Python 正则表达式：用匹配项的ID替换匹配项

我遇到了这种情况，我想知道我是否可以用正则表达式：

我有一个以下格式的字符串：

{{We all}} love {{stackoverflow}}.

我的问题是如何使用regex replace获得：

match1 love match2

---

试试这个

result = re.sub("([{]{2}[^}]+[}]{2})([^{]+)([{]{2}[^}]+[}]{2})", r"match1\2match2", subject)

解释

"""
(          # Match the regular expression below and capture its match into backreference number 1
   [{]        # Match the character “{”
      {2}        # Exactly 2 times
   [^}]       # Match any character that is NOT a “}”
      +          # Between one and unlimited times, as many times as possible, giving back as needed (greedy)
   [}]        # Match the character “}”
      {2}        # Exactly 2 times
)
(          # Match the regular expression below and capture its match into backreference number 2
   [^{]       # Match any character that is NOT a “{”
      +          # Between one and unlimited times, as many times as possible, giving back as needed (greedy)
)
(          # Match the regular expression below and capture its match into backreference number 3
   [{]        # Match the character “{”
      {2}        # Exactly 2 times
   [^}]       # Match any character that is NOT a “}”
      +          # Between one and unlimited times, as many times as possible, giving back as needed (greedy)
   [}]        # Match the character “}”
      {2}        # Exactly 2 times
)
"""

---

适用于任何数量的组。

假设在要替换的项目之外没有其他花括号，这应该是一件简单的事情。拿起一两本正则表达式教程，看看一些文档。我正在使用Python。将尝试查看一些建议的教程：）

s = '{{We all}} love {{stackoverflow}}.' #string to match
pat = re.compile(r'\{\{.*?\}\}') #pattern
#now replace each matched group by its index
for index,group in enumerate(re.findall(pat,s)):
    s = re.sub(group, 'match'+str(index+1), s)