本文实例讲述了Python实现的简单线性回归算法。分享给大家供大家参考,具体如下:
用python实现R的线性模型(lm)中一元线性回归的简单方法,使用R的women示例数据,R的运行结果:
> summary(fit)
Call:
lm(formula = weight ~ height, data = women)
Residuals:
Min 1Q Median 3Q Max
-1.7333 -1.1333 -0.3833 0.7417 3.1167
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -87.51667 5.93694 -14.74 1.71e-09 ***
height 3.45000 0.09114 37.85 1.09e-14 ***
---
Signif. codes: 0 ‘***' 0.001 ‘**' 0.01 ‘*' 0.05 ‘.' 0.1 ‘ ' 1
Residual standard error: 1.525 on 13 degrees of freedom
Multiple R-squared: 0.991, Adjusted R-squared: 0.9903
F-statistic: 1433 on 1 and 13 DF, p-value: 1.091e-14
python实现的功能包括:
- 计算pearson相关系数
- 使用最小二乘法计算回归系数
- 计算拟合优度判定系数R2R2
- 计算估计标准误差Se
- 计算显著性检验的F和P值
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import numpy as np
import scipy.stats as ss
class Lm:
"""简单一元线性模型,计算回归系数、拟合优度的判定系数和
估计标准误差,显著性水平"""
def __init__( self , data_source, separator):
self .beta = np.matrix(np.zeros( 2 ))
self .yhat = np.matrix(np.zeros( 2 ))
self .r2 = 0.0
self .se = 0.0
self .f = 0.0
self .msr = 0.0
self .mse = 0.0
self .p = 0.0
data_mat = np.genfromtxt(data_source, delimiter = separator)
self .xarr = data_mat[:, : - 1 ]
self .yarr = data_mat[:, - 1 ]
self .ybar = np.mean( self .yarr)
self .dfd = len ( self .yarr) - 2 # *度n-2
return
# 计算协方差
@staticmethod
def cov_custom(x, y):
result = sum ((x - np.mean(x)) * (y - np.mean(y))) / ( len (x) - 1 )
return result
# 计算相关系数
@staticmethod
def corr_custom(x, y):
return Lm.cov_custom(x, y) / (np.std(x, ddof = 1 ) * np.std(y, ddof = 1 ))
# 计算回归系数
def simple_regression( self ):
xmat = np.mat( self .xarr)
ymat = np.mat( self .yarr).T
xtx = xmat.T * xmat
if np.linalg.det(xtx) = = 0.0 :
print ( 'Can not resolve the problem' )
return
self .beta = np.linalg.solve(xtx, xmat.T * ymat) # xtx.I * (xmat.T * ymat)
self .yhat = (xmat * self .beta).flatten().A[ 0 ]
return
# 计算拟合优度的判定系数R方,即相关系数corr的平方
def r_square( self ):
y = np.mat( self .yarr)
ybar = np.mean(y)
self .r2 = np. sum (( self .yhat - ybar) * * 2 ) / np. sum ((y.A - ybar) * * 2 )
return
# 计算估计标准误差
def estimate_deviation( self ):
y = np.array( self .yarr)
self .se = np.sqrt(np. sum ((y - self .yhat) * * 2 ) / self .dfd)
return
# 显著性检验F
def sig_test( self ):
ybar = np.mean( self .yarr)
self .msr = np. sum (( self .yhat - ybar) * * 2 )
self .mse = np. sum (( self .yarr - self .yhat) * * 2 ) / self .dfd
self .f = self .msr / self .mse
self .p = ss.f.sf( self .f, 1 , self .dfd)
return
def summary( self ):
self .simple_regression()
corr_coe = Lm.corr_custom( self .xarr[:, - 1 ], self .yarr)
self .r_square()
self .estimate_deviation()
self .sig_test()
print ( 'The Pearson\'s correlation coefficient: %.3f' % corr_coe)
print ( 'The Regression Coefficient: %s' % self .beta.flatten().A[ 0 ])
print ( 'R square: %.3f' % self .r2)
print ( 'The standard error of estimate: %.3f' % self .se)
print ( 'F-statistic: %d on %s and %s DF, p-value: %.3e' % ( self .f, 1 , self .dfd, self .p))
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python执行结果:
The Regression Coefficient: [-87.51666667 3.45 ]
R square: 0.991
The standard error of estimate: 1.525
F-statistic: 1433 on 1 and 13 DF, p-value: 1.091e-14
其中求回归系数时用矩阵转置求逆再用numpy
内置的解线性方程组的方法是最快的:
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a = np.mat(women.xarr); b = np.mat(women.yarr).T
timeit (a.I * b)
99.9 µs ± 941 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
timeit ata.I * (a.T * b)
64.9 µs ± 717 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
timeit np.linalg.solve(ata, a.T * b)
15.1 µs ± 126 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
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希望本文所述对大家Python程序设计有所帮助。
原文链接:https://blog.csdn.net/qq_35753140/article/details/78699748