On 10/16/07, Julien Hillairet <[EMAIL PROTECTED]> wrote: > > 2007/10/16, Bill Baxter <[EMAIL PROTECTED]>: > > > > 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-2a810f7dccd3f7c700d1076f15078ad1fe3c6d0d > > > > --bb > > > > 2007/10/16, Charles R Harris < [EMAIL PROTECTED]>: > > > > > > > > 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 <[EMAIL PROTECTED]>: > > > > 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. -- . __ . |-\ . . [EMAIL PROTECTED]
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