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
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import numpy as np
import gaosi as gs
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代码
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"""
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)
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高斯模块
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# 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)
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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服务器之家。
原文链接：https://blog.csdn.net/weixin_45779228/article/details/109318170