Formula with dynamic number of variables

See ?as.formula, e.g.:

factors <- c("factor1", "factor2")
as.formula(paste("y~", paste(factors, collapse="+")))
# y ~ factor1 + factor2

where factors is a character vector containing the names of the factors you want to use in the model. This you can paste into an lm model, e.g.:

set.seed(0)
y <- rnorm(100)
factor1 <- rep(1:2, each=50)
factor2 <- rep(3:4, 50)
lm(as.formula(paste("y~", paste(factors, collapse="+"))))

# Call:
# lm(formula = as.formula(paste("y~", paste(factors, collapse = "+"))))

# Coefficients:
# (Intercept)      factor1      factor2  
#    0.542471    -0.002525    -0.147433

An oft forgotten function is reformulate. From ?reformulate:

reformulate creates a formula from a character vector.

A simple example:

listoffactors <- c("factor1","factor2")
reformulate(termlabels = listoffactors, response = 'y')

will yield this formula:

y ~ factor1 + factor2

Although not explicitly documented, you can also add interaction terms:

listofintfactors <- c("(factor3","factor4)^2")
reformulate(termlabels = c(listoffactors, listofintfactors), 
    response = 'y')

will yield:

y ~ factor1 + factor2 + (factor3 + factor4)^2

Another option could be to use a matrix in the formula:

Y = rnorm(10)
foo = matrix(rnorm(100),10,10)
factors=c(1,5,8)

lm(Y ~ foo[,factors])

You don't actually need a formula. This works:

lm(data_frame[c("Y", "factor1", "factor2")])

as does this:

v <- c("Y", "factor1", "factor2")
do.call("lm", list(bquote(data_frame[.(v)])))

I generally solve this by changing the name of my response column. It is easier to do dynamically, and possibly cleaner.

model_response <- "response_field_name"
setnames(model_data_train, c(model_response), "response") #if using data.table
model_gbm <- gbm(response ~ ., data=model_data_train, ...)