Counting opening parenthesis in pandas dataframe

Question

I am trying to count number of symbols with the string.punctuation module in python in a dataframe column but I can not find a way to have the opening parenthesis to be counted as python thinks does not consider it a string apparently.

I am working on linux + Jupyter notebook and python 3.8.

df = pd.DataFrame()
df['password'] = data
df['sign'] = 0
for i in string.punctuation:
    print(i)
    print(type(i))
    df['sign'] += df['password'].str.count(i)
    
df['sign'].iloc[:100]

This gives me:

!
<class 'str'>
"
<class 'str'>
#
<class 'str'>
$
<class 'str'>
%
<class 'str'>
&
<class 'str'>
'
<class 'str'>
(
<class 'str'>

and afterwards the exception:

/opt/conda/lib/python3.8/sre_parse.py in _parse(source, state, verbose, nested, first)
    834             p = _parse_sub(source, state, sub_verbose, nested + 1)
    835             if not source.match(")"):
--> 836                 raise source.error("missing ), unterminated subpattern",
    837                                    source.tell() - start)
    838             if group is not None:

error: missing ), unterminated subpattern at position 0

Thank you.

Answer 1

Example dataframe:

df = pd.DataFrame({'text': ['hel\\l\'o', 'hellO()world']})

Parentheses are part of the regex syntax so you need to escape them:

df['text'].str.count('\(')

To cover all of string.punctuation you can use:

df['text'].str.count(f'[{re.escape(string.punctuation)}]')