Your regex is matching on all - characters:

In [48]:
df_raw.replace(['-','\*'], ['0.00','0.00'], regex=True)

Out[48]:
       A          B
0   1.00        1.0
1  0.001  0.0045.00
2    NaN       0.00

If you put additional boundaries so that it only matches that single character with a termination then it works as expected:

In [47]:
df_raw.replace(['^-$'], ['0.00'], regex=True)

Out[47]:
      A       B
0  1.00     1.0
1    -1  -45.00
2   NaN    0.00

Here ^ means start of string and $ means end of string so it will only match on that single character.

Or you can just use replace which will only match on exact matches:

In [29]:

df_raw.replace('-',0)
Out[29]:
      A       B
0  1.00     1.0
1    -1  -45.00
2   NaN       0

You could use Series.str.replace:

import pandas as pd

df = pd.DataFrame(['$40,000*','$40000 conditions attached'], columns=['P'])
print(df)
#                             P
# 0                    $40,000*
# 1  $40000 conditions attached

df['P'] = df['P'].str.replace(r'\D+', '', regex=True).astype('int')
print(df)

yields

       P
0  40000
1  40000

since \D matches any character that is not a decimal digit.

When you don't use regex=True the method will look for the exact match of the replace_to value. When you use regex=True it will look for the sub strings too. So your code works when you use that parameter. Example

When you dont use the regex=True parameter, the replace method will look for the exact match of the replace_to value in the series.

df = pd.DataFrame()
df['k'] = ['YoYo.','mins.s.s','sos.s.','.','mama.mia']
df['k'].replace('.',',')

0       YoYo.
1    mins.s.s
2      sos.s.
3           ,
4    mama.mia

When you use regex = True it even matches the substrings of the series and changes the strings

df = pd.DataFrame()
df['k'] = ['YoYo.','mins.s.s','sos.s.','.','mama.mia']
df['k'].replace('\.',',',regex=True)

0       YoYo,
1    mins,s,s
2      sos,s,
3           ,
4    mama,mia

str.replace() ^v2|v3 does not recognize regular expressions.

To perform a substitution using a regular expression, use re.sub() ^v2|v3.

For example:

import re

line = re.sub(
           r"(?i)^.*interfaceOpDataFile.*$", 
           "interfaceOpDataFile %s" % fileIn, 
           line
       )

In a loop, it would be better to compile the regular expression first:

import re

regex = re.compile(r"^.*interfaceOpDataFile.*$", re.IGNORECASE)
for line in some_file:
    line = regex.sub("interfaceOpDataFile %s" % fileIn, line)
    # do something with the updated line

No. Regular expressions in Python are handled by the re module.

article = re.sub(r'(?is)</html>.+', '</html>', article)

In general:

str_output = re.sub(regex_search_term, regex_replacement, str_input)