Fitting a linear regression model in R

Question

I have a question regarding linear regression analysis in R:

I have several independent variables (about 20-30) and one dependent variable. To reach the best model, I try "all" relevant combinations of independent variables in order to maximize my adjusted R^2. However, this is a lot of work. So my question is: Is there a way to automatically fit a regression model in R, i.e. an automatic selection of these independent variables stored in a data frame, which yield the best description of the variation in the dependent variable?

Thank you for your help!

You should try stepwise regression, which selects the best combination by different methods - forward selection or backward selection. Moreover, you can try using regularization. — adomasb
This question deserves a long, detailed discussion of the merits of doing model selection and the best way to go about it. For a short answer check out the help to step(), the package leaps, and better still, the method called lasso regression (package glmnet) — ilir

Roman Luštrik Roman Luštrik · Accepted Answer · 2014-10-13T11:27:50

You can use step function, however analysis done with this approach may hit some bumps on the road if whoever is checking your work is against data dredging. Here is an example from step.

> summary(lm1 <- lm(Fertility ~ ., data = swiss))

Call:
lm(formula = Fertility ~ ., data = swiss)

Residuals:
     Min       1Q   Median       3Q      Max 
-15.2743  -5.2617   0.5032   4.1198  15.3213 

Coefficients:
                 Estimate Std. Error t value Pr(>|t|)    
(Intercept)      66.91518   10.70604   6.250 1.91e-07 ***
Agriculture      -0.17211    0.07030  -2.448  0.01873 *  
Examination      -0.25801    0.25388  -1.016  0.31546    
Education        -0.87094    0.18303  -4.758 2.43e-05 ***
Catholic          0.10412    0.03526   2.953  0.00519 ** 
Infant.Mortality  1.07705    0.38172   2.822  0.00734 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 7.165 on 41 degrees of freedom
Multiple R-squared:  0.7067,    Adjusted R-squared:  0.671 
F-statistic: 19.76 on 5 and 41 DF,  p-value: 5.594e-10

> slm1 <- step(lm1)
Start:  AIC=190.69
Fertility ~ Agriculture + Examination + Education + Catholic + 
    Infant.Mortality

                   Df Sum of Sq    RSS    AIC
- Examination       1     53.03 2158.1 189.86
<none>                          2105.0 190.69
- Agriculture       1    307.72 2412.8 195.10
- Infant.Mortality  1    408.75 2513.8 197.03
- Catholic          1    447.71 2552.8 197.75
- Education         1   1162.56 3267.6 209.36

Step:  AIC=189.86
Fertility ~ Agriculture + Education + Catholic + Infant.Mortality

                   Df Sum of Sq    RSS    AIC
<none>                          2158.1 189.86
- Agriculture       1    264.18 2422.2 193.29
- Infant.Mortality  1    409.81 2567.9 196.03
- Catholic          1    956.57 3114.6 205.10
- Education         1   2249.97 4408.0 221.43
> summary(slm1)

Call:
lm(formula = Fertility ~ Agriculture + Education + Catholic + 
    Infant.Mortality, data = swiss)

Residuals:
     Min       1Q   Median       3Q      Max 
-14.6765  -6.0522   0.7514   3.1664  16.1422 

Coefficients:
                 Estimate Std. Error t value Pr(>|t|)    
(Intercept)      62.10131    9.60489   6.466 8.49e-08 ***
Agriculture      -0.15462    0.06819  -2.267  0.02857 *  
Education        -0.98026    0.14814  -6.617 5.14e-08 ***
Catholic          0.12467    0.02889   4.315 9.50e-05 ***
Infant.Mortality  1.07844    0.38187   2.824  0.00722 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 7.168 on 42 degrees of freedom
Multiple R-squared:  0.6993,    Adjusted R-squared:  0.6707 
F-statistic: 24.42 on 4 and 42 DF,  p-value: 1.717e-10

Fitting a linear regression model in R

1 Answers