垂直堆叠对角线的5个2D阵列以构建整个2D阵列 [英] Stack vertically 5 2D-arrays in diagonal to build a whole 2d-array

问题描述

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]

对于大量使用的邻接矩阵，有效的numpy方法是什么?让我们将 n 个邻接矩阵堆叠在[n*20,n*20]

的新2D数组的对角线上

解决方案

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]])

l have 5 adjacency matrices (nump arrays) : A, B, C, D, E. each of dimension [20,20].

Given A, B, C, D, E, l would like to build F which stacks the 5 adjacency matrices. Since we have 5 2D arrays of [20,20] then F is of dimension [20*5,20*5] as follow :

F=np.zeros((100,100))

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

such that :

 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]

What is the efficient numpy way to do that for larage number of adjacency matrices ?. Let's, we have n adjacency matrices to stack in a diagonal of new 2D array of [n*20,n*20]

解决方案

You could use 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)

Sparse storage may be a good idea, anyway.

Alternatively, here is a more direct method in case you definitely want to use dense arrays:

>>> 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]])

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