85

Suppose, there is some data.frame foo_data_frame and one wants to find regression of the target column Y by some others columns. For that purpose usualy some formula and model are used. For example:

linear_model <- lm(Y ~ FACTOR_NAME_1 + FACTOR_NAME_2, foo_data_frame)

That does job well if the formula is coded statically. If it is desired to root over several models with the constant number of dependent variables (say, 2) it can be treated like that:

for (i in seq_len(factor_number)) {
  for (j in seq(i + 1, factor_number)) {
    linear_model <- lm(Y ~ F1 + F2, list(Y=foo_data_frame$Y,
                                         F1=foo_data_frame[[i]],
                                         F2=foo_data_frame[[j]]))
    # linear_model further analyzing...
  }
}

My question is how to do the same affect when the number of variables is changing dynamically during program running?

for (number_of_factors in seq_len(5)) {
   # Then root over subsets with #number_of_factors cardinality.
   for (factors_subset in all_subsets_with_fixed_cardinality) {
     # Here I want to fit model with factors from factors_subset.
     linear_model <- lm(Does R provide smth to write here?)
   }
}
Karolis Koncevičius
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Max
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    Thanks! your middle example made me realise I didn't need the solution to your question and could do something much simpler! – Mark Adamson Jan 29 '16 at 09:34

5 Answers5

108

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
Max Ghenis
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Joris Meys
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68

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

landroni
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mnel
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    @JorisMeys And it's so much nicer as it allows adding interaction terms! I've been looking for a similar solution for years.. – landroni Sep 22 '14 at 11:37
  • What if the x variables contain spaces? Say "factor 1" , "factor 2" etc.. – axiom Jan 22 '19 at 08:24
11

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])
Sacha Epskamp
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    +1, but be aware of the fact this doesn't allow to use interaction effects. For that one can construct a model matrix as well (see `?model.matrix` ) – Joris Meys Feb 09 '11 at 23:39
4

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)])))
G. Grothendieck
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1

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, ...)
bibzzzz
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