > in various places: > > > > > a = > > > > numpy.array([numpy.arange(5),None,numpy.nan,numpy.arange(6),Non > > > > e],dtype=numpy.object) > > > > a > > array([array([0, 1, 2, 3, 4]), None, nan, array([0, 1, 2, 3, 4, 5]), > None], dtype=object) > > > > numpy.equal(a,None) > > FutureWarning: comparison to `None` will result in an elementwise > object comparison in the future. > > > So far, I always ignored the warning, for lack of an idea how to > resolve it. > > Now, with Numpy 1.13, I have to resolve the issue, because it fails > with: > > ValueError: The truth value of an array with more than one element is > ambiguous. Use a.any() or a.all() > > It seem that the numpy.equal is applied to each inner array, > returning a Boolean array for each element, which cannot be coerced > to a single Boolean. > > The expression > > > > > numpy.vectorize(operator.is_)(a,None) > > gives the desired result, but feels a bit clumsy. > we could delay the deprecation probably. For a solutions, maybe we could add a ufunc for elementwise `is` on object arrays (dunno about the name, maybe `object_identity`. Just some quick thoughts. - Sebastian > Is there a cleaner, efficient way to do an element-wise (but shallow) > comparison? > > Thank you and best regards, > Martin Gfeller, Swisscom > > _______________________________________________ > NumPy-Discussion mailing list > [hidden email] > https://mail.python.org/mailman/listinfo/numpy-discussion > NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion signature.asc (817 bytes) Download Attachment |
In reply to this post by Martin.Gfeller
On Mon, Jul 17, 2017 at 2:13 AM, <[hidden email]> wrote:
> > Dear all > > I have object array of arrays, which I compare element-wise to None in various places: > > >>> a = numpy.array([numpy.arange(5),None,numpy.nan,numpy.arange(6),None],dtype=numpy.object) > >>> a > array([array([0, 1, 2, 3, 4]), None, nan, array([0, 1, 2, 3, 4, 5]), None], dtype=object) > >>> numpy.equal(a,None) > FutureWarning: comparison to `None` will result in an elementwise object comparison in the future. > > > So far, I always ignored the warning, for lack of an idea how to resolve it. > > Now, with Numpy 1.13, I have to resolve the issue, because it fails with: > > ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() > > It seem that the numpy.equal is applied to each inner array, returning a Boolean array for each element, which cannot be coerced to a single Boolean. > > The expression > > >>> numpy.vectorize(operator.is_)(a,None) > > gives the desired result, but feels a bit clumsy. Wrap the clumsiness up in a documented, tested utility function with a descriptive name and use that function everywhere instead. -- Robert Kern _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion |
Here’s a hack that lets you keep using ==:
Eric On Mon, 17 Jul 2017 at 17:45 Robert Kern <[hidden email]> wrote:
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On Mon, Jul 17, 2017 at 10:52 AM, Eric Wieser <[hidden email]> wrote:
Frankly, I'd stick with a well-named utility function. It's much more kind to those who have to read the code (e.g. you in 6 months). :-) Robert Kern
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