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

I've tweaked @aleskva's answer to better work with regex and wildcards:

import re

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

    return newString

replacenth('abdsahd124njhdasjk124ndjaksnd124ndjkas', '1.*?n', '15', 1)

This gives abdsahd15jhdasjk124ndjaksnd124ndjkas. Note the use of ? to make the query non-greedy.

I realise that the question explicitly states that they didn't want to use regex, however it may be useful to be able to use wildcards in a clear fashion (hence my answer).

Answer 2

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'

Answer 3

My two cents

a='01ab12ab23ab34ab45ab56ab67ab78ab89ab90';print('The original string: ', a)
sTar = 'ab';print('Look for: ', sTar)
n = 4; print('At occurence #:', n)
sSub = '***';print('Substitute with: ', sSub)

t = 0
for i in range(n):
   t = a.find(sTar,t)
   print(i+1, 'x occurence at', t)
   if t != -1: t+=1

t-=1   #reset, get the correct location
yy = a[:t] + a[t:].replace(sTar, sSub, 1)
print('New string is:', yy)

Output

The original string:  01ab12ab23ab34ab45ab56ab67ab78ab89ab90
Look for:  ab
At occurence #: 4
Substitute with:  ***
1 x occurence at 2
2 x occurence at 6
3 x occurence at 10
4 x occurence at 14
New string is: 01ab12ab23ab34***45ab56ab67ab78ab89ab90

Answer 4

I had a similar need, i.e to find the IPs in logs and replace only src IP or dst IP field selectively. This is how i achieved in a pythonic way;

import re

mystr = '203.23.48.0 DENIED 302 449 800 1.1 302 http d.flashresultats.fr  10.111.103.202 GET GET - 188.92.40.78 '
src = '1.1.1.1'
replace_nth = lambda mystr, pattern, sub, n: re.sub(re.findall(pattern, mystr)[n - 1], sub, mystr)
result = replace_nth(mystr, '\S*\d+\.\d+\.\d+\.\d+\S*', src, 2)
print(result)