Replace nth occurrence of substring in string

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 = 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 maybe it isn't the most Pythonic way, because the where variable construction needs importing re library. Maybe somebody will find even more Pythonic way.

Sources and some links in addition:

• where construction: How to find all occurrences of a substring?
• 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

I have come up with the below, which considers also options to replace all 'old' string occurrences to the left or to the right. Naturally, there is no option to replace all occurrences, as standard str.replace works perfect.

def nth_replace(string, old, new, n=1, option='only nth'):
    """
    This function replaces occurrences of string 'old' with string 'new'.
    There are three types of replacement of string 'old':
    1) 'only nth' replaces only nth occurrence (default).
    2) 'all left' replaces nth occurrence and all occurrences to the left.
    3) 'all right' replaces nth occurrence and all occurrences to the right.
    """
    if option == 'only nth':
        left_join = old
        right_join = old
    elif option == 'all left':
        left_join = new
        right_join = old
    elif option == 'all right':
        left_join = old
        right_join = new
    else:
        print("Invalid option. Please choose from: 'only nth' (default), 'all left' or 'all right'")
        return None
    groups = string.split(old)
    nth_split = [left_join.join(groups[:n]), right_join.join(groups[n:])]
    return new.join(nth_split)

You can use a while loop with str.find to find the nth occurrence if it exists and use that position to create the new string:

def nth_repl(s, sub, repl, n):
    find = s.find(sub)
    # If find is not -1 we have found at least one match for the substring
    i = find != -1
    # loop util we find the nth or we find no match
    while find != -1 and i != n:
        # find + 1 means we start searching from after the last match
        find = s.find(sub, find + 1)
        i += 1
    # If i is equal to n we found nth match so replace
    if i == n:
        return s[:find] + repl + s[find+len(sub):]
    return s

Example:

In [14]: s = "foobarfoofoobarbar"

In [15]: nth_repl(s, "bar","replaced",3)
Out[15]: 'foobarfoofoobarreplaced'

In [16]: nth_repl(s, "foo","replaced",3)
Out[16]: 'foobarfooreplacedbarbar'

In [17]: nth_repl(s, "foo","replaced",5)
Out[17]: 'foobarfoofoobarbar'

The last answer is nearly perfect - only one correction:

    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 = after.replace(sub, wanted)
        newString = before + after
        return newString

The after-string has to be stored in the this variable again after replacement. Thank you for the great solution!