Vectorisation
Table of Contents
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
Comparison operators, logical operators, and many functions are also vectorised:
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
Very important: the operator * gives you element-wise multiplication!
To do matrix multiplication, we need to use the %*% operator:
m %*% matrix(1, nrow=4, ncol=1)
[,1]
[1,] 22
[2,] 26
[3,] 30
matrix(1:4, nrow=1) %*% matrix(1:4, ncol=1)
[,1]
[1,] 30
For more on matrix algebra, see the Quick-R reference guide
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:
m ^ -1m * c(1, 0, -1)m > c(0, 20)m * c(1, 0, -1, 2)
Did you get the output you expected? If not, ask a helper!
Challenge solutions
Let's try this on the Fare column of the titanic dataset.
Make a new column in the titanic data frame that contains Fare 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
titanic$FareInteger <- round(titanic$Fare)
head(titanic)
PassengerId Survived Pclass
1 1 0 3
2 2 1 1
3 3 1 3
4 4 1 1
5 5 0 3
6 6 0 3
Name Sex Age SibSp
1 Braund, Mr. Owen Harris male 22 1
2 Cumings, Mrs. John Bradley (Florence Briggs Thayer) female 38 1
3 Heikkinen, Miss. Laina female 26 0
4 Futrelle, Mrs. Jacques Heath (Lily May Peel) female 35 1
5 Allen, Mr. William Henry male 35 0
6 Moran, Mr. James male NA 0
Parch Ticket Fare Cabin Embarked FareInteger
1 0 A/5 21171 7.2500 S 7
2 0 PC 17599 71.2833 C85 C 71
3 0 STON/O2. 3101282 7.9250 S 8
4 0 113803 53.1000 C123 S 53
5 0 373450 8.0500 S 8
6 0 330877 8.4583 Q 8
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:
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 0.08333333m * c(1, 0, -1)[,1] [,2] [,3] [,4] [1,] 1 4 7 10 [2,] 0 0 0 0 [3,] -3 -6 -9 -12m > c(0, 20)[,1] [,2] [,3] [,4] [1,] TRUE FALSE TRUE FALSE [2,] FALSE TRUE FALSE TRUE [3,] TRUE FALSE TRUE FALSE
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