Apply the ufunc op to all pairs (a, b) with a in A and b in B.
Let M = A.ndim, N = B.ndim. Then the result, C, of op.outer(A, B) is an array of dimension M + N such that:
![C[i_0, ..., i_{M-1}, j_0, ..., j_{N-1}] =
op(A[i_0, ..., i_{M-1}], B[j_0, ..., j_{N-1}])](../../_images/math/b7261a8987938a5f9cd3be2756aa0ce0fb559d32.png)
For A and B one-dimensional, this is equivalent to:
r = empty(len(A),len(B))
for i in xrange(len(A)):
    for j in xrange(len(B)):
        r[i,j] = op(A[i], B[j]) # op = ufunc in question
| Parameters : | A : array_like 
 B : array_like 
  | 
|---|---|
| Returns : | r : ndarray 
  | 
See also
Examples
>>> np.multiply.outer([1, 2, 3], [4, 5, 6])
array([[ 4,  5,  6],
       [ 8, 10, 12],
       [12, 15, 18]])
A multi-dimensional example:
>>> A = np.array([[1, 2, 3], [4, 5, 6]])
>>> A.shape
(2, 3)
>>> B = np.array([[1, 2, 3, 4]])
>>> B.shape
(1, 4)
>>> C = np.multiply.outer(A, B)
>>> C.shape; C
(2, 3, 1, 4)
array([[[[ 1,  2,  3,  4]],
        [[ 2,  4,  6,  8]],
        [[ 3,  6,  9, 12]]],
       [[[ 4,  8, 12, 16]],
        [[ 5, 10, 15, 20]],
        [[ 6, 12, 18, 24]]]])