如何限制正则表达式的 findall() 方法

问题描述

对于 findall 方法，是否有与 BeautifulSoup 的 limit=X 参数等效的正则表达式?我的意思是，如何找到有问题的前 X 个单词然后中断代码执行?谢谢

Is there a regex equivalent of BeautifulSoup's limit=X argument for the findall method? I mean, how to find the first X words in question and then break the code execution? thank you

推荐答案

您可以使用 re.finditer 因为它返回一个迭代器而不是一次生成所有值:

You can use re.finditer as it returns an iterator instead of generating all the values at once:

In [21]: strs="12345678"

In [22]: it=re.finditer("\d",strs)

In [23]: [next(it).group(0) for _ in xrange(4)] #returns only 4 mathces
Out[23]: ['1', '2', '3', '4']

虽然这可能会在限制大于匹配数时引发 StopIteration 错误.一个简单的解决方法是使用异常处理或使用 itertools.isclice :

Though this might raise StopIteration error when the limit is greater than the number of matches. A simple workaround is to use exception handling or use itertools.isclice :

In [26]: def limiter(strs,pattern,limit):
    it=re.finditer(pattern,strs)
    try:
        for _ in xrange(limit):
            yield next(it).group(0)
    except StopIteration:
        pass
   ....:

In [27]: list(limiter("12345","\d",3))
Out[27]: ['1', '2', '3']

In [28]: list(limiter("12345","\d",6))
Out[28]: ['1', '2', '3', '4', '5']

In [29]: list(limiter("12345","\d",10))
Out[29]: ['1', '2', '3', '4', '5']

关于 re.finditer 的帮助:

In [24]: re.finditer?
Type:       function
String Form:<function finditer at 0xb74114c4>
File:       /usr/lib/python2.7/re.py
Definition: re.finditer(pattern, string, flags=0)
Docstring:
Return an iterator over all non-overlapping matches in the
string.  For each match, the iterator returns a match object.

Empty matches are included in the result.

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