使用单词边界时如何忽略特殊字符

问题描述

s = '!sopa !sop !sopaa !sopii'

如何在使用词边界时忽略!

re.sub(r'\b\!sop\b', 'sopa', s)

输出:'!sopa !sop !sopaa !sopii'

解决方案

您似乎想要这样的东西.

>>>s = '!sopa !sop !sopaa !sopii'>>>re.sub(r'\B!sop\b', 'sopa', s)'!sopa sopa !sopaa !sopii'

您的正则表达式将失败，因为在 ! 符号之前实际上没有 \b 退出.也就是说，从上面你试图匹配 ! 符号，前提是它前面有一个非单词字符.\b 匹配单词 char 和非单词字符，反之亦然.\B 匹配两个单词和两个非单词字符.这里 \B 实际上存在于空格和 ! 之间，因为它们都是非单词字符.

s = '!sopa !sop !sopaa !sopii'

how to ignore ! when using word boundary

re.sub(r'\b\!sop\b', 'sopa', s)

output : '!sopa !sop !sopaa !sopii'

解决方案

Seems like you want something like this.

>>> s = '!sopa !sop !sopaa !sopii'
>>> re.sub(r'\B!sop\b', 'sopa', s)
'!sopa sopa !sopaa !sopii'

Your regex will fail because there isn't a \b actually exits before ! symbol. That is, from the above you're trying to match the ! symbol only if it's preceded by a non-word character. \b matches between a word char and a non-word character, vice versa. \B matches between two word and two non-word chars. Here \B is actaully exists between a space and !, since both are non-word characters.

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