1 Introduction

In programming languages loop structures, either with or without conditions, are used to repeat commands over multiple entities. For and while loops as well as if-else statements are also often used in R, but not as often as in many other programming languages. The reason for this is that many needs of the loops are addressed using vectorization or via apply functions.

This means that we can multiply all values in a vector in R by two by calling

vec.a <- c(1, 2, 3, 4)
vec.a * 2
## [1] 2 4 6 8

In many other and languages as well as in R, you can also create this with a loop instead

for (i in vec.a) {
  vec.a[i] <- vec.a[i] * 2
}

vec.a
## [1] 2 4 6 8

As you saw in the lecture, this is far less efficient and not by any means easier to type and we hence tend to avoid loops when possible.

After this exercise you should know:

  • What are the most common loop structures in R
  • Some common alternatives to using loops in R
  • How one can convert a short script to a function.
  • Use that new function in R.

2 Exercises

  1. Create a 100000 by 10 matrix with the numbers 1:1000000. Make a for-loop that calculates the sum for each row of the matrix. Verify that your results are consistent with what you obtain with the apply() function to calculate row sums as well as with the built-in rowSums() function. These functions were discussed in the lecture Elements of the programming language - part 2.

X <- matrix(1:1000000, nrow = 100000, ncol = 10)
for.sum <- vector()
# Note that this loop is much faster if you outside the loop create an empty vector of the right size.
# rwmeans <- vector('integer', 100000)
for (i in 1:nrow(X)) {
    for.sum[i] <- sum(X[i,])
}
head(for.sum)
## [1] 4500010 4500020 4500030 4500040 4500050 4500060
app.sum <- apply(X, MARGIN = 1, sum)
head(app.sum)
## [1] 4500010 4500020 4500030 4500040 4500050 4500060
rowSums.sum <- rowSums(X)
head(rowSums.sum)
## [1] 4500010 4500020 4500030 4500040 4500050 4500060
identical(for.sum, app.sum)
## [1] TRUE
identical(for.sum, rowSums.sum)
## [1] FALSE
identical(for.sum, as.integer(rowSums.sum))
## [1] TRUE
  1. Another common loop structure that is used is the while loop, which functions much like a for loop, but will only run as long as a test condition is TRUE. Modify your for loop from exercise 1 and make it into a while loop.

x <- 1
while.sum <- vector("integer", 100000)
while (x < 100000) {
  while.sum[x] <- sum(X[x,])
  x <- x + 1
}
head(while.sum)
## [1] 4500010 4500020 4500030 4500040 4500050 4500060
  1. Create a data frame with two numeric and one character vector. Write a loop that loops over the columns and reports the sum of the column values if it is numeric and the total number of characters if it is a character vector.

vector1 <- 1:10
vector2 <- c("Odd", "Loop", letters[1:8])
vector3 <- rnorm(10, sd = 10)
dfr1 <- data.frame(vector1, vector2, vector3, stringsAsFactors = FALSE)
sum.vec <- vector()
for(i in 1:ncol(dfr1)) {
  if (is.numeric(dfr1[,i])) {
      sum.vec[i] <- sum(dfr1[,i])
  } else {
      sum.vec[i] <- sum(nchar(dfr1[,i]))
  }
}
sum.vec
## [1]  55.00000  15.00000 -56.05122
  1. In question 3 you generated a loop to go over a data frame. Try to convert this code to a function in R. The function should take a single data frame name as argument.

dfr.info <- function(dfr) {
sum.vec <- vector()
for (i in 1:ncol(dfr)) {
    if (is.numeric(dfr[,i])) {
        sum.vec[i] <- mean(dfr[,i])
    } else {
        sum.vec[i] <- sum(nchar(dfr[,i]))
    }
}
sum.vec
}
  1. Read up on the if-else function in R. If possible use the if-else function to answer question 3.

  2. In all loops that we tried out we have created the variable where the output is saved outside the loop. Why is this?

  3. Advanced: At the lecture an approach to calculate factorials were implemented using recursion (function calling itself). Here we instead will have a go at generating Fibonacci numbers. A fibonacci number is part of a series of number with the following properties:

The first two numbers in the Fibonacci sequence are either 1 and 1, or 0 and 1, depending on the chosen starting point of the sequence, and each subsequent number is the sum of the previous two. Hence:

0, 1, 1, 2, 3, 5, 8, 13, 21, ...

or

1, 1, 2, 3, 5, 8, 13, 21, ...

Try to generate such a series using a recursive approach.