麻烦的
|
1
2
3
4
5
6
7
8
9
10
11
12
13
|
# TODO 构造增广矩阵，假设A，b行数相同def augmentMatrix(A, b): if(len(A) != len(b)): raise 'The number of rows is different' result = [] for i in range(len(A)): row = [] for j in range(len(A[i])): row.append(A[i][j]) for j in range(len(b[i])): row.append(b[i][j]) result.append(row) return result |
优化后
|
1
2
3
4
5
6
7
8
|
# TODO 构造增广矩阵，假设A，b行数相同def augmentMatrix(A, b): return [AA + bb for AA, bb in zip(A,b)] A = [[1,2,3],[4,5,6],[7,8,9]]b = [[1],[2],[3]]print augmentMatrix(A,b)[[1, 2, 3, 1], [4, 5, 6, 2], [7, 8, 9, 3]] |
注：解读一下AA+bb, 在 python中, [1, 2, 3] + [4]这样的表达式会返回[1, 2, 3, 4]
以上这篇在Python中构建增广矩阵的实现方法就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持服务器之家。
原文链接：https://blog.csdn.net/chuan403082010/article/details/79407431
