Replace Nth Occurrence of Substring in String

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

Python - replace every nth occurrence of string

The code you got from the previous question is a nice starting point, and only a minimal adaptation is required to have it change every nth occurence:

def nth_repl_all(s, sub, repl, nth):
    find = s.find(sub)
    # loop util we find no match
    i = 1
    while find != -1:
        # if i  is equal to nth we found nth matches so replace
        if i == nth:
            s = s[:find]+repl+s[find + len(sub):]
            i = 0
        # find + len(sub) + 1 means we start after the last match
        find = s.find(sub, find + len(sub) + 1)
        i += 1
    return s

Javascript: Replace nth occurrence of substring in multiline string

You regex doesn't match properly. . will not match newline characters so what I'll do is (?:.|\n) instead.

var someString="I have a cat, my cat is intelligent,\n my cat is very active";
someString= someString.replace(new RegExp("(?:(?:.|\n)*?my cat){2}"), function(x){return x.replace(RegExp("my cat$"), 'our cat abc')});
console.log(someString);

How to replace the Nth occurrence of a word in a string?

s = 'I have a banana and a banana and another banana'
word = 'banana'
replace_with = 'apple'
occurrance  = 2
word.join(s.split(word)[:occurrance]) + replace_with + word.join(s.split(word)[occurrance:])

'I have a banana and a apple and another banana'

If you just want to process the last occurance
then make occurrance = -1

Find nth occurrence of a character in a row and replace it in Python, in order to fix/replace a wrong/leftover ; separator in a text file

You should not replace that semi-colon with another character, but instead wrap the string with double quotes (and escape any double quotes that might already be part of the string by doubling them). This is how in CSV syntax you can include such delimiters. In fact, it is not bad practice to wrap such text with double quotes always.

Here is how I'd adapt your loop:

for row in listoflines:
    if row.count(';') > 13:  # maybe there are even two or more of them
        cells = row.split(';')
        # Escape double quotes and wrap in double quotes
        s = '"' + ';'.join(cells[12:-1]).replace('"', '""') + '"'
        # Replace the involved parts with that single part
        cells[12:-1] = [s]
        # ...and convert back to the string format
        row = ';'.join(cells)
      
    finaldocument.append(row)

Replace Nth occurrence of a character in a string with something else

You can replace ((?:\d+, ){3}\d), with \1\n

You basically capture everything till fourth comma in group1 and comma separately and replace it with \1\n which replaces matched text with group1 text and newline, giving you the intended results.

Regex Demo

R Code demo

gsub("((?:\\d+, ){3}\\d),", "\\1\n", "1, 2, 3, 4, 5, 6, 7, 8, 9, 10")

Prints,

[1] "1, 2, 3, 4\n 5, 6, 7, 8\n 9, 10"

Edit:

To generalize above solution to any text, we can change \d to [^,]

New R code demo

gsub("((?:[^,]+, ){3}[^,]+),", "\\1\n", "1, 2, 3, 4, 5, 6, 7, 8, 9, 10")
gsub("((?:[^,]+, ){3}[^,]+),", "\\1\n", "a, bb, ccc, dddd, 500, 600, 700, 800, 900, 1000")

Output,

[1] "1, 2, 3, 4\n 5, 6, 7, 8\n 9, 10"
[1] "a, bb, ccc, dddd\n 500, 600, 700, 800\n 900, 1000"

Replace String in between nth and n+1 occurrence of a character in Java

The better way - using regex

String input = "abc_1234_01233456_DC";
Matcher matcher = Pattern.compile("(_([^_]+)){2}").matcher(input);//{2} - number of occurrence
String result = matcher.find() ? new StringBuilder(input).replace(matcher.start(2), matcher.end(2), "78910").toString() : input; //abc_1234_78910_DC

That's it

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