Re: [Numpy-discussion] Should numpy.sqrt(-1) return 1j rather than nan?

[Scott S. <sin...@uk...>]

Fernando Perez wrote:
> Please note that I see a valid reason for scipy.foo != numpy.foo when
> the scipy version uses code with extra features, is faster, has
> additional options, etc.  But as I said in a previous message, I think
> that /for the same input/, we should really try to satisfy that
>
> numpy.foo(x) == scipy.foo(x) (which is NOT the same as 'numpy.foo is scipy.foo')
>
> within reason.

As far as I can tell this is exactly what happens. Consider the issue 
under discussion...

----------------------------------
 >>> import numpy as np
 >>> np.sqrt(-1)
-1.#IND
 >>> np.sqrt(-1+0j)
1j
 >>> a = complex(-1)
 >>> np.sqrt(a)
1j
 >>> import scipy as sp
 >>> sp.sqrt(-1)
-1.#IND
 >>> np.sqrt(-1+0j)
1j
 >>> sp.sqrt(a)
1j
 >>> np.__version__
'1.0rc1'
 >>> sp.__version__
'0.5.1'
 >>>

----------------------------------

I'm sure that this hasn't changed in the development versions.

Surely the point is that when your algorithm can potentially produce a 
complex result, the logical thing to do is to use a complex data type. 
In this case Numpy and Scipy behave in a way which is intuitive. If 
complex results are surprising and unexpected then the algorithm is 
probably in error or poorly understood ;-)

Cheers,
Scott