how can I replace the nth occurrences in a list of string without using the built in function

I'm kind of new in python and I have problem to write a script which take four element (ex str, Replacefrom, replaceto and n) find the characters and replace the nth occurrence.

Example:

>>> replaeceit("Mississippi", "s", "l", 2)
'Mislissippi'
>>> replae开发者_开发技巧ceit("Mississippi", "s", "l", 0)
'Mississippi'

the n is 2 so the code change the second s to l .. and when the n=0 then it dosen't do nything

honestly I don't know how to implement n to the equation this is my code so far without n

def replaceit(str,replacefrom,replaceto):
    new=""
    for letter in str:
        if letter== replacefrom:
            new=new+replaceto
        else: 
            new=new+letter
    return new

---

Okay, maybe now I understood what you are looking for:

def replaceit(st, remove, put, pos):
outs = ""
count = 0
for letter in st:
    if letter == remove:
        count += 1
        if count == pos:
            outs += put
        else:
            outs += letter
    else:
        outs += letter
return outs

Output:

In [84]: replaceit("Mississipi", "s", "l", 2)
Out[84]: 'Mislissipi'

Of course you can check that the arguments no2 and no3 are strings with len() of 1.

---

This is second first attempt at understanding your question:

def replaceit(s, replacefrom, replaceto, n):
  new_s, count = '', 0
  for letter in s:
    if letter == replacefrom:
      count += 1
      if count == n:
        new_s += replaceto
        continue
    new_s += letter
  return new_s

This matches your examples:

>>> replaceit("Mississippi", "s", "l", 2)
'Mislissippi'
>>> replaceit("Mississippi", "s", "l", 0)
'Mississippi'

If this is not what you want, please explain better.

You can also achieve the same with regular expressions:

def replaceit(s, replacefrom, replaceto, n):
  import re
  if n <= 0:
    return s
  return re.sub('(.*?%s)%s' % (('%s.*?' % replacefrom) * (n-1), replacefrom), r'\1%s' % replaceto, s)

---

Everyone always loves a generator expression.

from itertools import count
def replaceit(str, replacefrom, replaceto, n):
    c = count(1)
    return ''.join(replacefrom if l == replaceto and c.next() == n else l for l in str)