为什么 re.sub 替换整个模式，而不仅仅是其中的一个捕获组? [英] Why does re.sub replace the entire pattern, not just a capturing group within it?

问题描述

re.sub('a(b)','d','abc') 产生 dc，而不是 adc.

为什么re.sub要替换整个捕获组，而不是只替换捕获组'(b)'?

Why does re.sub replace the entire capturing group, instead of just capturing group'(b)'?

推荐答案

因为它应该替换整个模式:

Because it's supposed to replace the whole occurrence of the pattern:

返回通过替换 repl 替换 string 中最左边不重叠的模式所获得的字符串.

Return the string obtained by replacing the leftmost non-overlapping occurrences of the pattern in string by the replacement repl.

如果只替换某些子组，那么包含多个组的复杂正则表达式将不起作用.有几种可能的解决方案:

If it were to replace only some subgroup, then complex regexes with several groups wouldn't work. There are several possible solutions:

1. 完整指定模式:re.sub('ab', 'ad', 'abc') - 我最喜欢的，因为它非常易读和明确.
2. 捕获您想要保留的组，然后在模式中引用它们(注意它应该是原始字符串以避免转义):re.sub('(a)b', r'1d', 'abc')
3. 类似于上一个选项:提供一个回调函数作为 repl 参数，并使其处理 Match 对象并返回所需的结果.
4. 使用lookbehinds/lookaheds，它们不包含在匹配中，但会影响匹配:re.sub('(?<=a)b', r'd', 'abxb') 产生 adxb.组开头的 ?<= 表示这是一个前瞻".

1. Specify pattern in full: re.sub('ab', 'ad', 'abc') - my favorite, as it's very readable and explicit.
2. Capture groups which you want to preserve and then refer to them in the pattern (note that it should be raw string to avoid escaping): re.sub('(a)b', r'1d', 'abc')
3. Similar to previous option: provide a callback function as repl argument and make it process the Match object and return required result.
4. Use lookbehinds/lookaheds, which are not included in the match, but affect matching: re.sub('(?<=a)b', r'd', 'abxb') yields adxb. The ?<= in the beginning of the group says "it's a lookahead".

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