Python 正则表达式:对空字符串的模式匹配进行拆分 [英] Python regex: splitting on pattern match that is an empty string

问题描述

使用 re 模块，我似乎无法拆分空字符串的模式匹配:

<预><代码>>>>re.split(r'(?<!foo)(?=bar)', 'foobarbarbazbar')['foobarbarbazbar']

也就是说，即使找到匹配项，如果是空字符串，即使re.split也无法拆分字符串.

re.split 的文档 似乎支持我的结果.

对于这种特殊情况很容易找到解决方法":

<预><代码>>>>re.sub(r'(?<!foo)(?=bar)', 'qux', 'foobarbarbazbar').split('qux')['foobar', 'barbaz', 'bar']

但这是一种容易出错的方法，因为那样我必须注意已经包含我要拆分的子字符串的字符串:

<预><代码>>>>re.sub(r'(?<!foo)(?=bar)', 'qux', 'foobarbarquxbar').split('qux')['foobar', 'bar', '', 'bar']

有没有更好的方法来分割与 re 模块匹配的空模式?另外，为什么 re.split 一开始就不允许我这样做?我知道其他与正则表达式一起使用的拆分算法是可能的；例如，我可以使用 JavaScript 的内置 String.prototype.split().

解决方案

不幸的是 split 需要一个非零宽度匹配，但还没有被修复，因为很多不正确的代码取决于当前的行为，例如使用 [something]* 作为正则表达式.使用这样的模式现在将生成一个 FutureWarning 并且那些永远可以拆分任何东西，从 Python 3.5 开始抛出一个 ValueError :

<预><代码>>>>re.split(r'(?<!foo)(?=bar)', 'foobarbarbazbar')回溯(最近一次调用最后一次):文件<stdin>"，第 1 行，在 <module> 中文件/usr/lib/python3.6/re.py"，第 212 行，拆分返回 _compile(pattern, flags).split(string, maxsplit)ValueError: split() 需要一个非空的模式匹配.

这个想法是，经过一段时间的警告后，可以更改行为，以便您的正则表达式再次起作用.

<小时>

如果您不能使用regex模块，您可以使用re.finditer()编写自己的拆分函数:

def megasplit(pattern, string):splits = list((m.start(), m.end()) for m in re.finditer(pattern, string))开始 = [0] + [i[1] for i in splits]结束 = [i[0] for i in splits] + [len(string)]返回 [string[start:end] 开始，以 zip 结束(开始，结束)]打印(megasplit(r'(?<!foo)(?=bar)', 'foobarbarbazbar'))打印(megasplit(r'o'，'foobarbarbazbar'))

如果您确定匹配项仅为零宽度，则可以使用拆分的开头来简化代码:

导入重新def zerowidthsplit(模式，字符串):splits = list(m.start() for m in re.finditer(pattern, string))开始 = [0] + 分裂结束 = 拆分 + [ len(string) ]返回 [string[start:end] 开始，以 zip 结束(开始，结束)]打印(zerowidthsplit(r'(?<!foo)(?=bar)', 'foobarbarbazbar'))

With the re module, it seems that I am unable to split on pattern matches that are empty strings:

>>> re.split(r'(?<!foo)(?=bar)', 'foobarbarbazbar')
['foobarbarbazbar']

In other words, even if a match is found, if it's the empty string, even re.split cannot split the string.

The docs for re.split seem to support my results.

A "workaround" was easy enough to find for this particular case:

>>> re.sub(r'(?<!foo)(?=bar)', 'qux', 'foobarbarbazbar').split('qux')
['foobar', 'barbaz', 'bar']

But this is an error-prone way of doing it because then I have to beware of strings that already contain the substring that I'm splitting on:

>>> re.sub(r'(?<!foo)(?=bar)', 'qux', 'foobarbarquxbar').split('qux')
['foobar', 'bar', '', 'bar']

Is there any better way to split on an empty pattern match with the re module? Additionally, why does re.split not allow me to do this in the first place? I know it's possible with other split algorithms that work with regex; for example, I am able to do this with JavaScript's built-in String.prototype.split().

解决方案

It is unfortunate that the split requires a non-zero-width match, but it hasn't been to fixed yet, since quite a lot incorrect code depends on the current behaviour by using for example [something]*as the regex. Use of such patterns will now generate a FutureWarning and those that never can split anything, throw a ValueError from Python 3.5 onwards:

>>> re.split(r'(?<!foo)(?=bar)', 'foobarbarbazbar')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python3.6/re.py", line 212, in split
    return _compile(pattern, flags).split(string, maxsplit)
ValueError: split() requires a non-empty pattern match.

The idea is that after a certain period of warnings, the behaviour can be changed so that your regular expression would work again.

---

If you can't use the regex module, you can write your own split function using re.finditer():

def megasplit(pattern, string):
    splits = list((m.start(), m.end()) for m in re.finditer(pattern, string))
    starts = [0] + [i[1] for i in splits]
    ends = [i[0] for i in splits] + [len(string)]
    return [string[start:end] for start, end in zip(starts, ends)]

print(megasplit(r'(?<!foo)(?=bar)', 'foobarbarbazbar'))
print(megasplit(r'o', 'foobarbarbazbar'))

If you are sure that the matches are zero-width only, you can use the starts of the splits for easier code:

import re

def zerowidthsplit(pattern, string):
    splits = list(m.start() for m in re.finditer(pattern, string))
    starts = [0] + splits
    ends = splits + [ len(string) ]
    return [string[start:end] for start, end in zip(starts, ends)]

print(zerowidthsplit(r'(?<!foo)(?=bar)', 'foobarbarbazbar'))

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