Re[2]: [Numpy-discussion] Piecewise functions.

[Alan G I. <ai...@am...>]

On Thu, 22 Sep 2005, Andrea Riciputi apparently wrote:=20
> I've already tried something like this, but it doesn't work since f1=20
> and f2 return not valid values outside the range over they are=20
> defined. Perhaps an example could clarify; suppose that f1(x) =3D 1./=20
> sqrt(1 - x**2) for x <=3D 1, and f2(x) =3D 1./sqrt(x**2 - 1) for x > 1.=
=20

> Your suggestion, as the other I've tried, fails with a=20
> "OverflowError: math range error".=20

If you do it as I suggested, they should not I believe be=20
evaluated outside of their range.  So your function must be=20
generating an overflow error within this range.


>>> import math
>>> import random
>>> def f1(x): return math.sqrt(1-x**2)
...
>>> def f2(x): return 1./math.sqrt(x**2-1)
...
>>> def f(x): return x<=3D1 and f1(x) or f2(x)
...
>>> d =3D [random.uniform(0,2) for i in range(20)]
>>> fd =3D [f(x) for x in d]

Works fine.

Cheers,
Alan Isaac