替换字符串中除第一个以外的所有引用

str.partition 将字符串拆分为分隔符前面的部分、分隔符本身和后面的部分,如果分隔符不存在,则拆分为字符串和两个空字符串。归根结底是:

s = 'X did something. X found it to be good, and so X went home.'
before, first, after = s.partition('X')
result = before + first + after.replace('X', 'Y')

re.sub 使用一个函数:

import re


def repl(match, count=[0]):
    x, = count
    count[0] += 1
    if x > 0:
        return 'Y'
    return 'X'


print(re.sub('X', repl, 'X did something. X found it to be good, and so X went home.'))

X did something. Y found it to be good, and so Y went home.

这样做的目的是使用一个函数来保持数据的计数 X

另一种选择是查找第一个,并且仅在替换所有后进行 X

最后,从句首到句首

st = 'X did something. X found it to be good, and so X went home.'
first_found = st.find('X')
print (st[:first_found + 1] + st[first_found + 1:].replace('X', 'Y'))
# X did something. Y found it to be good, and so Y went home.

这是一个没有正则表达式的低技术解决方案。:)

>>> s = 'X did something. X found it to be good, and so X went home'
>>> s = s.replace('X', 'Y').replace('Y', 'X', 1)
>>> s
>>> 'X did something. Y found it to be good, and so Y went home'

解决方案如果 'Y' 可以存在于原始字符串中:

def replace_tail(s, target, replacement):
    try:
        pos = s.index(target)
    except ValueError:
        return s
    pos += len(target)
    head = s[:pos]
    tail = s[pos:]
    return head + tail.replace(target, replacement)

演示:

>>> s = 'Today YYY and XXX did something. XXX found it to be good, and so XXX went home without YYY.'
>>> replace_tail(s, 'XXX', 'YYY')
>>> 'Today YYY and XXX did something. YYY found it to be good, and so YYY went home without YYY.'

replace 如果可能的话。

切片 生成两个字符串:第一个字符串直到(包括)第一个元素,下一个片段包含其余元素。然后,我们可以在该零件上应用替换零件,并将这些零件合并回:

def replace_but_first(text, search, replace):
    try:
        idx = text.index(search) + len(search)
        return text[:idx] + text[idx:].replace(search, replace)
    except ValueError:  # we did not found a single match
        return text

>>> replace_but_first('X did something. X found it to be good, and so X went home.', 'X', 'Y')
'X did something. Y found it to be good, and so Y went home.'

如果您仍然对使用正则表达式操作感兴趣,可以使用 re.finditer() MatchObject 找到每个匹配案例的实例。将迭代器强制转换为列表允许您索引到 匹配对象 实例。在下面的函数中, [1:] 指示跳过第一个匹配。

def replace_rest(my_string, replacement):

    for match in list(re.finditer(r'(X)', my_string))[1:]:
        my_string = my_string[0:match.start()] + replacement + my_string[match.end():]

    return my_string

运行:

>>> my_string = "Person X did something. X found it to be good, and so Y went home."

输出:

>>> replace_rest(my_string, "Y")
'Person X did something. Y found it to be good, and so Y went home.'

:这对于忽略任何第n次出现的图案也很有用。