Evaluating the loop index as a variable name, rather than a string - r

Question

I would like to evaluate a string, eg x1, where x1 <- "disp", as the underlying value, i.e. disp, when x1 is the loop index.
A reproducible example, using the mtcars dataset as an example is below:
x1 <- "disp"
x2 <- "hp"
vars <- c("x1", "x2")
for (x in vars){
print(x)
}
Which gives me
#> [1] "x1"
#> [1] "x2"
Desired Outcome:
What I'm trying to get is a loop that runs these commands:
print(x1)
print(x2)
resulting in:
#> [1] "disp"
#> [1] "hp"
I recognise that the simplest solution would be to bypass x1 and x2 completely:
vars <- c("disp", "hp")
for (x in vars){
print(x)
}
But that's less helpful, as it will be very helpful to have x1, x2, etc, in my (unsimplified) problem.
Also, if purrr is a better way to do something like this, instead of a loop, I'd be very interested to understand that better.
If anyone has a suggestion on a better title for the question, I will also be very interested.
Deeper Question
I simplified my question above, hoping that would be enough to get what I needed, but for context, I'm trying to do something like this:
df <- mtcars
x1 <- "disp"
x2 <- "hp"
vars <- c("x1", "x2")
for (x in vars){
lm(mpg ~ x, data = mtcars)
}
Created on 2019-07-11 by the reprex package (v0.2.1)

Answer 1

The answer to your original question is to use get. However, since you want to do something beyond that and want to use vars as it is we can use get with as.formula
lst <- vector("list", length(vars))
for (x in seq_along(vars)) {
lst[[x]] <- lm(as.formula(paste0("mpg ~", get(vars[x]))), mtcars)
}
#[[1]]
#Call:
#lm(formula = as.formula(paste0("mpg ~", get(vars[.x]))), data = mtcars)
#Coefficients:
#(Intercept) disp
# 29.5999 -0.0412
#[[2]]
#Call:
#lm(formula = as.formula(paste0("mpg ~", get(vars[.x]))), data = mtcars)
#Coefficients:
#(Intercept) hp
# 30.0989 -0.0682
Using purrr you can do that with map
purrr::map(seq_along(vars), ~lm(as.formula(paste0("mpg ~", get(vars[.x]))), mtcars))

Answer 2

We can use lapply from base R and reformulate
lapply(mget(vars), function(x)
lm(reformulate(response = "mpg", termlabels = x), data = mtcars))
#$x1
#Call:
#lm(formula = reformulate(response = "mpg", termlabels = x), data = mtcars)
#Coefficients:
#(Intercept) disp
# 29.59985 -0.04122
#$x2
#Call:
#lm(formula = reformulate(response = "mpg", termlabels = x), data = mtcars)
#Coefficients:
#(Intercept) hp
# 30.09886 -0.06823

Answer 3

Already answered, but:
library(rlang)
library(tidyverse)
vars <- exprs(disp, hp) # without "character-quotes"
map(seq_along(vars), ~eval(expr(lm(mpg ~ !!vars[[.x]], mtcars))))
# or
vars <- c("disp", "hp")
map(vars, ~exec("lm", str_c("mpg ~ ", .x), data = mtcars))