Python Regex escape operator \ in substitutions & raw strings

Question

I don't understand the logic in the functioning of the scape operator \ in python regex together with r' of raw strings. Some help is appreciated.

code:

import re
text=' esto  .es  10  . er - 12 .23 with [  and.Other ] here is more ; puntuation'
print('text0=',text)
text1 = re.sub(r'(\s+)([;:\.\-])', r'\2', text)
text2 = re.sub(r'\s+\.', '\.', text)
text3 = re.sub(r'\s+\.', r'\.', text)
print('text1=',text1)
print('text2=',text2)
print('text3=',text3)

The theory says: backslash character ('\') to indicate special forms or to allow special characters to be used without invoking their special meaning.

And as far as the link provided at the end of this question explains, r' represents a raw string, i.e. there is no special meaning for symbols, it is as it stays.

so in the above regex I would expect text2 and text3 to be different, since the substitution text is '.' in text 2, i.e. a period, whereas (in principle) the substitution text in text 3 is r'.' which is a raw string, i.e. the string as it is should appear, backslash and period. But they result in the same:

The result is:

text0=  esto  .es  10  . er - 12 .23 with [  and.Other ] here is more ; puntuation
text1=  esto.es  10. er- 12.23 with [  and.Other ] here is more; puntuation
text2=  esto\.es  10\. er - 12\.23 with [  and.Other ] here is more ; puntuation
text3=  esto\.es  10\. er - 12\.23 with [  and.Other ] here is more ; puntuation
#text2=text3 but substitutions are not the same r'\.' vs '\.'

It looks to me that the r' does not work the same way in substitution part, nor the backslash. On the other hand my intuition tells me I am missing something here.

EDIT 1: Following @Wiktor Stribiżew comment. He pointed out that (following his link):

import re
print(re.sub(r'(.)(.)(.)(.)(.)(.)', 'a\6b', '123456'))
print(re.sub(r'(.)(.)(.)(.)(.)(.)', r'a\6b', '123456'))
# in my example the substitutions were not the same and the result were equal
# here indeed r' changes the results

which gives:

ab
a6b

that puzzles me even more.

Note: I read this stack overflow question about raw strings which is super complete. Nevertheless it does not speak about substitutions

Answer 1

A simple way to work around all these string escaping issues is to use a function/lambda as the repl argument, instead of a string. For example:

output = re.sub(
    pattern=find_pattern,
    repl=lambda _: replacement,
    string=input,
)

The replacement string won't be parsed at all, just substituted in place of the match.