[SciPy-dev] numpy.broadcast

[David Goldsmith]

--- On Wed, 8/5/09, Robert Kern <robert.kern at gmail.com> wrote:

> It is necessarily == D. Broadcasting is associative. 

Ah, that's the key I didn't understand!  That helps (me) a lot; I'm going to make a "Note" of it in numpy.broadcast's docstring.

DG

> 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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