mutate() for new columns in R Programming - Time & Space Complexity

Time Complexity: mutate() for new columns

O(n)

Understanding Time Complexity

We want to understand how the time needed to add new columns with mutate() changes as the data grows.

How does the work grow when the number of rows increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

library(dplyr)
data <- tibble(x = 1:1000)
data <- data %>% mutate(y = x * 2, z = y + 3)

This code creates two new columns y and z based on existing columns.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Going through each row to calculate new column values.
• How many times: Once for each row, calculating all new columns in one pass.

How Execution Grows With Input

As the number of rows grows, the work to add new columns grows too.

Input Size (n)	Approx. Operations
10	About 20 (2 columns x 10 rows)
100	About 200 (2 columns x 100 rows)
1000	About 2000 (2 columns x 1000 rows)

Pattern observation: The work grows directly with the number of rows and columns added.

Final Time Complexity

Time Complexity: O(n)

This means the time to add new columns grows in a straight line with the number of rows.

Common Mistake

[X] Wrong: "Adding multiple columns with mutate() takes much more time than adding one column because it repeats over the data multiple times."

[OK] Correct: Actually, mutate() processes all new columns in one pass over the data, so the time grows mostly with the number of rows, not the number of columns.

Interview Connect

Knowing how mutate() scales helps you write efficient data transformations and explain your choices clearly in real projects or interviews.

Self-Check

"What if we used mutate() inside a loop that runs for each row? How would the time complexity change?"

Practice

1. What does the mutate() function do in R's dplyr package?

easy

A. It sorts the data frame by a column.

B. It deletes rows from a data frame.

C. It merges two data frames together.

D. It adds new columns or modifies existing ones in a data frame.

Solution

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

   The mutate() function is designed to add new columns or change existing columns in a data frame.

2. Step 2: Compare with other functions

   Deleting rows is done by filter(), sorting by arrange(), and merging by join functions, not mutate().

3. Final Answer:

   It adds new columns or modifies existing ones in a data frame. -> Option D

4. Quick Check:

   mutate() = add/change columns [OK]

Hint: mutate() changes or adds columns, not rows or merges [OK]

Common Mistakes:

• Confusing mutate() with filter() for rows
• Thinking mutate() merges data frames
• Assuming mutate() sorts data

2. Which of the following is the correct syntax to add a new column total as the sum of columns a and b using mutate()?

easy

A. mutate(df, total = a + b)

B. mutate(df, total <- a + b)

C. mutate(df, total = sum(a, b))

D. mutate(df, total = a & b)

Solution

1. Step 1: Check assignment inside mutate()

   Inside mutate(), new columns are created with =, not with assignment operator <-.

2. Step 2: Understand how to sum columns element-wise

   To add columns element-wise, use a + b. Using sum(a, b) sums all values together, not element-wise. a & b is a logical AND, not addition.

3. Final Answer:

   mutate(df, total = a + b) -> Option A

4. Quick Check:

   Use = and + for element-wise sum [OK]

Hint: Use = and + inside mutate() for new column sums [OK]

Common Mistakes:

• Using <- instead of = inside mutate()
• Using sum() instead of + for element-wise addition
• Using & instead of + for addition

3. Given the data frame df <- data.frame(x = 1:3, y = 4:6), what will be the result of mutate(df, z = x * y)?

medium

A. A data frame with columns x, y, and z where z is the product of x and y.

B. A data frame with columns x and y only, no new column added.

C. An error because mutate() cannot multiply columns.

D. A data frame with column z only containing the product values.

Solution

1. Step 1: Understand mutate() behavior

   mutate() adds new columns without removing existing ones unless explicitly overwritten.

2. Step 2: Calculate new column values

   For each row, z = x * y means multiplying corresponding elements: 1*4=4, 2*5=10, 3*6=18.

3. Final Answer:

   A data frame with columns x, y, and z where z is the product of x and y. -> Option A

4. Quick Check:

   mutate() adds z = x * y element-wise [OK]

Hint: mutate() keeps old columns and adds new ones [OK]

Common Mistakes:

• Expecting mutate() to remove old columns
• Thinking mutate() cannot do arithmetic
• Assuming new column replaces all columns

4. What is wrong with this code snippet?

df <- data.frame(a = 1:3, b = 4:6)
df %>% mutate(c = a + d)

medium

A. The syntax of mutate() is incorrect; it needs a comma after c.

B. Column d does not exist in df, causing an error.

C. You cannot use the pipe operator %>% with mutate().

D. The code will run but c will be filled with NA values.

Solution

1. Step 1: Check column names in data frame

   The data frame df has columns a and b, but no column named d.

2. Step 2: Understand mutate() behavior with missing columns

   Trying to use a non-existent column d inside mutate() causes an error because R cannot find it.

3. Final Answer:

   Column d does not exist in df, causing an error. -> Option B

4. Quick Check:

   Using unknown columns in mutate() causes error [OK]

Hint: Check all columns exist before using in mutate() [OK]

Common Mistakes:

• Assuming mutate() creates missing columns automatically
• Thinking pipe operator %>% is invalid here
• Believing syntax error due to missing comma

5. You have a data frame df with columns score1 and score2. You want to add a new column average that is the average of these two scores, but only if both scores are greater than 0; otherwise, average should be NA. Which mutate() code correctly does this?

hard

A. mutate(df, average = (score1 + score2) / 2)

B. mutate(df, average = ifelse(score1 > 0 | score2 > 0, (score1 + score2) / 2, NA))

C. mutate(df, average = ifelse(score1 > 0 & score2 > 0, (score1 + score2) / 2, NA))

D. mutate(df, average = ifelse(score1 & score2, (score1 + score2) / 2, NA))

Solution

1. Step 1: Use ifelse() with correct condition

   The condition requires both score1 and score2 to be greater than 0, so use vectorized & inside ifelse().

2. Step 2: Calculate average or assign NA

   If condition is true, calculate average as (score1 + score2) / 2, else assign NA.

3. Final Answer:

   mutate(df, average = ifelse(score1 > 0 & score2 > 0, (score1 + score2) / 2, NA)) -> Option C

4. Quick Check:

   Use ifelse() with & for both conditions [OK]

Hint: Use ifelse() with & to check both conditions [OK]

Common Mistakes:

• Using | instead of & to require both conditions
• Omitting ifelse() and always calculating average
• Using score1 & score2 without comparison