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?

Don't know how to make it lowercase, but your should use `'([A-Z]){2,}'` instead of `'([A-Z]){2}'` to replace any instances. — khachik, Nov 10 '10 at 14:25

score 62 · Accepted Answer · edited May 23 '17 at 12:17

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'

score 9 · Answer 2 · answered Nov 10 '10 at 14:29

9

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'

answered Nov 10 '10 at 14:29

bobince

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Akash g krishnan · Answer 3 · 2020-06-20T07:18:07.513

1

def replace(s):
    return " ".join(re.findall(r"[A-Z]){2}", s)).lower()

I guess this is what you are looking for.

edited Jun 20 '20 at 07:18

answered Jun 16 '20 at 09:44

Akash g krishnan

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score 0 · Answer 4 · answered Nov 10 '10 at 14:27

0

You can do it with a regular expression, just pass a function as the replacement like the docs say. The problem is your pattern.

As it is, your pattern matches runs of any two capital letters. I'll leave the actual pattern to you, but it starts with AA|BB|CC|.

answered Nov 10 '10 at 14:27

Ignacio Vazquez-Abrams

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score 0 · Answer 5 · answered Nov 10 '10 at 14:30

The 'repl' parameter that identifies the replacement can be either a string (as you have it here) or a function. This will do what you wish:

import re

def toLowercase(matchobj):
   return matchobj.group(1).lower()

s = 'start TT end'
re.sub(r'([A-Z]){2}', toLowercase, s)
>>> 'start t end'

score 0 · Answer 6 · edited Nov 10 '10 at 14:35

0

Try this:

def tol(m):
   return m.group(0)[0].lower()

s = 'start TTT AAA end'
re.sub(r'([A-Z]){2,}', tol, s)

Note that this doesn't replace singe upper letters. If you want to do it, use r'([A-Z]){1,}'.

edited Nov 10 '10 at 14:35

SilentGhost

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answered Nov 10 '10 at 14:34

khachik

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OP says: *repeat themselves twice* – SilentGhost Nov 10 '10 at 14:35
@SilentGhost. My fault. the re should be as suggested by Ignacio, if single upper characters shouldn't be touched. – khachik Nov 10 '10 at 14:40
if you look and bobince's and jens's answers you see the shorter way to do this. – SilentGhost Nov 10 '10 at 14:44
I see, thanks. The question about single upper chars is still open. – khachik Nov 10 '10 at 14:56

score 0 · Answer 7 · answered Nov 10 '10 at 15:55

WARNING! This post has no re as requested. Continue with your own responsibility!

I do not know how possible are corner cases but this is how normal Python does my naive coding.

import string
s = 'start TT end AAA BBBBBBB'
for c in string.uppercase:
    s = s.replace(c+c,c.lower())
print s
""" Output:
start t end aA bbbB
"""

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

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