矩阵数组的克罗内克积

Question

我有两个矩阵数组 A 和 B，它们的形状相同： A.shape = B.shape = (M,N,P)

我想计算沿轴 0 的克罗内克积，因此：

KP[ii,:,:] = A[ii,:,:]⊗B[ii,:,:]

有没有办法在 numpy 中不使用 for 循环来做到这一点？

谢谢！

示例：

A = np.array([ [[1,0],
                [0,1]],
               [[1,0],
                [0,1]]
            ])

B = np.array([ [[1,0],
                [0,-1]],
               [[0,1],
                [1,0]]
            ])

KP = np.array( [
                [[1,0,0,0],
                 [0,-1,0,0],
                 [0,0,1,0],
                 [0,0,0,-1]],
                [[0,1,0,0],
                 [1,0,0,0],
                 [0,0,0,1],
                 [0,0,1,0]]
               ] )

相当于：

KP= np.zeros( (A.shape[0],
               A.shape[1]**2,
               A.shape[2]**2) )

for ii in range(A.shape[0]):

    KP[ii,:,:] = np.kron(A[ii,:,:],B[ii,:,:])

Answer 1

您可以使用 einsum，稍加练习就非常直观，或者使用经典的重塑和广播路线

A = np.array([ [[1,0],
                [0,1]],
               [[1,0],
                [0,1]]
            ])

B = np.array([ [[1,0],
                [0,-1]],
               [[0,1],
                [1,0]]
            ])

i,j,k = A.shape
i,l,m = B.shape
np.einsum("ijk,ilm->ijlkm",A,B).reshape(i,j*l,k*m)

# array([[[ 1,  0,  0,  0],
#         [ 0, -1,  0,  0],
#         [ 0,  0,  1,  0],
#         [ 0,  0,  0, -1]],
# 
#        [[ 0,  1,  0,  0],
#         [ 1,  0,  0,  0],
#         [ 0,  0,  0,  1],
#         [ 0,  0,  1,  0]]])

等效的非 einsum 表达式：

(A[:,:,None,:,None]*B[:,None,:,None,:]).reshape(i,j*l,k*m)

Answer 2

万一有人觉得这也有用（虽然可能效率不高）。

def give_kr_prod(matrices):
    #matrices list of 2 (or more in principle) matrices
    while len(matrices) != 1:
        sm, smf=[],[]
        for ind in range(len(matrices)):
            sm.append(matrices[ind])
            if ind%2==1 and ind>0:
                smf.append(np.kron(*sm))
                sm=[]
        matrices = smf
    return matrices[0]

matrices = np.random.randn(8,2,2)

矩阵数组的克罗内克积

Kronecker product of matrix array

numpy

python-3.x

numpy-ndarray

array-broadcasting