如何替换文本块中的多个正则表达式匹配

Question

我正在尝试将段落中的多个匹配项转换为链接，同时在最终输出中保留周围的文本。我匹配的模式让人想起 Markdown 的超链接语法，作为一种允许非技术用户定义他们希望在输入中链接的文本的方式（我通过 Sheets API/Python 访问的 Google Sheet）。我捕获的第一组是链接文本，第二组是查询字符串中键的值。

我已经能够成功匹配此模式的单个实例，但我的替换字符串替换了输出中的整个段落。

text = "2018 was a big year for my sourdough starter and me. Mostly 
we worked on developing this [tangy bread](19928) and these [chewy 
rolls] (9843). But we were also just content keeping each other 
company and inspired to bake."

def link_inline(text):
    # expand a proper link around recipe id
    ref = re.search(r"(\[.*?\]\(\d+\))", text, re.MULTILINE).group(1)
    if (len(ref) > 0):
        link = re.sub("\[(.*?)\]\((\d+)\)", r"<a href='https://www.foo.com/recipes?rid=\2'>\1</a>", ref)
        return text
    else:
        return "replacement failed"

目标是让此输出保持段落完整，并简单地将 \[(.*?)\]\((\d+)\) 模式匹配替换为以下字符串，包括组的反向引用：<a href="https://www.foo.com?bar=\2">\1</a>

所以它需要遍历文本来替换所有匹配项（大概是re.finditer?），并且还要在模式匹配之外维护原始文本。但我不确定如何正确定义循环并执行此替换，而不用我的替换字符串覆盖整个段落。

Answer 1

我使用了re.compile，而不是在整个组中放置括号，而是在.*? 周围放置一对，在\d+ 周围放置另一对，因为这两个部分代表我们要提取和放置的文本进入我们的网址。

import re

def link_inline(text):
    # expand a proper link around recipe id
    ref = re.compile("\[(.*?)\]\((\d+)\)")
    replacer = r'<a href="https://www.foo.com/recipes?rid=\2">\1</a>'
    return ref.sub(replacer, text)


text = """
2018 was a big year for my sourdough starter and me. Mostly we worked on
 developing this [tangy bread](19928) and
 these [chewy rolls](9843). But we were also just
 content keeping each other company and inspired to bake.
"""

print(link_inline(text))

输出：

2018 was a big year for my sourdough starter and me. Mostly we worked on
 developing this <a href="https://www.foo.com/recipes?rid=19928">tangy bread</a> and
 these <a href="https://www.foo.com/recipes?rid=9843">chewy rolls</a>. But we were also just
 content keeping each other company and inspired to bake.

作为健全性检查，我尝试在字符串text 中添加一些带括号和方括号的额外字符串，例如(this) here 和[this] here。一切仍然正常。