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.replace("substring", 2nd)

What is the simplest and most Pythonic way to achieve this?

Why not duplicate: I don't want to use regex for this approach and most of answers to similar questions I found are just regex stripping or really complex function. I really want as simple as possible and not regex solution.

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'

Answer 2

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

Answer 3

Probably one of the shortest solutions and simplest on here without any external library.

def replace_nth(sub,repl,txt,nth):
    arr=txt.split(sub)
    part1=sub.join(arr[:nth])
    part2=sub.join(arr[nth:])
    
    return part1+repl+part2

I did a couple of tests and it worked perfectly.

Answer 4

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)

Answer 5

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!

Answer 6

def replace_nth_occurance(some_str, original, replacement, n):
    """ Replace nth occurance of a string with another string
    """
    all_replaced = some_str.replace(original, replacement, n) # Replace all originals up to (including) nth occurance and assign it to the variable.
    for i in range(n):
        first_originals_back = all_replaced.replace(replacement, original, i) # Restore originals up to nth occurance (not including nth)
    return first_originals_back

Answer 7

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 8

Bit late to the party, but I would consider this way quite pythonian (as far as I understand the meaning of that) and it doesn't require a for loop or counter

def Nreplacer(string,srch,rplc,n):
    Sstring = string.split(srch)
    #first check if substring is even present n times
    #then paste the part before the nth substring to the part after the nth substring
    #, with the replacement inbetween
    if len(Sstring) > (n):
        return f'{srch.join(Sstring[:(n)])}{rplc}{srch.join(Sstring[n:])}' 
    else:
        return string

Answer 9

There is a simple bug in @Padraic Cunningham's answer when the occurrence n is only larger by only 1 than what allowed (n = maximum_occurances + 1).

So here is a corrected version of his code:

def nth_repl(s, old, new, n):
find = s.find(old)
# If find is not -1 we have found at least one match for the substring
i = find != -1
# loop until 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(old, find + 1)
    i += 1
# If i is equal to n we found nth match so replace
if i == n and i <= len(s.split(old))-1:
    return s[:find] + new + s[find+len(old):]
return s

Answer 10

Not pythonic and not efficient, but a one-liner is:

def replace_nth(base_str, find_str, replace_str, n):
    return base_str.replace(find_str, "xxxxx", n-1).replace(find_str, replace_str, 1).replace("xxxxx", find_str)

If you know that some "xxxxxx" placeholder doesn't exist in the string, you can replace the n-1 first ocurrences by a placeholder. Then replace the n-th ocurrence of the substring you're looking for, which at this point is the first ocurrence. Then replace all the placeholders back to the original substring.

Answer 11

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)

Answer 12

General solution: replace any specified instance(s) of a substring [pattern] with another string.

def replace(instring,pattern,replacement,n=[1]):
  """Replace specified instance(s) of pattern in string.

  Positional arguments
    instring - input string
     pattern - regular expression pattern to search for
 replacement - replacement

  Keyword arguments
           n - list of instances requested to be replaced [default [1]]

"""
  import re
  outstring=''
  i=0
  for j,m in enumerate(re.finditer(pattern,instring)):
    if j+1 in n: outstring+=instring[i:m.start()]+replacement
    else: outstring+=instring[i:m.end()]
    i=m.end()
  outstring+=instring[i:]
  return outstring

Answer 13

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 14

Elegant and short:

def replace_ocurrance(string,from,to,num)
     strange_char = “$&$@$$&”
     return string.replace(from,strange_char,num).replace(strange_char, from,num-1).replace(to, strange_char,1)

Answer 15

There's only a couple of non-Regex answers and wanted to provide my own solution which I think is simpler and easier to understand. Create a new string and count for the nth occurence of the character you want to replace.

def replace_nth_occurence(old_str, old_char, new_char, n):
    new_str = ""
    occurences = 0
    for s in old_str:
        if s == old_char:
            occurences += 1
            if occurences == n:
                new_str += new_char # append the new character instead of the old
            else:
                new_str += s
        else:
            new_str += s
    return new_str

replace_nth_occurence("Testing_One_Two_Three", "_", "?", 3)
>> Testing_One_Two?Three

Answer 16

I have a one-liner
if your regex is reg and you need to replace it with reg2:

"".join([reg + x if i != INDEX else reg2 + x for i, x in enumerate(YOUR_STRING.split(reg))])[len(reg):]

Answer 17

(This answer should be as a comment to @Padraic Cunningham, but I don't have the "points" yet to do so, so I can only post a new reply)

The solution provided by Padraic Cunningham is simple and works, except for the bug mentioned by @Haider.

To fix the bug, there is an easier solution than the one provided by Haider: Change the way the value of i is updated. It should not be incremented if the searched string has not been found.

Additionally, there is a bug if n is smaller than 1 (should not happen, but probably better to check).

The very slightly modified solution is thus:

def nth_repl(s, sub, repl, n):
    if n < 1:
        return s
    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 => This is wrong, as might not have found it
        i += find != -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