Understanding the Impact of Print Function in sapply()

The sapply() function is a versatile and powerful tool in R for applying a specified function to each element of a vector or list. However, one subtle aspect of its behavior can lead to unexpected results when using print statements within the function itself.

Background on sapply

For those unfamiliar with the basics of R’s sapply(), it is generally used to apply a function to each element of a vector or list, returning a vector or list containing the results. The syntax for sapply() is:

sapply(x, func)

Here, x is the input vector or list and func is the function to be applied.

The Question

The question at hand revolves around a specific example where print(x) within an anonymous function passed to sapply(). A user presents a scenario where:

data <- c("001","002","103","119","129")
n1 <- sapply(data, function(x){
    x <- gsub(pattern="(\\d+)(\\d\\d)$", "\\2", x)
    if(gsub("(\\d)(\\d)","\\1",x)=="0") 
         x <- gsub("(\\d)(\\d)","\\2",x)
})

n2 <- sapply(data, function(x){
    x <- gsub(pattern="(\\d+)(\\d\\d)$", "\\2", x)
    if(gsub("(\\d)(\\d)","\\1",x)=="0") 
         x <- gsub("(\\d)(\\d)","\\2",x)
     print(x)
}, USE.NAMES=FALSE)

The user asks why n2 yields a different result than n1. Specifically, they observe that n2 produces a vector of “1” “2” “3” “19” “29”, whereas n1 does not.

Debugging the Issue

Let’s dive into the code provided to understand why this happens.

n2 <- sapply(data, function(x) {
    x <- gsub(pattern = "(\\d+)(\\d\\d)$", "\\2", x)
    if (gsub("(\\d)(\\d)", "\\1", x) == "0") x <- gsub("(\\d)(\\d)", "\\2", x)
     print(x)
}, USE.NAMES=FALSE)

Here is the modified version of n2 with better indentation and added spaces:

n2 <- sapply(data, function(x) {
    x <- gsub(pattern = "(\\d+)(\\d\\d)$", "\\2", x)
    if (gsub("(\\d)(\\d)", "\\1", x) == "0") 
        x <- gsub("(\\d)(\\d)", "\\2", x)
    print(x)
}, USE.NAMES=FALSE)

The Secret to Understanding sapply’s Behavior

The key to understanding this behavior lies in how R handles the return value of functions inside sapply(). When you use an explicit <code>return</code> statement, R will return that specific output from the function. However, if no return statement is provided within the function, R returns the outcome of the last operation.

In this case, when we look at n1 and n2, there’s a critical difference in how they are executed:

n1 <- sapply(data, function(x){
    x <- gsub(pattern = "(\\d+)(\\d\\d)$", "\\2", x)
    if (gsub("(\\d)(\\d)", "\\1", x) == "0") 
         x <- gsub("(\\d)(\\d)","\\2",x)
})

n2 <- sapply(data, function(x){
    x <- gsub(pattern = "(\\d+)(\\d\\d)$", "\\2", x)
    if (gsub("(\\d)(\\d)", "\\1", x) == "0") 
         x <- gsub("(\\d)(\\d)","\\2",x)
     print(x)
}, USE.NAMES=FALSE)

Notice the difference in indentation between n1 and n2. Within n2, there’s an extra line of code that includes <code>print(x)</code>. This statement not only prints to the screen but also causes it to be returned from the function.

When no return statement is provided, R returns the outcome of the last operation. In this case, that would be x, which contains the result after executing both sub-operations of the if condition. Since there’s a non-zero string match in the first part of the if, it skips the second operation entirely.

However, since we have <code>print(x)</code> at the end, x has already been updated by that point and its value is what gets returned to n2.

Workaround

To avoid this unexpected behavior and get a consistent result, you can use an explicit return statement within your function:

print(x)

Alternatively, simply omit the <code>print(x)</code> line altogether.

The corrected versions of n1 and n2 would look like this:

n1 <- sapply(data, function(x){
    x <- gsub(pattern = "(\\d+)(\\d\\d)$", "\\2", x)
    if (gsub("(\\d)(\\d)", "\\1", x) == "0") 
         x <- gsub("(\\d)(\\d)","\\2",x)
})

n2 <- sapply(data, function(x){
    x <- gsub(pattern = "(\\d+)(\\d\\d)$", "\\2", x)
    if (gsub("(\\d)(\\d)", "\\1", x) == "0") 
         x <- gsub("(\\d)(\\d)","\\2",x)
})

This should ensure that both n1 and n2 produce identical results.

Conclusion

Understanding how R handles functions within sapply() can be crucial for debugging issues like this one. Remember to always check your function’s return value carefully, especially if you’re relying on its output in subsequent operations.

Last modified on 2023-07-12