On Tue, 27 Feb 2018 12:25:30 +1300, Gregory Ewing <greg.ew...@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. -- Seb -- https://mail.python.org/mailman/listinfo/python-list
