Subsetting Data

Overview

Teaching: 35 min
Exercises: 15 min

Questions

How can I work with subsets of data in R?

Objectives

To be able to subset vectors, factors, matrices, lists, and data frames

To be able to extract individual and multiple elements: by index, by name, using comparison operations

To be able to skip and remove elements from various data structures.

R has many powerful subset operators. Mastering them will allow you to easily perform complex operations on any kind of dataset.

There are six different ways we can subset any kind of object, and three different subsetting operators for the different data structures.

Let’s start with the workhorse of R: a simple numeric vector.

x <- c(5.4, 6.2, 7.1, 4.8, 7.5)
names(x) <- c('a', 'b', 'c', 'd', 'e')
x

  a   b   c   d   e 
5.4 6.2 7.1 4.8 7.5 

Atomic vectors

In R, simple vectors containing character strings, numbers, or logical values are called atomic vectors because they can’t be further simplified.

So now that we’ve created a dummy vector to play with, how do we get at its contents?

Accessing elements using their indices

To extract elements of a vector we can give their corresponding index, starting from one:

x[1]

  a 
5.4 

x[4]

  d 
4.8 

It may look different, but the square brackets operator is a function. For vectors (and matrices), it means “get me the nth element”.

We can ask for multiple elements at once:

x[c(1, 3)]

  a   c 
5.4 7.1 

Or slices of the vector:

x[1:4]

  a   b   c   d 
5.4 6.2 7.1 4.8 

the : operator creates a sequence of numbers from the left element to the right.

1:4

[1] 1 2 3 4

Just like if we had used the c function.

c(1, 2, 3, 4)

[1] 1 2 3 4

It can be used to quickly create sequences from any starting and stoping point.

-3:2

[1] -3 -2 -1  0  1  2

We can ask for the same element multiple times:

x[c(1,1,3)]

  a   a   c 
5.4 5.4 7.1 

If we ask for an index beyond the length of the vector, R will return a missing value:

x[6]

<NA> 
  NA 

This is a vector of length one containing an NA, whose name is also NA.

If we ask for the 0th element, we get an empty vector:

x[0]

named numeric(0)

Vector numbering in R starts at 1

In many programming languages (C and Python, for example), the first element of a vector has an index of 0. In R, the first element is 1.

Skipping and removing elements

If we use a negative number as the index of a vector, R will return every element except for the one specified:

x[-2]

  a   c   d   e 
5.4 7.1 4.8 7.5 

We can skip multiple elements:

x[c(-1, -5)]  # or x[-c(1,5)]

  b   c   d 
6.2 7.1 4.8 

Tip: Order of operations

A common trip up for novices occurs when trying to skip slices of a vector. It’s natural to to try to negate a sequence like so:
x[-1:3]
This gives a somewhat cryptic error:
Error in x[-1:3]: only 0's may be mixed with negative subscripts
But remember the order of operations. : is really a function. It takes its first argument as -1, and its second as 3, so generates the sequence of numbers: c(-1, 0, 1, 2, 3).

The correct solution is to wrap that function call in brackets, so that the - operator applies to the result:
x[-(1:3)]
  d   e 
4.8 7.5 

To remove elements from a vector, we need to assign the result back into the variable:

x <- x[-4]
x

  a   b   c   e 
5.4 6.2 7.1 7.5 

Challenge 1

Given the following code:
x <- c(5.4, 6.2, 7.1, 4.8, 7.5)
names(x) <- c('a', 'b', 'c', 'd', 'e')
print(x)
  a   b   c   d   e 
5.4 6.2 7.1 4.8 7.5 
Come up with at least 3 different commands that will produce the following output:
  b   c   d 
6.2 7.1 4.8 
After you find 3 different commands, compare notes with your neighbor. Did you have different strategies?
Solution to challenge 1

Any of the following are valid approaches:
x[2:4]
  b   c   d 
6.2 7.1 4.8 
x[-c(1,5)]
  b   c   d 
6.2 7.1 4.8 
x[c("b", "c", "d")]
  b   c   d 
6.2 7.1 4.8 
x[c(2,3,4)]
  b   c   d 
6.2 7.1 4.8 
Did you find any others that we didn’t include here?

Subsetting by name

We can extract elements by using their name, instead of extracting by index:

x <- c(a=5.4, b=6.2, c=7.1, d=4.8, e=7.5) # we can name a vector 'on the fly'
x[c("a", "c")]

  a   c 
5.4 7.1 

This is usually a much more reliable way to subset objects: the position of various elements can often change when chaining together subsetting operations, but the names will always remain the same!

Subsetting through other logical operations

We can also use any logical vector to subset:

x[c(FALSE, FALSE, TRUE, FALSE, TRUE)]

  c   e 
7.1 7.5 

Since comparison operators (e.g. >, <, ==) evaluate to logical vectors, we can also use them to succinctly subset vectors: the following statement gives the same result as the previous one.

x[x > 7]

  c   e 
7.1 7.5 

Breaking it down, this statement first evaluates x>7, generating a logical vector c(FALSE, FALSE, TRUE, FALSE, TRUE), and then selects the elements of x corresponding to the TRUE values.

We can use == to mimic the previous method of indexing by name (remember you have to use == rather than = for comparisons):

x[names(x) == "a"]

  a 
5.4 

Tip: Combining logical conditions

We often want to combine multiple logical criteria. For example, we might want to find all the countries that are located in Asia or Europe and have life expectancies within a certain range. Several operations for combining logical vectors exist in R:

&, the “logical AND” operator: returns TRUE if both the left and right are TRUE.

|, the “logical OR” operator: returns TRUE, if either the left or right (or both) are TRUE.

You may sometimes see && and || instead of & and |. These two-character operators only look at the first element of each vector and ignore the remaining elements. In general you should not use the two-character operators in data analysis; save them for programming, i.e. deciding whether to execute a statement.

!, the “logical NOT” operator: converts TRUE to FALSE and FALSE to TRUE. It can negate a single logical condition (eg !TRUE becomes FALSE), or a whole vector of conditions(eg !c(TRUE, FALSE) becomes c(FALSE, TRUE)).

Additionally, you can compare the elements within a single vector using the all function (which returns TRUE if every element of the vector is TRUE) and the any function (which returns TRUE if one or more elements of the vector are TRUE).

Challenge 2

Given the following code:
x <- c(5.4, 6.2, 7.1, 4.8, 7.5)
names(x) <- c('a', 'b', 'c', 'd', 'e')
print(x)
  a   b   c   d   e 
5.4 6.2 7.1 4.8 7.5 
Write a subsetting command to return the values in x that are greater than 4 and less than 7.
Solution to challenge 2
x_subset <- x[x<7 & x>4]
print(x_subset)
  a   b   d 
5.4 6.2 4.8 

Tip: Non-unique names

You should be aware that it is possible for multiple elements in a vector to have the same name. (For a data frame, columns can have the same name — although R tries to avoid this — but row names must be unique.) Consider these examples:
x <- 1:3
x
[1] 1 2 3
names(x) <- c('a', 'a', 'a')
x
a a a 
1 2 3 
x['a']  # only returns first value
a 
1 
x[names(x) == 'a']  # returns all three values
a a a 
1 2 3 

Tip: Getting help for operators

Remember you can search for help on operators by wrapping them in quotes: help("%in%") or ?"%in%".

Skipping named elements

Skipping or removing named elements is a little harder. If we try to skip one named element by negating the string, R complains (slightly obscurely) that it doesn’t know how to take the negative of a string:

x <- c(a=5.4, b=6.2, c=7.1, d=4.8, e=7.5) # we start again by naming a vector 'on the fly'
x[-"a"]

Error in -"a": invalid argument to unary operator

However, we can use the != (not-equals) operator to construct a logical vector that will do what we want:

x[names(x) != "a"]

  b   c   d   e 
6.2 7.1 4.8 7.5 

Skipping multiple named indices is a little bit harder still. Suppose we want to drop the "a" and "c" elements, so we try this:

x[names(x)!=c("a","c")]

Warning in names(x) != c("a", "c"): longer object length is not a multiple
of shorter object length

  b   c   d   e 
6.2 7.1 4.8 7.5 

R did something, but it gave us a warning that we ought to pay attention to - and it apparently gave us the wrong answer (the "c" element is still included in the vector)!

So what does != actually do in this case? That’s an excellent question.

Recycling

Let’s take a look at the comparison component of this code:

names(x) != c("a", "c")

Warning in names(x) != c("a", "c"): longer object length is not a multiple
of shorter object length

[1] FALSE  TRUE  TRUE  TRUE  TRUE

Why does R give FALSE as the third element of this vector, when names(x)[3] != "c" is obviously false? When you use !=, R tries to compare each element of the left argument with the corresponding element of its right argument. What happens when you compare vectors of different lengths?

Inequality testing

When one vector is shorter than the other, it gets recycled:

Inequality testing: results of recycling

In this case R repeats c("a", "c") as many times as necessary to match names(x), i.e. we get c("a","c","a","c","a"). Since the recycled "a" doesn’t match the third element of names(x), the value of != is FALSE. Because in this case the longer vector length (5) isn’t a multiple of the shorter vector length (2), R printed a warning message. If we had been unlucky and names(x) had contained six elements, R would silently have done the wrong thing (i.e., not what we intended it to do). This recycling rule can can introduce hard-to-find and subtle bugs!

The way to get R to do what we really want (match each element of the left argument with all of the elements of the right argument) it to use the %in% operator. The %in% operator goes through each element of its left argument, in this case the names of x, and asks, “Does this element occur in the second argument?”. Here, since we want to exclude values, we also need a ! operator to change “in” to “not in”:

x[! names(x) %in% c("a","c") ]

  b   d   e 
6.2 4.8 7.5 

Challenge 3

Selecting elements of a vector that match any of a list of components is a very common data analysis task. For example, the gapminder data set contains country and continent variables, but no information between these two scales. Suppose we want to pull out information from southeast Asia: how do we set up an operation to produce a logical vector that is TRUE for all of the countries in southeast Asia and FALSE otherwise?

Suppose you have the following data:
seAsia <- c("Myanmar","Thailand","Cambodia","Vietnam","Laos")

## read in the gapminder data that we downloaded in episode 2
gapminder <- read.csv("data/gapminder-FiveYearData.csv", header=TRUE)

## extract the `country` column from a data frame (we'll see this later);
## convert from a factor to a character;
## and get just the non-repeated elements
countries <- unique(as.character(gapminder$country))
There’s a wrong way (using only ==), which will give you a warning; a clunky way (using the logical operators == and |); and an elegant way (using %in%). See whether you can come up with all three and explain how they (don’t) work.
Solution to challenge 3
The wrong way to do this problem is:
(countries==seAsia)
This gives a warning:
"In countries == seAsia : longer object length is not a multiple of shorter object length"
and the wrong answer (a vector of all FALSE values), because none of the recycled values of seAsia happen to line up correctly with matching values in country.

The clunky (but technically correct) way to do this problem is
 (countries=="Myanmar" | countries=="Thailand" | countries=="Cambodia" | countries == "Vietnam" | countries=="Laos")
(or countries==seAsia[1] | countries==seAsia[2] | ...). This gives the correct values, but hopefully you can see how awkward it is (what if we wanted to select countries from a much longer list?).
The best way to do this problem is:
(countries %in% seAsia)
which is both correct and easy to type (and read).

Handling special values

At some point you will encounter functions in R that cannot handle missing, infinite, or undefined data.

There are a number of special functions you can use to filter out this data:

is.na will return all positions in a vector, matrix, or data.frame containing NA (or NaN)
likewise, is.nan, and is.infinite will do the same for NaN and Inf.
is.finite will return all positions in a vector, matrix, or data.frame that do not contain NA, NaN or Inf.
na.omit will filter out all missing values from a vector

Factor subsetting

Now that we’ve explored the different ways to subset vectors, how do we subset the other data structures?

Factor subsetting works the same way as vector subsetting.

We can subset by value:

f <- factor(c("a", "a", "b", "c", "c", "d"))
f[f == "a"]

[1] a a
Levels: a b c d

Using the %in% operator:

f[f %in% c("b", "c")]

[1] b c c
Levels: a b c d

Or we can subset by index as well:

f[1:3]

[1] a a b
Levels: a b c d

We can even exclude elements just like we did before. However, note that skipping elements will not remove the level even if no more of that category exists in the factor:

f[-3]

[1] a a c c d
Levels: a b c d

Matrix subsetting

Matrices are also subsetted using the [ function. In this case it takes two arguments: the first applying to the rows, the second to its columns:

set.seed(1)
m <- matrix(rnorm(6*4), ncol=4, nrow=6)
m[3:4, c(3,1)]

            [,1]       [,2]
[1,]  1.12493092 -0.8356286
[2,] -0.04493361  1.5952808

You can leave the first or second arguments blank to retrieve all the rows or columns respectively:

m[, c(3,4)]

            [,1]        [,2]
[1,] -0.62124058  0.82122120
[2,] -2.21469989  0.59390132
[3,]  1.12493092  0.91897737
[4,] -0.04493361  0.78213630
[5,] -0.01619026  0.07456498
[6,]  0.94383621 -1.98935170

If we only access one row or column, R will automatically convert the result to a vector:

m[3,]

[1] -0.8356286  0.5757814  1.1249309  0.9189774

If you want to keep the output as a matrix, you need to specify a third argument; drop = FALSE:

m[3, , drop=FALSE]

           [,1]      [,2]     [,3]      [,4]
[1,] -0.8356286 0.5757814 1.124931 0.9189774

Unlike vectors, if we try to access a row or column outside of the matrix, R will throw an error:

m[, c(3,6)]

Error in m[, c(3, 6)]: subscript out of bounds

Tip: Higher dimensional arrays

when dealing with multi-dimensional arrays, each argument to [ corresponds to a dimension. For example, a 3D array, the first three arguments correspond to the rows, columns, and depth dimension.

Because matrices are vectors, we can also subset using only one argument:

m[5]

[1] 0.3295078

This usually isn’t useful, and often confusing to read. However it is useful to note that matrices are laid out in column-major format by default. That is the elements of the vector are arranged column-wise:

matrix(1:6, nrow=2, ncol=3)

     [,1] [,2] [,3]
[1,]    1    3    5
[2,]    2    4    6

If you wish to populate the matrix by row, use byrow=TRUE:

matrix(1:6, nrow=2, ncol=3, byrow=TRUE)

     [,1] [,2] [,3]
[1,]    1    2    3
[2,]    4    5    6

Matrices can also be subsetted using their rownames and column names instead of their row and column indices.

Challenge 4

Given the following code:
m <- matrix(1:18, nrow=3, ncol=6)
print(m)
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    4    7   10   13   16
[2,]    2    5    8   11   14   17
[3,]    3    6    9   12   15   18
Which of the following commands will extract the values 11 and 14?

A. m[2,4,2,5]

B. m[2:5]

C. m[4:5,2]

D. m[2,c(4,5)]

Solution to challenge 4

From the above choices, D would give us the correct subset.

Note: you can use : and c interchangeably. For this problem, the command m[2, 4:5] would also give the correct subset.

List subsetting

Now we’ll introduce some new subsetting operators. There are three functions used to subset lists. We’ve already seen these when learning about atomic vectors and matrices: [, [[, and $.

Using [ will always return a list. If you want to subset a list, but not extract an element, then you will likely use [.

xlist <- list(a = "Software Carpentry", b = 1:10, data = head(iris))
xlist[1]

$a
[1] "Software Carpentry"

This returns a list with one element.

We can subset elements of a list exactly the same way as atomic vectors using [. Comparison operations however won’t work as they’re not recursive, they will try to condition on the data structures in each element of the list, not the individual elements within those data structures.

xlist[1:2]

$a
[1] "Software Carpentry"

$b
 [1]  1  2  3  4  5  6  7  8  9 10

To extract individual elements of a list, you need to use the double-square bracket function: [[.

xlist[[1]]

[1] "Software Carpentry"

Notice that now the result is a vector, not a list.

You can’t extract more than one element at once:

xlist[[1:2]]

Error in xlist[[1:2]]: subscript out of bounds

Nor use it to skip elements:

xlist[[-1]]

Error in xlist[[-1]]: attempt to select more than one element in get1index <real>

But you can use names to both subset and extract elements:

xlist[["a"]]

[1] "Software Carpentry"

The $ function is a shorthand way for extracting elements by name:

xlist$data

  Sepal.Length Sepal.Width Petal.Length Petal.Width Species
        5.1         3.5          1.4         0.2  setosa
        4.9         3.0          1.4         0.2  setosa
        4.7         3.2          1.3         0.2  setosa
        4.6         3.1          1.5         0.2  setosa
        5.0         3.6          1.4         0.2  setosa
        5.4         3.9          1.7         0.4  setosa

We can combine subsetting operators to further filter results:

xlist$data[1]

  Sepal.Length
        5.1
        4.9
        4.7
        4.6
        5.0
        5.4

xlist[[2]][2:4]

[1] 2 3 4

Challenge 5

Given the following list:
xlist <- list(a = "Software Carpentry", b = 1:10, data = head(iris))
Using your knowledge of both list and vector subsetting, extract the number 2 from xlist. Hint: the number 2 is contained within the “b” item in the list.
Solution to challenge 5
xlist$b[2]
[1] 2
xlist[[2]][2]
[1] 2
xlist[["b"]][2]
[1] 2

Challenge 6 - Advanced

To create a linear model showing the relationship between the response variable population and predictor variable lifeExp, we use the following command:
mod <- aov(pop ~ lifeExp, data=gapminder)
Extract the residual degrees of freedom (hint: attributes() will help you, don’t forget check out the help documentation!)
Solution to challenge 6

the attributes() command lists the attributes of the object mod:
attributes(mod)       ## `df.residual` is one of the names of `mod`
$names
 [1] "coefficients"  "residuals"     "effects"       "rank"          "fitted.values" "assign"       
 [7] "qr"            "df.residual"   "xlevels"       "call"          "terms"         "model"        

$class
[1] "aov" "lm" 
We can see that mod has a component called df.residual. We can access this value by using the $ operator.
mod$df.residual

Data frames

Remember the data frames are lists underneath the hood, so similar rules apply. However they are also two dimensional objects:

[ with one argument will act the same way as for lists, where each list element corresponds to a column. The resulting object will be a data frame:

head(gapminder[3])

Similarly, [[ will act to extract a single column:

head(gapminder[["lifeExp"]])

[1] 28.801 30.332 31.997 34.020 36.088 38.438

And $ provides a convenient shorthand to extract columns by name:

head(gapminder$year)

[1] 1952 1957 1962 1967 1972 1977

With two arguments, [ behaves the same way as for matrices:

gapminder[1:3,]

      country year      pop continent lifeExp gdpPercap
Afghanistan 1952  8425333      Asia  28.801  779.4453
Afghanistan 1957  9240934      Asia  30.332  820.8530
Afghanistan 1962 10267083      Asia  31.997  853.1007

If we subset a single row, the result will be a data frame (because the elements are mixed types):

gapminder[3,]

      country year      pop continent lifeExp gdpPercap
3 Afghanistan 1962 10267083      Asia  31.997  853.1007

But for a single column the result will be a vector (this can be changed with the third argument, drop = FALSE).

Challenge 7

Fix each of the following common data frame subsetting errors:
Extract observations collected for the year 1957
gapminder[gapminder$year = 1957,]
Extract all columns except 1 through to 4
gapminder[,-1:4]
Extract the rows where the life expectancy is longer the 80 years
gapminder[gapminder$lifeExp > 80]
Extract the first row, and the fourth and fifth columns (lifeExp and gdpPercap).
gapminder[1, 4, 5]
Advanced: extract rows that contain information for the years 2002 and 2007
gapminder[gapminder$year == 2002 | 2007,]
Solution to challenge 7

Fix each of the following common data frame subsetting errors:
Extract observations collected for the year 1957
# gapminder[gapminder$year = 1957,]
gapminder[gapminder$year == 1957,]
Extract all columns except 1 through to 4
# gapminder[,-1:4]
gapminder[,-c(1:4)]
Extract the rows where the life expectancy is longer the 80 years
# gapminder[gapminder$lifeExp > 80]
gapminder[gapminder$lifeExp > 80,]
Extract the first row, and the fourth and fifth columns (lifeExp and gdpPercap).
# gapminder[1, 4, 5]
gapminder[1, c(4, 5)]
Advanced: extract rows that contain information for the years 2002 and 2007
 # gapminder[gapminder$year == 2002 | 2007,]
 gapminder[gapminder$year == 2002 | gapminder$year == 2007,]
 gapminder[gapminder$year %in% c(2002, 2007),]

Challenge 8

Why does gapminder[1:20] return an error? How does it differ from gapminder[1:20, ]?

Create a new data.frame called gapminder_small that only contains rows 1 through 9 and 19 through 23. You can do this in one or two steps.
Solution to challenge 8

gapminder is a data.frame so needs to be subsetted on two dimensions. gapminder[1:20, ] subsets the data to give the first 20 rows and all columns.

To do this in one step, you can use the command:
gapminder_small <- gapminder[c(1:9, 19:23),]
There are many ways to subset in this way, you might have come up with a different approach.

Key Points

Indexing in R starts at 1, not 0.

Access individual values by location using [].

Access slices of data using [low:high].

Access arbitrary sets of data using [c(...)].

Use logical operations and logical vectors to access subsets of data.

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R for Reproducible Scientific Analysis

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Subsetting Data

Overview

Atomic vectors

Accessing elements using their indices

Vector numbering in R starts at 1

Skipping and removing elements

Tip: Order of operations

Challenge 1

Solution to challenge 1

Subsetting by name

Subsetting through other logical operations

Tip: Combining logical conditions

Challenge 2

Solution to challenge 2

Tip: Non-unique names

Tip: Getting help for operators

Skipping named elements

Recycling

Challenge 3

Solution to challenge 3

Handling special values

Factor subsetting

Matrix subsetting

Tip: Higher dimensional arrays

Challenge 4

Solution to challenge 4

List subsetting

Challenge 5

Solution to challenge 5

Challenge 6 - Advanced

Solution to challenge 6

Data frames

Challenge 7

Solution to challenge 7

Challenge 8

Solution to challenge 8

Key Points

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