Python正则表达式:使用量词进行前瞻 [英] Python regex: Following lookahead with quantifier

问题描述

只是想让我为这个问题着想，当我正积极地向前看时，这个问题就发生了.

Just trying to wrap my mind around this question, which occurred to me while I was messing around with positive lookaheads.

此正则表达式有意义吗?

Does this regex make any sense?

foo(?=bar)+

re.match()不会返回错误，但是如果'+'量词有任何意义，我无法弄清楚它会是什么. (FWIW，regex101.com给出错误先前的令牌不可量化" ...)

re.match() doesn't return an error, but if there's any sense to the '+' quantifier, I can't figure what it would be. (FWIW, regex101.com gives the error 'The preceding token is not quantifiable' ...)

谢谢.

/约翰

推荐答案

这里没有理由使用+量词.正则表达式先行查找和后退实际上不匹配任何文本，这意味着如果它们在某个位置匹配一次，则它们将连续匹配"无数次.似乎python很聪明，不会多次尝试匹配lookahead，因为它不会陷入无限循环中.

There is no reason to use the + quantifier here. Regular expression lookaheads and lookbehinds don't actually match any text, which means that if they match once in a position they will "match" an infinite number of times in a row. It seems python is smart enough not to try matching the lookahead more than once, as it does not get caught up in an infinite loop.

换句话说，只要坚持不合格的前瞻:foo(?=bar)最好.

In other words, just stick with unqualified lookaheads: foo(?=bar) is best.

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