效果挺好的！

不知道你们结果如何出来的!]本人的很正常!!
用对数的应变量,RMSE占平均数的14%, R2=68%
用原数,RMSE占平均数的41%,但R2=72%
!!

随机森林的结果:Type of random forest: regression
                     Number of trees: 3000
No. of variables tried at each split: 1

          Mean of squared residuals: 0.2355287
                    % Var explained: 68.3
线性回归结果:
Call:
lm(formula = lOzone ~ Solar.R + Wind + Temp, data = myda)

Residuals:
     Min       1Q   Median       3Q      Max
-2.06193 -0.29970 -0.00231  0.30756  1.23578

Coefficients:
              Estimate Std. Error t value Pr(>|t|)   
(Intercept) -0.2621323  0.5535669  -0.474 0.636798   
Solar.R      0.0025152  0.0005567   4.518 1.62e-05 ***
Wind        -0.0615625  0.0157130  -3.918 0.000158 ***
Temp         0.0491711  0.0060875   8.077 1.07e-12 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.5086 on 107 degrees of freedom
Multiple R-squared:  0.6644,    Adjusted R-squared:  0.655
F-statistic: 70.62 on 3 and 107 DF,  p-value: < 2.2e-16
二者结果是一致的.

最好的线性回归结果:与随机森林相当.对于此类简单的问题以用线性回归为好.

Call:
lm(formula = lOzone ~ Solar.R + Wind + Temp + I(Solar.R^2) +
    I(Wind^2), data = myda)

Residuals:
     Min       1Q   Median       3Q      Max
-1.85559 -0.25596  0.00228  0.31332  1.16265

Coefficients:
               Estimate Std. Error t value Pr(>|t|)   
(Intercept)   7.722e-01  6.357e-01   1.215 0.227208   
Solar.R       7.463e-03  2.323e-03   3.212 0.001749 **
Wind         -2.211e-01  5.885e-02  -3.757 0.000283 ***
Temp          4.194e-02  6.239e-03   6.722  9.6e-10 ***
I(Solar.R^2) -1.469e-05  6.736e-06  -2.181 0.031444 *  
I(Wind^2)     7.388e-03  2.589e-03   2.854 0.005208 **
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.4852 on 105 degrees of freedom
Multiple R-squared:  0.7002,    Adjusted R-squared:  0.686
F-statistic: 49.06 on 5 and 105 DF,  p-value: < 2.2e-16

请问这个结果的代码有吗？

% Var explained负数表示什么啊