Should numpy.sqrt(-1) return 1j rather than nan?

[Andrew Jaffe]

Tim Hochberg wrote:
> Travis Oliphant wrote:
>> Tim Hochberg wrote:
>>
>>   
>>> With python 2.5 out now, perhaps it's time to come up with a with 
>>> statement context manager. Something like:
>>>
>>>    a = numpy.arange(10)
>>>    a/a # ignores divide by zero
>>>    with errstate(divide='raise'):
>>>        a/a # raise exception on divide by zer
>>>    # Would ignore divide by zero again if we got here.
>>>
>>> -tim
>>>
>> This looks great.  I think most people aren't aware of the with 
>> statement and what it can do (I'm only aware because of your posts, for 
>> example). 
>>
>> So, what needs to be added to your example in order to just add it to 
>> numpy?
>>   
> As far as I know, just testing  and documentation -- however testing was 
> so minimal that I may find some other stuff. I'll try to clean it up 
> tomorrow so that I'm a little more confident that it works correctly and 
> I'll send another note out then.

For this particular application, why not a context manager which just 
substitutes in the appropriately-optimized version of sqrt? That is, 
don't change the error state, but actually change the value of the 
object pointed at by the name sqrt?

Andrew

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