How to match my string? Do I need negative lookbehind?-CodePudding

Let's assume we have the following string:

This thing costs $5000.

I'm trying to match up $5000 with negative lookbehind:

(?<!([:;]))\$?([0-9] )

So that it doesn't find a match if it has ";" or ':' behind $5000, eg. ;$5000 or ;5000.

First string:

This thing costs $5000 or 5000

Desired output:

$5000 or 5000

Second string:

This thing costs ;$5000 or ;5000

Desired output:

None

CodePudding user response：

Your implementation is great but there's a single flaw: you can match from the middle of the digits or from the $ sign.

Add \d and $ to the negative lookback and it'll work:

(?<![;:\d$])\$?([0-9] )

Examples:

>>> re.findall("(?<![;:\d$])\$?([0-9] )", "This thing costs ;$5000")
[]
>>> re.findall("(?<![;:\d$])\$?([0-9] )", "This thing costs $5000")
['5000']

Keep in mind I do suggest matching the number instead of dealing with negative lookbacks like so:

re.findall(r"\s\$?([0-9] )", "This thing costs ;$5000")

CodePudding user response：

Sometimes, matching what you do want rather than what you don't want is easier. As far as I can tell, what you're really looking for is an integer that optionally start with $. But, you still want to capture the $ if it's there. The ; and : are just red-herrings.

import re

values = ['This thing costs $5000.',  # $5000
          'This thing costs ;$5000.', # None
          'This thing costs 5000.',   # 5000
          'This thing costs ;5000.',  # None
          'This thing costs8000']     # None

pattern = r'.*\s(\$?\d )'

for value in values:
    # If a match is made, we want group 1 from that match.
    if match := re.match(pattern, value):
        print(match.group(1))
    else:
        print(match)

Output:

$5000
None
5000
None
None

See https://regexr.com/6sqbs for an in-depth explanation of my pattern .*\s(\$?\d )