python 实现方阵的对角线遍历示例
任务描述
对一个方阵矩阵,实现平行于主对角线方向的对角线元素遍历。
从矩阵索引入手:
[[ 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]]
上三角的索引遍历:
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
下三角的索引遍历:
1 0 2 1 3 2 4 3 2 0 3 1 4 2 3 0 4 1 4 0
代码
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)
上述代码中对于每条对角线的所有元素执行相同的赋值操作。
考虑将其中重复的部分封装成函数:
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
则代码变为:
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)
输出结果为:
[[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]]
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