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python實現簡單線性迴歸

用python實現R的線性模型(lm)中一元線性迴歸的簡單方法,使用R的women示例資料,R的執行結果:

> 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實現的功能包括:
1. 計算pearson相關係數
2. 使用最小二乘法計算迴歸係數
3. 計算擬合優度判定係數R2
4. 計算估計標準誤差Se
5. 計算顯著性檢驗的F和P值

import numpy as np
import scipy.stats as ss


class Lm:
    """簡單一元線性模型,計算迴歸係數、擬合優度的判定係數和
    估計標準誤差,顯著性水平"""

    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 # 計算協方差 @staticmethod def cov_custom(x, y): result = sum((x - np.mean(x)) * (y - np.mean(y))) / (len(x) - 1) return result # 計算相關係數 @staticmethod def corr_custom(x, y): return Lm.cov_custom(x, y) / (np.std(x, ddof=1) * np.std(y, ddof=1)) # 計算迴歸係數 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 # 計算擬合優度的判定係數R方,即相關係數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 # 計算估計標準誤差 def estimate_deviation(self): y = np.array(self.yarr) self.se = np.sqrt(np.sum((y - self.yhat) ** 2) / self.dfd) return # 顯著性檢驗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執行結果:

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

其中求迴歸係數時用矩陣轉置求逆再用numpy內建的解線性方程組的方法是最快的:

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)