# Vectorization

Last updated on 2024-05-12 | Edit this page

Estimated time: 25 minutes

## 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 8`

The 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 13`

Each 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 13`

Compare this to the output using vectorised operations.

### R

```
sum_xy <- x + y
sum_xy
```

### OUTPUT

`[1] 7 9 11 13`

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 14.880372
```

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 TRUE`

**Logical operators**

### R

```
a <- x > 3 # or, for clarity, a <- (x > 3)
a
```

### OUTPUT

`[1] FALSE FALSE FALSE TRUE`

Most functions also operate element-wise on vectors:

**Functions**

### R

```
x <- 1:4
log(x)
```

### OUTPUT

`[1] 0.0000000 0.6931472 1.0986123 1.3862944`

Vectorized 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 -12
```

### Tip: 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,] 30
```

### R

`matrix(1:4, nrow=1) %*% matrix(1:4, ncol=1)`

### OUTPUT

```
[,1]
[1,] 30
```

For 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 12
```

Write down what you think will happen when you run:

`m ^ -1`

`m * 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 12
```

Write 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 0.08333333
```

`m * c(1, 0, -1)`

### OUTPUT

```
[,1] [,2] [,3] [,4]
[1,] 1 4 7 10
[2,] 0 0 0 0
[3,] -3 -6 -9 -12
```

`m > c(0, 20)`

### OUTPUT

```
[,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:

### 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] 1.634984`

### R

`sum(1/(1:1e04)^2)`

### OUTPUT

`[1] 1.644834`

### R

```
n <- 10000
sum(1/(1:n)^2)
```

### OUTPUT

`[1] 1.644834`

We 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] 1.634984`

### R

`inverse_sum_of_squares(10000)`

### OUTPUT

`[1] 1.644834`

### R

```
n <- 10000
inverse_sum_of_squares(n)
```

### OUTPUT

`[1] 1.644834`

### Tip: 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
length
```

### OUTPUT

`[1] 2 4 6 5 7 9 8`

Vector `x`

was recycled to match the length of vector
`y`