Question

l有5个邻接矩阵（块数组）：A，B，C，D，E。每个矩阵的尺寸为[20,20]。

给出A，B，C，D，E，l要构建F，该F堆叠5个邻接矩阵。由于我们有5个[20,20]的2D数组，因此F的维数为[20 * 5,20 * 5]，如下所示：

F = np.zeros（（100,100））

    F=[
       [A,0,0,0,...,0],
       [0,...,B,...,0],
       [0,...,..,C,0],
       [0,.........D,..,0],
       [0,...........,E],
    ]

这样：

 A is indexed at F[0][:20]
 B is indexed at F[1][20:40]
 C is indexed at F[2][40:60]
 D is indexed at F[3][60:80]
 E is indexed at F[4][80:100]

对于大量邻接矩阵，有效的numpy方法是什么？让我们将 n 个邻接矩阵堆叠到[n*20,n*20]

的新2D数组的对角线上

Answer 1

您可以使用scipy.sparse.block_diag：

>>> AtoE = np.add.outer(np.arange(5, 10), np.zeros((3, 3), int))
>>> scipy.sparse.block_diag(AtoE).A
array([[5, 5, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [5, 5, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [5, 5, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 6, 6, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 6, 6, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 6, 6, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 7, 7, 7, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 7, 7, 7, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 7, 7, 7, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 8, 8, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 8, 8, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 8, 8, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 9, 9],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 9, 9],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 9, 9]], dtype=int64)

无论如何，稀疏存储可能是个好主意。

或者，如果您确实想使用密集数组，这是一种更直接的方法：

>>> A = AtoE[0]
>>> N, N = A.shape
>>> k = len(AtoE)
>>> out = np.zeros((k, N, k, N), A.dtype)
>>> np.einsum('ijik->ijk', out)[...] = AtoE
>>> out.reshape(k*N, k*N)
array([[5, 5, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [5, 5, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [5, 5, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 6, 6, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 6, 6, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 6, 6, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 7, 7, 7, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 7, 7, 7, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 7, 7, 7, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 8, 8, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 8, 8, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 8, 8, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 9, 9],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 9, 9],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 9, 9]])

以对角线垂直堆叠5个2D阵列以构建整个2D阵列

1 个答案: