Python实现的简单线性回归算法实例分析

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：

1. 计算pearson相关系数
2. 使用最小二乘法计算回归系数
3. 计算拟合优度判定系数r2r2
4. 计算估计标准误差se
5. 计算显著性检验的f和p值

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))

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：

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。