Numpy dot() function equivalent

Question

This question is just out of pure curiosity. Suppose I have 2 matrices a and b .

a=np.array([[1, 2],
            [2, 3],
            [4, 5]])

b=np.array([[1, 2, 3, 4],
            [2, 3, 4, 5]])

To find their dot product, I might use np.dot(a,b) . But is there any other way to do this? I am not asking for any other alias functions. But maybe another way to do this like np.sum(a*b, axis=1) (I know that doesn't work, it is just an example). And what if I have a 3-D matrix? Is there any other way to compute their dot product as well (without using any functions)?

Thanks in advance!

Answer 1

In [66]: a=np.array([[1, 2],
    ...:             [2, 3],
    ...:             [4, 5]])
    ...: 
    ...: b=np.array([[1, 2, 3, 4],
    ...:             [2, 3, 4, 5]])
    ...: 
    ...:             
In [67]: np.dot(a,b)
Out[67]: 
array([[ 5,  8, 11, 14],
       [ 8, 13, 18, 23],
       [14, 23, 32, 41]])
In [68]: a@b
Out[68]: 
array([[ 5,  8, 11, 14],
       [ 8, 13, 18, 23],
       [14, 23, 32, 41]])
In [69]: np.einsum('ij,jk',a,b)
Out[69]: 
array([[ 5,  8, 11, 14],
       [ 8, 13, 18, 23],
       [14, 23, 32, 41]])

Broadcasted multiply and sum:

In [71]: (a[:,:,None]*b[None,:,:]).sum(axis=1)
Out[71]: 
array([[ 5,  8, 11, 14],
       [ 8, 13, 18, 23],
       [14, 23, 32, 41]])
In [72]: (a[:,:,None]*b[None,:,:]).shape
Out[72]: (3, 2, 4)