Right-to-left string replace in Python?

Question

I want to do a string replace in Python, but only do the first instance going from right to left. In an ideal world I\'d have:

myStr = \"mississippi\"
print          


        

Answer 1

def rreplace(s, old, new):
    try:
        place = s.rindex(old)
        return ''.join((s[:place],new,s[place+len(old):]))
    except ValueError:
        return s

Answer 2

rsplit and join could be used to simulate the effects of an rreplace

>>> 'XXX'.join('mississippi'.rsplit('iss', 1))
'missXXXippi'

Answer 3

you may reverse a string like so:

myStr[::-1]

to replace just add the .replace:

print myStr[::-1].replace("iss","XXX",1)

however now your string is backwards, so re-reverse it:

myStr[::-1].replace("iss","XXX",1)[::-1]

and you're done. If your replace strings are static just reverse them in file to reduce overhead. If not, the same trick will work.

myStr[::-1].replace("iss"[::-1],"XXX"[::-1],1)[::-1]

Answer 4

It's kind of a dirty hack, but you could reverse the string and replace with also reversed strings.

"mississippi".reverse().replace('iss'.reverse(), 'XXX'.reverse(),1).reverse()

Answer 5

You could also use str.rpartition() which splits the string by the specified separator from right and returns a tuple:

myStr = "mississippi"

first, sep, last = myStr.rpartition('iss')
print(first + 'XXX' + last)
# missXXXippi

Answer 6

>>> re.sub(r'(.*)iss',r'\1XXX',myStr)
'missXXXippi'

The regex engine cosumes all the string and then starts backtracking untill iss is found. Then it replaces the found string with the needed pattern.

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Some speed tests

The solution with [::-1] turns out to be faster.

The solution with re was only faster for long strings (longer than 1 million symbols).