How to find and replace nth occurrence of word in a sentence using python regular expression?

Use negative lookahead like below.

>>> s = "cat goose  mouse horse pig cat cow"
>>> re.sub(r'^((?:(?!cat).)*cat(?:(?!cat).)*)cat', r'\1Bull', s)
'cat goose  mouse horse pig Bull cow'

DEMO

• ^ Asserts that we are at the start.
• (?:(?!cat).)* Matches any character but not of cat , zero or more times.
• cat matches the first cat substring.
• (?:(?!cat).)* Matches any character but not of cat , zero or more times.
• Now, enclose all the patterns inside a capturing group like ((?:(?!cat).)*cat(?:(?!cat).)*), so that we could refer those captured chars on later.
• cat now the following second cat string is matched.

OR

>>> s = "cat goose  mouse horse pig cat cow"
>>> re.sub(r'^(.*?(cat.*?){1})cat', r'\1Bull', s)
'cat goose  mouse horse pig Bull cow'

Change the number inside the {} to replace the first or second or nth occurrence of the string cat

To replace the third occurrence of the string cat, put 2 inside the curly braces ..

>>> re.sub(r'^(.*?(cat.*?){2})cat', r'\1Bull', "cat goose  mouse horse pig cat foo cat cow")
'cat goose  mouse horse pig cat foo Bull cow'

Play with the above regex on here ...

I use simple function, which lists all occurrences, picks the nth one's position and uses it to split original string into two substrings. Then it replaces first occurrence in the second substring and joins substrings back into the new string:

import re

def replacenth(string, sub, wanted, n)
    where = [m.start() for m in re.finditer(sub, string)][n-1]
    before = string[:where]
    after = string[where:]
    after.replace(sub, wanted, 1)
    newString = before + after
    print newString

For these variables:

string = 'ababababababababab'
sub = 'ab'
wanted = 'CD'
n = 5

outputs:

ababababCDabababab

Notes:

The where variable actually is a list of matches' positions, where you pick up the nth one. But list item index starts with 0 usually, not with 1. Therefore there is a n-1 index and n variable is the actual nth substring. My example finds 5th string. If you use n index and want to find 5th position, you'll need n to be 4. Which you use usually depends on the function, which generates our n.

This should be the simplest way, but it isn't regex only as you originally wanted.

Sources and some links in addition:

• where construction: Find all occurrences of a substring in Python
• string splitting: https://www.daniweb.com/programming/software-development/threads/452362/replace-nth-occurrence-of-any-sub-string-in-a-string
• similar question: Find the nth occurrence of substring in a string

Here's a way to do it without a regex:

def replaceNth(s, source, target, n):
    inds = [i for i in range(len(s) - len(source)+1) if s[i:i+len(source)]==source]
    if len(inds) < n:
        return  # or maybe raise an error
    s = list(s)  # can't assign to string slices. So, let's listify
    s[inds[n-1]:inds[n-1]+len(source)] = target  # do n-1 because we start from the first occurrence of the string, not the 0-th
    return ''.join(s)

Usage:

In [278]: s
Out[278]: 'cat goose  mouse horse pig cat cow'

In [279]: replaceNth(s, 'cat', 'Bull', 2)
Out[279]: 'cat goose  mouse horse pig Bull cow'

In [280]: print(replaceNth(s, 'cat', 'Bull', 3))
None