Substituting found regex based on rank

Question

In python how could we replace all matched cases of a pattern say '(AB)(.*?)(CD)' in text based on rank? for example reach from

bla bla blab ABftrCD bla bla ABgtCD bla blab ABnyhCD

to

bla bla blab ABftrCDn1 bla bla ABgtCDnn2 bla blab ABnyhCDnnn3

Answer 1

Using function in re.sub for replacement with a variable to keep track of replacement occurrence number

• used function attribute for variable but using a global variable is another option.

Code

import re

def func_replace(m):
    '''
        Replacement function used with re.sub
    '''
    func_replace.cnt += 1    # increment occurence count
    
    return f"{m.group(0)}n{func_replace.cnt}"

s = "bla bla blab ABftrCD bla bla ABgtCD bla blab ABnyhCD"

func_replace.cnt = 0   # initialize function cnt attribute (each time before calling re.sub)
print(re.sub(r'(AB)(.*?)(CD)', func_replace, s))

# Output: 'bla bla blab ABftrCDn1 bla bla ABgtCDn2 bla blab ABnyhCDn3'

Answer 2

Also you can use traditional loop for it:

def regex_number(string):
    
    finding = True
    l = 0
    k = 0
    while finding:

        i = string[l:].find('AB')

        if (i >= 0):

            j = string[l+i+2:].find('CD')

            if (j >= 0):

                k += 1
                sk = str(k)
                string = string[:l+i+j+4] + 'n'*k + sk + string[l+i+j+4:]
                l += i + j + k + len(sk) + 4

            if (j == -1):
                finding = False

        if (i == -1):
            finding = False

    return string