Fast Punctuation Removal with Pandas

Remove punctuations in pandas

Using Pandas str.replace and regex:

df["new_column"] = df['review'].str.replace('[^\w\s]','')

Fast punctuation removal with pandas

Setup

For the purpose of demonstration, let's consider this DataFrame.

df = pd.DataFrame({'text':['a..b?!??', '%hgh&12','abc123!!!', '$$$1234']})
df
        text
0   a..b?!??
1    %hgh&12
2  abc123!!!
3    $$$1234

Below, I list the alternatives, one by one, in increasing order of performance

str.replace

This option is included to establish the default method as a benchmark for comparing other, more performant solutions.

This uses pandas in-built str.replace function which performs regex-based replacement.

df['text'] = df['text'].str.replace(r'[^\w\s]+', '')

df
     text
0      ab
1   hgh12
2  abc123
3    1234

This is very easy to code, and is quite readable, but slow.

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regex.sub

This involves using the sub function from the re library. Pre-compile a regex pattern for performance, and call regex.sub inside a list comprehension. Convert df['text'] to a list beforehand if you can spare some memory, you'll get a nice little performance boost out of this.

import re
p = re.compile(r'[^\w\s]+')
df['text'] = [p.sub('', x) for x in df['text'].tolist()]

df
     text
0      ab
1   hgh12
2  abc123
3    1234

Note: If your data has NaN values, this (as well as the next method below) will not work as is. See the section on "Other Considerations".

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str.translate

python's str.translate function is implemented in C, and is therefore very fast.

How this works is:

1. First, join all your strings together to form one huge string using a single (or more) character separator that you choose. You must use a character/substring that you can guarantee will not belong inside your data.
2. Perform str.translate on the large string, removing punctuation (the separator from step 1 excluded).
3. Split the string on the separator that was used to join in step 1. The resultant list must have the same length as your initial column.

Here, in this example, we consider the pipe separator |. If your data contains the pipe, then you must choose another separator.

import string

punct = '!"#$%&\'()*+,-./:;<=>?@[\\]^_`{}~'   # `|` is not present here
transtab = str.maketrans(dict.fromkeys(punct, ''))

df['text'] = '|'.join(df['text'].tolist()).translate(transtab).split('|')

df
     text
0      ab
1   hgh12
2  abc123
3    1234

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Performance

str.translate performs the best, by far. Note that the graph below includes another variant Series.str.translate from MaxU's answer.

(Interestingly, I reran this a second time, and the results are slightly different from before. During the second run, it seems re.sub was winning out over str.translate for really small amounts of data.)

There is an inherent risk involved with using translate (particularly, the problem of automating the process of deciding which separator to use is non-trivial), but the trade-offs are worth the risk.

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Other Considerations

Handling NaNs with list comprehension methods; Note that this method (and the next) will only work as long as your data does not have NaNs. When handling NaNs, you will have to determine the indices of non-null values and replace those only. Try something like this:

df = pd.DataFrame({'text': [
    'a..b?!??', np.nan, '%hgh&12','abc123!!!', '$$$1234', np.nan]})

idx = np.flatnonzero(df['text'].notna())
col_idx = df.columns.get_loc('text')
df.iloc[idx,col_idx] = [
    p.sub('', x) for x in df.iloc[idx,col_idx].tolist()]

df
     text
0      ab
1     NaN
2   hgh12
3  abc123
4    1234
5     NaN

Dealing with DataFrames; If you are dealing with DataFrames, where every column requires replacement, the procedure is simple:

v = pd.Series(df.values.ravel())
df[:] = translate(v).values.reshape(df.shape)

Or,

v = df.stack()
v[:] = translate(v)
df = v.unstack()

Note that the translate function is defined below in with the benchmarking code.

Every solution has tradeoffs, so deciding what solution best fits your needs will depend on what you're willing to sacrifice. Two very common considerations are performance (which we've already seen), and memory usage. str.translate is a memory-hungry solution, so use with caution.

Another consideration is the complexity of your regex. Sometimes, you may want to remove anything that is not alphanumeric or whitespace. Othertimes, you will need to retain certain characters, such as hyphens, colons, and sentence terminators [.!?]. Specifying these explicitly add complexity to your regex, which may in turn impact the performance of these solutions. Make sure you test these solutions
on your data before deciding what to use.

Lastly, unicode characters will be removed with this solution. You may want to tweak your regex (if using a regex-based solution), or just go with str.translate otherwise.

For even more performance (for larger N), take a look at this answer by Paul Panzer.

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Appendix

Functions

def pd_replace(df):
    return df.assign(text=df['text'].str.replace(r'[^\w\s]+', ''))

def re_sub(df):
    p = re.compile(r'[^\w\s]+')
    return df.assign(text=[p.sub('', x) for x in df['text'].tolist()])

def translate(df):
    punct = string.punctuation.replace('|', '')
    transtab = str.maketrans(dict.fromkeys(punct, ''))

    return df.assign(
        text='|'.join(df['text'].tolist()).translate(transtab).split('|')
    )

# MaxU's version (https://stackoverflow.com/a/50444659/4909087)
def pd_translate(df):
    punct = string.punctuation.replace('|', '')
    transtab = str.maketrans(dict.fromkeys(punct, ''))

    return df.assign(text=df['text'].str.translate(transtab))

Performance Benchmarking Code

from timeit import timeit

import pandas as pd
import matplotlib.pyplot as plt

res = pd.DataFrame(
       index=['pd_replace', 're_sub', 'translate', 'pd_translate'],
       columns=[10, 50, 100, 500, 1000, 5000, 10000, 50000],
       dtype=float
)

for f in res.index: 
    for c in res.columns:
        l = ['a..b?!??', '%hgh&12','abc123!!!', '$$$1234'] * c
        df = pd.DataFrame({'text' : l})
        stmt = '{}(df)'.format(f)
        setp = 'from __main__ import df, {}'.format(f)
        res.at[f, c] = timeit(stmt, setp, number=30)

ax = res.div(res.min()).T.plot(loglog=True) 
ax.set_xlabel("N"); 
ax.set_ylabel("time (relative)");

plt.show()

Faster way to remove punctuations and special characters in pandas dataframe column

One way would be to keep only alphanumeric. Consider this dataframe

df=pd.DataFrame({'Text':['#^#346fetvx@!.,;:', 'fhfgd54@!#><?']})

    Text
0   #^#346fetvx@!.,;:
1   fhfgd54@!#><?

You can use

df['Text'] = df['Text'].str.extract('(\w+)', expand = False)

    Text
0   346fetvx
1   fhfgd54

Trying to remove punctuations from a column in Pandas

You should always try to avoid running pure Python code via apply() in Pandas. It's slow. Instead, use the special str property which exists on every Pandas string series:

In [9]: s = pd.Series(['hello', 'a,b,c', 'hmm...'])
In [10]: s.str.replace(r'[^\w\s]', '')
Out[10]: 
0    hello
1      abc
2      hmm
dtype: object

remove punctuation for each row in a pandas data frame

You need to iterate over the string in the dataframe, not over string.punctuation. You also need to build the string back up using .join().

df['cleaned'] = df['old'].apply(lambda x:''.join([i for i in x 
                                                  if i not in string.punctuation]))

When lambda expressions get long like that it can be more readable to write out the function definition separately, e.g. (thanks to @AndyHayden for the optimization tips):

def remove_punctuation(s):
    s = ''.join([i for i in s if i not in frozenset(string.punctuation)])
    return s

df['cleaned'] = df['old'].apply(remove_punctuation)

Remove all the punctuation from a dataframe, except some characters

Here's a way with string.punctuation:

>>> import re
>>> import string

>>> import pandas as pd

>>> df = pd.DataFrame({
...     'a': ['abc', 'de.$&$*f(@)<', '<g>hij<k>'],
...     'b': [1234, 5678, 91011],
...     'c': ['me <me@gmail.com>', '123 West-End Lane', '<<xyz>>']
... })

>>> punc = string.punctuation.replace('<', '').replace('>', '')

>>> pat = re.compile(f'[{punc}]')
>>> df.replace(pat, '')
           a      b                 c
0        abc   1234   me <megmailcom>
1       def<   5678  123 WestEnd Lane
2  <g>hij<k>  91011           <<xyz>>

You should double-check that this constant is inclusive of what you want:

String of ASCII characters which are considered punctuation characters
in the C locale.

Values:

>>> string.punctuation
'!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~'
>>> string.punctuation.replace('<', '').replace('>', '')
'!"#$%&\'()*+,-./:;=?@[\\]^_`{|}~'

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Notes:

• This solution uses an f-string (Python 3.6+)
• It encloses those literal characters in a character set to match any of them
• Note the difference between df.replace() and df[my_column_name].str.replace(). The signature for pd.DataFrame.replace() is DataFrame.replace(to_replace=None, value=None, inplace=False, limit=None, regex=False, method='pad'), where to_replace can be a regex.

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