Re: [Numpy-discussion] Re: Dot product threading?

[Bill B. <wb...@gm...>]

In Einstein summation notation, what numpy.dot() does now is:
c_riqk =3D a_rij * b_qjk

And you want:
c_[r]ik =3D a_[r]ij * b_[r]jk

where the brackets indicate a 'no summation' index.
Writing the ESN makes it clearer to me anyway.  :-)

--bb



On 5/18/06, Pau Gargallo <pau...@gm...> wrote:
>
> > I'm afraid I really don't understand the operation that you want.
>
> I think that the operation Angus wants is the following (at least I
> would like that one ;-)
>
> if you have two 2darrays of shapes:
>     a.shape =3D (n,k)
>     b.shape =3D (k,m)
> you get:
>     dot( a,b ).shape =3D=3D (n,m)
>
> Now, if you have higher dimensional arrays  (kind of "arrays of matrices"=
)
>     a.shape =3D I+(n,k)
>     b.shape =3D J+(k,m)
> where I and J are tuples, you get
>     dot( a,b ).shape =3D=3D I+J+(n,m)
>     dot( a,b )[ i,j ] =3D=3D dot( a[i],b[j] )          #i,j represent tup=
les
> That is the current behaviour, it computes the matrix product between
> every possible pair.
> For me that is similar to 'outer' but with matrix product.
>
> But sometimes it would be useful (at least for me) to have:
>      a.shape =3D I+(n,k)
>      b.shape =3D I+(k,m)
> and to get only:
>      dot2( a,b ).shape =3D=3D I+(n,m)
>      dot2( a,b )[i] =3D=3D dot2( a[i], b[i] )
>
> This would be a natural extension of the scalar product (a*b)[i] =3D=3D
> a[i]*b[i]
> If dot2 was a kind of ufunc, this will be the expected behaviour,
> while the current dot's behaviour will be obtained by dot2.outer(a,b).
>
> Does this make any sense?
>
> Cheers,
> pau
>
>