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I'm performing a linear regression using recipes to predict salary based on rank (assoc professor, assistant professor, and full professor), sex, discipline (applied or theoretical), years of service, and years since PhD. This data set is in the cars package.

I've created dummy variables, and transformed the dependent outcome variable into a more normal shape. I've standardized years of service and years since PhD into values between 0 and 1. 

salary.split <- initial_split(salary.df)

sal.train <- training(salary.split)
sal.test <- testing(salary.split)

sal.recipe <- recipe(salary ~ ., data = salary.df) %>% 
  step_log(salary) %>% 
  step_dummy(all_nominal()) %>% 
  step_range(yrs.since.phd) %>% 
  step_range(yrs.service)


sal.rec <- prep(sal.recipe, training = sal.train) %>% bake(new_data = sal.train)

sal.lm <- lm(sal.rec)

summary(sal.lm)

The results of the summary: 

Call:
lm(formula = sal.rec)

Residuals:
     Min       1Q   Median       3Q      Max 
-0.17727 -0.05780 -0.01406  0.04221  0.34499 

Coefficients:
                Estimate Std. Error t value Pr(>|t|)    
(Intercept)   -0.3052564  0.3240025  -0.942  0.34690    
yrs.service    0.8054404  0.0292577  27.529  < 2e-16 ***
salary         0.0375859  0.0285323   1.317  0.18877    
rank_AsstProf -0.0528260  0.0184926  -2.857  0.00459 ** 
rank_Prof      0.0740925  0.0174977   4.234 3.08e-05 ***
discipline_B  -0.0438070  0.0107863  -4.061 6.28e-05 ***
sex_Male       0.0006626  0.0165779   0.040  0.96815    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.08639 on 291 degrees of freedom
Multiple R-squared:  0.8656,    Adjusted R-squared:  0.8628 
F-statistic: 312.2 on 6 and 291 DF,  p-value: < 2.2e-16

When I look at the variable information (sal.recipe$var_info):

# A tibble: 6 x 4
  variable      type    role      source  
  <chr>         <chr>   <chr>     <chr>   
1 rank          nominal predictor original
2 discipline    nominal predictor original
3 yrs.since.phd numeric predictor original
4 yrs.service   numeric predictor original
5 sex           nominal predictor original
6 salary        numeric outcome   original

which shows salary as an outcome, not a predictor. Why is salary showing up as a coefficient when I look at the summary information for the linear model?

Winston Underwood
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    Because you haven't provided a formula for your model when using `lm()`. Passing an entire dataframe is interpreted as `lm(V1 ~ all + other + vars + in + dataframe)`. – Ritchie Sacramento May 18 '19 at 04:17

1 Answers1

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In this instance, sal.recipe is not a model object (like I thought) but instead is just a dataframe. Recipes are used to apply transformations to the data, and don't automatically construct the formula. In this case, passing sal.recipe to lm is simply passing a dataframe, which is what causes the confusion.

Winston Underwood
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