rreplace – How to replace the last occurrence of an expression in a string?

I’m not going to pretend that this is the most efficient way of doing it, but it’s a simple way. It reverses all the strings in question, performs an ordinary replacement using str.replace on the reversed strings, then reverses the result back the right way round:

>>> def rreplace(s, old, new, count):
...     return (s[::-1].replace(old[::-1], new[::-1], count))[::-1]
...
>>> rreplace('<div><div>Hello</div></div>', '</div>', '</bad>', 1)
'<div><div>Hello</div></bad>'

Here is a recursive solution to the problem:

def rreplace(s, old, new, occurence = 1):

    if occurence == 0:
        return s

    left, found, right = s.rpartition(old)

    if found == "":
        return right
    else:
        return rreplace(left, old, new, occurence - 1) + new + right

If you know that the ‘old’ string does not contain any special characters you can do it with a regex:

In [44]: s = '<div><div>Hello</div></div>'

In [45]: import re

In [46]: re.sub(r'(.*)</div>', r'1</bad>', s)
Out[46]: '<div><div>Hello</div></bad>'

Just reverse the string, replace first occurrence and reverse it again:

mystr = "Remove last occurrence of a BAD word. This is a last BAD word."

removal = "BAD"
reverse_removal = removal[::-1]

replacement = "GOOD"
reverse_replacement = replacement[::-1]

newstr = mystr[::-1].replace(reverse_removal, reverse_replacement, 1)[::-1]
print ("mystr:", mystr)
print ("newstr:", newstr)

Output:

mystr: Remove last occurence of a BAD word. This is a last BAD word.
newstr: Remove last occurence of a BAD word. This is a last GOOD word.

Here is a one-liner:

result = new.join(s.rsplit(old, maxreplace))

Return a copy of string s with all occurrences of substring old replaced by new. The first maxreplace occurrences are replaced.

and a full example of this in use:

s = 'mississipi'
old = 'iss'
new = 'XXX'
maxreplace = 1

result = new.join(s.rsplit(old, maxreplace))
>>> result
'missXXXipi'

Try this:

def replace_last(string, old, new):
    old_idx = string.rfind(old)
    return string[:old_idx] + new + string[old_idx+len(old):]

Similarly you can replace first occurrence by replacing string.rfind() with string.find().

I hope it helps.