Flatten a pandas dataframe column

Question

I currently have the following column:

0          [Joe]
1           John
2           Mary
3         [Joey]
4          Harry
5        [Susan]
6          Kevin

I can't seem to remove the [] with out making the rows with [] = NaN

To be clear I want the column to look like this:

0            Joe
1           John
2           Mary
3           Joey
4          Harry
5          Susan
6          Kevin

Can anyone help?

piRSquared · Accepted Answer · 2016-12-17 07:20:20Z

3

Your title seems to imply that some elements of your series are lists.

setup

s = pd.Series([['Joe'], 'John', 'Mary', ['Joey'], 'Harry', ['Susan'], 'Kevin'])
s

0      [Joe]
1       John
2       Mary
3     [Joey]
4      Harry
5    [Susan]
6      Kevin
dtype: object

option 1
apply with pd.Series

s.apply(pd.Series).squeeze()

0      Joe
1     John
2     Mary
3     Joey
4    Harry
5    Susan
6    Kevin
Name: 0, dtype: object

answered Dec 17, 2016 at 7:20

piRSquared

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Comments

Mohammad Yusuf · Accepted Answer · 2016-12-17 07:35:28Z

0

Try this:

df['column_name'] = df['column_name'].apply(lambda x: str(x).strip("'[]") if type(x) == list else x)

edited Dec 17, 2016 at 7:35

answered Dec 17, 2016 at 6:43

Mohammad Yusuf

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4 Comments

spies006 Over a year ago

AttributeError: 'list' object has no attribute 'strip' I think the lists within the column are causing issues. I need to address the multiple lists of length 1 within the column rather than just strip the []

spies006 Over a year ago

Yes that worked! It returned the column with single quotes i.e. '' where there was previously []. To remove those I added .strip("''") to the end of your solution so it ended up working with :

spies006 Over a year ago

df['column_name'] = df['column_name'].apply(lambda x: str(x).strip('[]').strip("''"))

Mohammad Yusuf Over a year ago

@firetheRookie Don't have to use an additional strip. You can do that in 1 strip. strip("'[]") Updated my answer.

Ted Petrou · Accepted Answer · 2016-12-17 18:34:52Z

0

Why not just do s.astype(str).str.strip ("'[]'") or

s.map(lambda x: x if type(x) != list else x [0])

edited Dec 17, 2016 at 18:34

answered Dec 17, 2016 at 18:27

Ted Petrou

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