R for reproducible scientific analysis
Subsetting data
Learning 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 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
The square brackets operator is just like any other 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 just creates a sequence of numbers from the left element to the right. I.e. x[1:4]
is equivalent to x[c(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)
But what about negative values?
Skipping and removing elements
If we use a negative number, 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
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
- Compare notes with your neighbour. Did you have different strategies?
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?”.
So why can’t we use ==
like before? That’s an excellent question.
Let’s take a look at just the comparison component:
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 FALSE FALSE FALSE
Obviously “c” is in the names of x
, so why didn’t this work? ==
works slightly differently to %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')
Warning in names(x) == c("a", "c", "e"): longer object length is not a
multiple of shorter object length
[1] TRUE FALSE FALSE 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 b e
5.4 6.2 7.5
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 c e
5.4 7.1 7.5
Since comparison operators evaluate to logical vectors, we can also use them to succinctly subset vectors:
x[x > 7]
c e
7.1 7.5
Challenge
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.
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 containingNA
.- 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
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
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
Because matrices are really just vectors underneath the hood, we can also subset using only one argument:
m[5]
[1] 0.3295078
This usually isn’t useful. 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
Matrices can also be subsetted using their rownames and column names instead of their row and column indices.
Challenge 2
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)]
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
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 3
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.
Challenge 4
Given a linear model:
mod <- aov(pop ~ lifeExp, data=gapminder)
Extract the residual degrees of freedom (hint: attributes()
will help you)
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])
pop
1 8425333
2 9240934
3 10267083
4 11537966
5 13079460
6 14880372
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$year)
[1] 1952 1957 1962 1967 1972 1977
With two arguments, [
behaves the same way as for matrices:
gapminder[1:3,]
country year pop continent lifeExp gdpPercap
1 Afghanistan 1952 8425333 Asia 28.801 779.4453
2 Afghanistan 1957 9240934 Asia 30.332 820.8530
3 Afghanistan 1962 10267083 Asia 31.997 853.1007
If we subset a single row, the result will be a data frame (because the elements are mixed types):
gapminder[3,]
country year pop continent lifeExp gdpPercap
3 Afghanistan 1962 10267083 Asia 31.997 853.1007
But for a single column the result will be a vector (this can be changed with the third argument, drop = FALSE
).
Challenge 5
Fix each of the following common data frame subsetting errors:
- Extract observations collected for the year 1957
gapminder[gapminder$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$lifeExp > 80]
- Extract the first row, and the fourth and fifth columns (
lifeExp
andgdpPercap
).
gapminder[1, 4, 5]
- Advanced: extract rows that contain information for the years 2002 and 2007
gapminder[gapminder$year == 2002 | 2007,]
Challenge 6
Why does
gapminder[1:20]
return an error? How does it differ fromgapminder[1:20, ]
?Create a new
data.frame
calledgapminder_small
that only contains rows 1 through 9 and 19 through 23. You can do this in one or two steps.
Challenge solutions
Solution to 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
x[2:4]
x[-c(1,5)]
x[c("b", "c", "d")]
x[c(2,3,4)]
Solution to challenge 2
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)]
Answer: D
Solution to challenge 3
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.
xlist$b[2]
xlist[[2]][2]
xlist[["b"]][2]
Solution to challenge 4
Given a linear model:
mod <- aov(pop ~ lifeExp, data=gapminder)
Extract the residual degrees of freedom (hint: attributes()
will help you)
attributes(mod) ## `df.residual` is one of the names of `mod`
mod$df.residual
Solution to challenge 5
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
andgdpPercap
).
# 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),]
Solution to challenge 6
- Why does
gapminder[1:20]
return an error? How does it differ fromgapminder[1:20, ]
?
Answer: 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.
- Create a new
data.frame
calledgapminder_small
that only contains rows 1 through 9 and 19 through 23. You can do this in one or two steps.
gapminder_small <- gapminder[c(1:9, 19:23),]