Use quasiquotation for formula syntax in a user-created function? - r

Question

When I run this code:
# Create example data
df <- tibble(age=rnorm(10),
income=rnorm(10))
make_model <- function(response_var, df){
# Create formula
form <- as.formula(response_var ~ .)
# Create model
model <- lm(form , data=df)
# Return coefficients
return(coef(model))
}
make_model(income, df)
I obtain the following error
Error in eval(predvars, data, env) : object 'income' not found
How can I make this function work using quasiquotation? I assume the logic is the same as how we can call library(dplyr) instead of library("dplyr").

Answer 1

Use blast() (to be included in rlang 0.5.0)
blast <- function(expr, env = caller_env()) {
eval_bare(enexpr(expr), env)
}
make_model <- function(data, column) {
f <- blast(!!enexpr(column) ~ .)
model <- lm(f, data = data)
coef(model)
}
df <- data.frame(
age = rnorm(10),
income = rnorm(10)
)
make_model(df, income)
#> (Intercept) age
#> -0.3563103 -0.2200773
Works flexibly:
blast(list(!!!1:3))
#> [[1]]
#> [1] 1
#>
#> [[2]]
#> [1] 2
#>
#> [[3]]
#> [1] 3

Answer 2

The following should work:
library(tidyverse)
# Your original function, modified
make_model <- function(df, column) {
column <- enexpr(column)
form <- as.formula(paste0(quo_text(column), " ~ ."))
model <- lm(form, data = df)
return(coef(model))
}
# Your original data and call
tibble(
age = rnorm(10),
income = rnorm(10)
) %>%
make_model(income)