How to reference objects in a different environment inside a formula - r

Question

I'm using fixest::feols() and I have a function I want to pass an argument to in order to subset the data using the subset = argument. However when keep getting the error: The argument 'subset' is a formula whose variables must be in the data set given in argument 'data'.
I have tried the following code:
library(fixest)
cars <- mtcars
my_fun <- function(data, hp.c.off) {
feols(mpg ~ disp + drat,
data = data,
subset = ~ hp > substitute(hp.c.off))
}
my_fun(data = cars, 150)
My expected outcome would be the same as if one typed:
feols(mpg ~ disp + drat,
data = cars,
subset = ~ hp > 150)
I know I have to replace the value of hp.c.off before passing it onto a formula. And one could do this by creating a string expression first and then using as.formula() however, I was wondering if there is a better way to do programmatically build the expression that didn't require creating a string expression first and then converting it into a formula.
Thanks!

Answer 1

1) Create the formula as a character string and then convert it to a formula.
my_fun <- function(data, hp.c.off) {
feols(mpg ~ disp + drat,
data = data,
subset = as.formula(paste("~ hp >", hp.c.off)))
}
2) or just don't use the subset= argument and instead use the data argument with subset.
my_fun <- function(data, hp.c.off) {
feols(mpg ~ disp + drat,
data = subset(data, hp > hp.c.off))
}
3) or use the fact that subset= can be a logical vector
my_fun <- function(data, hp.c.off) {
feols(mpg ~ disp + drat,
data = data,
subset = data$hp > hp.c.off)
}

Answer 2

You can use rlang::new_formula(), with rlang::expr() to quote the rhs and !!rlang::enexpr() to capture and inject the hp.c.off argument.
I don’t have fixest installed, but this demonstrates building the formula inside a function:
library(rlang)
cars <- mtcars
my_fun <- function(data, hp.c.off) {
new_formula(lhs = NULL, rhs = expr(hp > !!enexpr(hp.c.off)))
}
my_fun(data = cars, 150)
# ~hp > 150
# <environment: 0x1405e38>

Answer 3

Simple option is to pass an expression as argument to the function
my_fun <- function(data,expr = ~ hp > 150){
feols(mpg ~ disp + drat,
data = data,
subset = expr)
}
-testing
> my_fun(data = cars)
OLS estimation, Dep. Var.: mpg
Observations: 13
Standard-errors: IID
Estimate Std. Error t value Pr(>|t|)
(Intercept) 23.414923 8.019808 2.919636 0.015310 *
disp -0.021349 0.008284 -2.577276 0.027545 *
drat -0.201284 2.014207 -0.099932 0.922373
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
RMSE: 2.16851 Adj. R2: 0.300667