[SciPy-dev] numpy.broadcast

Robert Kern robert.kern@gmail....
Wed Aug 5 14:41:51 CDT 2009


On Wed, Aug 5, 2009 at 14:38, David Goldsmith<d_l_goldsmith@yahoo.com> wrote:
> --- On Wed, 8/5/09, Robert Kern <robert.kern@gmail.com> wrote:
>
>> wrote:
>> > I guess I don't really understand this too well - is
>> the below correct behavior, and if so, why?
>> >
>> >>>> b = np.broadcast(x, y, x, y)
>> >>>> b.nd # doesn't return what I'd expect
>> > 2
>
> Why isn't that 4?

Why would it be 4?

>> Why don't you expect this? It's the correct answer.
>> (x*y*x*y).shape == (3,3).

This is the example you need to pay attention to.

>> >>>> del b # maybe problem is that I have to
>> "clear" b first?
>> >>>> # or maybe it's that all args have to be
>> different?
>> > ...
>> >>>> b = np.broadcast(x, y, x * y)
>> >>>> b.nd
>> > 2
>
> Why isn't that 3?
>
> If x0, ..., xN are the arguments to `broadcast` and D = max(x0.nd, ..., xN.nd), is broadcast.nd necessarily <= D?  If so, then I think I'm on the road to understanding.

It is necessarily == D. Broadcasting is associative. The (x*y*z).shape
== (x*(y*z)).shape == ((x*y)*z).shape.

-- 
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless
enigma that is made terrible by our own mad attempt to interpret it as
though it had an underlying truth."
  -- Umberto Eco


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