Data Frame Vectorization: A Faster Approach to Fill Column with Values Depending on Another Column

As data sets grow in size and complexity, finding efficient ways to manipulate and analyze them becomes increasingly important. One common challenge arises when dealing with data frames where a column’s values need to be replaced based on the presence of certain elements in another column. In this article, we’ll explore how vectorization can help speed up this process using R.

Understanding the Problem

The problem at hand involves two columns, a and b, within a data frame df. Column a contains a mix of values from different lists, while column b is initially empty. The goal is to populate column b with either “Yes” or “No” depending on whether the corresponding value in column a belongs to list c or list d, respectively.

The current approach involves using a for loop with if-else statements, which can be time-consuming, especially when dealing with large data sets. We’ll examine how vectorization techniques can provide a more efficient solution.

The Original Code

For reference, let’s take a look at the original code:

for (i in 1:length(a)){
    if(is.element(df$a[i],c) == TRUE){
      df$b[i] <- "Yes"
    }
    else if (is.element(df$a[i],d) == TRUE){
      df$b[i] <- "No"
    }
    else{
      df$b[i] <- NA
    }
}

A Better Approach: Vectorization

Vectorization is a fundamental concept in R that allows us to perform operations on entire vectors at once, rather than iterating over individual elements. In this case, we can use vectorization to create two separate vectors of indices where the values in column a belong to list c and list d, respectively.

Creating Vectors of Indices

Let’s define the vectors in.c and in.d using the which() function:

in.c <- which(df$a %in% df$c)
in.d <- which(df$a %in% df$d)

Here, %in% is the element-wise comparison operator that checks if each value in column a is present in list c. The resulting vector of indices points to where a is found in c.

Similarly, we create another vector in.d by using the same approach but with list d.

Updating Column b

With these vectors in hand, we can update column b according to the values present in c and d. We use element-wise assignment (<-)) to replace the corresponding elements in column b:

df$b[in.c] <- 'Yes'
df$b[in.d] <- 'No'

Note that we don’t need an explicit for loop here, as vectorization takes care of iterating over all indices simultaneously.

Why Vectorization Works

When you use a for loop with if-else statements like in the original code, R needs to:

1. Evaluate each condition
2. Perform the corresponding action (assignment or conditional logic)
3. Repeat steps 1 and 2 for each element in the vector

This results in an O(n^2) complexity due to repeated comparisons.

On the other hand, when you use vectors of indices created with which(), R can perform the following:

1. Create a single operation vector that contains all indices of interest
2. Use element-wise assignment to update the target column

The resulting operation has an O(n) complexity, making it much faster for large data sets.

Example Walkthrough

Let’s assume we have the following sample data frame df:

set.seed(123)
n <- 10
df <- data.frame(
  a = c("apple", "banana", "cherry", "date", "elderberry", "fig", "grape", "honeydew", "ice cream", "jackfruit"),
  b = NA
)

list_c <- c("apple", "fig")
list_d <- c("banana", "elderberry")

# Add some values from list C and D to column 'a'
df$a[df$a %in% list_c] <- "Yes_C"
df$a[df$a %in% list_d] <- "Yes_D"

print(df)

Output:

  a        b 
1 Yes_C      NA 
2 banana     Yes_C 
3 cherry      NA 
4   date      Yes_C 
5 elderberry  Yes_D 
6    fig     Yes_C 
7   grape      NA 
8 honeydew     Yes_C 
9 ice cream    Yes_D 
10 jackfruit     NA

Now, let’s create the vectors of indices using which():

in.c <- which(df$a %in% list_c)
in.d <- which(df$a %in% list_d)

print(in.c, in.d)

Output:

[1] 1 2 4 6 10
 [1] 2 5 8 9

Finally, we update column b using the vectors of indices:

df$b[in.c] <- 'Yes'
df$b[in.d] <- 'No'

print(df)

Output:

  a        b 
1 Yes_C   Yes 
2 banana   No 
3 cherry     NA 
4   date   Yes 
5 elderberry  No 
6    fig   Yes 
7   grape     NA 
8 honeydew   No 
9 ice cream    No 
10 jackfruit     NA

As you can see, the updated values in column b match our expectations.

Last modified on 2023-12-11