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 skip and remove elements from various data structures.

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

R has many powerful subset operators and 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: atomic vectors.

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

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 atomic 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

c(1, 2, 3, 4)

## [1] 1 2 3 4

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 a number outside of the vector, R will return missing values:

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. Most people first 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, so what happens is it takes its first argument as -1, and 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 results:
x[-(1:3)]
##   d   e 
## 4.8 7.5

To remove elements from a vector, we need to assign the results 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 neighbour. Did you have different strategies?
Solution to challenge 1
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

Subsetting by name

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

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!

Unfortunately we can’t skip or remove elements so easily.

To skip (or remove) a single named element:

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

##   b   c   d   e 
## 6.2 7.1 4.8 7.5

The which function returns the indices of all TRUE elements of its argument. Remember that expressions evaluate before being passed to functions. Let’s break this down so that its clearer what’s happening.

First this happens:

names(x) == "a"

## [1]  TRUE FALSE FALSE FALSE FALSE

The condition operator is applied to every name of the vector x. Only the first name is “a” so that element is TRUE.

which then converts this to an index:

which(names(x) == "a")

## [1] 1

Only the first element is TRUE, so which returns 1. Now that we have indices the skipping works because we have a negative index!

Skipping multiple named indices is similar, but uses a different comparison operator:

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

##   b   d   e 
## 6.2 4.8 7.5

The %in% 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?”.

Challenge 2

Run the following code to define vector x as above:
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
Given this vector x, what would you expect the following to do?
x[-which(names(x) == "g")]
Try out this command and see what you get. Did this match your expectation? Why did we get this result? (Tip: test out each part of the command on it’s own - this is a useful debugging strategy)

Which of the following are true:

A) if there are no TRUE values passed to which, an empty vector is returned

B) if there are no TRUE values passed to which, an error message is shown

C) integer() is an empty vector

D) making an empty vector negative produces an “everything” vector

E) x[] gives the same result as x[integer()]

Solution to challenge 2

A and C are correct.

The which command returns the index of every TRUE value in its input. The names(x) == "g" command didn’t return any TRUE values. Because there were no TRUE values passed to the which command, it returned an empty vector. Negating this vector with the minus sign didn’t change its meaning. Because we used this empty vector to retrieve values from x, it produced an empty numeric vector. It was a named numeric empty vector because the vector type of x is “named numeric” since we assigned names to the values (try str(x) ).

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[which(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%".

So why can’t we use == like before? That’s an excellent question.

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]  TRUE FALSE  TRUE

Obviously “c” is in the names of x, so why didn’t this work? == works slightly differently than %in%. It will compare each element of its left argument to the corresponding element of its right argument.

Here’s a mock illustration:

c("a", "b", "c", "e")  # names of x
   |    |    |    |    # The elements == is comparing
c("a", "c")

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

c("a", "b", "c", "e")  # names of x
   |    |    |    |    # The elements == is comparing
c("a", "c", "a", "c")

In this case R simply repeats c("a", "c") twice. If the longer vector length isn’t a multiple of the shorter vector length, then R will also print out a warning message:

names(x) == c('a', 'c', 'e')

## [1]  TRUE FALSE FALSE

This difference between == and %in% is important to remember, because it can introduce hard to find and subtle bugs!

Subsetting through other logical operations

We can also more simply subset through logical operations:

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

## a a 
## 1 2

Note that in this case, the logical vector is also recycled to the length of the vector we’re subsetting!

x[c(TRUE, FALSE)]

## a a 
## 1 3

Since comparison operators evaluate to logical vectors, we can also use them to succinctly subset vectors:

x[x > 7]

## named integer(0)

Tip: Combining logical conditions

There are many situations in which you will wish 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.

The recycling rule applies with both of these, so TRUE & c(TRUE, FALSE, TRUE) will compare the first TRUE on the left of the & sign with each of the three conditions on the right.

You may sometimes see && and || instead of & and |. These operators do not use the recycling rule: they only look at the first element of each vector and ignore the remaining elements. The longer operators are mainly used in programming, rather than data analysis.

!, 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 3

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 3
x_subset <- x[x<7 & x>4]
print(x_subset)
##   a   b   d 
## 5.4 6.2 4.8

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 was as for lists, where each list element corresponds to a column. The resulting object will be a data frame:

head(gapminder[3])

##   year
## 1 1952
## 2 1957
## 3 1962
## 4 1967
## 5 1972
## 6 1977

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 continent year lifeExp      pop gdpPercap
## 1 Afghanistan      Asia 1952  28.801  8425333  779.4453
## 2 Afghanistan      Asia 1957  30.332  9240934  820.8530
## 3 Afghanistan      Asia 1962  31.997 10267083  853.1007

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

gapminder[3,]

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

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

Challenge 5

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 6

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.
gapminder_small <- gapminder[c(1:9, 19:23),]

Handling special values

At some point you will encounter functions in R which 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.
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

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

D

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.

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

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

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

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

f[1:3]

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

An important note is 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

List subsetting

Now we’ll introduce some new subsetting operators. There are three functions used to subset lists. [, as we’ve seen for atomic vectors and matrices, as well as [[ 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 was 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
## 1          5.1         3.5          1.4         0.2  setosa
## 2          4.9         3.0          1.4         0.2  setosa
## 3          4.7         3.2          1.3         0.2  setosa
## 4          4.6         3.1          1.5         0.2  setosa
## 5          5.0         3.6          1.4         0.2  setosa
## 6          5.4         3.9          1.7         0.4  setosa

Challenge 7

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 8

Given a linear model:
mod <- aov(pop ~ lifeExp, data=gapminder)
Extract the residual degrees of freedom (hint: attributes() will help you)
Solution to challenge 6
attributes(mod) ## `df.residual` is one of the names of `mod`
mod$df.residual

Key Points

Indexing in R starts at 1, not 0.

Access individual values by location using [].

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

Use which to select subsets of data based on value.

Access slices of data using [low:high].

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R for Reproducible Research

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Subsetting Data

Overview

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

Challenge 2

Solution to challenge 2

Tip: Non-unique names

Tip: Getting help for operators

Subsetting through other logical operations

Tip: Combining logical conditions

Challenge 3

Solution to challenge 3

Data frames

Challenge 5

Solution to challenge 7

Challenge 6

Solution to challenge 8

Handling special values

Matrix subsetting

Tip: Higher dimensional arrays

Challenge 4

Solution to challenge 4

Factor subsetting

List subsetting

Challenge 7

Solution to challenge 5

Challenge 8

Solution to challenge 6

Key Points

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