python 分别用梯度下降法和最小二乘法求线性回归

#coding=utf-8
 
import math
import matplotlib.pyplot as plt
import numpy as np
'''
author 
date          2013-12-20
description   分别用最小二乘法和梯度下降法求线性回归
'''
'''
梯度下降法
'''
def grediant(xset,yset,alpha,dgree):
    theta_0 = 0;
    theta_1 = 1;
    length = len(xset)
    jtheta = 0
    total = 0
    sum_total = 0
    for index in range(0,length):
        print index
        total = total + theta_0 + theta_1 * xset[index] - yset[index]
        tmp0 = theta_0  - alpha/length*(total)
        tmp1 = theta_1  -  alpha/length*(total)*xset[index]
        theta_0 = tmp0
        theta_1 = tmp1
        sum_total = sum_total + theta_0 + theta_1 * xset[index] - yset[index]
        jtheta_1 = 0.5 * length * math.pow(sum_total,2)
        comp = math.fabs(jtheta_1 - jtheta)
        print "%10.5f   %10.5f  %10.5f  %10.5f" %(comp,jtheta * dgree,theta_0,theta_1)
        if comp < jtheta * dgree:break
        jtheta = jtheta_1
    return theta_0,theta_1 
'''
最小二乘法
'''
def linefit( x,y):
    N = len(x)
    sx,sy,sxx,syy,sxy=0,0,0,0,0
    for i in range(0,N):
        sx  += x[i]
        sy  += y[i]
        sxx += x[i]*x[i]
        syy += y[i]*y[i]
        sxy += x[i]*y[i]
    a = (sy*sx/N -sxy)/( sx*sx/N -sxx)
    b = (sy - a*sx)/N
    r = abs(sy*sx/N-sxy)/math.sqrt( (sxx-sx*sx/N)*(syy-sy*sy/N))
    return a,b,r      
X=[1.5,2,1.5,2,3,3,3.5,3.5,4,4,5,5]
Y=[3,3.2,4,4.5,4,5,4.2,4.5,5,5.5,4.8,6.5]
#a,b = grediant(X,Y,0.0005,0.3)
a,b,r = linefit(X,Y)
print "y = %10.5fx + %10.5f" %(b,a)
x = np.linspace(0, 10, 10000)
#y = b * x + a
y = a * x + b
plt.plot(x,y)
plt.scatter(X, Y)
plt.show()