[Numpy-discussion] Curious behavior of __radd__

Travis Oliphant travis@continuum...
Wed Feb 1 23:58:58 CST 2012

This seems odd to me.   Unraveling what is going on (so far): 

	Let a = np.complex64(1j)  and b = A()

	* np.complex64.__add__ is calling np.add
	* np.add(a, b) needs to search for an "add" loop that matches the input types and it finds one with signature
		 ('O', 'O') -> 'O'
	* a is converted to an array via the equivalent of a1 = array(a,'O')
	* b is converted to an array in the same way b1 = array(b, 'O')

	Somehow a1 as an array of objects is an array of "complex" types instead of "complex64" types
	Then the equivalent of a1[()] + b1[()] is called. 

So, the conversion is being done by 

a1 = array(a, 'O')

I don't know why this is.   This seems like a regression, but I don't have an old version of NumPy to check.   



These should be the same type.   But they are not...



On Feb 1, 2012, at 1:26 PM, Andreas Kloeckner wrote:

> Hi all,
> here's something I don't understand. Consider the following code snippet:
> ---------------------------------------------------
> class A(object):
>    def __radd__(self, other):
>        print(type(other))
> import numpy as np
> np.complex64(1j) + A()
> ---------------------------------------------------
> In my world, this should print <type 'numpy.complex64'>.
> It does print <type 'complex'>.
> Who is casting my sized complex to a built-in complex, and why?
> It can be Python's type coercion, because the behavior is the same in
> Python 3.2. (And the docs say Python 3 doesn't support coercion.)
> (Please cc me.)
> Thanks,
> Andreas
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