CODEWITHSUNDEEP
regexpython-2.7
Ghoul Fool
Ghoul Fool

Reputation: 6949

Regex match, return remaining rest of string

Simple regex function that matches the start of a string "Bananas: " and returns the second part. I've done the regex, but it's not the way I expected it to work:

import re

def return_name(s):
  m = re.match(r"^Bananas:\s?(.*)", s)

  if m:
    # print m.group(0)
    # print m.group(1)
    return m.group(1)

somestring = "Bananas: Gwen Stefani" # Bananas: + name

print return_name(somestring) # Gwen Stefani - correct!

However, I'm convinced that you don't have identify the group with (.*) in order to get the same results. ie match first part of string - return the remaining part. But I'm not sure how to do that.

Also I read somewhere that you should be being cautious using .* in a regex.

Upvotes: 1

Views: 5618

Answers (2)

ndnenkov
ndnenkov

Reputation: 36101

You could use a lookbehind ((?<=)):

(?<=^Bananas:\s).*

Remember to use re.search instead of re.match as the latter will try to match at the start of the string (aka implicit ^).

As for the .* concerns - it can cause a lot of backtracking if you don't have a clear understanding of how regexes work, but in this case it is guaranteed to be a linear search.

Upvotes: 4

Aaron
Aaron

Reputation: 24802

Using the alternate regular expression module "regex" you could use perl's \K meta-character, which makes it able to discard previously matched content and only Keep the following.

I'm not really recommending this, I think your solution is good enough, and the lookbehind answer is also probably better than using another module just for that.

Upvotes: 1

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