The difference between [] and [[]] notations for accessing the elements of a list or dataframe



Answers

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.

Question

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

What is the difference between the two? In what situations should I use one over the other?




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.




[] 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"



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"



From Hadley Wickham:

Or using tidyverse / purrr:




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