Question

我从R和Python得到了多项式回归系数。

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中获得不同的回归系数？

Answer 1

在python中，^是按位OR运算符。您想要一个指数。尝试

Reg3 = ols("Yield ~ Nitrogen + I(Nitrogen**2)", data = df3)

使用R和Python的不同多项式回归系数

1 个答案: