R for reproducible scientific analysis

Control flow

Learning Objectives

  • Write conditional statements with if and else.
  • Write and understand for loops.

Often when we’re coding we want to control the flow of our actions. This can be done by setting actions to occur only if a condition or a set of conditions are met. Alternatively, we can also set an action to occur a particular number of times.

There are several ways you can control flow in R. For conditional statements, the most commonly used approaches are the constructs:

# 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
}

Say, for example, that we want R to print a message if a variable x has a particular value:

# sample a random number from a Poisson distribution
# with a mean (lambda) of 8

x <- rpois(1, lambda=8)

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

x
[1] 8

Note you may not get the same output as your neighbour because you may be sampling different random numbers from the same distribution.

Let’s set a seed so that we all generate the same ‘pseudo-random’ number, and then print more information:

x <- rpois(1, lambda=8)

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

Important: when R evaluates the condition inside if statements, it is looking for a logical element, i.e., TRUE or FALSE. This can cause some headaches for beginners. For example:

x  <-  4 == 3
if (x) {
  "4 equals 3"
}

As we can see, the message was not printed because the vector 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 years of birth from 1812 in the healthData dataset. Now do the same for 1910.

Did anyone get a warning message like this?

Warning in if (healthData$birthYear == 1812) {: the condition has length >
1 and only the first element will be used

If your condition evaluates to a vector with more than one logical element, the function if will still run, but will only evaluate the condition in the first element. Here you need to make sure your condition is of length 1.

Repeating operations

If you want to iterate over a set of values, when the order of iteration is important, and perform the same operation on each, a for loop will do the job. We saw for loops in the shell lessons earlier. This is the most flexible of looping operations, but therefore also the hardest to use correctly. Avoid using for loops unless the order of iteration is important: i.e. the calculation at each iteration depends on the results of previous iterations.

The basic structure of a for loop is:

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

For example:

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 bit creates a vector on the fly; you can iterate over any other vector as well.

We can use a for loop nested within another for loop to iterate over two things at once.

for (i in 1:5){
  for(j in c('a', 'b', 'c', 'd', 'e')){
    print(paste(i,j))
  }
}
[1] "1 a"
[1] "1 b"
[1] "1 c"
[1] "1 d"
[1] "1 e"
[1] "2 a"
[1] "2 b"
[1] "2 c"
[1] "2 d"
[1] "2 e"
[1] "3 a"
[1] "3 b"
[1] "3 c"
[1] "3 d"
[1] "3 e"
[1] "4 a"
[1] "4 b"
[1] "4 c"
[1] "4 d"
[1] "4 e"
[1] "5 a"
[1] "5 b"
[1] "5 c"
[1] "5 d"
[1] "5 e"

Rather than printing the results, we could write the 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"
[12] "3 b" "3 c" "3 d" "3 e" "4 a" "4 b" "4 c" "4 d" "4 e" "5 a" "5 b"
[23] "5 c" "5 d" "5 e"

This approach can be useful, but ‘growing your results’ (building the result object incrementally) is computationally inefficient, so avoid it when you are iterating through a lot of values.

A better way is to define your (empty) output object before filling in the values. For this example, it looks more involved, but is still more efficient.

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"
[12] "2 c" "3 c" "4 c" "5 c" "1 d" "2 d" "3 d" "4 d" "5 d" "1 e" "2 e"
[23] "3 e" "4 e" "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 3

Write a script that loops through the healthData data by illness level and prints out whether the mean health measure is smaller or larger than 8 units. Hint: you may want to check out the functions na.rm(), is.na() and unique()

Challenge 4

Modify the script from Challenge 4 to also loop over each study. This time print out whether the health measure is smaller than 5, between 5 and 8, or greater than 8.

Challenge solutions

Solution to challenge 1

Use an if statement to print a suitable message reporting whether there are any years of birth from 1812 in the healthData dataset. Now do the same for 1910.

if (any(healthData$birthYear == 1812)){
  print("There was at least one person born in 1812 in the dataset")
} else {
  print("There are no people in the dataset who were born in 1812")
}
[1] "There are no people in the dataset who were born in 1812"

if (any(healthData$birthYear == 1910)){
  print("There was at least one person born in 1910 in the dataset")
} else {
  print("There are no people in the dataset who were born in 1910")
}
[1] "There was at least one person born in 1910 in the dataset"

Solution to 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?

The rows and columns have been swapped between output_vector and output_vector2

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[j, i] <- temp_output
 }
}
output_vector2 <- as.vector(output_matrix)
output_vector2
[1] "1 a" "1 b" "1 c" "1 d" "1 e" "2 a" "2 b" "2 c" "2 d" "2 e" "3 a"
[12] "3 b" "3 c" "3 d" "3 e" "4 a" "4 b" "4 c" "4 d" "4 e" "5 a" "5 b"
[23] "5 c" "5 d" "5 e"

Solution to challenge 3

Write a script that loops through the healthData data by illness level and prints out whether the mean health measure is smaller or larger than 8 units.

for (illness in sort(unique(healthData$illnessReversed[!is.na(healthData$illnessReversed)]))){
  if ((mean(healthData$health[healthData$illnessReversed == illness], na.rm=T) > 8)){
  print(paste("The mean health measure for people with", illness, "illness is greater than 8 units"))
  } else {
  print(paste("The mean health measure for people with", illness, "illness is less than 8 units"))
  }
}
[1] "The mean health measure for people with 1 illness is less than 8 units"
[1] "The mean health measure for people with 2 illness is less than 8 units"
[1] "The mean health measure for people with 3 illness is less than 8 units"
[1] "The mean health measure for people with 4 illness is greater than 8 units"
[1] "The mean health measure for people with 5 illness is greater than 8 units"

Solution to challenge 4

Modify the script from Challenge 4 to also loop over each study. This time print out whether the health measure is smaller than 5, between 5 and 8, or greater than 8.

for (illness in sort(unique(healthData$illnessReversed[!is.na(healthData$illnessReversed)]))){
  for (group in unique(healthData$HIGroup)){
    if ((ans <- mean(healthData$health[healthData$illnessReversed == illness & healthData$HIGroup == group], na.rm=T)) > 8){
      print(paste("The mean health measure for people with", illness, "illness in group", group, "is greater than 8 units"))
    } else if (ans < 5){
      print(paste("The mean health measure for people with", illness, "illness in group", group, "is less than 5 units"))
    } else {
      print(paste("The mean health measure for people with", illness, "illness in group", group, "is between 5 and 8 units"))
    }
  }
}
[1] "The mean health measure for people with 1 illness in group Group 1 is less than 5 units"
[1] "The mean health measure for people with 1 illness in group Group 2 is less than 5 units"
[1] "The mean health measure for people with 2 illness in group Group 1 is between 5 and 8 units"
[1] "The mean health measure for people with 2 illness in group Group 2 is between 5 and 8 units"
[1] "The mean health measure for people with 3 illness in group Group 1 is between 5 and 8 units"
[1] "The mean health measure for people with 3 illness in group Group 2 is between 5 and 8 units"
[1] "The mean health measure for people with 4 illness in group Group 1 is greater than 8 units"
[1] "The mean health measure for people with 4 illness in group Group 2 is greater than 8 units"
[1] "The mean health measure for people with 5 illness in group Group 1 is greater than 8 units"
[1] "The mean health measure for people with 5 illness in group Group 2 is greater than 8 units"