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Control Flow in R

Adapted from Software Carpentry

Overview

Today we will learn to:

Make data-dependent choices with if...else statements
Use the ifelse() function for vectorized conditions
Write and understand for() loops
Repeat operations using while() loops

Questions

How can I make data-dependent choices in R?
How can I repeat operations in R?

Why Control Flow?

Often when coding, we want to control the flow of our actions:

Set actions to occur only if a condition is met
Set an action to occur a particular number of times
Make decisions based on data values

Control flow is essential for automating complex analyses!

Conditional Statements

The most common approaches for conditional statements:

# if
if (condition is true) {
  perform action
}

# if ... else
if (condition is true) {
  perform action
} else {  # that is, if the condition is false,
  perform alternative action
}

Simple if Statement

Print a message if a variable has a particular value:

x <- 8

if (x >= 10) {
  print("x is greater than or equal to 10")
}

x

[1] 8

The print statement doesn’t appear because x is not greater than 10.

Adding else

To print a message for numbers less than 10, add else:

x <- 8

if (x >= 10) {
  print("x is greater than or equal to 10")
} else {
  print("x is less than 10")
}

[1] "x is less than 10"

Multiple Conditions with else if

Test multiple conditions using else if:

x <- 8

if (x >= 10) {
  print("x is greater than or equal to 10")
} else if (x > 5) {
  print("x is greater than 5, but less than 10")
} else {
  print("x is less than 5")
}

[1] "x is greater than 5, but less than 10"

Logical Elements in Conditions

Important: R looks for a logical element (TRUE or FALSE) inside if() statements.

x <- 4 == 3
if (x) {
  "4 equals 3"
} else {
  "4 does not equal 3"
}

[1] "4 does not equal 3"

Understanding the Condition

The “not equal” message was printed because x is FALSE:

x <- 4 == 3
x

[1] FALSE

Challenge 1

Use an if() statement to print a suitable message reporting whether there are any records from 2002 in the gapminder dataset.

Now do the same for 2012.

Hint: Think about how to check if any values match a condition!

Challenge 1 Solution

First, let’s load the data:

gapminder <- read.csv(
  "https://raw.githubusercontent.com/swcarpentry/r-novice-gapminder/main/episodes/data/gapminder_data.csv"
)

Challenge 1 Solution (continued)

Check for records from 2002:

if (any(gapminder$year == 2002)) {
  print("Record(s) for the year 2002 found.")
}

[1] "Record(s) for the year 2002 found."

Challenge 1 Solution (for 2012)

Did you try this for 2012?

if (gapminder$year == 2012) {
  print("Record(s) for the year 2012 found.")
}

You may have received a warning or error!

Common Mistake with if()

The if() function only accepts singular (length 1) inputs:

Error in `if (gapminder$year == 2012) ...`:
! the condition has length > 1

The if() function will only evaluate the condition in the first element of the vector.

To use if(), make sure your input is singular (length 1)!

The ifelse() Function

R’s built-in ifelse() function accepts both singular and vector inputs:

# ifelse function
ifelse(condition is true, perform action, perform alternative action)

First argument: condition(s) to be met
Second argument: evaluated when TRUE
Third argument: evaluated when FALSE

ifelse() Example

y <- -3
ifelse(y < 0, "y is a negative number", "y is either positive or zero")

[1] "y is a negative number"

Works with vectors too!

values <- c(-3, 5, 0, -1, 8)
ifelse(values < 0, "negative", "non-negative")

[1] "negative"     "non-negative" "non-negative" "negative"     "non-negative"

any() and all() Functions

Two useful functions for checking conditions on vectors:

any() - returns TRUE if at least one TRUE value is found
all() - returns TRUE only if all values are TRUE

x <- c(TRUE, FALSE, TRUE)
any(x)  # TRUE

[1] TRUE

all(x)  # FALSE

[1] FALSE

Using any() with Data

Check if any records exist for a year:

any(gapminder$year == 2002)  # TRUE

[1] TRUE

any(gapminder$year == 2012)  # FALSE

[1] FALSE

Similar to the %in% operator!

Repeating Operations: for Loops

If you want to iterate over a set of values and perform the same operation on each, use a for() loop:

for (iterator in set of values) {
  do a thing
}

Basic for Loop

Print numbers 1 through 10:

for (i in 1:10) {
  print(i)
}

[1] 1
[1] 2
[1] 3
[1] 4
[1] 5
[1] 6
[1] 7
[1] 8
[1] 9
[1] 10

The 1:10 creates a vector on the fly.

Iterating Over Any Vector

You can iterate over any vector:

colors <- c("red", "green", "blue")
for (color in colors) {
  print(color)
}

[1] "red"
[1] "green"
[1] "blue"

Nested for Loops

Use a for() loop nested within another to iterate over two things:

for (i in 1:3) {
  for (j in c('a', 'b', 'c')) {
    print(paste(i, j))
  }
}

[1] "1 a"
[1] "1 b"
[1] "1 c"
[1] "2 a"
[1] "2 b"
[1] "2 c"
[1] "3 a"
[1] "3 b"
[1] "3 c"

Understanding Nested Loops

When the first index (i) is set to 1:

The second index (j) iterates through its full set
Once j is complete, i is incremented
Process continues until all indices are used

Storing Loop Results

Write loop output to a new object:

output_vector <- c()
for (i in 1:5) {
  for (j in c('a', 'b', 'c', 'd', 'e')) {
    temp_output <- paste(i, j)
    output_vector <- c(output_vector, temp_output)
  }
}
output_vector

 [1] "1 a" "1 b" "1 c" "1 d" "1 e" "2 a" "2 b" "2 c" "2 d" "2 e" "3 a" "3 b"
[13] "3 c" "3 d" "3 e" "4 a" "4 b" "4 c" "4 d" "4 e" "5 a" "5 b" "5 c" "5 d"
[25] "5 e"

Don’t Grow Your Results!

“Growing” results (building the result object incrementally) is computationally inefficient.

The problem:

Computers are bad at handling this
Calculations can quickly slow to a crawl

Better approach: Define an empty results object with appropriate dimensions before filling in values.

Better Approach: Pre-allocate

Define your output object before filling values:

output_matrix <- matrix(nrow = 5, ncol = 5)
j_vector <- c('a', 'b', 'c', 'd', 'e')
for (i in 1:5) {
  for (j in 1:5) {
    temp_j_value <- j_vector[j]
    temp_output <- paste(i, temp_j_value)
    output_matrix[i, j] <- temp_output
  }
}
output_vector2 <- as.vector(output_matrix)
output_vector2

 [1] "1 a" "2 a" "3 a" "4 a" "5 a" "1 b" "2 b" "3 b" "4 b" "5 b" "1 c" "2 c"
[13] "3 c" "4 c" "5 c" "1 d" "2 d" "3 d" "4 d" "5 d" "1 e" "2 e" "3 e" "4 e"
[25] "5 e"

Challenge 2

Compare the objects output_vector and output_vector2.

Are they the same?
If not, why not?
How would you change the last block of code to make output_vector2 the same as output_vector?

Challenge 2 Solution

Check if they’re identical:

all(output_vector == output_vector2)

[1] FALSE

They’re not the same! But all elements exist in both:

all(output_vector %in% output_vector2)

[1] TRUE

Challenge 2 Solution (continued)

The elements are sorted in different order because as.vector() outputs elements by column.

Fix: Transpose the output matrix:

output_vector2 <- as.vector(t(output_matrix))
all(output_vector == output_vector2)

[1] TRUE

while Loops

Sometimes you need to repeat an operation as long as a condition is met:

while (this condition is true) {
  do a thing
}

R interprets a condition being met as TRUE.

while Loop Example

Generate random numbers until you get one less than 0.1:

set.seed(42)  # For reproducibility
z <- 1
while (z > 0.1) {
  z <- runif(1)
  cat(z, "\n")
}

while Loop Warning

Be careful with while() loops!

Ensure your condition will eventually be FALSE
Otherwise you’ll be stuck in an infinite loop
The loop never terminates if the condition is always met

Challenge 3

Write a script that loops through the gapminder data by continent and prints out whether the mean life expectancy is smaller or larger than 50 years.

Hint: Use unique() to get continent names, then subset and calculate means.

Challenge 3 Solution

Step 1: Get unique continents:

unique(gapminder$continent)

[1] "Asia"     "Europe"   "Africa"   "Americas" "Oceania"

Challenge 3 Solution (continued)

Step 2: Loop and calculate:

thresholdValue <- 50

for (iContinent in unique(gapminder$continent)) {
  tmp <- mean(gapminder[gapminder$continent == iContinent, "lifeExp"])
  
  if (tmp < thresholdValue) {
    cat("Average Life Expectancy in", iContinent, "is less than", thresholdValue, "\n")
  } else {
    cat("Average Life Expectancy in", iContinent, "is greater than", thresholdValue, "\n")
  }
  rm(tmp)
}

Average Life Expectancy in Asia is greater than 50 
Average Life Expectancy in Europe is greater than 50 
Average Life Expectancy in Africa is less than 50 
Average Life Expectancy in Americas is greater than 50 
Average Life Expectancy in Oceania is greater than 50

Challenge 4

Modify the script from Challenge 3 to loop over each country.

This time print out whether the life expectancy is:

smaller than 50
between 50 and 70
greater than 70

Challenge 4 Solution

Add two thresholds and extend the if-else statements:

lowerThreshold <- 50
upperThreshold <- 70

for (iCountry in unique(gapminder$country)) {
  tmp <- mean(gapminder[gapminder$country == iCountry, "lifeExp"])
  
  if (tmp < lowerThreshold) {
    cat("Average Life Expectancy in", iCountry, "is less than", lowerThreshold, "\n")
  } else if (tmp > lowerThreshold && tmp < upperThreshold) {
    cat("Average Life Expectancy in", iCountry, "is between", lowerThreshold, "and", upperThreshold, "\n")
  } else {
    cat("Average Life Expectancy in", iCountry, "is greater than", upperThreshold, "\n")
  }
  rm(tmp)
}

Challenge 5 - Advanced

Write a script that loops over each country in the gapminder dataset:

Tests whether the country starts with a ‘B’
Graphs life expectancy against time as a line graph if the mean life expectancy is under 50 years

Hint: Use grep() with value = TRUE to find countries starting with “B”.

Challenge 5 Solution

Find countries starting with “B”:

grep("^B", unique(gapminder$country), value = TRUE)

 [1] "Bahrain"                "Bangladesh"             "Belgium"               
 [4] "Benin"                  "Bolivia"                "Bosnia and Herzegovina"
 [7] "Botswana"               "Brazil"                 "Bulgaria"              
[10] "Burkina Faso"           "Burundi"

Challenge 5 Solution (continued)

Complete solution:

thresholdValue <- 50
candidateCountries <- grep("^B", unique(gapminder$country), value = TRUE)

for (iCountry in candidateCountries) {
  tmp <- mean(gapminder[gapminder$country == iCountry, "lifeExp"])
  
  if (tmp < thresholdValue) {
    cat("Average Life Expectancy in", iCountry, "is less than", thresholdValue, 
        "- plotting life expectancy graph...\n")
    
    with(subset(gapminder, country == iCountry),
         plot(year, lifeExp,
              type = "o",
              main = paste("Life Expectancy in", iCountry, "over time"),
              ylab = "Life Expectancy",
              xlab = "Year"
         )
    )
  }
}

Average Life Expectancy in Bangladesh is less than 50 - plotting life expectancy graph...

Average Life Expectancy in Benin is less than 50 - plotting life expectancy graph...

Average Life Expectancy in Burkina Faso is less than 50 - plotting life expectancy graph...

Average Life Expectancy in Burundi is less than 50 - plotting life expectancy graph...

When to Use for Loops

The advice of many R users:

Learn about for() loops
But avoid them unless order of iteration is important
i.e., when calculations depend on previous iterations

If order doesn’t matter, consider vectorized alternatives like the purrr package for better computational efficiency.

Key Points

Use if and else to make choices
Use for to repeat operations
Use while for condition-based repetition
Pre-allocate results objects for efficiency
Consider vectorized alternatives when possible

Important Reminders

if() expects a single logical value (length 1)
Use any() or all() to summarize logical vectors
Use ifelse() for vectorized conditional operations
Avoid growing results inside loops
Be careful not to create infinite while loops

Resources

Software Carpentry: r-novice-gapminder
R Documentation: ?if, ?for, ?while
RStudio Cheatsheets