标签:odi encoding 位置 矩阵 print for range nbsp 数据
X = [[12,7,3],
[4 ,5,6],
[7 ,8,9]]
Y = [[5,8,1],
[6,7,3],
[4,5,9]]
#encoding=utf-8
z=[
[0,0,0],
[0,0,0],
[0,0,0]
]
x = [[12,7,3],
[4 ,5,6],
[7 ,8,9]]
y = [[5,8,1],
[6,7,3],
[4,5,9]]
for i in range(len(x)):
for j in range(len(x[0])):
z[i][j]=x[i][j]+y[i][j]
print z
[[x[i][j]+y[i][j] for j in range(len(x[0]))] for i in range(len(x))]
9.两个 3 行 3 列的矩阵,实现其对应位置的数据相加,并返回一个新矩阵
标签:odi encoding 位置 矩阵 print for range nbsp 数据
原文地址:https://www.cnblogs.com/luo25236240/p/9255404.html