Sorting with order() in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this R code using order()?

Consider the following R code that sorts a vector using order(). What will be the output?

R Programming

x <- c(5, 2, 9, 1)
sorted_x <- x[order(x)]
sorted_x

A [1] 9 5 2 1

B [1] 5 2 9 1

C [1] 1 2 5 9

D [1] 2 1 5 9

💡 Hint

The order() function returns the indices that would sort the vector.

The order(x) returns the indices that sort x in ascending order: 4, 2, 1, 3. Using these indices to subset x gives the sorted vector c(1, 2, 5, 9).

❓ Predict Output

intermediate

What is the output when ordering a data frame by two columns?

Given this data frame, what will be the output of ordering by age then score?

R Programming

df <- data.frame(name=c('Ann', 'Bob', 'Cara', 'Dan'), age=c(25, 25, 22, 25), score=c(90, 85, 88, 85))
ordered_df <- df[order(df$age, df$score), ]
ordered_df$name

A [1] "Cara" "Ann" "Bob" "Dan"

B [1] "Cara" "Bob" "Dan" "Ann"

C [1] "Dan" "Bob" "Ann" "Cara"

D [1] "Bob" "Ann" "Dan" "Cara"

💡 Hint

First sort by age ascending, then by score ascending.

The data frame rows sorted by age then score are: Cara (22,88), Bob (25,85), Dan (25,85), Ann (25,90). Since Bob and Dan have the same age and score, their original order is preserved.

🔧 Debug

advanced

Why does this order() call produce an error?

Examine the code below. Why does it produce an error?

R Programming

x <- c(3, 1, 4)
order(x > 2, x)

A No error; returns indices ordering by x > 2 then x

B Error because order() expects vectors of the same length and type

C Error because logical vector cannot be used directly with numeric vector in order()

D Error because > operator is not vectorized

💡 Hint

Check if logical vectors can be used in order().

The order() function can handle logical vectors because they are treated as numeric (FALSE=0, TRUE=1). So no error occurs and it returns indices ordering by x > 2 then x.

📝 Syntax

advanced

Which option has correct syntax to sort a vector in descending order using order()?

Choose the correct R code to sort vector v in descending order using order().

R Programming

v <- c(10, 5, 8, 3)

A v[order(-1 * v)]

B v[order(v, decreasing=TRUE)]

C v[order(v, desc=TRUE)]

D v[order(-v)]

💡 Hint

Remember order() does not have a decreasing argument; use negative values to reverse order.

order() does not support decreasing argument. Using -v reverses the order. Option D and D are similar but D uses multiplication which works but is less idiomatic. Option D is invalid because order() has no decreasing argument. Option D uses a non-existent argument desc.

🚀 Application

expert

How many items are in the sorted vector after removing duplicates?

Given the vector v <- c(4, 2, 4, 3, 2, 1), how many items are in the vector after sorting with order() and removing duplicates?

R Programming

v <- c(4, 2, 4, 3, 2, 1)
unique_sorted_v <- unique(v[order(v)])
length(unique_sorted_v)

A 4

B 5

C 6

D 3

💡 Hint

First sort the vector, then remove duplicates, then count the length.

Sorting v gives c(1, 2, 2, 3, 4, 4). Removing duplicates leaves c(1, 2, 3, 4), which has length 4.

Practice

1. What does the order() function in R do?

easy

A. Sorts the vector and returns the sorted values directly

B. Removes duplicates from a vector

C. Returns the indices that would sort a vector in ascending order

D. Calculates the sum of all elements in a vector

Solution

1. Step 1: Understand the purpose of order()

   The order() function returns the positions (indices) that would sort the vector, not the sorted vector itself.

2. Step 2: Differentiate from sort()

   Unlike sort(), which returns sorted values, order() returns the order of indices to rearrange the vector.

3. Final Answer:

   Returns the indices that would sort a vector in ascending order -> Option C

4. Quick Check:

   order() = indices for sorting [OK]

Hint: Remember: order() gives positions, sort() gives sorted values [OK]

Common Mistakes:

• Confusing order() with sort()
• Expecting order() to return sorted values
• Using order() to remove duplicates

2. Which of the following is the correct syntax to sort a vector v in decreasing order using order()?

easy

A. v[sort(v, decreasing = TRUE)]

B. order(v, decreasing = TRUE)

C. sort(order(v), decreasing = TRUE)

D. v[order(v, decreasing = TRUE)]

Solution

1. Step 1: Use order() with decreasing = TRUE

   To sort vector v in decreasing order, use order(v, decreasing = TRUE) to get indices for sorting from largest to smallest.

2. Step 2: Index vector v with order()

   Use these indices to reorder v as v[order(v, decreasing = TRUE)].

3. Final Answer:

   v[order(v, decreasing = TRUE)] -> Option D

4. Quick Check:

   v[order(v, decreasing=TRUE)] sorts descending [OK]

Hint: Index vector with order() for sorting, add decreasing=TRUE for reverse [OK]

Common Mistakes:

• Using order() without indexing the vector
• Trying to sort order() output directly
• Using sort() on order() output incorrectly

3. What is the output of the following R code?

v <- c(4, 2, 9, 7)
v[order(v)]

medium

A. [1] 2 4 7 9

B. [1] 4 2 9 7

C. [1] 9 7 4 2

D. [1] 2 7 4 9

Solution

1. Step 1: Calculate order(v)

   For v = c(4, 2, 9, 7), order(v) returns indices that sort v: positions of values in ascending order are 2 (2), 1 (4), 4 (7), 3 (9).

2. Step 2: Index v with order(v)

   Using these indices: v[c(2,1,4,3)] gives 2 4 7 9.

3. Final Answer:

   [1] 2 4 7 9 -> Option A

4. Quick Check:

   v[order(v)] = sorted vector ascending [OK]

Hint: order() gives indices to sort; index vector to get sorted values [OK]

Common Mistakes:

• Expecting order() to return sorted values directly
• Mixing up indices and values
• Incorrect indexing order

4. Identify the error in the following R code snippet:

v <- c(5, 3, 8, 1)
sorted_v <- order(v)
print(sorted_v)

medium

A. order() returns indices, not sorted values; should index v with order(v)

B. Missing parentheses in order function call

C. Vector v is not defined before order()

D. print() cannot display vectors

Solution

1. Step 1: Understand what order() returns

   order(v) returns the indices that would sort v, not the sorted vector itself.

2. Step 2: Correct usage to get sorted vector

   To get sorted values, use v[order(v)]. The code assigns indices to sorted_v, so printing it shows positions, not sorted values.

3. Final Answer:

   order() returns indices, not sorted values; should index v with order(v) -> Option A

4. Quick Check:

   order() = indices, index vector to sort [OK]

Hint: Use v[order(v)] to get sorted vector, not order(v) alone [OK]

Common Mistakes:

• Expecting order() to return sorted vector
• Printing order() output directly
• Confusing order() with sort()

5. Given two related vectors names <- c("Anna", "Bob", "Cara") and scores <- c(88, 95, 90), how do you sort names by scores in descending order using order()?

hard

A. order(names, decreasing = TRUE)[scores]

B. names[order(scores, decreasing = TRUE)]

C. scores[order(names, decreasing = TRUE)]

D. sort(names, by = scores, decreasing = TRUE)

Solution

1. Step 1: Use order() on scores with decreasing = TRUE

   To sort names by scores descending, get indices with order(scores, decreasing = TRUE).

2. Step 2: Index names with these order indices

   Use these indices to reorder names: names[order(scores, decreasing = TRUE)].

3. Final Answer:

   names[order(scores, decreasing = TRUE)] -> Option B

4. Quick Check:

   Index names by order(scores, decreasing=TRUE) [OK]

Hint: Sort related vectors by ordering one, index the other [OK]

Common Mistakes:

• Trying to sort names directly by scores without order()
• Using sort() with non-existent 'by' argument
• Mixing order() arguments or indexing incorrectly