Substitution for comma match

Question

For the input string as shown below, I am trying to substitute UK1/ after Street and every , and skip hyphen to create expected output shown below.

Input = Street1-2,4,6,8-10

Expected output = StreetUK/1-2,UK/4,UK/6,UK/8-10

With the regex pattern I have trouble substitute for each captured group. How can I capture all the required groups for every , and sub required string.

    replacements = []
    pattern = r"(Street)?(?:\d+)(((,)?(?:\d+))*[-]?(?:\d+))*"
    
    def replacement(x):
        replacements.append(f"{x.group(1)}{'UK'}/")
    
    input = 'Street1-2,4,6,8-10'
    m = re.sub(pattern, replacement, input)
    print(m, [''.join(x) for x in replacements] )
    

The above code just prints ['StreetUK/'] but not as expected.

Answer 1

You can search using this regex:

(Street|,)(?=\d)

and replace with:

\1UK/

RegEx Demo

Code:

import re
input = 'Street1-2,4,6,8-10'
print(re.sub(r'(Street|,)(?=\d)', r'\1UK/', input))

Output:

StreetUK/1-2,UK/4,UK/6,UK/8-10

If you want to do these replacements only for input starting with Street then use:

print(re.sub(r'(Street|,)(?=\d)', r'\1UK/', input) if input.startswith("Street") else input)

RegEx Explanation:

• (Street|,): Match Street or comma in capture group #1
• (?=\d): Make sure there is a digit after the current position
• \1UK/: is replacement to put back-reference of group #1 back followed by UK/ in the substituted string