How to modify a pandas DataFrame in a function so that changes are seen by the caller?

Question

I find myself doing repetitive tasks to various [pandas][1] DataFrames, so I made a function to do the processing. How do I modify df in the function process_df(df) so that the caller sees all changes (without assigning a return value)?

A simplified version of the code:

def process_df(df): df.columns = map(str.lower, df.columns) df = pd.DataFrame({'A': [1], 'B': [2]}) process_df(df) print df

 A B 0 1 2

EDIT new code:

def process_df(df): df = df.loc[:, 'A'] df = pd.DataFrame({'A': [1], 'B': [2]}) process_df(df) print df

 A B 0 1 2

IIUC your code is working, after function process_df column names becomes [a, b] — Anton Protopopov
– Anton Protopopov, Commented Feb 2, 2016 at 5:16
Indeed it is. My bad. In the process of simplifying I left out the part that does not work. I will repost as a new question. — ChaimG
– ChaimG, Commented Feb 2, 2016 at 5:27

Igor Raush · Accepted Answer · 2016-02-02 05:55:10Z

Indexing a DataFrame using ix, loc, iloc, etc. returns a view of the underlying data (it is a read operation). In order to modify the contents of the frame you will need to use in-place transforms. For example,

def process_df(df): # drop all columns except for A df.drop(df.columns[df.columns != 'A'], axis=1, inplace=True) df = DataFrame({'A':[1,2,3], 'B':[1,2,3]}) process_df(df)

To change the order of columns, you can do something like this:

def process_df(df): # swap A and B df.columns = ['B', 'A'] df[['B', 'A']] = df[['A', 'B']]

How would I go about re-arranging the order of multiple columns? All the examples I have seen to do this begin with df = ... which wouldn't be an in-place transform.
Added an example above. It's not pretty. Out of curiosity, why the restriction that you can't use the return value of process_df?
Code (and error) reduction. The function is always used to modify the existing DataFrame.

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