numpy matrix multiplication row-wise

1
votes

I have one matrix, like

a = np.array([[1, 2, 3], [2, 3, 4], [3, 4, 5], [4, 5, 6]])

and I want to get a new matrix, where each element is the matrix product of the row of a with itself:

np.array([
    np.dot(np.array([a[0]]).T, np.array([a[0]])),
    np.dot(np.array([a[1]]).T, np.array([a[1]])),
    np.dot(np.array([a[2]]).T, np.array([a[2]])),
    np.dot(np.array([a[3]]).T, np.array([a[3]])),
])

which will be a 4x4 matrix with each element a 3x3 matrix. After this I can sum over the 0 axis to get a new 3x3 matrix.

Is there any more elegant way to implement this except using loop?

2
pythonnumpymatrixmultiplication

2 Answers

2
votes

Use NumPy broadcasting to keep the first axis aligned and perform outer product between the second one -

a[:,:,None]*a[:,None,:] # or a[...,None]*a[:,None]

With np.einsum, translates to -

np.einsum('ij,ik->ijk',a,a)
0
votes

I might be missing something but isn't this just matrix multiplication ? 

>>> a.T @ a
array([[30, 40, 50],
       [40, 54, 68],
       [50, 68, 86]])

>>> np.sum(np.array([
      np.dot(np.array([a[0]]).T, np.array([a[0]])),
      np.dot(np.array([a[1]]).T, np.array([a[1]])),
      np.dot(np.array([a[2]]).T, np.array([a[2]])),
      np.dot(np.array([a[3]]).T, np.array([a[3]])),
    ]), axis=0)
array([[30, 40, 50],
       [40, 54, 68],
       [50, 68, 86]])