R for reproducible scientific analysis
Control flow
Learning Objectives
- Write conditional statements with
if
andelse
. - 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 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
andall
The
any
function will return TRUE if at least one TRUE value is found within a vector, otherwise it will returnFALSE
. This can be used in a similar way to the%in%
operator. The functionall
, as the name suggests, will only returnTRUE
if all values in the vector areTRUE
.
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 gapminder
data by continent and prints out whether the mean life expectancy is smaller or larger than 50 years.
Challenge 4
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 5 - 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.