Re: [Numpy-discussion] How to compare an array of arrays elementwise to None in Numpy 1.13 (was easy before)?

Here’s a hack that lets you keep using ==: class IsCompare: __array_priority__ = 999999 # needed to make it work on either side of `==` def __init__(self, val): self._val = val def __eq__(self, other): return other is self._val def __neq__(self, other): return other is not self._val a == IsCompare(None) # a is None a == np.array(IsCompare(None)) # broadcasted a is None Eric ​ On Mon, 17 Jul 2017 at 17:45 Robert Kern wrote:

On Mon, Jul 17, 2017 at 2:13 AM, wrote:

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Dear all

I have object array of arrays, which I compare element-wise to None in

various places:

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...

...

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a =

numpy.array([numpy.arange(5),None,numpy.nan,numpy.arange(6),None],dtype=numpy.object)

...

...

...

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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

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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 NumPy-Discussion@python.org https://mail.python.org/mailman/listinfo/numpy-discussion