基于python numpy库实现的二次型标准化的算法
寒假在家，闲来无事，摆烂之余，写了一个基于python numpy的矩阵标准化算法，适用于任意对称阵的标准化（由于规范化很简单就做）。
没用用到任何高级的东西，就是遍历循环按照矩阵的初等变换来标准化，可能效率不是很高，但是除非大型矩阵（几万几百万那种的），一般能见到的矩阵就是秒出结果，当数据较少或较小时算法间的差别并不大（基本没有什么区别）
代码如下：
1 import numpy as np 2 3 a = np.array(eval(input('请输入矩阵：')), dtype=np.float16) 4 x = len(a) 5 print('原矩阵：\n',a) 6 for n in range(x): 7 if a[n, n] == 0 and n != x - 1: 8 v = 0 9 for i in range(x): 10 v = v + a[n, i] 11 if v == 0: 12 continue 13 else: 14 for i in range(x): 15 if a[n, i] == 0: 16 continue 17 else: 18 for p in range(x): 19 a[int(f'{p}'), int(f'{n}')] -= a[int(f'{p}'), int(f'{i}')] 20 for p in range(x): 21 a[int(f'{n}'), int(f'{p}')] -= a[int(f'{i}'), int(f'{p}')] 22 break 23 for i in range(n + 1, x): 24 xi_shu = a[n, i] / a[n, n] 25 for p in range(x): 26 a[int(f'{p}'), int(f'{i}')] -= a[int(f'{p}'), int(f'{n}')] * xi_shu 27 for p in range(x): 28 a[int(f'{i}'), int(f'{p}')] -= a[int(f'{n}'), int(f'{p}')] * xi_shu 29 else: 30 for i in range(n + 1, x): 31 xi_shu = a[n, i] / a[n, n] 32 for p in range(x): 33 a[int(f'{p}'), int(f'{i}')] -= a[int(f'{p}'), int(f'{n}')] * xi_shu 34 for p in range(x): 35 a[int(f'{i}'), int(f'{p}')] -= a[int(f'{n}'), int(f'{p}')] * xi_shu 36 print('标准化矩阵：\n',a)
