scipy.linalg具有block_diag函数以自动执行此操作
>>> a1 = np.array([[1,1,1],[1,1,1],[1,1,1]])
>>> a2 = np.array([[2,2,2],[2,2,2],[2,2,2]])
>>> a3 = np.array([[3,3,3],[3,3,3],[3,3,3]])
>>> import scipy.linalg
>>> scipy.linalg.block_diag(a1, a2, a3)
array([[1, 1, 1, 0, 0, 0, 0, 0, 0],
[1, 1, 1, 0, 0, 0, 0, 0, 0],
[1, 1, 1, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 2, 2, 2, 0, 0, 0],
[0, 0, 0, 2, 2, 2, 0, 0, 0],
[0, 0, 0, 2, 2, 2, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 3, 3, 3],
[0, 0, 0, 0, 0, 0, 3, 3, 3],
[0, 0, 0, 0, 0, 0, 3, 3, 3]])
>>> r = np.array([[1,1,1,0,0,0,0,0,0],[1,1,1,0,0,0,0,0,0],[1,1,1,0,0,0,0,0,0], [0,0,0,2,2,2,0,0,0],[0,0,0,2,2,2,0,0,0],[0,0,0,2,2,2,0,0,0],[0,0,0,0,0,0,3,3,3],[0,0,0,0,0,0,3,3,3],[0,0,0,0,0,0,3,3,3]])
>>> (scipy.linalg.block_diag(a1, a2, a3) == r).all()
True
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