Replace nth occurrence of substring in string

Question

I want to replace the n\'th occurrence of a substring in a string.

There\'s got to be something equivalent to what I WANT to do which is

mystring.repl

Answer 1

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'