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