How to dynamically reference datasets in function call of linear regression

1
votes

Let's say I have a function like this:

data("mtcars")
ncol(mtcars)

test <- function(string){
      fit <- lm(mpg ~ cyl,
                     data = string)
      return(fit)
}

I'd like to be able to have the "string" variable evaluated as the dataset for a linear regression like so:

test("mtcars")

However, I get an error:

Error in eval(predvars, data, env) : invalid 'envir' argument of type 'character'

I've tried using combinations of eval and parse, but to no avail. Any ideas?

2
rlinear-regression

2 Answers

2
votes

You can use get() to search by name for an object.

test <- function(string){
  fit <- lm(mpg ~ cyl, data = get(string))
  return(fit)
}

test("mtcars")

# Call:
# lm(formula = mpg ~ cyl, data = get(string))
# 
# Coefficients:
# (Intercept)          cyl  
#      37.885       -2.876

You can add one more line to make the output look better. Notice the change of the Call part in the output. It turns from data = get(string) to data = mtcars.

test <- function(string){
  fit <- lm(mpg ~ cyl, data = get(string))
  fit$call$data <- as.name(string)
  return(fit)
}

test("mtcars")

# Call:
# lm(formula = mpg ~ cyl, data = mtcars)
# 
# Coefficients:
# (Intercept)          cyl  
#      37.885       -2.876
1
votes

Try this slight change to your code:

#Code
test <- function(string){
  fit <- lm(mpg ~ cyl,
            data = eval(parse(text=string)))
  return(fit)
}
#Apply
test("mtcars")

Output:

Call:
lm(formula = mpg ~ cyl, data = eval(parse(text = string)))

Coefficients:
(Intercept)          cyl  
     37.885       -2.876