将3D numpy数组转换成块对角矩阵 [英] 3D numpy array into block diagonal matrix
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问题描述
我正在寻找一种将nXaXb numpy数组转换为块对角矩阵的方法。我已经碰到过 scipy.linalg。 block_diag ,其缺点(对于我而言)是要求矩阵的每个块分别给出。但是,这在n很高时很有挑战性,因此要使情况更清楚,可以说我有
I am looking for a way to convert a nXaXb numpy array into a block diagonal matrix. I have already came across scipy.linalg.block_diag, the down side of which (for my case) is it requires each blocks of the matrix to be given separately. However, this is challenging when n is very high, so to make things more clear lets say I have a
import numpy as np
a = np.random.rand(3,2,2)
array([[[ 0.33599705, 0.92803544],
[ 0.6087729 , 0.8557143 ]],
[[ 0.81496749, 0.15694689],
[ 0.87476697, 0.67761456]],
[[ 0.11375185, 0.32927167],
[ 0.3456032 , 0.48672131]]])
我要实现的目标与
from scipy.linalg import block_diag
block_diag(a[0], a[1],a[2])
array([[ 0.33599705, 0.92803544, 0. , 0. , 0. , 0. ],
[ 0.6087729 , 0.8557143 , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0.81496749, 0.15694689, 0. , 0. ],
[ 0. , 0. , 0.87476697, 0.67761456, 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0.11375185, 0.32927167],
[ 0. , 0. , 0. , 0. , 0.3456032 , 0.48672131]])
这实际上是一个示例,其中a具有数百个元素。
This is just as an example in actual case a has hundreds of elements.
推荐答案
尝试使用 block_diag(* a)
。参见下面的示例:
Try using block_diag(*a)
. See example below:
In [9]: paste
import numpy as np
a = np.random.rand(3,2,2)
from scipy.linalg import block_diag
b = block_diag(a[0], a[1],a[2])
c = block_diag(*a)
b == c
## -- End pasted text --
Out[9]:
array([[ True, True, True, True, True, True],
[ True, True, True, True, True, True],
[ True, True, True, True, True, True],
[ True, True, True, True, True, True],
[ True, True, True, True, True, True],
[ True, True, True, True, True, True]], dtype=bool)
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