在python中利用最小二乘拟合二次抛物线函数的方法

#!/usr/bin/env python
# coding:utf-8


import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import leastsq


# 待拟合的数据
X = np.array([1,2,3,4,5,6])
Y=np.array([9.1,18.3,32,47,69.5,94.8])


# 二次函数的标准形式
def func(params, x):
 a, b, c = params
 return a * x * x + b * x + c


# 误差函数，即拟合曲线所求的值与实际值的差
def error(params, x, y):
 return func(params, x) - y


# 对参数求解
def slovePara():
 p0 = [10, 10, 10]

 Para = leastsq(error, p0, args=(X, Y))
 return Para


# 输出最后的结果
def solution():
 Para = slovePara()
 a, b, c = Para[0]
 print "a=",a," b=",b," c=",c
 print "cost:" + str(Para[1])
 print "求解的曲线是:"
 print("y="+str(round(a,2))+"x*x+"+str(round(b,2))+"x+"+str(c))

 plt.figure(figsize=(8,6))
 plt.scatter(X, Y, color="green", label="sample data", linewidth=2)

 # 画拟合直线
 x=np.linspace(0,12,100) ##在0-15直接画100个连续点
 y=a*x*x+b*x+c ##函数式
 plt.plot(x,y,color="red",label="solution line",linewidth=2)
 plt.legend() #绘制图例
 plt.show()


solution()

Minimize the sum of squares of a set of equations.
 x = arg min(sum(func(y)**2,axis=0))
   y

a= 2.06607141425 b= 2.5975001036 c= 4.68999985496
cost:1
求解的曲线是:
y=2.07x*x+2.6x+4.68999985496

def square_wave(x,p):
 a, b, c, T = p
 y = np.where(np.mod(x-b,T)<T/2, 1+c/a, 0)
 y = np.where(np.mod(x-b,T)>T/2, -1+c/a, y)
 return a*y

def gaussian_wave(x,p):
 a, b, c, d= p
 return a*np.exp(-(x-b)**2/(2*c**2))+d