R for reproducible scientific analysis

Vectorisation

Learning Objectives

  • To understand vectorised operations in R.

Most of R’s functions are vectorised, meaning that the function will operate on all elements of a vector without needing to loop through and act on each element one at a time. This makes writing code more concise, easy to read, and less error prone.

x <- 1:4
x * 2
[1] 2 4 6 8

The multiplication happened to each element of the vector.

We can also add two vectors together:

y <- 6:9
x + y
[1]  7  9 11 13

Each element of x was added to its corresponding element of y:

x:  1  2  3  4
    +  +  +  +
y:  6  7  8  9
---------------
    7  9 11 13

Challenge 1

Let’s try this on the health column of the healthData dataset.

Make a new column in the healthData data frame that contains health rounded to the nearest integer. Check the head or tail of the data frame to make sure it worked.

Hint: R has a round() function

Comparison operators, logical operators, and many functions are also vectorized:

Comparison operators

x > 2
[1] FALSE FALSE  TRUE  TRUE

Logical operators

a <- x > 3  # or, for clarity, a <- (x > 3)
a
[1] FALSE FALSE FALSE  TRUE

Most functions also operate element-wise on vectors:

Functions

x <- 1:4
log(x)
[1] 0.0000000 0.6931472 1.0986123 1.3862944

Vectorised operations work element-wise on matrices:

m <- matrix(1:12, nrow=3, ncol=4)
m * -1  
     [,1] [,2] [,3] [,4]
[1,]   -1   -4   -7  -10
[2,]   -2   -5   -8  -11
[3,]   -3   -6   -9  -12

Challenge 2

Given the following matrix:

m <- matrix(1:12, nrow=3, ncol=4)
m
     [,1] [,2] [,3] [,4]
[1,]    1    4    7   10
[2,]    2    5    8   11
[3,]    3    6    9   12

Write down what you think will happen when you run:

  1. m ^ -1
  2. m * c(1, 0, -1)
  3. m > c(0, 20)
  4. m * c(1, 0, -1, 2)

Did you get the output you expected? If not, ask a helper!

Challenge 3

We’re interested in looking at the sum of the following sequence of fractions:

 x = 1/(1^2) + 1/(2^2) + 1/(3^2) + ... + 1/(n^2)

This would be tedious to type out, and impossible for high values of n. Use vectorisation to compute x when n=100. What is the sum when n=10,000?

Challenge solutions

Solution to challenge 1

Let’s try this on the health column of the healthData dataset.

Make a new column in the healthData data frame that contains health rounded to the nearest integer. Check the head or tail of the data frame to make sure it worked.

Hint: R has a round() function

healthData$healthInteger <- round(healthData$health)
head(healthData)
  id conscientiousness extraversion intellect agreeableness neuroticism
1  3             5.825        3.986     6.044         4.613       3.649
2  4             7.732        7.016     6.821         6.649       6.299
3  7             6.498        2.697     5.527         3.087       4.091
4  8             5.881        2.504     4.234         4.613       3.649
5 10             4.254        5.147     4.751         3.850       3.208
6 12             7.508        3.535     6.821         4.613       5.415
   sex selfRatedHealth mentalAdjustment illnessReversed health
1 Male               4                2               3   6.74
2 Male               5                3               5  11.96
3 Male               3                3               4   8.05
4 Male               3                2               4   6.48
5 Male               4                2               3   6.74
6 Male               4                2               5   9.01
  alcoholUseInYoungAdulthood education birthYear HIGroup healthInteger
1                          2         9      1909 Group 1             7
2                          3         8      1905 Group 1            12
3                          2         6      1910 Group 1             8
4                          1         8      1905 Group 1             6
5                          2         9      1910 Group 1             7
6                          2         4      1911 Group 1             9

Solution to challenge 2

Given the following matrix:

m <- matrix(1:12, nrow=3, ncol=4)
m
     [,1] [,2] [,3] [,4]
[1,]    1    4    7   10
[2,]    2    5    8   11
[3,]    3    6    9   12

Write down what you think will happen when you run:

  1. m ^ -1
          [,1]      [,2]      [,3]       [,4]
[1,] 1.0000000 0.2500000 0.1428571 0.10000000
[2,] 0.5000000 0.2000000 0.1250000 0.09090909
[3,] 0.3333333 0.1666667 0.1111111 0.08333333
  1. m * c(1, 0, -1)
     [,1] [,2] [,3] [,4]
[1,]    1    4    7   10
[2,]    0    0    0    0
[3,]   -3   -6   -9  -12
  1. m > c(0, 20)
      [,1]  [,2]  [,3]  [,4]
[1,]  TRUE FALSE  TRUE FALSE
[2,] FALSE  TRUE FALSE  TRUE
[3,]  TRUE FALSE  TRUE FALSE

Challenge 3

We’re interested in looking at the sum of the following sequence of fractions:

 x = 1/(1^2) + 1/(2^2) + 1/(3^2) + ... + 1/(n^2)

This would be tedious to type out, and impossible for high values of n. Can you use vectorisation to compute x, when n=100? How about when n=10,000?

sum(1/(1:100)^2)
[1] 1.634984

sum(1/(1:1e04)^2)
[1] 1.644834

n <- 10000
sum(1/(1:n)^2)
[1] 1.644834

We can also obtain the same results using a function:

inverse_sum_of_squares <- function(n) {
  sum(1/(1:n)^2)
}
inverse_sum_of_squares(100)
[1] 1.634984

inverse_sum_of_squares(10000)
[1] 1.644834

n <- 10000
inverse_sum_of_squares(n)
[1] 1.644834