Subsetting Data
Last updated on 2024-03-12 | 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. Why so many? Well, this gives us a lot of flexibility that can come in useful. At first, though, it may seem overwhelming. Don’t worry!
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, or their numbered place in the vector 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
. Here NA
stands for “Not
Available”, and is a common way to represent a missing value.
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 (you have to use ==
rather than =
for
comparisons, as R has another use for =
):
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 plots that are located in Casco Bay or Penobscot Bay and have urchin densities 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 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: 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
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:
R
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
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:
R
x[names(x) != "a"]
OUTPUT
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:
R
x[names(x)!=c("a","c")]
WARNING
Warning in names(x) != c("a", "c"): longer object length is not a multiple of
shorter object length
OUTPUT
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:
R
names(x) != c("a", "c")
WARNING
Warning in names(x) != c("a", "c"): longer object length is not a multiple of
shorter object length
OUTPUT
[1] FALSE TRUE TRUE TRUE TRUE
Why does R give TRUE
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?
When one vector is shorter than the other, it gets recycled.
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 TRUE
. 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”:
R
x[! names(x) %in% c("a","c") ]
OUTPUT
b d e
6.2 4.8 7.5
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
Data frames are two dimensional objects (under the hood they are structured as lists in R, but we aren’t going to go into detail on that). Data frames have some useful subsetting operators.
Let’s look at the Casco Bay DMR kelp-urchin data again.
R
casco_dmr <- read.csv("data/casco_kelp_urchin.csv") # if you don't already have the data loaded
[
with one argument will extract a column - each element
of the data frame corresponds to a column. The resulting object will be
a data frame:
R
head(casco_dmr[3])
OUTPUT
exposure.code
1 2
2 2
3 3
4 2
5 2
6 1
R
class(casco_dmr[3])
OUTPUT
[1] "data.frame"
Similarly, [[
will act to extract a single
column:
R
head(casco_dmr[["exposure.code"]])
OUTPUT
[1] 2 2 3 2 2 1
And $
provides a convenient shorthand to extract columns
by name:
R
head(casco_dmr$exposure.code)
OUTPUT
[1] 2 2 3 2 2 1
To select specific rows and/or columns, you can provide two arguments
to [
. The first identifies the rows to subset, and the
second the columns.
R
casco_dmr[1:3, ]
OUTPUT
year region exposure.code coastal.code latitude longitude depth crust
1 2001 Casco Bay 2 2 43.72766 -70.10721 5 6.1
2 2001 Casco Bay 2 2 43.76509 -69.96087 5 31.5
3 2001 Casco Bay 3 3 43.75199 -69.93420 5 31.5
understory kelp urchin month day survey site
1 38.5 92.5 0 6 15 dmr 66
2 74.0 59.0 0 6 15 dmr 71
3 96.5 7.7 0 6 15 dmr 70
If we subset a single row, the result will be a data frame (because the elements are mixed types):
R
casco_dmr[3, ]
OUTPUT
year region exposure.code coastal.code latitude longitude depth crust
3 2001 Casco Bay 3 3 43.75199 -69.9342 5 31.5
understory kelp urchin month day survey site
3 96.5 7.7 0 6 15 dmr 70
But for a single column the result will be a vector (this can be
changed with the argument, drop = FALSE
).
R
str(casco_dmr[, 3])
OUTPUT
int [1:90] 2 2 3 2 2 1 2 3 3 5 ...
R
str(casco_dmr[, 3, drop = FALSE])
OUTPUT
'data.frame': 90 obs. of 1 variable:
$ exposure.code: int 2 2 3 2 2 1 2 3 3 5 ...
Challenge 3
Fix each of the following common data frame subsetting errors:
- Extract observations collected for the year 2011
- Extract all columns except 1 through to 4
R
casco_dmr[, -1:4]
- Extract the rows where the kelp percent cover is greater than 80%
R
casco_dmr[casco_dmr$kelp > 80]
- Extract the first row, and the fifth and sixth columns
(
latitude
andlongitude
).
R
casco_dmr[1, 5, 6]
- Advanced: extract rows that contain information for the years 2002 or 2007
R
casco_dmr[casco_dmr$year == 2002 | 2007,]
Fix each of the following common data frame subsetting errors:
- Extract observations collected for the year 2011
R
# casco_dmr[casco_dmr$year = 2011, ]
casco_dmr[casco_dmr$year == 2011, ]
- Extract all columns except 1 through to 4
R
# casco_dmr[, -1:4]
casco_dmr[,-c(1:4)]
- Extract the rows where the kelp cover is greater than 80%
R
# casco_dmr[casco_dmr$kelp > 80]
casco_dmr[casco_dmr$kelp > 80,]
- Extract the first row, and the fifth and sixth columns
(
latitude
andlongitude
).
R
# casco_dmr[1, 5, 6]
casco_dmr[1, c(5, 6)]
- Advanced: extract rows that contain information for the years 2002 and 2007
R
# casco_dmr[casco_dmr$year == 2002 | 2007,]
casco_dmr[casco_dmr$year == 2002 | casco_dmr$year == 2007,]
casco_dmr[casco_dmr$year %in% c(2002, 2007),]
casco_dmr
is a data.frame so it needs to be subsetted on two dimensions.casco_dmr[1:20, ]
subsets the data to give the first 20 rows and all columns.
R
casco_dmr_small <- casco_dmr[c(1:9, 19:23),]