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R provides two different methods for accessing the elements of a list or data.frame: [] and [[]].

What is the difference between the two, and when should I use one over the other?

MrFlick
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Sharpie
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11 Answers11

414

The R Language Definition is handy for answering these types of questions:

R has three basic indexing operators, with syntax displayed by the following examples

    x[i]
    x[i, j]
    x[[i]]
    x[[i, j]]
    x$a
    x$"a"

For vectors and matrices the [[ forms are rarely used, although they have some slight semantic differences from the [ form (e.g. it drops any names or dimnames attribute, and that partial matching is used for character indices). When indexing multi-dimensional structures with a single index, x[[i]] or x[i] will return the ith sequential element of x.

For lists, one generally uses [[ to select any single element, whereas [ returns a list of the selected elements.

The [[ form allows only a single element to be selected using integer or character indices, whereas [ allows indexing by vectors. Note though that for a list, the index can be a vector and each element of the vector is applied in turn to the list, the selected component, the selected component of that component, and so on. The result is still a single element.

Mus
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ars
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    What is the reasoning behind using [[ vs [ to index with a single number vs vector? Why not just use [ for both? I guess you can use the [[ to get back a single entry, and [ with one index returns a list of length 1...but why not just make [ return a single entry with one index instead of a list? Why might you ever want a length-1 list returned? – wordsforthewise Nov 17 '17 at 18:19
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    @wordsforthewise, when programming you can have a vector of undefined length that you want to use for indexing. Having `[` always return a list means that you get the same output class for `x[v]` regardless of the length of `v`. For example, one might want to `lapply` over a subset of a list: `lapply(x[v], fun)`. If `[` would drop the list for vectors of length one, this would return an error whenever `v` has length one. – Axeman Dec 15 '17 at 10:48
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    I think this explains more clearly, http://adv-r.had.co.nz/Subsetting.html – Nate Anderson Sep 12 '19 at 21:24
206

The significant differences between the two methods are the class of the objects they return when used for extraction and whether they may accept a range of values, or just a single value during assignment.

Consider the case of data extraction on the following list:

foo <- list( str='R', vec=c(1,2,3), bool=TRUE )

Say we would like to extract the value stored by bool from foo and use it inside an if() statement. This will illustrate the differences between the return values of [] and [[]] when they are used for data extraction. The [] method returns objects of class list (or data.frame if foo was a data.frame) while the [[]] method returns objects whose class is determined by the type of their values.

So, using the [] method results in the following:

if( foo[ 'bool' ] ){ print("Hi!") }
Error in if (foo["bool"]) { : argument is not interpretable as logical

class( foo[ 'bool' ] )
[1] "list"

This is because the [] method returned a list and a list is not valid object to pass directly into an if() statement. In this case we need to use [[]] because it will return the "bare" object stored in 'bool' which will have the appropriate class:

if( foo[[ 'bool' ]] ){ print("Hi!") }
[1] "Hi!"

class( foo[[ 'bool' ]] )
[1] "logical"

The second difference is that the [] operator may be used to access a range of slots in a list or columns in a data frame while the [[]] operator is limited to accessing a single slot or column. Consider the case of value assignment using a second list, bar():

bar <- list( mat=matrix(0,nrow=2,ncol=2), rand=rnorm(1) )

Say we want to overwrite the last two slots of foo with the data contained in bar. If we try to use the [[]] operator, this is what happens:

foo[[ 2:3 ]] <- bar
Error in foo[[2:3]] <- bar : 
more elements supplied than there are to replace

This is because [[]] is limited to accessing a single element. We need to use []:

foo[ 2:3 ] <- bar
print( foo )

$str
[1] "R"

$vec
     [,1] [,2]
[1,]    0    0
[2,]    0    0

$bool
[1] -0.6291121

Note that while the assignment was successful, the slots in foo kept their original names.

Brad Gilbert
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Sharpie
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143

Double brackets accesses a list element, while a single bracket gives you back a list with a single element.

lst <- list('one','two','three')

a <- lst[1]
class(a)
## returns "list"

a <- lst[[1]]
class(a)
## returns "character"
Thomas
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medriscoll
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126

From Hadley Wickham:

From Hadley Wickham

My (crappy looking) modification to show using tidyverse / purrr:

enter image description here

jzadra
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61

[] extracts a list, [[]] extracts elements within the list

alist <- list(c("a", "b", "c"), c(1,2,3,4), c(8e6, 5.2e9, -9.3e7))

str(alist[[1]])
 chr [1:3] "a" "b" "c"

str(alist[1])
List of 1
 $ : chr [1:3] "a" "b" "c"

str(alist[[1]][1])
 chr "a"
Brad Gilbert
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24

Just adding here that [[ also is equipped for recursive indexing.

This was hinted at in the answer by @JijoMatthew but not explored.

As noted in ?"[[", syntax like x[[y]], where length(y) > 1, is interpreted as:

x[[ y[1] ]][[ y[2] ]][[ y[3] ]] ... [[ y[length(y)] ]]

Note that this doesn't change what should be your main takeaway on the difference between [ and [[ -- namely, that the former is used for subsetting, and the latter is used for extracting single list elements.

For example,

x <- list(list(list(1), 2), list(list(list(3), 4), 5), 6)
x
# [[1]]
# [[1]][[1]]
# [[1]][[1]][[1]]
# [1] 1
#
# [[1]][[2]]
# [1] 2
#
# [[2]]
# [[2]][[1]]
# [[2]][[1]][[1]]
# [[2]][[1]][[1]][[1]]
# [1] 3
#
# [[2]][[1]][[2]]
# [1] 4
#
# [[2]][[2]]
# [1] 5
#
# [[3]]
# [1] 6

To get the value 3, we can do:

x[[c(2, 1, 1, 1)]]
# [1] 3

Getting back to @JijoMatthew's answer above, recall r:

r <- list(1:10, foo=1, far=2)

In particular, this explains the errors we tend to get when mis-using [[, namely:

r[[1:3]]

Error in r[[1:3]] : recursive indexing failed at level 2

Since this code actually tried to evaluate r[[1]][[2]][[3]], and the nesting of r stops at level one, the attempt to extract through recursive indexing failed at [[2]], i.e., at level 2.

Error in r[[c("foo", "far")]] : subscript out of bounds

Here, R was looking for r[["foo"]][["far"]], which doesn't exist, so we get the subscript out of bounds error.

It probably would be a bit more helpful/consistent if both of these errors gave the same message.

Bhargav Rao
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MichaelChirico
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Being terminological, [[ operator extracts the element from a list whereas [ operator takes subset of a list.

submartingale
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16

To help newbies navigate through the manual fog, it might be helpful to see the [[ ... ]] notation as a collapsing function - in other words, it is when you just want to 'get the data' from a named vector, list or data frame. It is good to do this if you want to use data from these objects for calculations. These simple examples will illustrate.

(x <- c(x=1, y=2)); x[1]; x[[1]]
(x <- list(x=1, y=2, z=3)); x[1]; x[[1]]
(x <- data.frame(x=1, y=2, z=3)); x[1]; x[[1]]

So from the third example:

> 2 * x[1]
  x
1 2
> 2 * x[[1]]
[1] 2
Gavin Simpson
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Redfoot
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    As a newbie, I found it helpful in the 3 assignments to x (using "<-") to replace x=1 with w=1 to avoid confusion with the x that is the target of "<-" – user36800 Jul 06 '15 at 15:53
  • Although very simple, I really like this explanation. Another simple demonstration: `iris[[1]]` returns a vector, whereas `iris[1]` returns a data.frame – stevec Jun 09 '20 at 14:38
16

Both of them are ways of subsetting. The single bracket will return a subset of the list, which in itself will be a list. i.e., It may or may not contain more than one elements. On the other hand, a double bracket will return just a single element from the list.

-Single bracket will give us a list. We can also use single bracket if we wish to return multiple elements from the list. Consider the following list:

>r<-list(c(1:10),foo=1,far=2);

Now, please note the way the list is returned when I try to display it. I type r and press enter.

>r

#the result is:-

[[1]]

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

$foo

[1] 1

$far

[1] 2

Now we will see the magic of single bracket:

>r[c(1,2,3)]

#the above command will return a list with all three elements of the actual list r as below

[[1]]

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

$foo

[1] 1


$far

[1] 2

which is exactly the same as when we tried to display value of r on screen, which means the usage of single bracket has returned a list, where at index 1 we have a vector of 10 elements, then we have two more elements with names foo and far. We may also choose to give a single index or element name as input to the single bracket. e.g.,:

> r[1]

[[1]]

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

In this example, we gave one index "1" and in return got a list with one element(which is an array of 10 numbers)

> r[2]

$foo

[1] 1

In the above example, we gave one index "2" and in return got a list with one element:

> r["foo"];

$foo

[1] 1

In this example, we passed the name of one element and in return a list was returned with one element.

You may also pass a vector of element names like:

> x<-c("foo","far")

> r[x];

$foo

[1] 1

$far
[1] 2

In this example, we passed an vector with two element names "foo" and "far".

In return we got a list with two elements.

In short, a single bracket will always return you another list with number of elements equal to the number of elements or number of indices you pass into the single bracket.

In contrast, a double bracket will always return only one element. Before moving to double bracket a note to be kept in mind. NOTE:THE MAJOR DIFFERENCE BETWEEN THE TWO IS THAT SINGLE BRACKET RETURNS YOU A LIST WITH AS MANY ELEMENTS AS YOU WISH WHILE A DOUBLE BRACKET WILL NEVER RETURN A LIST. RATHER A DOUBLE BRACKET WILL RETURN ONLY A SINGLE ELEMENT FROM THE LIST.

I will site a few examples. Please keep a note of the words in bold and come back to it after you are done with the examples below:

Double bracket will return you the actual value at the index.(It will NOT return a list)

  > r[[1]]

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


  >r[["foo"]]

    [1] 1

for double brackets if we try to view more than one elements by passing a vector it will result in an error just because it was not built to cater to that need, but just to return a single element.

Consider the following

> r[[c(1:3)]]
Error in r[[c(1:3)]] : recursive indexing failed at level 2
> r[[c(1,2,3)]]
Error in r[[c(1, 2, 3)]] : recursive indexing failed at level 2
> r[[c("foo","far")]]
Error in r[[c("foo", "far")]] : subscript out of bounds
AndrewGB
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Jijo Mathew
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    Downvoted because "passing a vector... will result in an error just because it was not built to cater to that need" is incorrect; see my new answer. – MichaelChirico Apr 30 '16 at 20:31
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    Downvoted because it makes strong claims like "WHILE A DOUBLE BRACKET WILL NEVER RETURN A LIST". It is not true - if we have an object that is list of lists, double bracket will return another list. – dabsingh Jan 03 '19 at 23:39
  • The fact that `[]` returns a list class even if it is a single digit is very unintuitive. They should've created another syntax like `([])` for the list and the `[[]]` to access the actual element is fine. I prefer to think of `[[]]` as the raw value like in other languages. – TokyoToo Jun 11 '20 at 09:14
  • This is objectively incorrect, as others have stated `[[` will happily return you a list if that is the selected element. The correct answer is that `[` returns the selected item as a subset of its parent, while `[[` returns the raw selected item in and of itself free of its parent object. – D3SL Jul 26 '21 at 10:39
8

For yet another concrete use case, use double brackets when you want to select a data frame created by the split() function. If you don't know, split() groups a list/data frame into subsets based on a key field. It's useful if when you want to operate on multiple groups, plot them, etc.

> class(data)
[1] "data.frame"

> dsplit<-split(data, data$id)
> class(dsplit)
[1] "list"

> class(dsplit['ID-1'])
[1] "list"

> class(dsplit[['ID-1']])
[1] "data.frame"
Gavin Simpson
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Peter
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-1

Please refer the below-detailed explanation.

I have used Built-in data frame in R, called mtcars.

> mtcars 
               mpg cyl disp  hp drat   wt ... 
Mazda RX4     21.0   6  160 110 3.90 2.62 ... 
Mazda RX4 Wag 21.0   6  160 110 3.90 2.88 ... 
Datsun 710    22.8   4  108  93 3.85 2.32 ... 
           ............

The top line of the table is called the header which contains the column names. Each horizontal line afterward denotes a data row, which begins with the name of the row, and then followed by the actual data. Each data member of a row is called a cell.

single square bracket "[]" operator

To retrieve data in a cell, we would enter its row and column coordinates in the single square bracket "[]" operator. The two coordinates are separated by a comma. In other words, the coordinates begin with row position, then followed by a comma, and ends with the column position. The order is important.

Eg 1:- Here is the cell value from the first row, second column of mtcars.

> mtcars[1, 2] 
[1] 6

Eg 2:- Furthermore, we can use the row and column names instead of the numeric coordinates.

> mtcars["Mazda RX4", "cyl"] 
[1] 6 

Double square bracket "[[]]" operator

We reference a data frame column with the double square bracket "[[]]" operator.

Eg 1:- To retrieve the ninth column vector of the built-in data set mtcars, we write mtcars[[9]].

mtcars[[9]] [1] 1 1 1 0 0 0 0 0 0 0 0 ...

Eg 2:- We can retrieve the same column vector by its name.

mtcars[["am"]] [1] 1 1 1 0 0 0 0 0 0 0 0 ...