CODEWITHSUNDEEP
rfunctionevaluation
BioBroo
BioBroo

Reputation: 683

Generate a list of functions in R

I would like to generate a list of functions in R. Basically, I'd like to generate a list of functions of the form f <- function(x){a*x} where I specify a vector of a values. Is this possible?

Or is it possible to specify a value of a and then have R interpret the function with a evaluated already? To clarify, if I specify a <- 2, can I make R think of f as function(x){2*x} instead of function(x){a*x}?

Upvotes: 1

Views: 199

Answers (3)

BioBroo
BioBroo

Reputation: 683

Here's another solution, which I tried but thought had failed. We can use a closure, which is a function that returns a function.

cl <- function(a){
    f <- function(x){a*x}
return(f)
}

cl(1)(1)
cl(5)(1)

Upvotes: 0

Rich Scriven
Rich Scriven

Reputation: 99341

Something like this may also be handy. You can alter the body of a function with body and get

> f <- function(x) a*x
> f
function(x) a*x

> a <- 2
> body(f)[[2]] <- get('a')
> f
function (x)
2 * x

> f(1:5)
# [1]  2  4  6  8 10

The body of a function is a list, and it's elements can be accessed just like any other list.

> body(f)
2 * x
> as.list(body(f))
# [[1]]
# `*`

# [[2]]
# [1] 2

# [[3]]
# x

Upvotes: 2

MrFlick
MrFlick

Reputation: 206263

Yes you can do that. Here's an example

a<-1:5

fn<-lapply(a, function(i) {
    force(i)
    function(x) {
        x * i
    }
})


fn[[1]](1)
fn[[5]](1)

The only "trick" here is the force() function. Because R uses lazy evaluation, it doesn't decode the i value until you actually need it. If you don't put the force in there, when you go to use a function, it will finally resolve i and i will just be the last value it was in the lapply which is 5 in this case.

Upvotes: 4

Related Questions