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 npimport scipy.stats as ssclass 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 # 自由度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：
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a = np.mat(women.xarr); b = np.mat(women.yarr).Ttimeit (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) |
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
