Function to regress chosen variable against all others - r

Question

In my dataset I have 6 variables(x1,x2,x3,x4,x5,x6), i wish to create a function that allows me to input one variable and it will do the formula with the rest of the variables in the data set.
For instance,
fitRegression <- function(data, dependentVariable) {
fit = lm(formula = x1 ~., data = data1)
return(fit)
}
fitRegression(x2)
However, this function only returns me with results of x1. My desire result will be inputting whatever variables and will automatically do the formula with the rest of the variables.
For Example:
fitRegression(x2)
should subtract x2 from the variable list therefore we only compare x2 with x1,x3,x4,x5,x6.
and if:
fitRegression(x3)
should subtract x3 from the comparable list, therefore we compare x3 with x1,x2,x4,x5,x6.
Is there any ways to express this into my function, or even a better function.

Answer 1

You can do it like this:
# sample data
sampleData <- data.frame(matrix(rnorm(500),100,5))
colnames(sampleData) <- c("A","B","C","D","E")
# function
fitRegression <- function(mydata, dependentVariable) {
# select your independent and dependent variables
dependentVariableIndex<-which(colnames(mydata)==dependentVariable)
independentVariableIndices<-which(colnames(mydata)!=dependentVariable)
fit = lm(formula = as.formula(paste(colnames(mydata)[dependentVariableIndex], "~", paste(colnames(mydata)[independentVariableIndices], collapse = "+"), sep = "" )), data = mydata)
return(fit)
}
# ground truth
lm(formula = A~B+C+D+E, data = sampleData)
# reconcile results
fitRegression(sampleData, "A")

Answer 2

You want to select the Y variable in your argument. The main difficulty is to pass this argument without any quotes in your function (it is apparently the expected result in your code). Therefore you can use this method, using the combination deparse(substitute(...)):
fitRegression <- function(data, dependentVariable) {
formula <- as.formula(paste0(deparse(substitute(dependentVariable)), "~."))
return(lm(formula, data) )
}
fitRegression(mtcars, disp)
That will return the model.

Answer 3

The below function uses "purrr" and "caret" it produces a list of models.
df <-mtcars
library(purrr);library(caret)
#create training set
vect <- createDataPartition(1:nrow(df), p=0.8, list = FALSE)
#build model list
ModList <- 1:length(df) %>%
map(function(col) train(y= df[vect,col], x= df[vect,-col], method="lm"))