In Pandas Dataframe error: nothing to repeat at position 17217 - pandas

Question

I am trying to use str.contains to identify elements of one column of a dataframe in another column of another dataframe. Here is the code:
pattern = fr"(?:{'|'.join(strategic_accounts['Account Name'])})"
all_leads['in_strategic_list'] = all_leads['Company'].str.contains(pattern).astype(int)
Here are the heads of both dataframes as well as position 17271 of the all_leads dataframe. I don't understand the error because it looks like there isn't anything abnormal at position 17217. Also, all related errors online seem to refer to error nothing to repeat at position 0 which seems like it would be a different error since mine came up at loc 17217. Any insights appreciated! Thanks!
This mock example works perfectly with the same code:
df1 = pd.DataFrame({'name': ['Marc', 'Jake', 'Sam', 'Brad', 'SpongeBob']})
df2 = pd.DataFrame({'IDs': ['Jake', 'John', 'Marc', 'Tony', 'Bob']})
pattern = fr"(?:{'|'.join(df2['IDs'])})"
df1['In_df2'] = df1['name'].str.contains(pattern).astype(int)
Update:
I have managed to figure out that the error is referring to loc 17217 in pattern not in strategic_accounts df. Printing the loc 17217 in pattern returns '*'. I have tried to apply this function to pattern before inserting it into the str.contains and I can't seem to get it to remove.
import re
pattern = fr"(?:{'|'.join(strategic_accounts['Account Name'])})"
def esc_spec_char(pattern):
for p in pattern:
if p == '\*':
re.sub('*', '1', p)
else:
continue
return pattern
pattern = esc_spec_char(pattern)
pattern[17217]
New_Update:
I have applied #LiamFiddler's method of turning the string into a re.Pattern object and run it on a dummy df and while it does seem to escape the * it doesn't seem to find the N. Not sure if I made some mistake. Here is the code:
sries = pd.Series(['x','y','$','%','^','N','*'])
ac = '|'.join(sries)
p = re.compile(re.escape(ac))
df1 = pd.DataFrame(data = {'Id' : [123, 232, 344, 455, 566, 377],
'col2' : ["N", "X", "Y", '*', "W", "Z"]})
df1['col2'].str.contains(p, regex=True).astype(int)

Answer 1

EDIT
I realized that re.escape() also escapes the | delimiter, so I think the appropriate solution is to map re.escape() to the series before joining the names:
strategic_accounts['Escaped Accounts'] = strategic_accounts['Account Name'].apply(lambda x: re.escape(x))
pattern = re.compile('|'.join(strategic_accounts['Escaped Accounts']))
Then you can proceed as below with using Series.str.contains(). On your sample dataframe, here is what I get:
sries = pd.Series(['x','y','$','%','^','N','*'])
ac = sries.apply(lambda x: re.escape(x))
p = re.compile('|'.join(ac))
df1 = pd.DataFrame(data = {'Id' : [123, 232, 344, 455, 566, 377],
'col2' : ["N", "X", "Y", '*', "W", "Z"]})
df1['col2'].str.contains(p, regex=True).astype(int)
Out:
0 1
1 0
2 0
3 1
4 0
5 0
Original
Ok, so based on the discovery of the special character, I think this is your solution:
First, we need to escape the special characters in the strings so that they don't mess up the regex. Fortunately, Python's re module has an .escape() method specifically for escaping special characters.
import re
accounts = '|'.join(strategic_accounts['Account Name'])
pattern = re.compile(re.escape(accounts))
Now we can proceed as before:
all_leads['in_strategic_list'] = all_leads['Company'].str.contains(pattern, regex=True).astype(int)