Python正则表达式：拆分为空字符串的模式匹配

Question

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() .

Answer 1

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'))

Answer 2

import regex
x="bazbarbarfoobar"
print regex.split(r"(?<!baz)(?=bar)",x,flags=regex.VERSION1)

You can use regex module here for this.

or

(.+?(?<!foo))(?=bar|$)|(.+?foo)$

Use re.findall .

See demo