When selecting a single column from a pandas DataFrame(say df.iloc[:, 0], df['A'], or df.A, etc), the resulting vector is automatically converted to a Series instead of a single-column DataFrame. However, I am writing some functions that takes a DataFrame as an input argument. Therefore, I prefer to deal with single-column DataFrame instead of Series so that the function can assume say df.columns is accessible. Right now I have to explicitly convert the Series into a DataFrame by using something like pd.DataFrame(df.iloc[:, 0]). This doesn’t seem like the most clean method. Is there a more elegant way to index from a DataFrame directly so that the result is a single-column DataFrame instead of Series?

As @Jeff mentions there are a few ways to do this, but I recommend using loc/iloc to be more explicit (and raise errors early if you’re trying something ambiguous):

In [10]: df = pd.DataFrame([[1, 2], [3, 4]], columns=['A', 'B'])

In [11]: df
   A  B
0  1  2
1  3  4

In [12]: df[['A']]

In [13]: df[[0]]

In [14]: df.loc[:, ['A']]

In [15]: df.iloc[:, [0]]

Out[12-15]:  # they all return the same thing:
0  1
1  3

The latter two choices remove ambiguity in the case of integer column names (precisely why loc/iloc were created). For example:

In [16]: df = pd.DataFrame([[1, 2], [3, 4]], columns=['A', 0])

In [17]: df
   A  0
0  1  2
1  3  4

In [18]: df[[0]]  # ambiguous
0  1
1  3

As Andy Hayden recommends, utilizing .iloc/.loc to index out (single-columned) dataframe is the way to go; another point to note is how to express the index positions.
Use a listed Index labels/positions whilst specifying the argument values to index out as Dataframe; failure to do so will return a ‘pandas.core.series.Series’


    A_1 = train_data.loc[:,'Fraudster']
    print('A_1 is of type', type(A_1))
    A_2 = train_data.loc[:, ['Fraudster']]
    print('A_2 is of type', type(A_2))
    A_3 = train_data.iloc[:,12]
    print('A_3 is of type', type(A_3))
    A_4 = train_data.iloc[:,[12]]
    print('A_4 is of type', type(A_4))


    A_1 is of type <class 'pandas.core.series.Series'>
    A_2 is of type <class 'pandas.core.frame.DataFrame'>
    A_3 is of type <class 'pandas.core.series.Series'>
    A_4 is of type <class 'pandas.core.frame.DataFrame'>

These three approaches have been mentioned:

pd.DataFrame(df.loc[:, 'A'])  # Approach of the original post
df.loc[:,[['A']]              # Approach 2 (note: use iloc for positional indexing)
df[['A']]                     # Approach 3

pd.Series.to_frame() is another approach.

Because it is a method, it can be used in situations where the second and third approaches above do not apply. In particular, it is useful when applying some method to a column in your dataframe and you want to convert the output into a dataframe instead of a series. For instance, in a Jupyter Notebook a series will not have pretty output, but a dataframe will.

# Basic use case: 

# Use case 2 (this will give you pretty output in a Jupyter Notebook): 

# Use case 3: 

# Use case 4: 
def some_function(num): 


You can use df.iloc[:, 0:1], in this case the resulting vector will be a DataFrame and not series.

As you can see:

enter image description here

(Talking about pandas 1.3.4)

I’d like to add a little more context to the answers involving .to_frame(). If you select a single row of a data frame and execute .to_frame() on that, then the index will be made up of the original column names and you’ll get numeric column names. You can just tack on a .T to the end to transpose that back into the original data frame’s format (see below).

import pandas as pd
print(pd.__version__)  #1.3.4

df = pd.DataFrame({
    "col1": ["a", "b", "c"],
    "col2": [1, 2, 3]

# series
df.loc[0, ["col1", "col2"]]

# dataframe (column names are along the index; not what I wanted)
df.loc[0, ["col1", "col2"]].to_frame()
    #       0
    # col1  a
    # col2  1

# looks like an actual single-row dataframe.
# To me, this is the true answer to the question
# because the output matches the format of the
# original dataframe.
df.loc[0, ["col1", "col2"]].to_frame().T
    #   col1 col2
    # 0    a    1

# this works really well with .to_dict(orient="records") which is 
# what I'm ultimately after by selecting a single row
df.loc[0, ["col1", "col2"]].to_frame().T.to_dict(orient="records")
    # [{'col1': 'a', 'col2': 1}]