• 计算方法 | 曲线拟合的最小二乘法


    就是找个破多项式,让每个点和真实数据的差值的平方最小,原理就是这么简单。

    1.写正则方程组(法方程组)

    2.写结果向量

    3.解方程组

    丢代码:

     1 import numpy as np
     2 import math
     3 # data = eval(input("请输入数据:"))
     4 data = [(0.6,7.05),(1.3,12.2),(1.64,14.4),(1.8,15.2),(2.1,17.4),(2.3,19.6),(2.44,20.2)]
     5 n = 2
     6 m = len(data)
     7 fa_mat = [[0 for j in range(n)] for i in range(n)]
     8 for i in range(len(fa_mat)):
     9     for j in range(len(fa_mat[0])):
    10         _sum = 0
    11         for k in range(m):
    12             _sum += data[k][0] ** (i+j)
    13         fa_mat[i][j] = _sum
    14 print("法方程组如下:")
    15 print(np.array(fa_mat))
    16 # 计算结果向量
    17 b = [[0] for i in range(n)]
    18 for i in range(n):
    19     _sum = 0
    20     for k in range(m):
    21         _sum += ((data[k][0] ** i) * data[k][1])
    22         # print("_sum += (({} ** {}) * {})".format(data[k][0],i,data[k][1]))
    23     b[i][0] = _sum
    24 print("结果向量如下:")
    25 print(np.array(b))
    26 print("下面是解方程组:
    
    ")
    27 # 解线性方程组 使用三角分解法 ======================================
    28 mat = fa_mat[:]
    29 b = b[:]
    30 height = len(mat)
    31 width = len(mat[0])
    32 
    33 # 1 直接分解矩阵
    34 # 初始化LU
    35 U = [[0 for i in range(width)] for j in range(height)]
    36 L = [[0 if i!=j else 1 for i in range(width) ] for j in range(height)]
    37 
    38 print(L,U)
    39 
    40 # 从U第一行开始,对于L来说是从第一列开始
    41 for i in range(height):
    42     # 依次计算LU的行、列
    43     for j in range(i, width):
    44         _sum = 0
    45         for k in range(i):
    46             _sum += L[i][k] * U[k][j]
    47             print("U{}{}_sum += L[{}][{}] * U[{}][{}] = {} * {} = {}".format(i,j,i,k,k,j,L[i][k],U[k][j],L[i][k] * U[k][j]))
    48         U[i][j] = mat[i][j] - _sum
    49     # L的i.j是反着的
    50     for j in range(i, width):
    51         _sum = 0
    52         if j > i:
    53             for k in range(i):
    54                 _sum += L[j][k] * U[k][i]
    55                 print("L{}{}_sum += L[{}][{}] * U[{}][{}] = {} * {} = {}".format(j,i,j,k,k,i,L[j][k],U[k][i],L[j][k] * U[k][i]))
    56             L[j][i] = (mat[j][i] - _sum) / U[i][i]
    57     print("L = 
    {}".format(np.array(L)))
    58     print("U = 
    {}".format(np.array(U)))
    59     
    60 print("分解完毕,开始计算Ly=b")
    61 # 分解完毕 解Ly =b
    62 y = [[0] for i in range(height)]
    63 # print(y)
    64 for i in range(height):
    65     _sum = 0
    66     for k in range(i):
    67         _sum += L[i][k] * y[k][0]
    68     y[i][0] = (b[i][0] - _sum)          # 这里不用除以系数因为对角线为1
    69     print(y)
    70 print("计算完毕,开始计算Ux=y")
    71 x = [[0] for i in range(height)]
    72 for i in range(height-1,-1,-1):
    73     _sum = 0
    74     for k in range(height-1,i,-1):
    75         _sum += U[i][k] * x[k][0]
    76     x[i][0] = (y[i][0] - _sum) / U[i][i]
    77     print(x)
    78 print("最终结果x: 
    {}".format(np.array(x)))

    over.

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  • 原文地址:https://www.cnblogs.com/Mz1-rc/p/13881006.html
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