Seb wrote: > 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.
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 -- https://mail.python.org/mailman/listinfo/python-list
