rreplace-如何替换字符串中表达式的最后一次出现?

问题:rreplace-如何替换字符串中表达式的最后一次出现?

Python中是否有一种快速的方法来替换字符串,而不是像从头开始那样replace从头开始呢?例如:

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

>>> '<div><div>Hello</div></div>'.rreplace('</div>','</bad>',1)
>>> '<div><div>Hello</div></bad>'

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('</div>','</bad>',1)
>>> '<div><div>Hello</div></bad>'

回答 0

>>> 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'
>>> 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'

回答 1

我不会假装这是最有效的方法,但这是一种简单的方法。它反转所有有问题的字符串,对反转的字符串执行普通替换str.replace,然后以正确的方式将结果反向反转:

>>> 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>'

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>'

回答 2

这里是单线:

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

返回字符串s的副本,其中所有出现的子字符串oldnew替换。替换第一个maxreplace事件。

以及使用中的完整示例:

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

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

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'

回答 3

只需反转字符串,替换第一次出现的字符串,然后再次反转它:

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)

输出:

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.

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.

回答 4

如果您知道“旧”字符串不包含任何特殊字符,则可以使用正则表达式进行操作:

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>'

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>'

回答 5

这是该问题的递归解决方案:

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

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