1

I need to create a diagonal matrix in python from an X matrix which is repeated 3 times. In matlab I do it in the following way:

  X=[1 2 3;
     4 5 6;
     7 8 9]

for i=1:1:3
  Brep{i}=X;      
end    
Mdiag=blkdiag(Brep{:})

Mdiag =

 1     2     3     0     0     0     0     0     0
 4     5     6     0     0     0     0     0     0
 7     8     9     0     0     0     0     0     0
 0     0     0     1     2     3     0     0     0
 0     0     0     4     5     6     0     0     0
 0     0     0     7     8     9     0     0     0
 0     0     0     0     0     0     1     2     3
 0     0     0     0     0     0     4     5     6
 0     0     0     0     0     0     7     8     9

But I don't know how to do this in Pyhton.

I would appreciate any help.

--------------------------------- ANSWER:--------------------------------------

Taking into account the answer provided by Fabrizio Bernini, I modified his code in order to do it more general. The code can now repeat the matrix X (any dimension), n times on the diagonal. For example:

import numpy as np

n=5

x = np.array([[1, 2, 3],[4, 5, 6],[7, 8, 9]])

Mdiag = np.zeros((n*x.shape[0], n*x.shape[1]))

for i in range(n):

Mdiag[x.shape[0]*i:x.shape[0]*i+x.shape[0],x.shape[1]*i:x.shape[1]*i+x.shape[1]]= x

Mdiag =[1., 2., 3., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.],
       [4., 5., 6., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.],
       [7., 8., 9., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 1., 2., 3., 0., 0., 0., 0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 4., 5., 6., 0., 0., 0., 0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 7., 8., 9., 0., 0., 0., 0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0., 1., 2., 3., 0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0., 4., 5., 6., 0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0., 7., 8., 9., 0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0., 0., 0., 0., 1., 2., 3., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0., 0., 0., 0., 4., 5., 6., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0., 0., 0., 0., 7., 8., 9., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 1., 2., 3.],
       [0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 4., 5., 6.],
       [0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 7., 8., 9.]

Thanks you all for the answers provided.

4

5 回答 5

5

你可以使用scipy.linalg.block_diag：

from scipy.linalg import block_diag
X=[[1, 2, 3],
     [4,5,6],
     [7, 8, 9]]
block_diag(X, X, X)

输出：

array([[1, 2, 3, 0, 0, 0, 0, 0, 0],
       [4, 5, 6, 0, 0, 0, 0, 0, 0],
       [7, 8, 9, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 1, 2, 3, 0, 0, 0],
       [0, 0, 0, 4, 5, 6, 0, 0, 0],
       [0, 0, 0, 7, 8, 9, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 1, 2, 3],
       [0, 0, 0, 0, 0, 0, 4, 5, 6],
       [0, 0, 0, 0, 0, 0, 7, 8, 9]], dtype=int32)

np.bmat使用和np.asmatrix的另一个选项np.asarray可能是这样的：

import numpy as np

X=np.asmatrix([[1, 2, 3],
     [4,5,6],
     [7, 8, 9]])
y = np.asmatrix(np.zeros((3, 3)))

r=np.asarray(np.bmat('X, y, y; y, X, y; y, y, X'))
print(r)
于 2020-07-05T08:51:48.797 回答
3

一种方法是创建一个大零矩阵并直接粘贴x到对角线位置：

N = 3
y = np.zeros((x.shape[0] * N, x.shape[1] * N))
for i in range(N):
    y[i * x.shape[0] : (i + 1) * x.shape[0], 
      i * x.shape[1] : (i + 1) * x.shape[1]] = x

另一种解决方案：根据需要创建一个水平“带”x和尽可能多的 0。然后将那条腰带的三个副本堆叠起来，移动：

belt = np.hstack([x, np.zeros((x.shape[0], (x.shape[1] - 1) * N))])
np.vstack([np.roll(belt, i) for i in range(0, x.shape[0] * N, x.shape[0])])
于 2020-07-05T08:41:45.760 回答
1

您没有标记任何“高级”模块（scipy，numpy，...）所以我们开始：

您可以创建一个自己执行此操作的函数。矩阵在 python 中显示为列表列表（除非你进入 numpy 等）：

def diag_size(what, size = 4):
    """Create a diagonal matrix of size len(what)*size where the list of list
    'what' is on the diagonal of the embiggended result"""
    rv = [[]]
    s = len(what) # quadratic => len(what) == len(what[0])
    
    for row in range(size*s):
        for col in range(size*s):
            if row//s == col//s:
                rv[-1].append(base[row%s][col%s])
            else:
                rv[-1].append(0)
        rv.append([])

    return rv[:-1]

n = 3 # size of the initial "matrix"
# create base case
base = [list(range(i+1,i+n+1)) for i in range(0,n*n,n)]

# create the diagonal one
print(*diag_size(base),sep="\n")

输出：

# with size=4 you get a 4 wide diagonal matrix
[1, 2, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[4, 5, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[7, 8, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 1, 2, 3, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 4, 5, 6, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 7, 8, 9, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 1, 2, 3, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 4, 5, 6, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 7, 8, 9, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 5, 6]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 7, 8, 9]
于 2020-07-05T08:53:53.693 回答
0

你可以使用 numpy 做这样的事情：

import numpy
x = np.array([[1, 2, 3],[4, 5, 6],[7, 8, 9]])
Mdiag = np.zeros((9, 9))
for i in range(3):
    Mdiag[3*i:3*i+3, 3*i:3*i+3] = x

您还可以使用 np.reshape 生成初始矩阵：

x = np.arange(1, 10)
x = x.reshape((3, 3))

更一般地说，如果你有一个 nxn 子矩阵并且想要创建一个 MxM 对角块矩阵，其中 M = c*n，你可以这样做：

import numpy
x = ... define here your nxn matrix
Mdiag = np.zeros((M, M))
for i in range(n):
    Mdiag[n*i:n*i+n, n*i:n*i+n] = x

有 numpy 内置函数来复制块矩阵和生成块矩阵，但我不知道有一个只用一个命令生成对角块矩阵的函数。另请参阅此内容以供参考：

[1]：https ://numpy.org/doc/stable/reference/generated/numpy.matlib.repmat.html#numpy.matlib.repmat和 [2]：https ://numpy.org/doc/stable/参考/生成/numpy.block.html?highlight=block%20matrix

于 2020-07-05T09:11:52.153 回答
-1

试试这个代码。

import numpy as np
X = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
M = np.zeros(shape = (9, 9))
for i in range(3):
  M[3*i:3*(i+1), 3*i:3*(i+1)] = X
于 2020-07-05T08:41:26.773 回答