python 实现方阵的对角线遍历示例

[[ 1 2 3 4 5]
 [ 6 7 8 9 10]
 [11 12 13 14 15]
 [16 17 18 19 20]
 [21 22 23 24 25]]

0 0
1 1
2 2
3 3
4 4

0 1
1 2
2 3
3 4

0 2
1 3
2 4

0 3
1 4

0 4

1 0
2 1
3 2
4 3

2 0
3 1
4 2

3 0
4 1

4 0

import numpy as np

A = np.arange(25)+1
A = np.mat(A.reshape([5, 5]))
print(A)

"""
[[ 1 2 3 4 5]
 [ 6 7 8 9 10]
 [11 12 13 14 15]
 [16 17 18 19 20]
 [21 22 23 24 25]]
"""

Num_element = A.shape[0]
c = int((Num_element-1)/2)
# print(c)


R = np.zeros_like(A)
# print(R)

for j in range(Num_element):
 print()
 i = 0
 # print(i, j)
 while np.max([i, j])<Num_element:
 print(i, j)
 if np.abs(i-j)%2==0:
 R[i, j] = A[c-int((j-i)/2), c+int((j-i)/2)]
 else:
 R[i, j] = (A[c-int((j-i-1)/2), c+int((j-i+1)/2)]+A[c-int((j-i+1)/2), c+int((j-i-1)/2)])/2

 i=i+1
 j=j+1

# print(R)

for k in range(1, Num_element):
 print()
 i = 0
 # print(i, j)
 while np.max([k, i])<Num_element:
 print(k, i)
 if np.abs(k-i)%2==0:
 R[k, i] = A[c-int((i-k)/2), c+int((i-k)/2)]
 else:
 R[k, i] = (A[c-int((i-k-1)/2), c+int((i-k+1)/2)]+A[c-int((i-k+1)/2), c+int((i-k-1)/2)])/2

 k=k+1
 i=i+1

print(R)

def diag_opreation(k, i, Num_element, R, A):
 c = int((Num_element-1)/2)
 while np.max([k, i])<Num_element:
 print(k, i)
 if np.abs(k-i)%2==0:
 R[k, i] = A[c-int((i-k)/2), c+int((i-k)/2)]
 else:
 R[k, i] = (A[c-int((i-k-1)/2), c+int((i-k+1)/2)]+A[c-int((i-k+1)/2), c+int((i-k-1)/2)])/2
 
 k=k+1
 i=i+1
 return R

for j in range(Num_element):
 print()
 i = 0
 # print(i, j)
 R = diag_opreation(i, j, Num_element, R, A)

# print(R)

for k in range(1, Num_element):
 print()
 i = 0
 # print(i, j)
 R = diag_opreation(k, i, Num_element, R, A)

print(R)

[[13 11 9 7 5]
 [15 13 11 9 7]
 [17 15 13 11 9]
 [19 17 15 13 11]
 [21 19 17 15 13]]