How can I use `I()` with `paste0`? - r

Question

I have multiple dataframes and I would like to evaluate (multiple) different models on each. MWE
df1 <- data.frame(A3 = c(-5, 5, 1),
B3 = c(0, 10, 1))
df2 <- data.frame(A4 = c(5, 15, 1))
B4 = c(10, 20, 1))
myfun <- function(arg1, arg2){ # arg1 =1 or 2
if (arg2 == 1){
eqn <- paste0("A", arg1+2) ~ paste0("B", arg1+2) + I(as.name(paste0("B", arg1+2))^2)
} else {
eqn <- paste0("A", arg1+2) ~ paste0("B", arg1+2) + I(as.name(paste0("B", arg1+2))^2) +I(as.name(paste0("B", arg1+2))^3)
}
return (lm(formula = eqn, data = eval(as.name(paste0("df", arg1)))
)
)
}
For example if I run myfun(1,2) I would like to get lm(A4 ~ B4 + I(B4^2) + I(B4^3), data = df2). But whatever I try I get the following error message Error in (paste0("B", arg1 +2))^2 : non-numeric argument to binary operator. From what I read in ?I, I imagine this is because R isolates whatever is passed into I(), so it doesn't realize I am trying to transform a variable: is that what is going on, and is it something I can fix? Also, is there a better way to estimate multiple models quickly? All the similar questions I found used the same data.frame across models, while I have to account for the response (and predictor) variables coming from different dataframes for different models.

Answer 1

Maybe this is what you are looking for:
The issue is that your are doing a math operation on a string, i.e with (paste0("B", arg1 +2))^2 you try to square a string, that's why you get the error. Inytead you can simply glue you formula together as a string an d converted it to a formula via as.formula:
df1 <- data.frame(A3 = c(-5, 5, 1),
B3 = c(0, 10, 1))
df2 <- data.frame(A4 = c(5, 15, 1))
B4 = c(10, 20, 1)
myfun <- function(arg1, arg2){ # arg1 =1 or 2
if (arg2 == 1){
eqn <- paste0("A", arg1+2, " ~ B", arg1+2," + I(B", arg1+2, "^2)")
} else {
eqn <- paste0("A", arg1+2, " ~ B", arg1+2," + I(B", arg1+2, "^2) + I(B", arg1+2, "^3)")
}
return (lm(formula = as.formula(eqn), data = eval(as.name(paste0("df", arg1)))
)
)
}
myfun(2, 1)
#>
#> Call:
#> lm(formula = as.formula(eqn), data = eval(as.name(paste0("df",
#> arg1))))
#>
#> Coefficients:
#> (Intercept) B4 I(B4^2)
#> 0.84795 0.12281 0.02924

Answer 2

An option is also to construct the formula with glue
myfun <- function(arg1, arg2){
eqn <- switch(arg2,
`1` = glue::glue("A{arg1 + 2}~ B{arg1+2} + I(B{arg1+2}^2)"),
glue::glue("A{arg1 + 2}~ B{arg1+2}",
"+ I(B{arg1+2}^2) + I(B{arg1+2}^3)")
)
model <- lm(eqn, data = get(paste0('df', arg1), envir = .GlobalEnv))
model$call <- as.formula(eqn)
return(model)
}
myfun(2, 1)
#Call:
#A4 ~ B4 + I(B4^2)
#Coefficients:
#(Intercept) B4 I(B4^2)
# 0.84795 0.12281 0.02924