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')