On 10/16/07, Julien Hillairet <julien.hillairet@gmail.com> wrote:
2007/10/16, Bill Baxter <wbaxter@gmail.com>:
dot() also serves as Numpy's matrix multiply function. So it's trying to interpret that as a (3,N) matrix times a (3,N) matrix.
See examples here:
http://www.scipy.org/Numpy_Example_List_With_Doc#head-2a810f7dccd3f7c700d107...
--bb
2007/10/16, Charles R Harris < charlesr.harris@gmail.com>:
Dot is matrix multiplication, not the "dot" product you were expecting. It is also a bit ambiguous, as you see with the 1-D vectors, where you got what you expected.
Chuck
2007/10/16, Robert Kern <robert.kern@gmail.com>:
When given two 2-D arrays, dot() essentially does matrix multiplication. The last dimension of the first argument is matched with the next-to-last dimension of the second argument.
-- Robert Kern
Thank you for your answers. So, is there a "proper" solution to do the dot product as I had expected it ?
You might try tensordot. Without thinking it through too much: numpy.tensordot(a0, a1, axes=[-1,-1]) seems to do what you want. -- . __ . |-\ . . tim.hochberg@ieee.org