regex - re.sub in python : verbose mode does not work with replacement pattern? -

is there way around limitation of re.sub? not functional verbose mode (with reference here) in replace pattern; not eliminate whitespace or comments (yet interpret backreferences properly).

import remport re  ft1=r"""(?p<test>[0-9]+)""" ft2=r"""\g<test>and then: \g<test> #this remains"""  print re.sub(ft1,ft2,"front 1234 back",flags=re.verbose) #does not work  #result: front 1234and then: 1234 #this remains 

re.verbose not apply replacement pattern... there work-around? (simpler working groups after re.match.)

here way have found "compile" re replace expression sub. there few constraints: both spaces , newlines have written spaces written re match expression (in square brackets: [ ] , [\n\n\n]) , whole replace expression should have verbose newline @ beginning.

an example: searches string , detects word repeated after /ins/ , /del/, replaces occurrences single occurrence of word in front of .

both match , replace expressions complex, why want verbose version of replace expression.

===========================

import re  test = "<p>le petit <ins>homme à</ins> <del>homme en</del> ressorts</p>"   find=r"""     <ins>     (?p<front>[^<]+)          #there added matches      (?p<delim1>[ .!,;:]+)     #get delimiter     (?p<back1>[^<]*?)     </ins>     [ ]     <del>     (?p=front)     (?p<delim2>[ .!,;:]+)     (?p<back2>[^<]*?)     </del> """ replace = r"""     <<<<<\g<front>>>>>         #pop out in front matching thing     <ins>     \g<delim1>     \g<back1>     </ins>     [ ]          <del>         \g<delim2>             #put delimiters , backend     \g<back2>     </del> """  flatreplace = r"""<<<<<\g<front>>>>><ins>\g<delim1>\g<back1></ins> <del>\g<delim2>\g<back2></del>"""   def compilerepl(instring):      outstring=instring     #get space @ front of line     outstring=re.sub(r"\n\s+","\n",outstring)     #get space @ end of line     outstring=re.sub(r"\s+\n","",outstring)      #get rid of comments     outstring=re.sub(r"\s*#[^\n]*\n","\n",outstring)     #preserve space in brackets, , eliminate brackets     outstring=re.sub(r"(?<!\[)\[(\s+)\](?!\[)",r"\1",outstring)     # rid of newlines not in brackets     outstring=re.sub(r"(?<!\[)(\n)+(?!\])","",outstring)     #get rid of brackets around newlines     outstring=re.sub(r"\[((\\n)+)\]",r"\1",outstring)     #trim brackets         outstring=re.sub(r"\[\[(.*?)\]\]","[\\1]",outstring)     return outstring   assert(flatreplace == compilerepl(replace))   print test print compilerepl(replace) print re.sub(find,compilerepl(replace),test, flags=re.verbose)  #<p>le petit <ins>homme à</ins> <del>homme en</del> ressorts</p> #<<<<<\g<front>>>>><ins>\g<delim1>\g<back1></ins> <del>\g<delim2>\g<back2></del> #<p>le petit <<<<<homme>>>><ins> à</ins> <del> en</del> ressorts</p>