Subsetting Data
Last updated on 20230815  Edit this page
Overview
Questions
 How can I work with subsets of data in R?
Objectives
 To be able to subset vectors 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.
R
x < c(5.4, 6.2, 7.1, 4.8, 7.5)
names(x) < c('a', 'b', 'c', 'd', 'e')
x
OUTPUT
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:
R
x[1]
OUTPUT
a
5.4
R
x[4]
OUTPUT
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:
R
x[c(1, 3)]
OUTPUT
a c
5.4 7.1
Or slices of the vector:
R
x[1:4]
OUTPUT
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.
R
1:4
OUTPUT
[1] 1 2 3 4
R
c(1, 2, 3, 4)
OUTPUT
[1] 1 2 3 4
We can ask for the same element multiple times:
R
x[c(1, 1, 3)]
OUTPUT
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:
R
x[6]
OUTPUT
<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:
R
x[0]
OUTPUT
named numeric(0)
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:
R
x[2]
OUTPUT
a c d e
5.4 7.1 4.8 7.5
We can skip multiple elements:
R
x[c(1, 5)] # or x[c(1,5)]
OUTPUT
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:
R
x[1:3]
This gives a somewhat cryptic error:
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:
R
x[(1:3)]
OUTPUT
d e
4.8 7.5
To remove elements from a vector, we need to assign the result back into the variable:
R
x < x[4]
x
OUTPUT
a b c e
5.4 6.2 7.1 7.5
Challenge 1
Given the following code:
R
x < c(5.4, 6.2, 7.1, 4.8, 7.5)
names(x) < c('a', 'b', 'c', 'd', 'e')
print(x)
OUTPUT
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:
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?
R
x[2:4]
OUTPUT
b c d
6.2 7.1 4.8
R
x[c(1,5)]
OUTPUT
b c d
6.2 7.1 4.8
R
x[c("b", "c", "d")]
OUTPUT
b c d
6.2 7.1 4.8
R
x[c(2,3,4)]
OUTPUT
b c d
6.2 7.1 4.8
Subsetting by name
We can extract elements by using their name, instead of extracting by index:
R
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")]
OUTPUT
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:
R
x[c(FALSE, FALSE, TRUE, FALSE, TRUE)]
OUTPUT
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.
R
x[x > 7]
OUTPUT
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):
R
x[names(x) == "a"]
OUTPUT
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: returnsTRUE
if both the left and right areTRUE
. 

, the “logical OR” operator: returnsTRUE
, if either the left or right (or both) areTRUE
.
You may sometimes see &&
and 
instead of &
and 
. These twocharacter
operators only look at the first element of each vector and ignore the
remaining elements. In general you should not use the twocharacter
operators in data analysis; save them for programming, i.e. deciding
whether to execute a statement.

!
, the “logical NOT” operator: convertsTRUE
toFALSE
andFALSE
toTRUE
. It can negate a single logical condition (eg!TRUE
becomesFALSE
), or a whole vector of conditions(eg!c(TRUE, FALSE)
becomesc(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
).
R
x_subset < x[x<7 & x>4]
print(x_subset)
OUTPUT
a b d
5.4 6.2 4.8
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 containingNA
(orNaN
)  likewise,
is.nan
, andis.infinite
will do the same forNaN
andInf
. 
is.finite
will return all positions in a vector, matrix, or data.frame that do not containNA
,NaN
orInf
. 
na.omit
will filter out all missing values from a vector
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:
R
head(gapminder[3])
OUTPUT
pop
1 8425333
2 9240934
3 10267083
4 11537966
5 13079460
6 14880372
Similarly, [[
will act to extract a single
column:
R
head(gapminder[["lifeExp"]])
OUTPUT
[1] 28.801 30.332 31.997 34.020 36.088 38.438
And $
provides a convenient shorthand to extract columns
by name:
R
head(gapminder$year)
OUTPUT
[1] 1952 1957 1962 1967 1972 1977
To select specific rows and/or columns, you can provide two arguments
to [
R
gapminder[1:3, ]
OUTPUT
country year pop continent lifeExp gdpPercap
1 Afghanistan 1952 8425333 Asia 28.801 779.4453
2 Afghanistan 1957 9240934 Asia 30.332 820.8530
3 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):
R
gapminder[3, ]
OUTPUT
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 3
Fix each of the following common data frame subsetting errors:
 Extract observations collected for the year 1957
R
$year = 1957, ] gapminder[gapminder
 Extract all columns except 1 through to 4
R
gapminder[, 1:4]
 Extract the rows where the life expectancy is longer the 80 years
R
gapminder[gapminder$lifeExp > 80]
 Extract the first row, and the fourth and fifth columns
(
lifeExp
andgdpPercap
).
R
gapminder[1, 4, 5]
 Advanced: extract rows that contain information for the years 2002 and 2007
R
gapminder[gapminder$year == 2002  2007,]
Fix each of the following common data frame subsetting errors:
 Extract observations collected for the year 1957
R
# gapminder[gapminder$year = 1957, ]
gapminder[gapminder$year == 1957, ]
 Extract all columns except 1 through to 4
R
# gapminder[, 1:4]
gapminder[,c(1:4)]
 Extract the rows where the life expectancy is longer the 80 years
R
# gapminder[gapminder$lifeExp > 80]
gapminder[gapminder$lifeExp > 80,]
 Extract the first row, and the fourth and fifth columns
(
lifeExp
andgdpPercap
).
R
# gapminder[1, 4, 5]
gapminder[1, c(4, 5)]
 Advanced: extract rows that contain information for the years 2002 and 2007
R
# gapminder[gapminder$year == 2002  2007,]
gapminder[gapminder$year == 2002  gapminder$year == 2007,]
gapminder[gapminder$year %in% c(2002, 2007),]
gapminder
is a data.frame so it needs to be subsetted on two dimensions.gapminder[1:20, ]
subsets the data to give the first 20 rows and all columns.
R
gapminder_small < gapminder[c(1:9, 19:23),]