Environment and closures in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a closure capturing a variable

What is the output of this R code?

make_counter <- function() {
  count <- 0
  function() {
    count <<- count + 1
    count
  }
}
c <- make_counter()
c()
c()

R Programming

make_counter <- function() {
  count <- 0
  function() {
    count <<- count + 1
    count
  }
}
c <- make_counter()
c()
c()

A

[1] 2
[1] 3

B

[1] 0
[1] 1

C

[1] 1
[1] 1

D

[1] 1
[1] 2

💡 Hint

Think about how the <<- operator changes the variable in the parent environment.

The inner function increments 'count' in the parent environment each time it is called, so the first call returns 1, the second returns 2.

❓ Predict Output

intermediate

Value of variable after nested function calls

What is the value of x after running this code?

x <- 5
f <- function() {
  x <- 10
  g <- function() {
    x <<- 20
  }
  g()
  x
}
f()
x

R Programming

x <- 5
f <- function() {
  x <- 10
  g <- function() {
    x <<- 20
  }
  g()
  x
}
f()
x

A 5

B 10

C 20

D Error

💡 Hint

The <<- operator modifies the variable in the nearest parent environment where it exists.

The <<- operator in g() modifies the global x, changing it from 5 to 20. The local x inside f() remains 10, but the global x is 20 after f() runs.

🔧 Debug

advanced

Why does this closure not remember the updated value?

Consider this code:

make_funcs <- function() {
  funcs <- list()
  for (i in 1:3) {
    funcs[[i]] <- function() i
  }
  funcs
}
fs <- make_funcs()
fs[[1]]()
fs[[2]]()
fs[[3]]()

What is the output of the three calls to fs[[1]](), fs[[2]](), and fs[[3]]()?

R Programming

make_funcs <- function() {
  funcs <- list()
  for (i in 1:3) {
    funcs[[i]] <- function() i
  }
  funcs
}
fs <- make_funcs()
fs[[1]]()
fs[[2]]()
fs[[3]]()

A

[1] 1
[1] 2
[1] 3

B

[1] 3
[1] 3
[1] 3

C

[1] 1
[1] 1
[1] 1

D Error in function calls

💡 Hint

Think about when the variable i is looked up by the functions.

All functions capture the same variable i, which after the loop ends has the value 3. So all functions return 3.

📝 Syntax

advanced

Identify the error in this closure code

What error does this code produce?

make_adder <- function(x) {
  function(y) {
    x + y
  }
}
add5 <- make_adder(5)
add5("3")

R Programming

make_adder <- function(x) {
  function(y) {
    x + y
  }
}
add5 <- make_adder(5)
add5("3")

A Error: non-numeric argument to binary operator

B Error: object 'y' not found

C No error, returns 8

D Error: unexpected string constant

💡 Hint

Check the types of the arguments used in the addition.

Adding a number and a string causes a type error in R, resulting in 'non-numeric argument to binary operator'.

🚀 Application

expert

How many unique environments are created?

Consider this code:

make_funcs <- function() {
  funcs <- list()
  for (i in 1:3) {
    funcs[[i]] <- local({
      x <- i
      function() x
    })
  }
  funcs
}
fs <- make_funcs()
length(unique(sapply(fs, environment)))

What is the output of the last line?

R Programming

make_funcs <- function() {
  funcs <- list()
  for (i in 1:3) {
    funcs[[i]] <- local({
      x <- i
      function() x
    })
  }
  funcs
}
fs <- make_funcs()
length(unique(sapply(fs, environment)))

A 3

B 1

C 0

D Error

💡 Hint

Each call to local creates a new environment for the function.

Using local inside the loop creates a new environment for each function, so there are 3 unique environments.