paste formula in function (c constant?) - r

Question

I'm doing a function pasting a formula and then returning a feols result. But, I get a c at the beginning. How can I solve this?
library(dplyr)
library(fixest)
data(base_did)
base_did = base_did %>% mutate(D = 5*rnorm(1080),
x2 = 10*rnorm(1080),
rand_wei = abs(rnorm(1080)))
f <- function(data, arg=NULL){
arg = enexpr(arg)
if (length(arg) == 0) {
formula = "D ~ 1"
}
else {
formula = paste(arg, collapse = " + ")
formula = paste("D ~ ", formula, sep = "")
}
formula = paste(formula, " | id + period", sep = "")
denom.lm <- feols(as.formula(formula), data = data,
weights = abs(data$rand_wei))
return(denom.lm)
}
f(base_did, arg = c(x1,x2))
#Error in feols(as.formula(formula), data = data, weights = abs(data$rand_wei)) :
# Evaluation of the right-hand-side of the formula raises an error:
# In NULL: Evaluation of .Primitive("c") returns an object of length 1
#while the data set has 1080 rows.
If I return(formula) at the end. I get [1] "D ~ c + x1 + x2 | id + period".
But I need only D ~ x1 + x2 | id + period.

Answer 1

Perhaps one option to make your function work would be to pass the arguments via ... so that c is not needed and which would prevent the c to be added to your formula. To make this work you also have switch to enexprs inside your function.
Note: I slightly adjusted your function for the reprex to return just the formula.
library(dplyr, warn = FALSE)
library(fixest)
data(base_did)
base_did = base_did %>% mutate(D = 5*rnorm(1080),
x2 = 10*rnorm(1080),
rand_wei = abs(rnorm(1080)))
f <- function(data, ...){
arg = enexprs(...)
if (length(arg) == 0) {
formula = "D ~ 1"
}
else {
formula = paste(arg, collapse = " + ")
formula = paste("D ~ ", formula, sep = "")
}
formula = paste(formula, " | id + period", sep = "")
as.formula(formula)
}
f(base_did, x1, x2)
#> D ~ x1 + x2 | id + period
#> <environment: 0x7fe8f3567618>
f(base_did)
#> D ~ 1 | id + period
#> <environment: 0x7fe8f366f848>
UPDATE There is probably a better approach but after some research a possible option would be:
Note: When passing multiple arguments via c enexpr will return a call object which behaves like a list and where the first element contains the function name, i.e. c. That's why you get the c added to your formula.
f <- function(data, arg = NULL) {
arg <- enexpr(arg)
if (length(arg) == 0) {
formula <- "D ~ 1"
} else {
if (length(arg) > 1) arg <- vapply(as.list(arg[-1]), rlang::as_string, FUN.VALUE = character(1))
formula <- paste(arg, collapse = " + ")
formula <- paste("D ~ ", formula, sep = "")
}
formula <- paste(formula, " | id + period", sep = "")
as.formula(formula)
}
f(base_did, c(x1, x2))
#> D ~ x1 + x2 | id + period
#> <environment: 0x7fa763431388>
f(base_did, x1)
#> D ~ x1 | id + period
#> <environment: 0x7fa763538c40>
f(base_did)
#> D ~ 1 | id + period
#> <environment: 0x7fa765e22028>

Answer 2

You can tremendously simplify your function if you use fixest's built-in formula manipulation tools (see here). In particular the dot-square-bracket operator:
library(fixest)
data(base_did)
n = 1080
base = within(base_did, {
D = 5 * rnorm(n)
x2 = 10 *rnorm(n)
rand_wei = abs(rnorm(n))
})
f = function(data, ctrl = "1"){
feols(D ~ .[ctrl] | id + period,
data = data, weights = ~abs(rand_wei))
}
est1 = f(base)
est2 = f(base, ~x1) # with a formula
est3 = f(base, c("x1", "x2")) # with a character vector
etable(est1, est2, est3)
#> est1 est2 est3
#> Dependent Var.: D D D
#>
#> x1 0.0816 (0.0619) 0.0791 (0.0618)
#> x2 -0.0157 (0.0186)
#> Fixed-Effects: -------- --------------- ----------------
#> id Yes Yes Yes
#> period Yes Yes Yes
#> _______________ ________ _______________ ________________
#> S.E. type -- by: id by: id
#> Observations 1,080 1,080 1,080
#> R2 0.12810 0.13005 0.13094
#> Within R2 -- 0.00224 0.00326
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
EDIT: note that the formula thing (est2) only works with version >= 0.11.0.