[Numpy-discussion] Possible new multiplication operators for Python

Grégory Lielens gregory.lielens@fft...
Tue Aug 19 03:10:40 CDT 2008

On Tue, 2008-08-19 at 09:49 +0200, Grégory Lielens wrote:

> Using __call__ as matmul:
>    b = a.I - ( (a.I)(u) / (c.I + (v/a)(u)) )(v) / a

oups, of course you do not have right-divide in this case, it would thus

   b = a.I - (a.I) (u) ( ( c.I + (v)(a.I)(u) ).I ) (v) (a.I) 

hum, given the amount of re-read I had to do, even with
I think that visual parsing for complex formula was the biggest
drawback ;-)

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