我从 R 和 Python 得到不同的多项式回归系数。
R
X <- c(0,0, 10, 10, 20, 20)
Y <- c(5, 7, 15, 17, 9, 11)
fm1 <- lm(Y~X+I(X^2))
summary(fm1)
Call:
lm(formula = Y ~ X + I(X^2))
Residuals:
1 2 3 4 5 6
-1 1 -1 1 -1 1
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 6.00000 1.00000 6.000 0.00927 **
X 1.80000 0.25495 7.060 0.00584 **
I(X^2) -0.08000 0.01225 -6.532 0.00729 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.414 on 3 degrees of freedom
Multiple R-squared: 0.9441, Adjusted R-squared: 0.9068
F-statistic: 25.33 on 2 and 3 DF, p-value: 0.01322
anova(fm1)
Analysis of Variance Table
Response: Y
Df Sum Sq Mean Sq F value Pr(>F)
X 1 16.000 16.000 8.000 0.066276 .
I(X^2) 1 85.333 85.333 42.667 0.007292 **
Residuals 3 6.000 2.000
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
python
Nitro = [0, 0, 10, 10, 20, 20]
Yield = [5, 7, 15, 17, 9, 11]
import pandas as pd
df3 = pd.DataFrame(
{
"Nitrogen": Nitro,
"Yield": Yield
}
)
from statsmodels.formula.api import ols
from statsmodels.stats.anova import anova_lm
Reg3 = ols("Yield ~ Nitrogen + I(Nitrogen^2)", data = df3)
Fit3 = Reg3.fit()
print(Fit3.summary())
OLS Regression Results
==============================================================================
Dep. Variable: Yield R-squared: 0.944
Model: OLS Adj. R-squared: 0.907
Method: Least Squares F-statistic: 25.33
Date: Fri, 27 Jul 2018 Prob (F-statistic): 0.0132
Time: 19:25:22 Log-Likelihood: -8.5136
No. Observations: 6 AIC: 23.03
Df Residuals: 3 BIC: 22.40
Df Model: 2
Covariance Type: nonrobust
===================================================================================
coef std err t P>|t| [0.025 0.975]
-----------------------------------------------------------------------------------
Intercept 10.0000 0.935 10.690 0.002 7.023 12.977
Nitrogen 2.2000 0.314 7.001 0.006 1.200 3.200
I(Nitrogen ^ 2) -2.0000 0.306 -6.532 0.007 -2.974 -1.026
==============================================================================
Omnibus: nan Durbin-Watson: 3.333
Prob(Omnibus): nan Jarque-Bera (JB): 1.000
Skew: 0.000 Prob(JB): 0.607
Kurtosis: 1.000 Cond. No. 30.4
==============================================================================
print(anova_lm(Fit3))
df sum_sq mean_sq F PR(>F)
Nitrogen 1.0 16.000000 16.000000 8.000000 0.066276
I(Nitrogen ^ 2) 1.0 85.333333 85.333333 42.666667 0.007292
Residual 3.0 6.000000 2.000000 NaN NaN
问题
- 为什么在 R 和 Python 中得到不同的回归系数？
请您参考如下方法:
在 python 中 ^ 是按位或运算符。你想要一个指数。尝试
Reg3 = ols("Yield ~ Nitrogen + I(Nitrogen**2)", data = df3)
