How to compare an array of arrays elementwise to None in Numpy 1.13 (was easy before)?
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
On Mon, 2017-07-17 at 09:13 +0000, Martin.Gfeller@swisscom.com 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.
Yes, I guess ones bug is someone elses feature :(, if it is very bad, 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 NumPy-Discussion@python.org https://mail.python.org/mailman/listinfo/numpy-discussion
On Mon, Jul 17, 2017 at 2:13 AM,
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
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
On Mon, Jul 17, 2017 at 2:13 AM,
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 NumPy-Discussion@python.org https://mail.python.org/mailman/listinfo/numpy-discussion
On Mon, Jul 17, 2017 at 10:52 AM, Eric Wieser
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
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
участники (4)
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Eric Wieser
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Martin.Gfeller@swisscom.com
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Robert Kern
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Sebastian Berg