On Sat, Sep 26, 2009 at 18:17, Erik Tollerud wrote:
>> I'm sure you mean np.multiply.reduce().
> Yes, sorry - typo.
>
>>> Or, if there's a better way to just start with the first 3 1d
>>> vectorsand jump straight to the broadcast product (basically, an outer
>>> product over arbitrary number of di
> I'm sure you mean np.multiply.reduce().
Yes, sorry - typo.
>> Or, if there's a better way to just start with the first 3 1d
>> vectorsand jump straight to the broadcast product (basically, an outer
>> product over arbitrary number of dimensions...)?
>
> Well, numpy doesn't support arbitrary numb
On Sat, Sep 26, 2009 at 17:17, Erik Tollerud wrote:
> I'm encountering behavior that I think makes sense, but I'm not sure
> if there's some numpy function I'm unaware of that might speed up this
> operation.
>
> I have a (potentially very long) sequence of vectors, but for
> examples' sake, I'll
I'm encountering behavior that I think makes sense, but I'm not sure
if there's some numpy function I'm unaware of that might speed up this
operation.
I have a (potentially very long) sequence of vectors, but for
examples' sake, I'll stick with three: [A,B,C] with lengths na,nb, and
nc. To get th
