Using a regular expression to replace upper case repeated letters in python with a single lowercase letter


Question

I am trying to replace any instances of uppercase letters that repeat themselves twice in a string with a single instance of that letter in a lower case. I am using the following regular expression and it is able to match the repeated upper case letters, but I am unsure as how to make the letter that is being replaced lower case.

import re
s = 'start TT end'
re.sub(r'([A-Z]){2}', r"\1", s)
>>> 'start T end'

How can I make the "\1" lower case? Should I not be using a regular expression to do this?

1
27
11/10/2010 2:21:46 PM

Accepted Answer

Pass a function as the repl argument. The MatchObject is passed to this function and .group(1) gives the first parenthesized subgroup:

import re
s = 'start TT end'
callback = lambda pat: pat.group(1).lower()
re.sub(r'([A-Z]){2}', callback, s)

EDIT
And yes, you should use ([A-Z])\1 instead of ([A-Z]){2} in order to not match e.g. AZ. (See @bobince's answer.)

import re
s = 'start TT end'
re.sub(r'([A-Z])\1', lambda pat: pat.group(1).lower(), s) # Inline

Gives:

'start t end'
43
5/23/2017 12:17:29 PM

You can't change case in a replacement string. You would need a replacement function:

>>> def replacement(match):
...     return match.group(1).lower()
... 
>>> re.sub(r'([A-Z])\1', replacement, 'start TT end')
'start t end'

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