In [498]: A = np.arange(12).reshape(4,3)
In [499]: B = np.arange(4,10).reshape(3,2)
In [500]: A
Out[500]:
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11]])
In [501]: B
Out[501]:
array([[4, 5],
[6, 7],
[8, 9]])
Справочное итеративное решение:
In [504]: res = np.zeros((A.shape[0],B.shape[1]), A.dtype)
...: for i,row in enumerate(A):
...: for j,col in enumerate(B.T):
...: res[i,j] = np.min(row+col)
...:
In [505]: res
Out[505]:
array([[ 4, 5],
[ 7, 8],
[10, 11],
[13, 14]])
Более быстрая версия с использованием трансляции:
In [506]: np.min(A[:,:,None]+B[None,:,:], axis=1)
Out[506]:
array([[ 4, 5],
[ 7, 8],
[10, 11],
[13, 14]])
===
Продемонстрировать эквивалентность матричного произведения:
In [507]: np.dot(A,B)
Out[507]:
array([[ 22, 25],
[ 76, 88],
[130, 151],
[184, 214]])
In [508]: np.sum(A[:,:,None]*B[None,:,:], axis=1)
Out[508]:
array([[ 22, 25],
[ 76, 88],
[130, 151],
[184, 214]])