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Programming with R

Control Flow

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Overview

Teaching: 20 min
Exercises: 10 min
Questions

  • How can I make data-dependent choices in R?

  • How can I repeat operations in R?

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.

Conditional Statements

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.

x <- rpois(1, lambda=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"

x <- rpois(1, lambda=8)

if (x >= 10) {
  print("x is greater than or equal to 10")
} else if (10 > x > 6) {
  print("x is between 10 and 6")
} else {
  print("x is less than or equal to 6")
}

[1] "x is greater than or equal to 10"

Tip: pseudo-random numbers

In the above case, the function rpois generates a random number following a Poisson distribution with a mean (i.e. lambda) of 8. The function set.seed guarantees that all machines will generate the exact same ‘pseudo-random’ number (more about pseudo-random numbers). So if we set.seed(10), we see that x takes the value 8. You should get the exact same number.

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. Then write a similar statement that reports if there are both 2002 and 2012 records.

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.

While loops

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
}

As an example, here’s a while loop 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.

z <- 1
while(z > 0.1){
  z <- runif(1)
  print(z)
}

while loops will not always be appropriate. You have to be particularly careful that you don’t end up in an infinite loop because your condition is never met.

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_matrix <- matrix(nrow=5, ncol=5)
for(i in 1:5){
  for(j in c('a', 'b', 'c', 'd', 'e')){
    temp_output <- paste(i, j)
    output_matrix[i, j] <- 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"

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.

Tip: don’t grow your results

One of the biggest things that trips up novices and experienced R users alike, is building a results object (vector, list, matrix, data frame) as your for loop progresses. Computers are very bad at handling this, so your calculations can very quickly slow to a crawl. It’s much better to define an empty results object before hand of the appropriate dimensions. So if you know the end result will be stored in a matrix like above, create an empty matrix with 5 row and 5 columns, then at each iteration store the results in the appropriate location.

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

Write a script that loops through the gapminder data by continent and prints out the continent, the mean life expectancy on that continent, and whether or not that life expectancy is larger than 65 years. Hints: If x is a numeric vector, mean(x) returns the mean of x. For any vector x, unique(x) returns a vector with the unique values of x. Finally cat("x is",6) prints x is 6.

Challenge 3

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

Key Points

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