Should numpy.sqrt(-1) return 1j rather than nan?
oliphant.travis at ieee.org
Wed Oct 11 23:02:21 CDT 2006
Greg Willden wrote:
> On 10/11/06, *Travis Oliphant* <oliphant at ee.byu.edu
> <mailto:oliphant at ee.byu.edu>> wrote:
> Stefan van der Walt wrote:
> >Further, if I understand correctly, changing sqrt and power to give
> >the right answer by default will slow things down somewhat. But
> is it
> >worth sacrificing intuitive usage for speed?
> For NumPy, yes.
> This is one reason that NumPy by itself is not a MATLAB replacement.
> This is not about being a Matlab replacement.
> This is about correctness.
I disagree. NumPy does the "correct" thing when you realize that sqrt
is a function that returns the same type as it's input. The field
over-which the operation takes place is defined by the input data-type
and not the input "values". Either way can be considered correct
mathematically. As Paul said it was a design decision not to go
searching through the array to determine whether or not there are
negative numbers in the array.
Of course you can do that if you want and that's what scipy.sqrt does.
> Numpy purports to handle complex numbers.
> Mathematically, sqrt(-1) is a complex number.
Or, maybe it's undefined if you are in the field of real numbers. It
> Therefore Numpy *must* return a complex number.
Only if the input is complex. That is a reasonable alternative to your
> If Numpy doesn't return a complex number then it shouldn't pretend to
> support complex numbers.
Of course it supports complex numbers, it just doesn't support automatic
conversion to complex numbers. It supports complex numbers the same
way Python supports them (i.e. you have to use cmath to get sqrt(-1) == 1j)
People can look at this many ways without calling the other way of
looking at it unreasonable.
I don't see a pressing need to change this in NumPy, and in fact see
many reasons to leave it the way it is. This discussion should move to
the scipy list because that is the only place where a change could occur.
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