The difference between bracket and double bracket for accessing the elements of a list or dataframe

Question

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

User · Answer

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

User · Answer

From Hadley Wickham     My  crappy looking  modification to show using tidyverse   purrr

User · Answer

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   gt  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  lt - 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  lt - 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

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Both of them are ways of subsetting  The single bracket will return a subset of the list  which in itself will be a list  ie 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 -   gt r lt -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   gt 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 -   gt 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  eg    gt  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    gt  r 2    foo   1  1   In the above example we gave one index  2  and in return got a list with one element   gt  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 -   gt  x lt -c  foo   far     gt  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 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      gt  r  1          1   1  2  3  4  5  6  7  8  9 10      gt 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   gt  r  c 1 3    Error in r  c 1 3      recursive indexing failed at level 2  gt  r  c 1 2 3    Error in r  c 1  2  3      recursive indexing failed at level 2  gt  r  c  foo   far     Error in r  c  foo    far       subscript out of bounds

User · Answer

In addition   Following the L I N K of the A N S W E R here    Here is a little example addressing the following point   x i  j   vs  x  i  j    df1    lt - data frame a   1 3  df1 b  lt - list 4 5  6 7  8 9   df1  1 2   df1 1 2   str df1  1 2    str df1 1 2

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Double brackets accesses a list element  while a single bracket gives you back a list with a single element   lst  lt - list  one   two   three    a  lt - lst 1  class a     returns  list   a  lt - lst  1   class a     returns  character

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Being terminological     operator extracts the element from a list whereas   operator takes subset of a list

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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  lt - 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  lt - 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     lt - bar Error in foo  2 3    lt - 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    lt - 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

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Please refer the below-detailed explanation   I have used Built-in data frame in R  called mtcars    gt  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    gt  mtcars 1  2    1  6   Eg 2 - Furthermore  we can use the row and column names instead of the numeric coordinates    gt  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

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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    gt  class data   1   data frame    gt  dsplit lt -split data  data id   gt  class dsplit   1   list    gt  class dsplit  ID-1     1   list    gt  class dsplit   ID-1      1   data frame

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The R Language Definition is handy for answering these types of questions    http   cran r-project org doc manuals R-lang html Indexing     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

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extracts a list       extracts elements within the list  alist  lt - 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

[r] The difference between bracket [ ] and double bracket [[ ]] for accessing the elements of a list or dataframe

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