I have three same-size square matrices in NumPy. I would like to combine these to a block-diagonal matrix.
Example:
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]])
What is the best way to do this?
To build a block of matrix, use the numpy. block() method in Python Numpy. Blocks in the innermost lists are concatenated along the last dimension (-1), then these are concatenated along the secondlast dimension (-2), and so on until the outermost list is reached.
Step 1: Import numpy. Step 2: Take dimensions as input from the user. Step 3: Print the identity matrix using numpy. identity() function.
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
Since these answers, numpy
has added a block
function
In [695]: A=np.arange(1,10).reshape(3,3)
In [696]: B=np.arange(10,14).reshape(2,2)
In [698]: C = np.zeros((3,2),int)
In [699]: np.block([[A,C],[C.T,B]])
Out[699]:
array([[ 1, 2, 3, 0, 0],
[ 4, 5, 6, 0, 0],
[ 7, 8, 9, 0, 0],
[ 0, 0, 0, 10, 11],
[ 0, 0, 0, 12, 13]])
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