matrix multiplication

[Ian Kelly]

On Tue, Feb 27, 2018 at 4:08 AM, Peter Otten <__peter__ at web.de> wrote:
> Seb wrote:
>
>> On Tue, 27 Feb 2018 12:25:30 +1300,
>> Gregory Ewing <greg.ewing at canterbury.ac.nz> wrote:
>>
>>> Seb wrote:
>>>> I was wondering is whether there's a faster way of multiplying each
>>>> row (1x3) of a matrix by another matrix (3x3), compared to looping
>>>> through the matrix row by row as shown in the code.
>>
>>> Just multiply the two matrices together.
>>
>>> If A is an nx3 matrix and B is a 3x3 matrix, then C = A @ B is an nx3
>>> matrix where C[i] = A[i] @ B.
>>
>>> (This is a property of matrix multiplication in general, nothing
>>> special about numpy.)
>>
>> I think that's only true if B is the same for every row in A.  In the
>> code I posted, B varies by row of A.
>
> Yeah, you would have to substitute the N 3x3 matrices with an Nx3x3 tensor,
> though I don't know if numpy provides an op such that
>
> Nx3 op Nx3x3 --> desired result
>
> or
>
> op(Nx3, Nx3x3) --> desired result

Nx1x3 @ Nx3x3 ought to do it, with the result being Nx1x3.