R for reproducible scientific analysis

Control flow

Learning objectives

  • Write conditional statements with if and else.
  • Write and understand while and 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 records from 2002 in the gapminder dataset. Now do the same for 2012.

Did anyone get a warning message like this?

Warning in if (gapminder$year == 2012) {: 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.

Tip: any and all

The any function will return TRUE if at least one TRUE value is found within a vector, otherwise it will return FALSE. This can be used in a similar way to the %in% operator. The function all, as the name suggests, will only return TRUE if all values in the vector are TRUE.

Repeating operations

Sometimes you will find yourself needing to repeat an operation until a certain condition is met. You can do this with a while loop.

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

Let’s try an example, shall we? We’ll try to come up with some simple code that generates random numbers from a uniform distribution (the runif function) between 0 and 1 until it gets one that’s less than 0.1.

while(z > 0.1){
  z <- runif(1)
  print(z)
}
Error: object 'z' not found

But wait, that doesn’t work! What’s the problem?

The problem is that we haven’t defined z, and so the very first time the while loop’s condition is checked (z > 0.1), while just says “Okay, that’s not true so I’m not going to execute this block of code”. The same thing would have happened if we defined z to be anything less than 0.1. Let’s fix it.

z <- 1
while(z > 0.1){
  z <- runif(1)
  print(z)
}
[1] 0.4269077
[1] 0.6931021
[1] 0.08513597

Challenge 2

Use a while loop to construct a vector called ‘pet_list’ with the value: ‘cat’, ‘dog’, ‘dog’, ‘dog’, ‘dog’ (N.B. using a loop may not be the most efficient way to do this, but it illustrates the principle!)

while loops will not always be appropriate. 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 3

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 4

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.

Challenge 5

Modify the script from Challenge 4 to also loop over each country. This time print out whether the life expectancy is smaller than 50, between 50 and 70, or greater than 70.

Challenge 6 - Advanced

Write a script that loops over each country in the gapminder dataset, tests whether the country starts with a ‘B’, and graphs life expectancy against time as a line graph if the mean life expectancy is under 50 years.

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