[SOLVED] How to skip, if starts with, but match other strings

How to skip, if starts with, but match other strings

I want to match and substitute for strings as shown in the example below, but not for some strings which start with test or !!. I have used negative lookahead to skip matching unwanted strings but (Street|,)(?=\d) matching for Street & comma replacing group 1 with UK/ is not working as expected.

import re
input = [ 'Street1-2,4,6,8-10', 
          '!! Street4/31/2',
          'test Street4' ]
pattern = r'(^(?!test\s|!!\s).*(Street|,)(?=\d))'
output = [re.sub(pattern, r'\g<1>UK/', line) for line in input ]

Actual output:

['Street1-2,4,6,UK/8-10', '!! Street4/31/2', 'test Street4']

Expected output:

['StreetUK/1-2,UK/4,UK/6,UK/8-10', '!! Street4/31/2', 'test Street4']

Solution

You could change the pattern to use 2 capture groups, and then use a callback with re.sub.

The callback checks if there is a group 1 value. If there is, use it in the replacement, else use group 2 followed by UK/

^((?:!!|test)\s.*)|(Street|,)(?=\d)

The regex matches

^((?:!!|test)\s.*) Capture either !! or test at the start of the string followed by a whitespace char and then the rest of the line in group 1
| Or
(Street|,)(?=\d) Capture either Street or , in group 2 while asserting a digit to the right

See a regex101 demo

import re

lst = ['Street1-2,4,6,8-10',
       '!! Street4/31/2',
       'test Street4']

pattern = r'^((?:!!|test)\s.*)|(Street|,)(?=\d)'

output = [re.sub(pattern, lambda m: m.group(1) or m.group(2) + 'UK/', line) for line in lst]

print(output)

Output

['StreetUK/1-2,UK/4,UK/6,UK/8-10', '!! Street4/31/2', 'test Street4']