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Detailed explanation of regular greedy and non-greedy characteristics of Python
Detailed explanation of regular greedy and non-greedy characteristics of Python
这篇文章主要介绍了Python正则表达式中贪婪/非贪婪特性的相关资料，文中通过示例代码介绍的很详细，对大家具有一定的参考价值，需要的朋友下面来一起看看吧。
之前已经简单介绍了Python正则表达式的基础与捕获，那么在这一篇文章里，我将总结一下正则表达式的贪婪/非贪婪特性。
贪婪
默认情况下，正则表达式将进行贪婪匹配。所谓“贪婪”，其实就是在多种长度的匹配字符串中，选择较长的那一个。例如，如下正则表达式本意是选出人物所说的话，但是却由于“贪婪”特性，出现了匹配不当：
>>> sentence = """You said "why?" and I say "I don't know".""" >>> re.findall(r'"(.*)"', sentence) ['why?" and I say "I don\'t know']
再比如，如下的几个例子都说明了正则表达式“贪婪”的特性：
>>> re.findall('hi*', 'hiiiii')
['hiiiii']
>>> re.findall('hi{2,}', 'hiiiii')
['hiiiii']
>>> re.findall('hi{1,3}', 'hiiiii')
['hiii']非贪婪
当我们期望正则表达式“非贪婪”地进行匹配时，需要通过语法明确说明：
{2,5}? 捕获2-5次，但是优先次数少的匹配
在这里，问号?可能会有些让人犯晕，因为之前他已经有了自己的含义：前面的匹配出现0次或1次。其实，只要记住，当问号出现在表现不定次数的正则表达式部分之后时，就表示非贪婪匹配。
还是上面的那几个例子，用非贪婪匹配，则结果如下：
>>> re.findall('hi*?', 'hiiiii')
['h']
>>> re.findall('hi{2,}?', 'hiiiii')
['hii']
>>> re.findall('hi{1,3}?', 'hiiiii')
['hi']另外一个例子中，使用非贪婪匹配，结果如下：
>>> sentence = """You said "why?" and I say "I don't know".""" >>> re.findall(r'"(.*?)"', sentence) ['why?', "I don't know"]
捕获与非贪婪
严格来说，这一部分并不是非贪婪特性。但是由于其行为与非贪婪类似，所以为了方便记忆，就将其放在一起了。
(?=abc) 捕获，但不消耗字符，且匹配abc
(?!abc) 捕获，不消耗，且不匹配abc
在正则表达式匹配的过程中，其实存在“消耗字符”的过程，也就是说，一旦一个字符在匹配过程中被检索（消耗）过，后面的匹配就不会再检索这一字符了。
知道这个特性有什么用呢？还是用例子说明。比如，我们想找出字符串中出现过1次以上的单词：
>>> sentence = "Oh what a day, what a lovely day!" >>> re.findall(r'\b(\w+)\b.*\b\1\b', sentence) ['what']
这样的正则表达式显然无法完成任务。为什么呢？原因就是，在第一个(\w+)匹配到what，并且其后的\1也匹配到第二个what的时候，“Oh what a day, what”这一段子串都已经被正则表达式消耗了，所以之后的匹配，将直接从第二个what之后开始。自然地，这里只能找出一个出现了两次的单词。
那么解决方案，就和上面提到的(?=abc)语法相关了。这样的语法可以在分组匹配的同时，不消耗字符串！所以，正确的书写方式应该是：
>>> re.findall(r'\b(\w+)\b(?=.*\b\1\b)', sentence) ['what', 'a', 'day']
如果我们需要匹配一个至少包含两个不同字母的单词，则可以使用(?!abc)的语法：
>>> re.search(r'([a-z]).*(?!\1)[a-z]', 'aa', re.IGNORECASE) >>> re.search(r'([a-z]).*(?!\1)[a-z]', 'ab', re.IGNORECASE) <_sre.SRE_Match object; span=(0, 2), match='ab'>
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