The difference between double brace `[[…]]` and single brace `[..]` indexing in Pandas

Question

I\'m confused about the syntax regarding the following line of code:

x_values = dataframe[[\'Brains\']]

The dataframe object consists of 2

Answer 1

There is no special syntax in Python for [[ and ]]. Rather, a list is being created, and then that list is being passed as an argument to the DataFrame indexing function.

As per @MaxU's answer, if you pass a single string to a DataFrame a series that represents that one column is returned. If you pass a list of strings, then a DataFrame that contains the given columns is returned.

So, when you do the following

# Print "Brains" column as Series
print(df['Brains'])
# Return a DataFrame with only one column called "Brains"
print(df[['Brains']])

It is equivalent to the following

# Print "Brains" column as Series
column_to_get = 'Brains'
print(df[column_to_get])
# Return a DataFrame with only one column called "Brains"
subset_of_columns_to_get = ['Brains']
print(df[subset_of_columns_to_get])

In both cases, the DataFrame is being indexed with the [] operator.

Python uses the [] operator for both indexing and for constructing list literals, and ultimately I believe this is your confusion. The outer [ and ] in df[['Brains']] is performing the indexing, and the inner is creating a list.

>>> some_list = ['Brains']
>>> some_list_of_lists = [['Brains']]
>>> ['Brains'] == [['Brains']][0]
True
>>> 'Brains' == [['Brains']][0][0] == [['Brains'][0]][0]
True

What I am illustrating above is that at no point does Python ever see [[ and interpret it specially. In the last convoluted example ([['Brains'][0]][0]) there is no special ][ operator or ]][ operator... what happens is

• A single-element list is created (['Brains'])
• The first element of that list is indexed (['Brains'][0] => 'Brains')
• That is placed into another list ([['Brains'][0]] => ['Brains'])
• And then the first element of that list is indexed ([['Brains'][0]][0] => 'Brains')

Answer 2

Other solutions demonstrate the difference between a series and a dataframe. For the Mathematically minded, you may wish to consider the dimensions of your input and output. Here's a summary:

Object                                Series          DataFrame
Dimensions (obj.ndim)                      1                  2
Syntax arg dim                             0                  1
Syntax                             df['col']        df[['col']]
Max indexing dim                           1                  2
Label indexing              df['col'].loc[x]   df.loc[x, 'col']
Label indexing (scalar)      df['col'].at[x]    df.at[x, 'col']
Integer indexing           df['col'].iloc[x]  df.iloc[x, 'col']
Integer indexing (scalar)   df['col'].iat[x]   dfi.at[x, 'col']

When you specify a scalar or list argument to pd.DataFrame.__getitem__, for which [] is syntactic sugar, the dimension of your argument is one less than the dimension of your result. So a scalar (0-dimensional) gives a 1-dimensional series. A list (1-dimensional) gives a 2-dimensional dataframe. This makes sense since the additional dimension is the dataframe index, i.e. rows. This is the case even if your dataframe happens to have no rows.

Answer 3

Consider this:

Source DF:

In [79]: df
Out[79]:
   Brains  Bodies
0      42      34
1      32      23

Selecting one column - results in Pandas.Series:

In [80]: df['Brains']
Out[80]:
0    42
1    32
Name: Brains, dtype: int64

In [81]: type(df['Brains'])
Out[81]: pandas.core.series.Series

Selecting subset of DataFrame - results in DataFrame:

In [82]: df[['Brains']]
Out[82]:
   Brains
0      42
1      32

In [83]: type(df[['Brains']])
Out[83]: pandas.core.frame.DataFrame

Conclusion: the second approach allows us to select multiple columns from the DataFrame. The first one just for selecting single column...

Demo:

In [84]: df = pd.DataFrame(np.random.rand(5,6), columns=list('abcdef'))

In [85]: df
Out[85]:
          a         b         c         d         e         f
0  0.065196  0.257422  0.273534  0.831993  0.487693  0.660252
1  0.641677  0.462979  0.207757  0.597599  0.117029  0.429324
2  0.345314  0.053551  0.634602  0.143417  0.946373  0.770590
3  0.860276  0.223166  0.001615  0.212880  0.907163  0.437295
4  0.670969  0.218909  0.382810  0.275696  0.012626  0.347549

In [86]: df[['e','a','c']]
Out[86]:
          e         a         c
0  0.487693  0.065196  0.273534
1  0.117029  0.641677  0.207757
2  0.946373  0.345314  0.634602
3  0.907163  0.860276  0.001615
4  0.012626  0.670969  0.382810

and if we specify only one column in the list we will get a DataFrame with one column:

In [87]: df[['e']]
Out[87]:
          e
0  0.487693
1  0.117029
2  0.946373
3  0.907163
4  0.012626