Multiply 2D NumPy arrays element-wise and sum

问题:

I am wondering if there is a quicker way/dedicated NumPy function to perform element-wise multiplication of 2D NumPy arrays and then sum all the elements. I currently use np.sum(np.multiply(A, B)) where A, B are NumPy arrays of equal dimension m x n.

回答1:

You can use np.tensordot -

np.tensordot(A,B, axes=((0,1),(0,1)))

Another way with np.dot after flattening the inputs -

A.ravel().dot(B.ravel())

Another with np.einsum -

np.einsum('ij,ij',A,B)

Sample run -

In [14]: m,n = 4,5

In [15]: A = np.random.rand(m,n)

In [16]: B = np.random.rand(m,n)

In [17]: np.sum(np.multiply(A, B))
Out[17]: 5.1783176986341335

In [18]: np.tensordot(A,B, axes=((0,1),(0,1)))
Out[18]: array(5.1783176986341335)

In [22]: A.ravel().dot(B.ravel())
Out[22]: 5.1783176986341335

In [21]: np.einsum('ij,ij',A,B)
Out[21]: 5.1783176986341326

Runtime test

In [23]: m,n = 5000,5000

In [24]: A = np.random.rand(m,n)
    ...: B = np.random.rand(m,n)
    ...: 

In [25]: %timeit np.sum(np.multiply(A, B))
    ...: %timeit np.tensordot(A,B, axes=((0,1),(0,1)))
    ...: %timeit A.ravel().dot(B.ravel())
    ...: %timeit np.einsum('ij,ij',A,B)
    ...: 
10 loops, best of 3: 52.2 ms per loop
100 loops, best of 3: 19.5 ms per loop
100 loops, best of 3: 19.5 ms per loop
100 loops, best of 3: 19 ms per loop