# Vectorization

## Overview

Teaching:10 min

Exercises:15 minQuestions

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.

```
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

`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.## Solution to challenge 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.`gapminder$pop_millions <- gapminder$pop / 1e6 head(gapminder)`

`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`

## Challenge 2

On a single graph, plot population, in millions, against year, for all countries. Don’t worry about identifying which country is which.

Repeat the exercise, graphing only for China, India, and Indonesia. Again, don’t worry about which is which.

## Solution to challenge 2

Refresh your plotting skills by plotting population in millions against year.

`ggplot(gapminder, aes(x = year, y = pop_millions)) + geom_point()`

`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**

```
x > 2
```

```
[1] FALSE FALSE TRUE TRUE
```

**Logical operators**

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

```
[1] FALSE FALSE FALSE TRUE
```

## Tip: some useful functions for logical vectors

`any()`

will return`TRUE`

ifanyelement of a vector is`TRUE`

`all()`

will return`TRUE`

ifallelements of a vector are`TRUE`

Most functions also operate element-wise on vectors:

**Functions**

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

```
[1] 0.0000000 0.6931472 1.0986123 1.3862944
```

Vectorized 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
```

## 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:`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

## Challenge 3

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`

`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!

## Solution to challenge 3

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 0.08333333`

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

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

`m > c(0, 20)`

`[,1] [,2] [,3] [,4] [1,] TRUE FALSE TRUE FALSE [2,] FALSE TRUE FALSE TRUE [3,] TRUE FALSE TRUE FALSE`

## Challenge 4

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 4

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`

## Key Points

Use vectorized operations instead of loops.