User-defined function with lapply function

Question

I'm attempting to establish a user-defined function that inputs predetermined variables (independent and dependent) from the active data frame. Let's take the example data frame df below looking at a coin toss outcome as a result of other recorded variables:

> df
  outcome toss    person  hand age
1       H    1      Mary  Left  18
2       T    2     Allen  Left  12
3       T    3       Dom  Left  25
4       T    4 Francesca  Left  42
5       H    5      Mary Right  18
6       H    6     Allen Right  12
7       H    7       Dom Right  25
8       T    8 Francesca Right  42

The df data frame has a binomial response outcome being either heads or tails and I am going to look at how person , hand , and age might affect this categorical outcome. I plan to use a forward-selection approach which will test one variable against toss and then progress to add more.

As to keep things simple, I want to be able to identify the response/dependent (eg, outcome ) and predictor/independent (eg, person , hand ) variables before my user-defined function as such:

> independent<-c('person','hand','age')
> dependent<-'outcome'

Then create my function using the lapply and glm functions:

> test.func<-function(some_data,the_response,the_predictors)
+ {
+     lapply(the_predictors,function(a)
+         {
+         glm(substitute(as.name(the_response)~i,list(i=as.name(a))),data=some_data,family=binomial)
+     })
+ }

Yet, when I attempt to run the function with the predetermined vectors, this occurs:

> test.func(df,dependent,independent)
Error in as.name(the_response) : object 'the_response' not found

My expected response would be the following:

models<-lapply(independent,function(x)
+ {
+     glm(substitute(outcome~i,list(i=as.name(x))),data=df,family=binomial)
+ })
> models
[[1]]

Call:  glm(formula = substitute(outcome ~ i, list(i = as.name(x))), 
    family = binomial, data = df)

Coefficients:
    (Intercept)        personDom  personFrancesca       personMary  
      1.489e-16       -1.799e-16        1.957e+01       -1.957e+01  

Degrees of Freedom: 7 Total (i.e. Null);  4 Residual
Null Deviance:      11.09 
Residual Deviance: 5.545    AIC: 13.55

[[2]]

Call:  glm(formula = substitute(outcome ~ i, list(i = as.name(x))), 
    family = binomial, data = df)

**End Snippet**

As you can tell, using lapply and glm , I have created 3 simple models without all of the extra work doing it individually. You may be asking why create a user-defined function when you have simple code right there? I plan to run a while or repeat loop and it will decrease clutter.

Thank you for your assistance

Answer 1

I know code only answers are deprecated but I thought you were almost there and could just use the nudge to use the formula function (and to include 'the_response in the substitution):

 test.func<-function(some_data,the_response,the_predictors)
 {
     lapply(the_predictors,function(a)
         {print(   form<- formula(substitute(resp~i,
                                             list(resp=as.name(the_response), i=as.name(a)))))
         glm(form, data=some_data,family=binomial)
     })
 }

Test:

> test.func(df,dependent,independent)
outcome ~ person
<environment: 0x7f91a1ba5588>
outcome ~ hand
<environment: 0x7f91a2b38098>
outcome ~ age
<environment: 0x7f91a3fad468>
[[1]]

Call:  glm(formula = form, family = binomial, data = some_data)

Coefficients:
    (Intercept)        personDom  personFrancesca       personMary  
      8.996e-17       -1.540e-16        1.957e+01       -1.957e+01  

Degrees of Freedom: 7 Total (i.e. Null);  4 Residual
Null Deviance:      11.09 
Residual Deviance: 5.545    AIC: 13.55

[[2]]

Call:  glm(formula = form, family = binomial, data = some_data)

#snipped