Python - Matrix Outer Product
Given two matrices A: m * r B: n * r I want to generate another matrix C: m * n, with each entry C_ij being a matrix calculated by the outer product of A_i and B_j. For example, A
Solution 1:
The Einstein notation expresses this problem nicely
In [85]: np.einsum('ac,bd->abcd',A,B)
Out[85]:
array([[[[ 3, 1],
[ 6, 2]],
[[ 1, 2],
[ 2, 4]]],
[[[ 9, 3],
[12, 4]],
[[ 3, 6],
[ 4, 8]]]])
Solution 2:
temp = numpy.multiply.outer(A, B)
C = numpy.swapaxes(temp, 1, 2)
NumPy ufuncs, such as multiply
, have an outer
method that almost does what you want. The following:
temp = numpy.multiply.outer(A, B)
produces a result such that temp[a, b, c, d] == A[a, b] * B[c, d]
. You want C[a, b, c, d] == A[a, c] * B[b, d]
. The swapaxes
call rearranges temp
to put it in the order you want.
Solution 3:
Use numpy;
In [1]: import numpy as np
In [2]: A = np.array([[1, 2], [3, 4]])
In [3]: B = np.array([[3, 1], [1, 2]])
In [4]: C = np.outer(A, B)
In [5]: C
Out[5]:
array([[ 3, 1, 1, 2],
[ 6, 2, 2, 4],
[ 9, 3, 3, 6],
[12, 4, 4, 8]])
Once you have the desired result, you can use numpy.reshape()
to mold it in almost any shape you want;
In [6]: C.reshape([4,2,2])
Out[6]:
array([[[ 3, 1],
[ 1, 2]],
[[ 6, 2],
[ 2, 4]],
[[ 9, 3],
[ 3, 6]],
[[12, 4],
[ 4, 8]]])
Solution 4:
Simple Solution with Numpy Array Broadcasting
Since, you want C_ij = A_i * B_j
, this can be achieved simply by numpy broadcasting on element-wise-product of column-vector-A and row-vector-B, as shown below:
# import numpy as np# A = [[1, 2], [3, 4]]# B = [[3, 1], [1, 2]]
A, B = np.array(A), np.array(B)
C = A.reshape(-1,1) * B.reshape(1,-1)
# same as: # C = np.einsum('i,j->ij', A.flatten(), B.flatten())print(C)
Output:
array([[ 3, 1, 1, 2],
[ 6, 2, 2, 4],
[ 9, 3, 3, 6],
[12, 4, 4, 8]])
You could then get your desired four sub-matrices by using numpy.dsplit()
or numpy.array_split()
as follows:
np.dsplit(C.reshape(2, 2, 4), 2)
# same as:# np.array_split(C.reshape(2,2,4), 2, axis=2)
Output:
[array([[[ 3, 1],
[ 6, 2]],
[[ 9, 3],
[12, 4]]]),
array([[[1, 2],
[2, 4]],
[[3, 6],
[4, 8]]])]
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