Python实现的简单线性回归算法实例分析

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))

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)