lapply function to pass single and + arguments to LM

Question

I am stuck trying to pass "+" arguments to lm.

My 2 lines of code below work fine for single arguments like:

model_combinations=c('.', 'Long', 'Lat', 'Elev')


lm_models = lapply(model_combinations, function(x) {
                               lm(substitute(Y ~ i, list(i=as.name(x))), data=climatol_ann)})

But same code fails if I add 'Lat+Elev' at end of list of model_combinations as in:

model_combinations=c('.', 'Long', 'Lat', 'Elev', 'Lat+Elev') 

Error in eval(expr, envir, enclos) : object 'Lat+Elev' not found

I've scanned posts but am unable to find solution.

Answer 1

I've generally found it more robust/easier to understand to use reformulate to construct formulas via string manipulations rather than trying to use substitute() to modify an expression, e.g.

model_combinations <- c('.', 'Long', 'Lat', 'Elev', 'Lat+Elev')
model_formulas <- lapply(model_combinations,reformulate,
                         response="Y")
lm_models <- lapply(model_formulas,lm,data=climatol_ann)

Because reformulate works at a string level, it doesn't have a problem if the elements are themselves non-atomic (e.g. Lat+Elev). The only tricky situation here is if your data argument or variables are constructed in some environment that can't easily be found, but passing an explicit data argument usually avoids problems.

(You can also use as.formula(paste(...)) or as.formula(sprintf(...)); reformulate() is just a convenient wrapper.)