[python] Strip / trim all strings of a dataframe

Cleaning the values of a multitype data frame in python/pandas, I want to trim the strings. I am currently doing it in two instructions :

import pandas as pd

df = pd.DataFrame([['  a  ', 10], ['  c  ', 5]])

df.replace('^\s+', '', regex=True, inplace=True) #front
df.replace('\s+$', '', regex=True, inplace=True) #end

df.values

This is quite slow, what could I improve ?

You can try:

df[0] = df[0].str.strip()

or more specifically for all string columns

non_numeric_columns = list(set(df.columns)-set(df._get_numeric_data().columns))
df[non_numeric_columns] = df[non_numeric_columns].apply(lambda x : str(x).strip())

def trim(x):
    if x.dtype == object:
        x = x.str.split(' ').str[0]
    return(x)

df = df.apply(trim)

Money Shot

Here's a compact version of using applymap with a straightforward lambda expression to call strip only when the value is of a string type:

df.applymap(lambda x: x.strip() if isinstance(x, str) else x)

Full Example

A more complete example:

import pandas as pd


def trim_all_columns(df):
    """
    Trim whitespace from ends of each value across all series in dataframe
    """
    trim_strings = lambda x: x.strip() if isinstance(x, str) else x
    return df.applymap(trim_strings)


# simple example of trimming whitespace from data elements
df = pd.DataFrame([['  a  ', 10], ['  c  ', 5]])
df = trim_all_columns(df)
print(df)


>>>
   0   1
0  a  10
1  c   5

Working Example

Here's a working example hosted by trinket: https://trinket.io/python3/e6ab7fb4ab

You can use the apply function of the Series object:

>>> df = pd.DataFrame([['  a  ', 10], ['  c  ', 5]])
>>> df[0][0]
'  a  '
>>> df[0] = df[0].apply(lambda x: x.strip())
>>> df[0][0]
'a'

Note the usage of strip and not the regex which is much faster

Another option - use the apply function of the DataFrame object:

>>> df = pd.DataFrame([['  a  ', 10], ['  c  ', 5]])
>>> df.apply(lambda x: x.apply(lambda y: y.strip() if type(y) == type('') else y), axis=0)

   0   1
0  a  10
1  c   5

If you really want to use regex, then

>>> df.replace('(^\s+|\s+$)', '', regex=True, inplace=True)
>>> df
   0   1
0  a  10
1  c   5

But it should be faster to do it like this:

>>> df[0] = df[0].str.strip()