How fill a regex string with parameters

Question

I would like to fill regex variables with string.

import re

hReg = re.compile("/robert/(?P<action>([a-zA-Z0-9]*))/$")
hMatch = hReg.match("/robert/delete/")
args = hMatch.groupdict()

args variable is now a dict with {"action":"delete"}.

How i can reverse this process ? With args dict and regex pattern, how i can obtain the string "/robert/delete/" ?

it's possible to have a function just like this ?

def reverse(pattern, dictArgs):

Thank you

Answer 1

This function should do it

def reverse(regex, dict):
    replacer_regex = re.compile('''
        \(\?P\<         # Match the opening
            (.+?)       # Match the group name into group 1
        \>\(.*?\)\)     # Match the rest
        '''
        , re.VERBOSE)

    return replacer_regex.sub(lambda m : dict[m.group(1)], regex)

You basically match the (\?P...) block and replace it with a value from the dict.

EDIT: regex is the regex string in my exmple. You can get it from patter by

regex_compiled.pattern

EDIT2: verbose regex added