Dynamically insert variables into a fable model using rlang - r

Question

I am trying to dynamically insert variables into a fable model.
Data
library(dplyr)
library(fable)
library(stringr)
df <- tsibbledata::aus_retail %>%
filter(State == "Victoria", Industry == "Food retailing") %>%
mutate(reg_test = rnorm(441, 5, 2),
reg_test2 = rnorm(441, 5, 2))
Note that there can be an undetermined number of regressors included in the tsibble, but in this example, I have only two (reg_test and reg_test2). All regressor columns will start with reg_
Problem Function
I have a function where I want to dynamically put the regressor columns into an ARIMA model using the fable package.
test_f <- function(df) {
var_names <- str_subset(names(df), "reg_") %>%
paste0(collapse = "+")
test <- enquo(var_names)
df %>%
model(ARIMA(Turnover ~ !!test))
}
test_f(df)
# A mable: 1 x 3
# Key: State, Industry [1]
State Industry `ARIMA(Turnover ~ ~"reg_test+reg_tes~
<chr> <chr> <model>
1 Victoria Food retaili~ <NULL model>
Warning message:
1 error encountered for ARIMA(Turnover ~ ~"reg_test+reg_test2")
[1] invalid model formula in ExtractVars
I know that it is just putting the string var_names into the formula, which does not work, but I can't figure out how to create var_names in such a way that I can enquo() it correctly.
I read through the Quasiquotation section here
I searched SO but have not found the answer yet.
This question with pasre_expr() seemed to get closer, but still not what I wanted.
I know that I can use sym() if I have one variable, but I don't know how many reg_ variables there will be and I want to include them all.
Expected Output
By putting in the variables manually, I can show the output that I expect.
test <- df %>%
model(ARIMA(Turnover ~ reg_test + reg_test2))
test$`ARIMA(Turnover ~ reg_test + reg_test2)`[[1]]
Series: Turnover
Model: LM w/ ARIMA(2,1,0)(0,1,2)[12] errors
Coefficients:
ar1 ar2 sma1 sma2 reg_test reg_test2
-0.6472 -0.3541 -0.4115 -0.0793 -0.0296 -0.6143
s.e. 0.0473 0.0479 0.0520 0.0446 0.5045 0.5273
sigma^2 estimated as 884.9: log likelihood=-2058.04
AIC=4130.08 AICc=4130.35 BIC=4158.5
I also imagine that there is a better way for me to make the formula in the ARIMA function. If this can fix my problem as well, that will work too.
I appreciate any help!

Answer 1

You're possibly making this a bit more complicated than it needs to be. You can convert a string to a formula by doing as.formula(string), so simply build your formula as a string, convert it to a formula, then feed it to ARIMA. Here's a reprex:
library(dplyr)
library(fable)
library(stringr)
df <- tsibbledata::aus_retail %>%
filter(State == "Victoria", Industry == "Food retailing") %>%
mutate(reg_test = rnorm(441, 5, 2),
reg_test2 = rnorm(441, 5, 2))
test_f <- function(df) {
var_names <- paste0(str_subset(names(df), "reg_"), collapse = " + ")
mod <- model(df, ARIMA(as.formula(paste("Turnover ~", var_names))))
unclass(mod[1, 3][[1]])[[1]]
}
test_f(df)
#> Series: Turnover
#> Model: LM w/ ARIMA(2,1,0)(0,1,1)[12] errors
#>
#> Coefficients:
#> ar1 ar2 sma1 reg_test reg_test2
#> -0.6689 -0.376 -0.4765 0.3363 1.0194
#> s.e. 0.0448 0.045 0.0426 0.4978 0.5436
#>
#> sigma^2 estimated as 883.1: log likelihood=-2058.28
#> AIC=4128.56 AICc=4128.76 BIC=4152.91
Created on 2020-04-23 by the reprex package (v0.3.0)