Using the outer product you get a matrix that has the right size and contents but it is m*n by p*q Was that a misprint in your post?

...

...

x array([1, 2, 3, 4, 5, 6]) x.shape=(3,2) y = 10*transpose(x) y array([[10, 30, 50], [20, 40, 60]]) z = outerproduct(x.flat, y.flat) z array([[ 10, 30, 50, 20, 40, 60], [ 20, 60, 100, 40, 80, 120], [ 30, 90, 150, 60, 120, 180], [ 40, 120, 200, 80, 160, 240], [ 50, 150, 250, 100, 200, 300], [ 60, 180, 300, 120, 240, 360]])

-----Original Message----- From: numpy-discussion-admin@lists.sourceforge.net [mailto:numpy-discussion-admin@lists.sourceforge.net]On Behalf Of Nils Wagner Sent: Wednesday, July 18, 2001 4:10 AM To: numpy-discussion@lists.sourceforge.net Subject: [Numpy-discussion] Kronecker product Hi, I would appreciate it, if Numpy could handle the Kronecker-product of two matrices X, Y. kron(X,Y) is the Kronecker tensor product of X and Y. The result is a large matrix formed by taking all possible products between the elements of X and those of Y. If X is m-by-n and Y is p-by-q, then kron(X,Y) is m*p-by-n*q. Nils _______________________________________________ Numpy-discussion mailing list Numpy-discussion@lists.sourceforge.net http://lists.sourceforge.net/lists/listinfo/numpy-discussion

I would appreciate it, if Numpy could handle the Kronecker-product of two matrices X, Y.

kron(X,Y) is the Kronecker tensor product of X and Y.

import Numeric def kron(x, y): return Numeric.multiply.outer(x, y)

The result is a large matrix formed by taking all possible products between the elements of X and those of Y. If X is m-by-n and Y is p-by-q, then kron(X,Y) is m*p-by-n*q.

OK, there's one difference: the function shown about returns an array of shape (m, n, p, q). If the input arrays are always 2D, the following will do what you need: import Numeric def kron(x, y): z = Numeric.transpose(Numeric.multiply.outer(x, y), [0, 2, 1, 3]) z.shape = (z.shape[0]*z.shape[1], z.shape[2]*z.shape[3]) return z Konrad. -- ------------------------------------------------------------------------------- Konrad Hinsen | E-Mail: hinsen@cnrs-orleans.fr Centre de Biophysique Moleculaire (CNRS) | Tel.: +33-2.38.25.56.24 Rue Charles Sadron | Fax: +33-2.38.63.15.17 45071 Orleans Cedex 2 | Deutsch/Esperanto/English/ France | Nederlands/Francais -------------------------------------------------------------------------------

Using the outer product you get a matrix that has the right size and contents but it is m*n by p*q Was that a misprint in your post?

...

...

x array([1, 2, 3, 4, 5, 6]) x.shape=(3,2) y = 10*transpose(x) y array([[10, 30, 50], [20, 40, 60]]) z = outerproduct(x.flat, y.flat) z array([[ 10, 30, 50, 20, 40, 60], [ 20, 60, 100, 40, 80, 120], [ 30, 90, 150, 60, 120, 180], [ 40, 120, 200, 80, 160, 240], [ 50, 150, 250, 100, 200, 300], [ 60, 180, 300, 120, 240, 360]])

-----Original Message----- From: numpy-discussion-admin@lists.sourceforge.net [mailto:numpy-discussion-admin@lists.sourceforge.net]On Behalf Of Nils Wagner Sent: Wednesday, July 18, 2001 4:10 AM To: numpy-discussion@lists.sourceforge.net Subject: [Numpy-discussion] Kronecker product Hi, I would appreciate it, if Numpy could handle the Kronecker-product of two matrices X, Y. kron(X,Y) is the Kronecker tensor product of X and Y. The result is a large matrix formed by taking all possible products between the elements of X and those of Y. If X is m-by-n and Y is p-by-q, then kron(X,Y) is m*p-by-n*q. Nils _______________________________________________ Numpy-discussion mailing list Numpy-discussion@lists.sourceforge.net http://lists.sourceforge.net/lists/listinfo/numpy-discussion

I would appreciate it, if Numpy could handle the Kronecker-product of two matrices X, Y.

kron(X,Y) is the Kronecker tensor product of X and Y.

import Numeric def kron(x, y): return Numeric.multiply.outer(x, y)

The result is a large matrix formed by taking all possible products between the elements of X and those of Y. If X is m-by-n and Y is p-by-q, then kron(X,Y) is m*p-by-n*q.

OK, there's one difference: the function shown about returns an array of shape (m, n, p, q). If the input arrays are always 2D, the following will do what you need: import Numeric def kron(x, y): z = Numeric.transpose(Numeric.multiply.outer(x, y), [0, 2, 1, 3]) z.shape = (z.shape[0]*z.shape[1], z.shape[2]*z.shape[3]) return z Konrad. -- ------------------------------------------------------------------------------- Konrad Hinsen | E-Mail: hinsen@cnrs-orleans.fr Centre de Biophysique Moleculaire (CNRS) | Tel.: +33-2.38.25.56.24 Rue Charles Sadron | Fax: +33-2.38.63.15.17 45071 Orleans Cedex 2 | Deutsch/Esperanto/English/ France | Nederlands/Francais -------------------------------------------------------------------------------