Vectorization
Last updated on 2024-11-05 | Edit this page
Overview
Questions
- How can I operate on all the elements of a vector at once?
Objectives
- To understand vectorized operations in R.
Most of R’s functions are vectorized, 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.
R
x <- 1:4
x * 2
OUTPUT
[1] 2 4 6 8The multiplication happened to each element of the vector.
We can also add two vectors together:
R
y <- 6:9
x + y
OUTPUT
[1] 7 9 11 13Each element of x was added to its corresponding element
of y:
Here is how we would add two vectors together using a for loop:
R
output_vector <- c()
for (i in 1:4) {
output_vector[i] <- x[i] + y[i]
}
output_vector
OUTPUT
[1] 7 9 11 13Compare this to the output using vectorised operations.
R
sum_xy <- x + y
sum_xy
OUTPUT
[1] 7 9 11 13Challenge 1
Let’s try this on the pop column of the
gapminder dataset.
Make a new column in the gapminder data frame that
contains population in units of millions of people. Check the head or
tail of the data frame to make sure it worked.
Let’s try this on the pop column of the
gapminder dataset.
Make a new column in the gapminder data frame that
contains population in units of millions of people. Check the head or
tail of the data frame to make sure it worked.
R
gapminder$pop_millions <- gapminder$pop / 1e6
head(gapminder)
OUTPUT
country year pop continent lifeExp gdpPercap pop_millions
1 Afghanistan 1952 8425333 Asia 28.801 779.4453 8.425333
2 Afghanistan 1957 9240934 Asia 30.332 820.8530 9.240934
3 Afghanistan 1962 10267083 Asia 31.997 853.1007 10.267083
4 Afghanistan 1967 11537966 Asia 34.020 836.1971 11.537966
5 Afghanistan 1972 13079460 Asia 36.088 739.9811 13.079460
6 Afghanistan 1977 14880372 Asia 38.438 786.1134 14.880372Challenge 2
On a single graph, plot population, in millions, against year, for all countries. Do not worry about identifying which country is which.
Repeat the exercise, graphing only for China, India, and Indonesia. Again, do not worry about which is which.
Refresh your plotting skills by plotting population in millions against year.
R
ggplot(gapminder, aes(x = year, y = pop_millions)) +
geom_point()
R
countryset <- c("China","India","Indonesia")
ggplot(gapminder[gapminder$country %in% countryset,],
aes(x = year, y = pop_millions)) +
geom_point()
Comparison operators, logical operators, and many functions are also vectorized:
Comparison operators
R
x > 2
OUTPUT
[1] FALSE FALSE TRUE TRUELogical operators
R
a <- x > 3 # or, for clarity, a <- (x > 3)
a
OUTPUT
[1] FALSE FALSE FALSE TRUETip: some useful functions for logical vectors
any() will return TRUE if any
element of a vector is TRUE.all() will return TRUE if all
elements of a vector are TRUE.
Most functions also operate element-wise on vectors:
Functions
R
x <- 1:4
log(x)
OUTPUT
[1] 0.0000000 0.6931472 1.0986123 1.3862944Vectorized operations work element-wise on matrices:
R
m <- matrix(1:12, nrow=3, ncol=4)
m * -1
OUTPUT
[,1] [,2] [,3] [,4]
[1,] -1 -4 -7 -10
[2,] -2 -5 -8 -11
[3,] -3 -6 -9 -12Tip: element-wise vs. matrix multiplication
Very important: the operator * gives you element-wise
multiplication! To do matrix multiplication, we need to use the
%*% operator:
R
m %*% matrix(1, nrow=4, ncol=1)
OUTPUT
[,1]
[1,] 22
[2,] 26
[3,] 30R
matrix(1:4, nrow=1) %*% matrix(1:4, ncol=1)
OUTPUT
[,1]
[1,] 30For more on matrix algebra, see the Quick-R reference guide
Challenge 3
Given the following matrix:
R
m <- matrix(1:12, nrow=3, ncol=4)
m
OUTPUT
[,1] [,2] [,3] [,4]
[1,] 1 4 7 10
[2,] 2 5 8 11
[3,] 3 6 9 12Write 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!
Given the following matrix:
R
m <- matrix(1:12, nrow=3, ncol=4)
m
OUTPUT
[,1] [,2] [,3] [,4]
[1,] 1 4 7 10
[2,] 2 5 8 11
[3,] 3 6 9 12Write down what you think will happen when you run:
m ^ -1
OUTPUT
[,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)
OUTPUT
[,1] [,2] [,3] [,4]
[1,] 1 4 7 10
[2,] 0 0 0 0
[3,] -3 -6 -9 -12m > c(0, 20)
OUTPUT
[,1] [,2] [,3] [,4]
[1,] TRUE FALSE TRUE FALSE
[2,] FALSE TRUE FALSE TRUE
[3,] TRUE FALSE TRUE FALSEChallenge 4
We’re interested in looking at the sum of the following sequence of fractions:
R
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?
We’re interested in looking at the sum of the following sequence of fractions:
R
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?
R
sum(1/(1:100)^2)
OUTPUT
[1] 1.634984R
sum(1/(1:1e04)^2)
OUTPUT
[1] 1.644834R
n <- 10000
sum(1/(1:n)^2)
OUTPUT
[1] 1.644834We can also obtain the same results using a function:
R
inverse_sum_of_squares <- function(n) {
sum(1/(1:n)^2)
}
inverse_sum_of_squares(100)
OUTPUT
[1] 1.634984R
inverse_sum_of_squares(10000)
OUTPUT
[1] 1.644834R
n <- 10000
inverse_sum_of_squares(n)
OUTPUT
[1] 1.644834Tip: Operations on vectors of unequal length
Operations can also be performed on vectors of unequal length, through a process known as recycling. This process automatically repeats the smaller vector until it matches the length of the larger vector. R will provide a warning if the larger vector is not a multiple of the smaller vector.
R
x <- c(1, 2, 3)
y <- c(1, 2, 3, 4, 5, 6, 7)
x + y
WARNING
Warning in x + y: longer object length is not a multiple of shorter object
lengthOUTPUT
[1] 2 4 6 5 7 9 8Vector x was recycled to match the length of vector
y
Key Points
- Use vectorized operations instead of loops.