Pass expression as variable to curve

Question

I am trying to use a variable that stores a function name as a character string to draw a curve. Something like the following:

f1 <- function(x) 0*x
f2 <- function(x) 1 + 0*x
f3 <- function(x) 2 + 0*x

fn <- "f1"

plot.new()
plot.window(xlim = c(0, 1), ylim = c(-1, 3))

curve(get(fn), add = TRUE)
Error in curve(get(fn), add = TRUE) : 
  'expr' must be a function, or a call or an expression containing 'x'

curve(f1, add = TRUE)
curve(f2, add = TRUE)
curve(f3, add = TRUE)

I know I am missing something with how curve handles the expressions with substitute, or how get is passing the function. I am particularly confused because class(get(fn)) returns "function".

Answer 1

Well, the issue is that curve() doesn't evaluate the parameter you pass in. It looks at what you passed in to the parameter, not the results of evaluating the parameter. When you run

curve(fn)

what you're passing in is a "name" (or "symbol"). That triggers the function to find the function with that name. When you run

curve(get("f1"))  #error

you've passed in a call. Again, that call isn't executed by curve to see that it returns a function. As the error message says, if you pass in a call, that expression should contain x. For example

curve(f1(x))

is an example of the call syntax with the x variable.

If you really need to specify the function as a character vector, you can convert the character to a name object and then build the call to curve with do.call. For example

do.call("curve",list(as.name(fn)))

or you can modify your call to trigger the resulting function with the special x variable

curve(get(fn)(x))