Ordered factors in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of ordered factor comparison

What is the output of this R code snippet?

R Programming

levels <- c("low", "medium", "high")
ord_factor <- factor(c("medium", "low", "high"), levels = levels, ordered = TRUE)
ord_factor[1] > ord_factor[2]

A Error: comparison not possible

B [1] FALSE

C [1] NA

D [1] TRUE

💡 Hint

Remember that ordered factors allow comparison based on their levels.

The factor is ordered with levels low < medium < high. Comparing 'medium' > 'low' returns TRUE.

❓ Predict Output

intermediate

Number of levels in an ordered factor

How many levels does this ordered factor have?

R Programming

ord_factor <- ordered(c("small", "medium", "large", "medium"), levels = c("small", "medium", "large"))
length(levels(ord_factor))

A Error: levels not found

B 3

C 2

D 4

💡 Hint

Levels are defined explicitly in the factor creation.

The factor has three levels: small, medium, and large.

🔧 Debug

advanced

Why does this ordered factor comparison fail?

What error does this code produce and why?

R Programming

f1 <- factor(c("a", "b"), ordered = TRUE)
f2 <- factor(c("a", "b"), levels = c("b", "a"), ordered = TRUE)
f1 > f2

A Error: comparison not possible between factors with different levels

B [1] TRUE TRUE

C [1] FALSE FALSE

D Error: factors must have the same levels

💡 Hint

Check if the factors have the same levels and order.

Even though both factors have the same values, their levels are set in different orders ('a' < 'b' vs. 'b' < 'a'), causing comparison to fail.

🧠 Conceptual

advanced

Effect of changing levels order on ordered factor comparison

Given this code, what is the output of the comparison?

R Programming

ord_factor <- ordered(c("low", "medium", "high"), levels = c("high", "medium", "low"))
ord_factor[1] < ord_factor[3]

A [1] NA

B [1] TRUE

C [1] FALSE

D Error: invalid comparison

💡 Hint

The order of levels defines the comparison order.

Levels are reversed: high < medium < low. 'low' is last, 'high' is first, so 'low' < 'high' is FALSE.

📝 Syntax

expert

Identify the syntax error in ordered factor creation

Which option contains the correct syntax to create an ordered factor with levels "small", "medium", "large"?

A factor(c("small", "medium", "large"), levels = c("small", "medium", "large"), ordered = TRUE)

B factor(c("small", "medium", "large"), ordered = TRUE, levels = c("small", "medium", "large"))

C ordered(c("small", "medium", "large"), c("small", "medium", "large"))

D ordered(c("small", "medium", "large"), levels = c("small", "medium", "large"))

💡 Hint

Check the correct argument names and their order.

Option A correctly uses factor() with levels and ordered=TRUE. Option A correctly uses ordered() with named levels argument. Option A has argument order wrong but still works in R. Option A misses argument name for levels causing error.