Numpy中的特殊矩阵

Question

我想制作一个看起来像这样的numpy数组:

m = [1, 1, 1, 0, 0, 0, 0, 0, 0
     0, 0, 0, 1, 1, 1, 0, 0, 0 
     0, 0, 0, 0, 0, 0, 1, 1, 1]

我已经看到这个答案在Numpy中制作特殊的对角矩阵,我有这个:

a = np.zeros(3,9)
a[0, 0] = 1
a[0, 1] = 1
a[0, 2] = 1
a[1, 3] = 1
a[1, 4] = 1
a[1, 5] = 1
a[2, 6] = 1
a[2, 7] = 1
a[2, 8] = 1

但是我想使用'for'cicle,如何有效地填充对角线？

Answer 1

一种方法是简单地水平拉伸身份阵列;

> np.repeat(np.identity(3, dtype=int), 3, axis=1)

array([[1, 1, 1, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 1, 1, 1, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 1, 1, 1]])

Answer 2

如果m表示1一行中的sn数和行数，可以列出两种方法来解决它。

使用np.kron它很简单，就像这样 -

def kron_based(m,n):
    return np.kron(np.eye(n,dtype=int), np.ones(m,dtype=int))

使用零初始化和填充将是 -

def initialization_based(m,n):
    A = np.zeros((n,n*m),dtype=int)
    A.reshape(n,n,m)[np.eye(n,dtype=bool)] = 1
    return A

样品运行 -

In [54]: m = 4 # Number of 1s in a row. Note that this is 3 for your case
    ...: n = 3 # Number of rows
    ...: 

In [55]: initialization_based(m,n)
Out[55]: 
array([[1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1]])

In [56]: kron_based(m,n)
Out[56]: 
array([[1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1]])