`error: unbalanced parenthesis` while checking if an item presents in a pandas dataframe

问题

df=pd.DataFrame({"A":["one","two","three"],"B":["fopur","give","six"]})

when I do,

df.B.str.contains("six").any()
out[2]=True

when I do,

df.B.str.contains("six)").any()

I am getting the below error,

C:\ProgramData\Anaconda3\lib\sre_parse.py in parse(str, flags, pattern)
    868     if source.next is not None:
    869         assert source.next == ")"
--> 870         raise source.error("unbalanced parenthesis")
    871 
    872     if flags & SRE_FLAG_DEBUG:

error: unbalanced parenthesis at position 3

Please help!

回答1:

You can set regex=False in in pandas.Series.str.contains:

df.B.str.contains("six)", regex=False).any()

If you want to match irrespective of case,

df.B.str.contains("Six)", case=False, regex=False).any() 
out[]: True

https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.str.contains.html

Info:

Parenthesis are special characters in regular expressions that need to be "escaped", see for example here or here.

回答2:

You need escape ) by \ because special regex character:

df.B.str.contains("six\)").any()

More general:

import re

df.B.str.contains(re.escape("six)")).any()

来源：https://stackoverflow.com/questions/48699907/error-unbalanced-parenthesis-while-checking-if-an-item-presents-in-a-pandas-d