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

开发者 https://www.devze.com 2022-12-25 12:44 出处:网络
Is there a quick way in Python to replace strings but, instead of starting from the beginning as replace does, starting from the end? For example:

Is there a quick way in Python to replace strings but, instead of starting from the beginning as replace does, starting from the end? For example:

>>> def rreplace(old, new, occurrence)
>>>     ... # Code to replace the last occurrences of old by new

>>> '<div><div>Hello</div></div>'.rreplace('<开发者_开发问答/div>','</bad>',1)
>>> '<div><div>Hello</div></bad>'


>>> def rreplace(s, old, new, occurrence):
...  li = s.rsplit(old, occurrence)
...  return new.join(li)
... 
>>> s
'1232425'
>>> rreplace(s, '2', ' ', 2)
'123 4 5'
>>> rreplace(s, '2', ' ', 3)
'1 3 4 5'
>>> rreplace(s, '2', ' ', 4)
'1 3 4 5'
>>> rreplace(s, '2', ' ', 0)
'1232425'


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'


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>'


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.


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>'


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


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.


If you have a list of strings you can use list comprehension and string slicing in a one liner to cover the whole list.. No need to use a function;

myList = [x[::-1].replace('<div>'[::-1],'<bad>'[::-1],1)[::-1] if x.endswith('<div>') else x for x in myList]

I use if else to keep the items in the list that don't meet the criteria for replacement otherwise your list would just be the items that do meet the criteria.

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