Should numpy.sqrt(-1) return 1j rather than nan?
Travis Oliphant
oliphant at ee.byu.edu
Wed Oct 11 16:02:50 CDT 2006
Sven Schreiber wrote:
>>This is user adjustable. You change the error mode to raise on
>>'invalid' instead of pass silently which is now the default.
>>
>>-Travis
>>
>>
>>
>
>Could you please explain how this adjustment is done, or point to the
>relevant documentation.
>
>
numpy.sqrt(-1)
old = seterr(invalid='raise')
numpy.sqrt(-1) # should raise an error
seterr(**old) # restores error-modes for current thread
numpy.sqrt(-1)
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