欢迎您访问程序员文章站本站旨在为大家提供分享程序员计算机编程知识!
您现在的位置是: 首页  >  IT编程

Python实现曲线拟合的最小二乘法

程序员文章站 2022-05-09 08:49:21
本文实例为大家分享了python曲线拟合的最小二乘法,供大家参考,具体内容如下模块导入import numpy as npimport gaosi as gs代码"""本函数通过创建增广矩阵,并调用高...

本文实例为大家分享了python曲线拟合的最小二乘法,供大家参考,具体内容如下

模块导入

import numpy as np
import gaosi as gs

代码

"""
本函数通过创建增广矩阵,并调用高斯列主元消去法模块进行求解。

"""
import numpy as np
import gaosi as gs

shape = int(input('请输入拟合函数的次数:'))

x = np.array([0.6,1.3,1.64,1.8,2.1,2.3,2.44])
y = np.array([7.05,12.2,14.4,15.2,17.4,19.6,20.2])
data = []
for i in range(shape*2+1):
 if i != 0:
 data.append(np.sum(x**i))
 else:
 data.append(len(x))
b = []
for i in range(shape+1):
 if i != 0:
 b.append(np.sum(y*x**i))
 else:
 b.append(np.sum(y))
b = np.array(b).reshape(shape+1,1)
n = np.zeros([shape+1,shape+1])
for i in range(shape+1):
 for j in range(shape+1):
 n[i][j] = data[i+j]
result = gs.handle(n,b)
if not result:
 print('增广矩阵求解失败!')
 exit()
fun='f(x) = '
for i in range(len(result)):
 if type(result[i]) == type(''):
 print('存在*变量!')
 fun = fun + str(result[i])
 elif i == 0:
 fun = fun + '{:.3f}'.format(result[i])
 else:
 fun = fun + '+{0:.3f}*x^{1}'.format(result[i],i)
print('求得{0}次拟合函数为:'.format(shape))
print(fun)

高斯模块

# 导入 numpy 模块
import numpy as np


# 行交换
def swap_row(matrix, i, j):
 m, n = matrix.shape
 if i >= m or j >= m:
 print('错误! : 行交换超出范围 ...')
 else:
 matrix[i],matrix[j] = matrix[j].copy(),matrix[i].copy()
 return matrix


# 变成阶梯矩阵
def matrix_change(matrix):
 m, n = matrix.shape
 main_factor = []
 main_col = main_row = 0
 while main_row < m and main_col < n:
 # 选择进行下一次主元查找的列
 main_row = len(main_factor)
 # 寻找列中非零的元素
 not_zeros = np.where(abs(matrix[main_row:,main_col]) > 0)[0]
 # 如果该列向下全部数据为零,则直接跳过列
 if len(not_zeros) == 0:
 main_col += 1
 continue
 else:
 # 将主元列号保存在列表中
 main_factor.append(main_col)
 # 将第一个非零行交换至最前
 if not_zeros[0] != [0]:
 matrix = swap_row(matrix,main_row,main_row+not_zeros[0])
 # 将该列主元下方所有元素变为零
 if main_row < m-1:
 for k in range(main_row+1,m):
 a = float(matrix[k, main_col] / matrix[main_row, main_col])
 matrix[k] = matrix[k] - matrix[main_row] * matrix[k, main_col] / matrix[main_row, main_col]
 main_col += 1
 return matrix,main_factor


# 回代求解
def back_solve(matrix, main_factor):
 # 判断是否有解
 if len(main_factor) == 0:
 print('主元错误,无主元! ...')
 return none
 m, n = matrix.shape
 if main_factor[-1] == n - 1:
 print('无解! ...')
 return none
 # 把所有的主元元素上方的元素变成0
 for i in range(len(main_factor) - 1, -1, -1):
 factor = matrix[i, main_factor[i]]
 matrix[i] = matrix[i] / float(factor)
 for j in range(i):
 times = matrix[j, main_factor[i]]
 matrix[j] = matrix[j] - float(times) * matrix[i]
 # 先看看结果对不对
 return matrix


# 结果打印
def print_result(matrix, main_factor):
 if matrix is none:
 print('阶梯矩阵为空! ...')
 return none
 m, n = matrix.shape
 result = [''] * (n - 1)
 main_factor = list(main_factor)
 for i in range(n - 1):
 # 如果不是主元列,则为*变量
 if i not in main_factor:
 result[i] = '(free var)'
 # 否则是主元变量,从对应的行,将主元变量表示成非主元变量的线性组合
 else:
 # row_of_main表示该主元所在的行
 row_of_main = main_factor.index(i)
 result[i] = matrix[row_of_main, -1]
 return result


# 得到简化的阶梯矩阵和主元列
def handle(matrix_a, matrix_b):
 # 拼接成增广矩阵
 matrix_01 = np.hstack([matrix_a, matrix_b])
 matrix_01, main_factor = matrix_change(matrix_01)
 matrix_01 = back_solve(matrix_01, main_factor)
 result = print_result(matrix_01, main_factor)
 return result


if __name__ == '__main__':
 a = np.array([[2, 1, 1], [3, 1, 2], [1, 2, 2]],dtype=float)
 b = np.array([[4],[6],[5]],dtype=float)
 a = handle(a, b)

以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持。