Special operators (%in%, %*%) in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of %in% operator with factors

What is the output of this R code?

f1 <- factor(c("apple", "banana", "cherry"))
f2 <- factor(c("banana", "date", "apple"))
f1 %in% f2

R Programming

f1 <- factor(c("apple", "banana", "cherry"))
f2 <- factor(c("banana", "date", "apple"))
f1 %in% f2

A [1] TRUE TRUE FALSE

B [1] FALSE TRUE TRUE

C [1] TRUE FALSE TRUE

D [1] FALSE FALSE FALSE

💡 Hint

Remember that %in% checks if elements of the left vector are present in the right vector, comparing levels for factors.

The %in% operator returns a logical vector indicating if each element of f1 is found in f2. 'apple' and 'banana' are in f2, but 'cherry' is not.

❓ Predict Output

intermediate

Matrix multiplication with %*%

What is the result of this R code?

A <- matrix(c(1, 2, 3, 4), nrow=2)
B <- matrix(c(5, 6, 7, 8), nrow=2)
A %*% B

R Programming

A <- matrix(c(1, 2, 3, 4), nrow=2)
B <- matrix(c(5, 6, 7, 8), nrow=2)
A %*% B

A

     [,1] [,2]
[1,]   23   31
[2,]   34   46

B

     [,1] [,2]
[1,]   23   34
[2,]   31   46

C

     [,1] [,2]
[1,]   26   30
[2,]   38   44

D

     [,1] [,2]
[1,]   17   20
[2,]   39   46

💡 Hint

Recall matrix multiplication rules: multiply rows of A by columns of B and sum.

The product matrix is calculated as: element (1,1) = 1*5 + 3*6 = 23, element (1,2) = 1*7 + 3*8 = 31, element (2,1) = 2*5 + 4*6 = 34, element (2,2) = 2*7 + 4*8 = 46.

❓ Predict Output

advanced

Behavior of %in% with NA values

What is the output of this R code?

x <- c(1, 2, NA, 4)
y <- c(2, NA, 5)
x %in% y

R Programming

x <- c(1, 2, NA, 4)
y <- c(2, NA, 5)
x %in% y

A [1] FALSE TRUE FALSE FALSE

B [1] FALSE TRUE TRUE FALSE

C [1] FALSE TRUE NA FALSE

D [1] FALSE TRUE NA NA

💡 Hint

Think about how NA values are treated in logical comparisons with %in%.

The %in% operator returns NA when comparing NA to NA because NA means unknown, so it cannot confirm presence or absence. So the third element is NA.

❓ Predict Output

advanced

Matrix multiplication dimension mismatch error

What error does this R code produce?

A <- matrix(1:6, nrow=2)
B <- matrix(1:6, nrow=2)
A %*% B

R Programming

A <- matrix(1:6, nrow=2)
B <- matrix(1:6, nrow=2)
A %*% B

A Error in A %*% B : missing values in matrix

B Error in A %*% B : object 'B' not found

C Error in A %*% B : invalid matrix multiplication

D Error in A %*% B : non-conformable arguments

💡 Hint

Check the dimensions of matrices for multiplication compatibility.

Matrix multiplication requires the number of columns in A to equal the number of rows in B. A is 2x3, B is 2x3, so columns of A (3) does not match rows of B (2), producing 'Error in A %*% B : non-conformable arguments'.

❓ Predict Output

expert

Complex use of %in% with lists and vectors

What is the output of this R code?

lst <- list(a = 1:3, b = 4:6)
v <- c(2, 5, 7)
sapply(lst, function(x) v %in% x)

R Programming

lst <- list(a = 1:3, b = 4:6)
v <- c(2, 5, 7)
sapply(lst, function(x) v %in% x)

A

       a     b
[1,] TRUE TRUE
[2,] FALSE TRUE
[3,] FALSE FALSE

B

       a     b
[1,] TRUE FALSE
[2,] FALSE TRUE
[3,] FALSE FALSE

C

       a     b
[1,] TRUE FALSE
[2,] TRUE TRUE
[3,] FALSE FALSE

D

       a     b
[1,] TRUE FALSE
[2,] FALSE TRUE
[3,] TRUE FALSE

💡 Hint

Check each element of v against each list element using %in%, then sapply returns a matrix.

For list element 'a' (1:3), 2 is in it (TRUE), 5 and 7 are not (FALSE). For 'b' (4:6), 5 is in it (TRUE), 2 and 7 are not (FALSE). So the matrix shows TRUE/FALSE accordingly.