最简单的方法可能就是做以下事情：

import numpy as np

n = 10
a = np.zeros(n)

a[0] = 3
a[1] = -1
a[len(a)-1] = 3
a[len(a)-2] = -1

>>print(a)
output: [ 3. -1.  0.  0.  0.  0.  0.  0.  -1. 3.]

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希望这能有所帮助;）

In [97]: n=10                                                                   
In [98]: arr = np.zeros(n,int)                                                  
In [99]: arr[[0,-1]]=3; arr[[1,-2]]=-1                                          
In [100]: arr                                                                   
Out[100]: array([ 3, -1,  0,  0,  0,  0,  0,  0, -1,  3])

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很容易变成（n，1）：

In [101]: arr[:,None]                                                           
Out[101]: 
array([[ 3],
       [-1],
       [ 0],
       [ 0],
       [ 0],
       [ 0],
       [ 0],
       [ 0],
       [-1],
       [ 3]])

型

N = 100 # you can set to whatever number you want
a = np.zeros((N,1)) # you must specify the dimension as a tuple if you want exactly Nx1 array.
a[0]=3
a[1]=-1
a[-1]=3
a[-2]=-1

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我想有效的方法是：

import numpy as np

n = 100

I = np.arange(n)

matrix = np.row_stack(0*I)
matrix[0]=3
matrix[1]=-1
matrix[n-2]=-1
matrix[n-1]=3

print("\Matrix is \n",matrix)

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的数据