How to use regular expressions do reverse search?

问题

For example:
My string is: 123456789 nn nn oo nn nn mlm nn203.
My target is: nn.

Then, I match string from the end to the beginning and return the first match result and its postion.
In this examlpe, the result is nn start in [-5] end in [-3].
I wrote the simple funcitonto do this process, but how to use regular expressions to do this job?

回答1:

For the string itself, just do a findall and use the last one:

import re

st='123456 nn1 nn2 nn3 nn4 mlm nn5 mlm'

print re.findall(r'(nn\d+)',st)[-1]

Prints nn5

You can also do the same thing using finditer which makes it easier finding the relevant indexes:

print [(m.group(),m.start(),m.end()) for m in re.finditer(r'(nn\d+)',st)][-1]

Prints ('nn5', 27, 30)

If you have a lot of matches and you only want the last, sometimes it makes sense to simply reverse the string and pattern:

m=re.search(r'(\d+nn)',st[::-1])
offset=m.start(1)
print st[-m.start(1)-len(m.group(1)):-m.start(1)]

Prints nn5

回答2:

First, if you're not looking for a regular expression, string.rfind is a lot easier to get right.

You can use a regular expression by using a negative lookahead, see the documentation of re:

import re
s = "123456789 nn nn oo nn nn mlm nn203"
match = re.search("(nn)(?!.*nn.*)", s)

# for your negative numbers:
print (match.start()-len(s), match.end()-len(s))
# (-5, -3)

回答3:

Idea:

find reversed regexp (in your case irrelevant) in reversed string
resulting indexes convert to negative numbers + switch start<->end

Example:

>>> import re
>>> s = "123456789 nn nn oo nn nn mlm nn203"
>>> m = re.search("(nn)", s[::-1])
>>> -m.end(), -m.start()
(-5, -3)

来源：https://stackoverflow.com/questions/16510017/how-to-use-regular-expressions-do-reverse-search

标签

python

regex

reverse