Replace string with other with two possible patterns in python

I would like to replace everything in String untill "err" or "error" appear for empty string.
So:
"abc err efg" -> "efg",
"abc error efg" -> "efg.

How to do it with one pattern using re.sub?

I tried this:

lines_input = ['some line err hello', 'some line error hello']
rep = {r'^.*?err': '', r'^.*?error': ''}
dict((re.escape(k), v) for k, v in rep.items())
pattern = re.compile("|".join(rep.keys()))
for line in lines_input:
     print pattern.sub(lambda m: rep[re.escape(m.group(0))], line)

and had KeyError. Would like to have:

hello
hello

You don't need multiple regular expressions. The two patterns are the same except for the optional or at the end of error, so use an optional group.

line = re.sub(r'.*err(or)?\s*', '', line)

With your shown samples please try following Python code. Written and tested in Python3.10. Its using re library and split function using regex ( err(?:or)?(?: |$)).

Also if an element of list is not having error OR err then it will not print anything in it.

##Importing re library of Python here.
import re

##Input:
lines_input = ['some line err hello', 'some line error hello', 'hi there!!!'] 

##python:
[val2[-1] for val2 in [re.split('( err(?:or)?(?: |$))',val1) for val1 in lines_input] if len(val2)>=2]

You can achieve that with a single pattern

import re

lines_input = ['some line err hello', 'some line error hello']

pattern = re.compile(r'^.*?\b(?:err|error)\b')

for line in lines_input:
    # Replace the matched portion with an empty string
    modified_line = pattern.sub('', line)
    print(modified_line.strip())

Output

hello
hello