pandas dataframe : add & remove prefix/suffix from all cell values of entire dataframe

Question

To add a prefix/suffix to a dataframe, I usually do the following..

For instance, to add a suffix '@',

df = df.astype(str) + '@'

This has basically appended a '@' to all cell values.

I would like to know how to remove this suffix. Is there a method available with the pandas.DataFrame class directly that removes a particular prefix/suffix character from the entire DataFrame ?

I've tried iterating through the rows (as series) while using rstrip('@') as follows:

for index in range(df.shape[0]):
    row = df.iloc[index]
    row = row.str.rstrip('@')

Now, in order to make dataframe out of this series,

new_df = pd.DataFrame(columns=list(df))
new_df = new_df.append(row)

However, this doesn't work. Gives empty dataframe.

Is there something really basic that I am missing?

Answer 1

You could use applymap to apply your string method to each element:

df = df.applymap(lambda x: str(x).rstrip('@'))

Note: I wouldn't expect this to be as fast as the vectorized approach: pd.Series.str.rstrip i.e. transforming each column separately

Answer 2

You could make this real easy and just use pandas.DataFrame.replace() method to replace all "@" with a "":

df.replace("@", "")

If you are worried about "@" being replaced not just at the end of your values, you could use regex:

df.replace("@$", "", regex=True) 

Answer 3

You can use apply and the str.strip method of pd.Series:

In [13]: df
Out[13]:
       a       b      c
0    dog   quick    the
1   lazy    lazy    fox
2  brown   quick    dog
3  quick     the   over
4  brown    over   lazy
5    fox   brown  quick
6  quick     fox    the
7    dog  jumped    the
8   lazy   brown    the
9    dog    lazy    the

In [14]: df = df + "@"

In [15]: df
Out[15]:
        a        b       c
0    dog@   quick@    the@
1   lazy@    lazy@    fox@
2  brown@   quick@    dog@
3  quick@     the@   over@
4  brown@    over@   lazy@
5    fox@   brown@  quick@
6  quick@     fox@    the@
7    dog@  jumped@    the@
8   lazy@   brown@    the@
9    dog@    lazy@    the@

In [16]: df = df.apply(lambda S:S.str.strip('@'))

In [17]: df
Out[17]:
       a       b      c
0    dog   quick    the
1   lazy    lazy    fox
2  brown   quick    dog
3  quick     the   over
4  brown    over   lazy
5    fox   brown  quick
6  quick     fox    the
7    dog  jumped    the
8   lazy   brown    the
9    dog    lazy    the

Note, your approach doesn't work because when you do the following assignment in your for-loop:

row = row.str.rstrip('@')

This merely assigns the result of row.str.strip to the name row without mutating the DataFrame. This is the same behavior for all python objects and simple name assignment:

In [18]: rows = [[1,2,3],[4,5,6],[7,8,9]]

In [19]: print(rows)
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

In [20]: for row in rows:
    ...:     row = ['look','at','me']
    ...:

In [21]: print(rows)
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

To actually change the underlying data structure you need to use a mutator method:

In [22]: rows
Out[22]: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

In [23]: for row in rows:
    ...:     row.append("LOOKATME")
    ...:

In [24]: rows
Out[24]: [[1, 2, 3, 'LOOKATME'], [4, 5, 6, 'LOOKATME'], [7, 8, 9, 'LOOKATME']]

Note that slice-assignment is just syntactic sugar for a mutator method:

In [26]: rows
Out[26]: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

In [27]: for row in rows:
    ...:     row[:] = ['look','at','me']
    ...:
    ...:

In [28]: rows
Out[28]: [['look', 'at', 'me'], ['look', 'at', 'me'], ['look', 'at', 'me']]

This is analogous to pandas loc or iloc based assignment.