在R中向量化'deparse（substitute（d））'吗？

Question

I'm wondering why, when running my below using: bb(d = c(dnorm, dcauchy) ) I get an error saying: object 'c(dnorm, dcauchy)' not found ?

PS But as I show below, the function has no problem with bb(d = c(dnorm)) .

bb <- function(d){

 d <- if(is.character(d)) d else deparse(substitute(d))

  h <- numeric(length(d))
for(i in 1:length(d)){
  h[i] <- get(d[i])(1)  ## is there something about `get` that I'm missing?
    }
  h
}
# Two Examples of Use:
bb(d = dnorm)                # Works OK 
bb(d = c(dnorm, dcauchy) )   # Error: object 'c(dnorm, dcauchy)' not found

# But if you run:
bb(d = c("dnorm", "dcauchy"))# Works OK

Answer 1

Try this alternative where you pass the functions directly to your function

bb <- function(d){
  if (!is.list(d)) d <- list(d)
  sapply(d, function(x) x(1))  
}

bb(d = list(dnorm, dcauchy))
bb(d = dnorm)

The c() function is meant to combine vectors, it's not a magic "array" function or anything. If you have collections of simple atomic types, you can join them with c() , but for more complicated objects like functions, you need to collect those in a list, not a vector.