Question

我正在研究python项目并利用numpy。我经常需要通过单位矩阵计算矩阵的Kronecker乘积。这些是我的代码中相当大的瓶颈，所以我想优化它们。我必须采取两种产品。第一个是：

np.kron(np.eye(N), A)

只需使用scipy.linalg.block_diag即可轻松优化此版本。该产品相当于：

la.block_diag(*[A]*N)

大约快10倍。但是，我不确定如何优化第二类产品：

np.kron(A, np.eye(N))

我可以使用类似的技巧吗？

Answer 1

一种方法是初始化4D的输出数组，然后从A为其分配值。这样的任务将广播价值，这是我们在NumPy中获得效率的地方。

因此，解决方案就是这样 -

# Get shape of A
m,n = A.shape

# Initialize output array as 4D
out = np.zeros((m,N,n,N))

# Get range array for indexing into the second and fourth axes 
r = np.arange(N)

# Index into the second and fourth axes and selecting all elements along
# the rest to assign values from A. The values are broadcasted.
out[:,r,:,r] = A

# Finally reshape back to 2D
out.shape = (m*N,n*N)

作为一个功能 -

def kron_A_N(A, N):  # Simulates np.kron(A, np.eye(N))
    m,n = A.shape
    out = np.zeros((m,N,n,N),dtype=A.dtype)
    r = np.arange(N)
    out[:,r,:,r] = A
    out.shape = (m*N,n*N)
    return out

要模拟np.kron(np.eye(N), A)，只需按第一和第二轴交换操作，类似地换第三和第四轴 -

def kron_N_A(A, N):  # Simulates np.kron(np.eye(N), A)
    m,n = A.shape
    out = np.zeros((N,m,N,n),dtype=A.dtype)
    r = np.arange(N)
    out[r,:,r,:] = A
    out.shape = (m*N,n*N)
    return out

计时 -

In [174]: N = 100
     ...: A = np.random.rand(100,100)
     ...: 

In [175]: np.allclose(np.kron(A, np.eye(N)), kron_A_N(A,N))
Out[175]: True

In [176]: %timeit np.kron(A, np.eye(N))
1 loops, best of 3: 458 ms per loop

In [177]: %timeit kron_A_N(A, N)
10 loops, best of 3: 58.4 ms per loop

In [178]: 458/58.4
Out[178]: 7.842465753424658

高效的Kronecker产品，带有单位矩阵和常规矩阵 - NumPy / Python

1 个答案: