List of strings between regex matches

Question

How do I find all the strings between a regex pattern? For example,

>>> s="123 asd 12 456 sfd g 789"
>>> reg=re.compile("\d{3}")
>>> reg.findall(s)
['123', '456', '789']

I want to find:

[' asd 12 ', ' sfd g ']

Answer 1

Use the .split() method instead of .findall() :

>>> reg.split(s)
['', ' asd 12 ', ' sfd g ', '']

It includes all results in between the matches, including the empty strings at the start and end. You can filter those out:

>>> filter(None, reg.split(s))
[' asd 12 ', ' sfd g ']

although on Python 3 you'd need to use list(filter(None, reg.split(s))) , or iterate over the result of filter() .

Answer 2

使用re.split而不是re.findall 。

Answer 3

You could try something like:

>>> reg = re.compile(r'(?:\d{3})?(.*?)\d{3}')
>>> reg.findall("123 asd 12 456 sfd g 789")
[' asd 12 ', ' sfd g ']

Since .findall() won't find overlapping matches, you need to specify the first group of numbers as being an optional match. In the end, it might be better to take a different approach than regexes alone for a more robust solution.

Answer 4

>>> s = "123 asd 12 456 sfd g 789"
>>> filter(None, re.compile("\d{3}").split(s))
[' asd 12 ', ' sfd g ']