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Case insensitive replace

What's the easiest way to do a case-insensitive string replacement in Python?


B
Brian Moeskau

The string type doesn't support this. You're probably best off using the regular expression sub method with the re.IGNORECASE option.

>>> import re
>>> insensitive_hippo = re.compile(re.escape('hippo'), re.IGNORECASE)
>>> insensitive_hippo.sub('giraffe', 'I want a hIPpo for my birthday')
'I want a giraffe for my birthday'

If you're only doing a single replace, or want to save lines of code, it's more efficient to use a single substitution with re.sub and the (?i) flag: re.sub('(?i)' + re.escape('hippo'), 'giraffe', 'I want a hIPpo for my birthday')
Why re.escape for a string of letters only? Thanks.
@Elena, it's not needed for 'hippo', but would be useful if the to-replace value was passed into a function, so it's really more of a good example than anything else.
Besides having to re.escape your needle, there's another trap here which this answer fails to avoid, noted in stackoverflow.com/a/15831118/1709587: since re.sub processes escape sequences, as noted in docs.python.org/library/re.html#re.sub, you need to either escape all backslashes in your replacement string or use a lambda.
This doesn't work for replacing r'A\BC' with r'D\EF' in r'xxxA\BCxxxA\BCxxx') - The correct answer is blow, the one from johv
U
Unknown
import re
pattern = re.compile("hello", re.IGNORECASE)
pattern.sub("bye", "hello HeLLo HELLO")
# 'bye bye bye'

Or one-liner: re.sub('hello', 'bye', 'hello HeLLo HELLO', flags=re.IGNORECASE)
Note that re.sub only supports this flag since Python 2.7.
B
Bill the Lizard

In a single line:

import re
re.sub("(?i)hello","bye", "hello HeLLo HELLO") #'bye bye bye'
re.sub("(?i)he\.llo","bye", "he.llo He.LLo HE.LLO") #'bye bye bye'

Or, use the optional "flags" argument:

import re
re.sub("hello", "bye", "hello HeLLo HELLO", flags=re.I) #'bye bye bye'
re.sub("he\.llo", "bye", "he.llo He.LLo HE.LLO", flags=re.I) #'bye bye bye'

E
Excelsior_07

Continuing on bFloch's answer, this function will change not one, but all occurrences of old with new - in a case insensitive fashion.

def ireplace(old, new, text):
    idx = 0
    while idx < len(text):
        index_l = text.lower().find(old.lower(), idx)
        if index_l == -1:
            return text
        text = text[:index_l] + new + text[index_l + len(old):]
        idx = index_l + len(new) 
    return text

Very well done. Much better than regex; it handles all kinds of characters, whereas the regex is very fussy about anything non-alphanumeric. Preferred answer IMHO.
All you have to do is escape the regex: the accepted answer is much shorter and easier to read than this.
Escape only works for matching, backslashes in the destination can mess things up still.
Possibly the fastest method for a case-insensitive replace, tested against both using an arrayed string and using regex.
b
bFloch

This doesn't require RegularExp

def ireplace(old, new, text):
    """ 
    Replace case insensitive
    Raises ValueError if string not found
    """
    index_l = text.lower().index(old.lower())
    return text[:index_l] + new + text[index_l + len(old):] 

Good one, however this does not change all occurrences of old with new, but only the first occurrence.
It's less readable than the regex version. No need to reinvent the wheel here.
It would be interesting to do a performance comparison between this and the upvoted versions, it might be faster, which matters for some applications. Or it might be slower because it does more work in interpreted Python.
M
Mark Amery

Like Blair Conrad says string.replace doesn't support this.

Use the regex re.sub, but remember to escape the replacement string first. Note that there's no flags-option in 2.6 for re.sub, so you'll have to use the embedded modifier '(?i)' (or a RE-object, see Blair Conrad's answer). Also, another pitfall is that sub will process backslash escapes in the replacement text, if a string is given. To avoid this one can instead pass in a lambda.

Here's a function:

import re
def ireplace(old, repl, text):
    return re.sub('(?i)'+re.escape(old), lambda m: repl, text)

>>> ireplace('hippo?', 'giraffe!?', 'You want a hiPPO?')
'You want a giraffe!?'
>>> ireplace(r'[binfolder]', r'C:\Temp\bin', r'[BinFolder]\test.exe')
'C:\\Temp\\bin\\test.exe'

N
Nico Bako

This function uses both the str.replace() and re.findall() functions. It will replace all occurences of pattern in string with repl in a case-insensitive way.

def replace_all(pattern, repl, string) -> str:
   occurences = re.findall(pattern, string, re.IGNORECASE)
   for occurence in occurences:
       string = string.replace(occurence, repl)
       return string

M
Murray

An interesting observation about syntax details and options:

Python 3.7.2 (tags/v3.7.2:9a3ffc0492, Dec 23 2018, 23:09:28) [MSC v.1916 64 bit (AMD64)] on win32

import re
old = "TREEROOT treeroot TREerOot"
re.sub(r'(?i)treeroot', 'grassroot', old)

'grassroot grassroot grassroot'

re.sub(r'treeroot', 'grassroot', old)

'TREEROOT grassroot TREerOot'

re.sub(r'treeroot', 'grassroot', old, flags=re.I)

'grassroot grassroot grassroot'

re.sub(r'treeroot', 'grassroot', old, re.I)

'TREEROOT grassroot TREerOot'

So the (?i) prefix in the match expression or adding "flags=re.I" as a fourth argument will result in a case-insensitive match. BUT, using just "re.I" as the fourth argument does not result in case-insensitive match.

For comparison,

re.findall(r'treeroot', old, re.I)

['TREEROOT', 'treeroot', 'TREerOot']

re.findall(r'treeroot', old)

['treeroot']


This does not provide an answer to the question. please edit your answer to ensure that it improves upon other answers already present in this question.
From the re.sub docs it 5 parameters: re.sub(pattern, repl, string, count=0, flags=0) which is why flags=re.I works but trying to pass it as a positional parameter fails, it's in the wrong position.
N
Nimantha

I was having \t being converted to the escape sequences (scroll a bit down), so I noted that re.sub converts backslashed escaped characters to escape sequences.

To prevent that I wrote the following:

Replace case insensitive.

import re
    def ireplace(findtxt, replacetxt, data):
        return replacetxt.join(  re.compile(findtxt, flags=re.I).split(data)  )

Also, if you want it to replace with the escape characters, like the other answers here that are getting the special meaning bashslash characters converted to escape sequences, just decode your find and, or replace string. In Python 3, might have to do something like .decode("unicode_escape") # python3

findtxt = findtxt.decode('string_escape') # python2
replacetxt = replacetxt.decode('string_escape') # python2
data = ireplace(findtxt, replacetxt, data)

Tested in Python 2.7.8


N
Nimantha
i='I want a hIPpo for my birthday'
key='hippo'
swp='giraffe'

o=(i.lower().split(key))
c=0
p=0
for w in o:
    o[c]=i[p:p+len(w)]
    p=p+len(key+w)
    c+=1
print(swp.join(o))

For learning: generally when you do a search and replace on a string, it's better to not have to turn it into an array first. That's why the first answer is probably the best. While it's using an external module, it's treating the string as one whole string. It's also a bit clearer what's happening in the process.
For learning: its very difficult for a developer with no context to read this code and decipher what its doing :)