如何将块转换为块对角矩阵(NumPy) [英] How can I transform blocks into a blockdiagonal matrix (NumPy)

问题描述

在NumPy中，我有三个相同大小的正方形矩阵.我想将它们组合成块对角矩阵.

I have three same-size square matrices in NumPy. I would like to combine these to a block-diagonal matrix.

示例:

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]])

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]])

做到这一点的最佳方法是什么?

What is the best way to do this?

推荐答案

scipy.linalg具有block_diag函数以自动执行此操作

scipy.linalg has a block_diag function to do this automatically

>>> 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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