Python实现最小二乘法 Python数据拟合实现最小二乘法示例解析

#文件名core.py
import numpy as np
def leastSquare(x,y):
    if len(x)==2:
    #此时x为自然序列
        sx = 0.5*(x[1]-x[0]+1)*(x[1]+x[0])
        ex = sx/(x[1]-x[0]+1)
        sx2 = ((x[1]*(x[1]+1)*(2*x[1]+1))
              -(x[0]*(x[0]-1)*(2*x[0]-1)))/6
        x = np.array(range(x[0],x[1]+1))
    else:
        sx = sum(x)
        ex = sx/len(x)
        sx2 = sum(x**2)    
    sxy = sum(x*y)
    ey = np.mean(y)
    a = (sxy-ey*sx)/(sx2-ex*sx)
    b = (ey*sx2-sxy*ex)/(sx2-ex*sx)
    return a,b

>>> x = np.arange(25)
>>> y = x*15+20+np.random.randn(len(x))*5	#randn生成正态分布噪声
>>> a,b = core.leastSquare(x,y)				
>>> plt.scatter(x,y)						#原始数据散点图
<matplotlib.collections.PathCollection object at 0x00000218DEBBEDC8>
>>> plt.plot(x,a*x+b)						#拟合直线
[<matplotlib.lines.Line2D object at 0x00000218E0314FC8>]
>>> plt.show()

#传入参数格式为np.array,n为阶数
def leastSquareMulti(x,y,n):
    X = [np.sum(x**i) for i in range(2*n+1)]
    Y = np.array([[np.sum(y*x**i)] for i in range(n+1)])
    S = np.array([X[i:i+n+1] for i in range(n+1)])
    return np.linalg.solve(S,Y)		#

>>> x = np.arange(25)
>>> y = x**3+3*x**2+2*x+12
>>> import core
>>> core.leastSquareMulti(x,y,3)
array([[12.],		#此为常数项
       [ 2.],
       [ 3.],
       [ 1.]])

def expFit(x,y):
    y0 = y[0:-3]
    y1 = y[1:-2]
    y2 = y[2:-1]
    B,C = leastSquare(y2/y0,y1/y0)
    b1 = np.log((B-np.sqrt(B**2+4*C))/2)
    b2 = np.log((B+np.sqrt(B**2+4*C))/2)
    X = np.exp(b1-b2)*x
    Y = y/np.exp(b2*x)
    a1,a2 = leastSquare(X,Y)
    return a1,a2,b1,b2