R: Evaluate an expression in a data frame with arguments that are passed as an object -

i want write function evaluates expression in data frame, 1 using expressions may or may not contain user-defined objects. think magic word "non-standard evaluation", cannot quite figure out yet.

one simple example (yet realistic purposes): say, want evaluate lm() call variables found in data frame.

mydf <- data.frame(x=1:10, y=1:10) 

a function can written follows:

f <- function(df, expr){   expr <- substitute(expr)   pf <- parent.frame()   eval(expr, df, pf) } 

such want using following command.

f(mydf, lm(y~x))  # call: # lm(formula = y ~ x) #  # coefficients: # (intercept)            x   #    1.12e-15     1.00e+00   

nice. however, there cases in more convenient save model equation in object before calling lm(). unfortunately above function no longer it.

fml <- y~x  f(mydf, lm(fml)) # error in eval(expr, envir, enclos): object 'y' not found 

can explain why second call doesn't work? how function altered, such both calls lead desired results? (desired=fitted model)

cheers!

from ?lm, re data argument:

if not found in data, variables taken environment(formula)

in first case, formula created in eval(expr, df, pf) call, environment of formula environment based on df. in second case, formula created in global environment, why doesn't work.

because formulas come own environment, can tricky handle in nse.

you try:

with(mydf,   {     print(lm(y~x))     fml <- y~x     print(lm(fml))   } ) 

but isn't ideal you. short of checking whether names in captured parameter resolve formulas, , re-assigning environments, you'll have trouble. worse, isn't obvious re-assigning environment right thing do. in many cases, want in formula environment.

there loosely related discussion on issue on r chat:

• ben bolker outlines issue
• josh o'brien points old references