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.

Practice

1. What is the main purpose of creating an ordered factor in R?

easy

A. To store numeric data more efficiently

B. To keep categories in a specific order and allow comparisons like greater than or less than

C. To convert character strings into numeric values automatically

D. To remove duplicate values from a vector

Solution

1. Step 1: Understand what ordered factors do

   Ordered factors keep categories in a specific order defined by the user.

2. Step 2: Recognize the comparison feature

   They allow comparisons like greater than or less than between categories.

3. Final Answer:

   To keep categories in a specific order and allow comparisons like greater than or less than -> Option B

4. Quick Check:

   Ordered factors = ordered categories with comparison [OK]

Hint: Ordered factors let you compare categories by order [OK]

Common Mistakes:

• Thinking ordered factors store numeric data
• Confusing ordered factors with numeric conversion
• Believing ordered factors remove duplicates

2. Which of the following is the correct way to create an ordered factor f with levels "low" < "medium" < "high" in R?

easy

A. f <- factor(c("low", "high", "medium"), ordered=TRUE, levels=c("low", "medium", "high"))

B. f <- ordered(c("low", "high", "medium"), levels=c("high", "medium", "low"))

C. f <- factor(c("low", "high", "medium"), levels=c("high", "medium", "low"), ordered=TRUE)

D. f <- ordered(c("low", "high", "medium"))

Solution

1. Step 1: Check the function and parameters

   Using factor() with ordered=TRUE and specifying levels sets the order.

2. Step 2: Verify the levels order

   The levels must be in ascending order: "low", "medium", "high".

3. Final Answer:

   f <- factor(c("low", "high", "medium"), ordered=TRUE, levels=c("low", "medium", "high")) -> Option A

4. Quick Check:

   Correct syntax with ordered=TRUE and levels order [OK]

Hint: Use factor(..., ordered=TRUE, levels=correct_order) [OK]

Common Mistakes:

• Reversing the levels order
• Using ordered() without levels
• Setting ordered=TRUE but wrong levels order

3. What will be the output of the following R code?

f <- factor(c("small", "medium", "large"), ordered=TRUE, levels=c("small", "medium", "large"))
f > "medium"

medium

A. [FALSE FALSE FALSE]

B. [FALSE TRUE TRUE]

C. [FALSE FALSE TRUE]

D. [TRUE FALSE TRUE]

Solution

1. Step 1: Understand the factor levels and order

   The levels are "small" < "medium" < "large".

2. Step 2: Compare each element to "medium"

   "small" > "medium" is FALSE, "medium" > "medium" is FALSE, "large" > "medium" is TRUE.

3. Final Answer:

   [FALSE FALSE TRUE] -> Option C

4. Quick Check:

   Only "large" is greater than "medium" [OK]

Hint: Compare ordered factor elements to level with > operator [OK]

Common Mistakes:

• Confusing the order of levels
• Assuming equality returns TRUE
• Mixing up TRUE/FALSE positions

4. Identify the error in this R code that tries to create an ordered factor:

levels <- c("low", "medium", "high")
f <- factor(c("low", "high", "medium"), ordered=TRUE, levels=levels)

medium

A. The ordered=TRUE argument is missing

B. The levels vector is not in alphabetical order

C. The factor vector contains values not in levels

D. The variable name 'levels' conflicts with the function argument name

Solution

1. Step 1: Check variable and argument names

   Using 'levels' as a variable name shadows the function argument 'levels'.

2. Step 2: Understand the conflict

   This can cause unexpected behavior or errors because R confuses the variable with the argument.

3. Final Answer:

   The variable name 'levels' conflicts with the function argument name -> Option D

4. Quick Check:

   Variable name shadows argument name [OK]

Hint: Avoid naming variables 'levels' to prevent conflicts [OK]

Common Mistakes:

• Ignoring variable name conflicts
• Assuming alphabetical order is required
• Missing ordered=TRUE argument

5. You have a survey with responses: "Agree", "Neutral", "Disagree". You want to create an ordered factor response where "Disagree" < "Neutral" < "Agree". Which R code correctly creates this ordered factor and allows you to check if a response is at least "Neutral"?

hard

A. response <- factor(c("Agree", "Disagree", "Neutral"), ordered=TRUE, levels=c("Disagree", "Neutral", "Agree"))
response >= "Neutral"

B. response <- ordered(c("Agree", "Disagree", "Neutral"), levels=c("Agree", "Neutral", "Disagree"))
response >= "Neutral"

C. response <- factor(c("Agree", "Disagree", "Neutral"), levels=c("Disagree", "Neutral", "Agree"))
response >= "Neutral"

D. response <- ordered(c("Agree", "Disagree", "Neutral"))
response >= "Neutral"

Solution

1. Step 1: Set levels in correct order

   The levels must be "Disagree" < "Neutral" < "Agree" to reflect the survey order.

2. Step 2: Create ordered factor and compare

   Using factor(..., ordered=TRUE, levels=...) creates the ordered factor. Then, response >= "Neutral" checks if responses are at least "Neutral".

3. Final Answer:

   response <- factor(c("Agree", "Disagree", "Neutral"), ordered=TRUE, levels=c("Disagree", "Neutral", "Agree")) response >= "Neutral" -> Option A

4. Quick Check:

   Correct order and comparison with ordered factor [OK]

Hint: Set levels in order and use ordered=TRUE for comparisons [OK]

Common Mistakes:

• Wrong levels order
• Missing ordered=TRUE
• Using ordered() without levels