Question

我只是在学习pyhton并希望定义一个返回nxn方阵的函数，其中主对角线（i = j），上对角线（j = i + 1）和下对角线（j = i）的预定义值-1）和所有其他元素相等0.

任何帮助将不胜感激，

感谢

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Answer 1

如果a,b,c分别是上对角线，主对角线和下对角线的三个列表，您可以按如下方式编写：

import numpy as np
a=[1,2,3,4]
b=[5,6,7,8,9]
c=[10,11,12,13]
n=len(b)
m=np.zeros((n,n))
for i in range(0,n-1):
    m[i,i+1]=a[i]
    m[i,i]=b[i]
    m[i+1,i]=c[i]
m[n-1,n-1]=b[n-1]
print(m)

在上面的代码中，您初始化零矩阵，然后根据您的列表仅更新上，下和主对角线条目。输出是

[[  5.   1.   0.   0.   0.]
 [ 10.   6.   2.   0.   0.]
 [  0.  11.   7.   3.   0.]
 [  0.   0.  12.   8.   4.]
 [  0.   0.   0.  13.   9.]]

编辑：@hpaulj建议的缩短方式是

m=np.diag(a,1)+np.diag(b,0)+np.diag(c,-1)

np.diag(r,k)创建一个矩阵，其中主对角线上方的k'对角线（如果k下方为负值）为r，其余条目为0

请参阅此处的文档： https://docs.scipy.org/doc/numpy/reference/generated/numpy.diag.html

Answer 2

这是一个方法，它使用numpy.diag接受在对角线上添加的项目列表，并利用k参数指定我们想要修改的对角线。

import numpy as np

def create_diag(n, l):
    arr = np.zeros((n, n))

    # check that l contains an odd number of elements
    if len(l) % 2 != 1:
        return arr

    # check that we have at most the number of diagonals in the matrix
    if len(l) >= 2*n:
        return arr

    # set limits of diagonals to set. setting 3 diagonals means limits=[-1,0,1]. setting 1 diagonal means limits=[0]
    limit = int((len(l) - 1) / 2)
    limits = range(-limit, limit + 1)

    # maps k-->list of values on the diagonal
    diag_map = dict(zip(limits, l))
    for k, v in diag_map.items():
        diag = [v] * (n - abs(k))
        arr += np.diag(diag, k=k)
    return arr

# elements to put on the diagonal, centered about the main diagonal (i=j)
l = [-1,1,2]
n = 6

diag_arr = create_diag(n, l)

# [[ 1.  2.  0.  0.  0.  0.]
#  [-1.  1.  2.  0.  0.  0.]
#  [ 0. -1.  1.  2.  0.  0.]
#  [ 0.  0. -1.  1.  2.  0.]
#  [ 0.  0.  0. -1.  1.  2.]
#  [ 0.  0.  0.  0. -1.  1.]]

定义一个返回nxn方阵的函数

2 个答案: