Wsrtjtyk

I need a simple numpy function do this multiplication without a for-loop and more time efficient.

Actually I want a function that multiplies each row of a to b

a=np.arange(2,12).reshape(5,2)

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

c=np.array([[a[i,:]@b] for i in range(a.shape[0])])

With numpy.tensordot:

c = np.tensordot(a, b, axes=1)

If you insist that shape will be the same:

c.reshape(5,1,2)

Using numpy einsum you could do (edited to reshape the array based on @dobkind's answer):

c = np.einsum('ki,ij->kj', a, b).reshape(5,1,2)

which should be faster.

%timeit np.einsum('ki,ij->kj', a, b).reshape(5,1,2)

1.87 µs ± 10.5 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

versus (using the @ matrix multiplication operator, which works in Python 3)

%timeit np.array([[a[i,:]@b] for i in range(a.shape[0])])

10.2 µs ± 36.7 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

To use @, make a a 3d array (5,1,2) with pairs well with (2,2) (or (1,2,2) by automatic broadcasting).

In [448]: np.array([[a[i,:]@b] for i in range(a.shape[0])])

Out[448]: 

array([[[11, 16]],



       [[19, 28]],



       [[27, 40]],



       [[35, 52]],



       [[43, 64]]])



In [450]: a[:,None,:]@b

Out[450]: 

array([[[11, 16]],



       [[19, 28]],



       [[27, 40]],



       [[35, 52]],



       [[43, 64]]])

This is actually a bit faster than the einsum solution - though with a example this small I wouldn't make a big deal about timings.

Use matmul function from numpy to multiply two matrices.
Let me know whether this helps. Thank you.

c = np.matmul(a,b)