Should numpy.sqrt(-1) return 1j rather than nan?
oliphant at ee.byu.edu
Wed Oct 11 15:41:25 CDT 2006
pearu at cens.ioc.ee wrote:
>I have recieved the following note from a user:
>In SciPy 0.3.x the ufuncs were overloaded by more "intelligent" versions.
>A very attractive feature was that sqrt(-1) would yield 1j as in Matlab.
>Then you can program formulas directly (e.g., roots of a 2nd order
>polynomial) and the right answer is always achieved. In the Matlab-Python
>battle in mathematics education, this feature is important.
>Now in SciPy 0.5.x sqrt(-1) yields nan. A lot of code we have, especially
>for introductory numerics and physics courses, is now broken.
>This has already made my colleagues at the University skeptical to
>Python as "this lack of backward compatibility would never happen in Matlab".
This was a consequence of moving scipy_base into NumPy but not exposing
the scimath library in NumPy. It would be a very easy thing to put
from numpy.lib.scimath import *
into the scipy name-space.
I'm supportive of that as a backward-compatibility measure.
>Another problem related to Numeric and numpy is that in these courses we
>use ScientificPython several places, which applies Numeric and will
>continue to do so. You then easily get a mix of numpy and Numeric
>in scripts, which may cause problems and at least extra overhead.
>Just converting to numpy in your own scripts isn't enough if you call
>up libraries using and returning Numeric.
>I wonder, what are the reasons that numpy.sqrt(-1) returns nan?
Because that is backwards compatible. You have to construct a
function-wrapper in order to handle the negative case correctly. The
function wrapper is going to be slower. Thus, it is placed in a
>Could sqrt(-1) made to return 1j again?
Not in NumPy. But, in scipy it could.
>If not, shouldn't
>numpy.sqrt(-1) raise a ValueError instead of returning silently nan?
This is user adjustable. You change the error mode to raise on
'invalid' instead of pass silently which is now the default.
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