Find two of the same character in a string with regular expressions

This is in reference to a question I asked before here

I received a solution to the problem in that question but ended up needing to go with regex for this particular part.

I need a regular expression to search and replace a string for instances of two vowels in a row that are the same, so the "oo" in "took", or the "ee" in "bees" and replace it with the one of the letters that was replaced and a :.

Some examples of expe开发者_如何学Pythoncted behavior:

"took" should become "to:k"

"waaeek" should become "wa:e:k"

"raaag" should become "ra:ag"

Thank you for the help.

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Try this:

re.sub(r'([aeiou])\1', r'\1:', str)

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Search for ([aeiou])\1 and replace it with \1:

I don't know about python, but you should be able to make the regex case insensitive and global with something like /([aeiou])\1/gi

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What NOT to do: As noted, this will match any two vowels together. Leaving this answer as an example of what NOT to do. The correct answer (in this case) is to use backreferences as mentioned in numerous other answers.

import re

data = ["took","waaeek","raaag"]

for s in data:
  print re.sub(r'([aeiou]){2}',r'\1:',s)

This matches exactly two occurrences {2} of any member of the set [aeiou]. and replaces it with the vowel, captured with the parens () and placed in the sub string by the \1 followed by a ':'

Output:

to:k
wa:e:k
ra:ag

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You'll need to use a back reference in your search expression. Try something like: ([a-z])+\1 (or ([a-z])\1 for just a double).