Should numpy.sqrt(-1) return 1j rather than nan?
pearu at cens.ioc.ee
pearu at cens.ioc.ee
Wed Oct 11 19:39:57 EDT 2006
On Wed, 11 Oct 2006, Travis Oliphant wrote:
> >Interestingly, in worst cases numpy.sqrt is approximately ~3 times slower
> >than scipy.sqrt on negative input but ~2 times faster on positive input:
> >
> >In [47]: pos_input = numpy.arange(1,100,0.001)
> >
> >In [48]: %timeit -n 1000 b=numpy.sqrt(pos_input)
> >1000 loops, best of 3: 4.68 ms per loop
> >
> >In [49]: %timeit -n 1000 b=scipy.sqrt(pos_input)
> >1000 loops, best of 3: 10 ms per loop
> >
> >
>
> This is the one that concerns me. Slowing everybody down who knows they
> have positive values just for people that don't seems problematic.
I think the code in scipy.sqrt can be optimized from
def _fix_real_lt_zero(x):
x = asarray(x)
if any(isreal(x) & (x<0)):
x = _tocomplex(x)
return x
def sqrt(x):
x = _fix_real_lt_zero(x)
return nx.sqrt(x)
to (untested)
def _fix_real_lt_zero(x):
x = asarray(x)
if not isinstance(x,(nt.csingle,nt.cdouble)) and any(x<0):
x = _tocomplex(x)
return x
def sqrt(x):
x = _fix_real_lt_zero(x)
return nx.sqrt(x)
or
def sqrt(x):
old = nx.seterr(invalid='raises')
try:
r = nx.sqrt(x)
except FloatingPointError:
x = _tocomplex(x)
r = nx.sqrt(x)
nx.seterr(**old)
return r
I haven't timed these cases yet..
Pearu
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