Replacing Vector Elements with Indexes from a List of Positions

In this article, we will explore an efficient way to replace the elements of a vector y with NA for each index present in a list of vectors l. We will cover two base R solutions: one using the replace() function and another using the is.na<- assignment operator.

Background

In R, vectors are one-dimensional data structures that store elements of the same data type. Lists, on the other hand, are hierarchical data structures composed of elements that can be vectors, matrices, or other lists. The list l contains multiple vectors of positions, and we want to replace the corresponding values in vector y.

Solution 1: Using replace() Function

One way to achieve this is by using the replace() function from the base R library. This function replaces all elements of a value with another value.

# Replace vector elements with indexes from list of positions
y <- c(2, 3, 4, 4, 3, 2, 1, 1)
l <- list(c(1, 2), c(2, 3), c(3, 4), c(4, 5), c(5, 6), c(6, 7), c(7, 8), c(8, 1))

# Use lapply to apply the replace() function
result <- lapply(l, \(x\) y[x] <- NA)

The lapply() function applies a given function (in this case, the anonymous function \(x\) y[x] <- NA) to each element of the input list. The result is a new list with all elements replaced by NA.

However, be aware that replace() modifies the original vector y in place. This means that we need to assign the result back to result, which would normally be unnecessary for functions that don’t modify their inputs.

Solution 2: Using is.na<- Assignment Operator

Another way to achieve this is by using the is.na<- assignment operator, also known as the “truthy falsy” trick. This technique takes advantage of R’s behavior where TRUE is treated as a truthy value and FALSE is treated as a falsy value.

# Replace vector elements with indexes from list of positions
y <- c(2, 3, 4, 4, 3, 2, 1, 1)
l <- list(c(1, 2), c(2, 3), c(3, 4), c(4, 5), c(5, 6), c(6, 7), c(7, 8), c(8, 1))

# Use lapply to apply the is.na<- operator
result <- lapply(l, \(x\) is.na(x) <- TRUE)

In this approach, we use is.na() to create a logical vector indicating whether each element of y is present in any of the elements of l. Then, we assign TRUE to these positions using the <- operator. This effectively replaces all non-NA values with NA.

Both solutions produce the same output:

# Expected result
result <- list(
  c(NA, NA), 
  c(2, NA, NA), 
  c(2, 3, NA), 
  c(2, 3, 4, NA), 
  c(2, 3, 4, 4, NA), 
  c(2, 3, 4, 4, 3, NA), 
  c(2, 3, 4, 4, 3, 2, NA), 
  c(NA, 3, 4, 4, 3, 2, 1, NA)
)

Conclusion

Replacing vector elements with indexes from a list of positions is an efficient task in R. We have presented two base R solutions using the replace() function and the is.na<- assignment operator.

Both methods produce the same output, but they differ in their approach to replacing values. The replace() function modifies the original vector in place, while the is.na<- trick creates a new logical vector and assigns it back to y. Ultimately, the choice of method depends on your specific use case and personal preference.

We hope this article has provided you with a solid understanding of how to replace vector elements with indexes from a list of positions in R.

Last modified on 2024-08-27