R: subsetting within a function - r

Question

Suppose I have a data frame in the environment, mydata, with three columns, A, B, C.
mydata = data.frame(A=c(1,2,3),
B=c(4,5,6),
C=c(7,8,9))
I can create a linear model with
lm(C ~ A, data=mydata)
I want a function to generalize this, to regress B or C on A, given just the name of the column, i.e.,
f = function(x){
lm(x ~ A, data=mydata)
}
f(B)
f(C)
or
g = function(x){
lm(mydata$x ~ mydata$A)
}
g(B)
g(C)
These solutions don't work. I know there is something wrong with the evaluation, and I have tried permutations of quo() and enquo() and !!, but no success.
This is a simplified example, but the idea is, when I have dozens of similar models to build, each fairly complicated, with only one variable changing, I want to do so without repeating the entire formula each time.

Answer 1

If we want to pass unquoted column name, and option is {{}} from tidyverse. With select, it can take both string and unquoted
library(dplyr)
printcol2 <- function(data, x) {
data %>%
select({{x}})
}
printcol2(mydata, A)
# A
#1 1
#2 2
#3 3
printcol2(mydata, 'A')
# A
#1 1
#2 2
#3 3
If the OP wanted to pass unquoted column name to be passed in lm
f1 <- function(x){
rsp <- deparse(substitute(x))
fmla <- reformulate("A", response = rsp)
out <- lm(fmla, data=mydata)
out$call <- as.symbol(paste0("lm(", deparse(fmla), ", data = mydata)"))
out
}
f1(B)
#Call:
#lm(B ~ A, data = mydata)
#Coefficients:
#(Intercept) A
# 3 1
f1(C)
#Call:
#lm(C ~ A, data = mydata)
#Coefficients:
#(Intercept) A
# 6 1

Answer 2

Maybe you are looking for deparse(substitute(.)). It accepts arguments quoted or not quoted.
f = function(x, data = mydata){
y <- deparse(substitute(x))
fmla <- paste(y, 'Species', sep = '~')
lm(as.formula(fmla), data = data)
}
mydata <- iris
f(Sepal.Length)
#
#Call:
#lm(formula = as.formula(fmla), data = data)
#
#Coefficients:
# (Intercept) Speciesversicolor Speciesvirginica
# 5.006 0.930 1.582
f(Petal.Width)
#
#Call:
#lm(formula = as.formula(fmla), data = data)
#
#Coefficients:
# (Intercept) Speciesversicolor Speciesvirginica
# 0.246 1.080 1.780

Answer 3

I think generally, you might be looking for:
printcol <- function(x){
print(x)
}
printcol(mydata$A)
This doesn't involve any fancy evaluation, you just need to specify the variable you'd like to subset in your function call.
This gives us:
[1] 1 2 3
Note that you're only printing the vector A, and not actually subsetting column A from mydata.