CODEWITHSUNDEEP
rscoping
Abishek
Abishek

Reputation: 33

How does scoping work in R?

a <- 10
f <- function(){
    a <<- 3
    function() {
      a <<- 15
      a <- 5
      return(a)
    }
}
g <- f()
g()
a

When I run the above code I get the following output:

5

15

Can someone please explain how the second output is 15 instead of 3? And the same code returns 10 for a when I do not use "<<-" in the function f.

Upvotes: 1

Views: 52

Answers (1)

Rui Barradas
Rui Barradas

Reputation: 76402

Just see what your function f returns.

f()
function() {
      a <<- 15
      a <- 5
      return(a)
    }
<environment: 0x000000001f1c7898>

So when you assign this result, a function, to g it becomes that function in the output above. The code line a <<- 3 is never executed when you call g.
In other words, the second output is explained in the same fashion the first output is explained, g sets the variable a that exists in the globalenv to 15 and returns a value of variable a <- 5 created in its own environment.

EDIT.
Note that if, like Dason says in his comment, print a before calling g its value has changed to 3 like you expect it in your question. The code line a <<- 3 was indeed executed when f was called.

g <- f()   # call 'f'

a          # print 'a', the value has changed
[1] 3

g()        # now call 'g'
[1] 5

a          # and the value of 'a' changed again
[1] 15

Upvotes: 2

Related Questions