python中两个单词之间的正则表达式

Question

I am trying to get value

l1 = [u'/worldcup/archive/southafrica2010/index.html', u'/worldcup/archive/germany2006/index.html', u'/worldcup/archive/edition=4395/index.html', u'/worldcup/archive/edition=1013/index.html', u'/worldcup/archive/edition=84/index.html', u'/worldcup/archive/edition=76/index.html', u'/worldcup/archive/edition=68/index.html', u'/worldcup/archive/edition=59/index.html', u'/worldcup/archive/edition=50/index.html', u'/worldcup/archive/edition=39/index.html', u'/worldcup/archive/edition=32/index.html', u'/worldcup/archive/edition=26/index.html', u'/worldcup/archive/edition=21/index.html', u'/worldcup/archive/edition=15/index.html', u'/worldcup/archive/edition=9/index.html', u'/worldcup/archive/edition=7/index.html', u'/worldcup/archive/edition=5/index.html', u'/worldcup/archive/edition=3/index.html', u'/worldcup/archive/edition=1/index.html']

I'm trying to do regular expression starting off with something like this below

m = re.search(r"\d+", l)
print m.group()

but I want value between "archive/" and "/index.html"
I goggled and have tried something like (?<=archive/\\/index.html).*(?=\\/index.html:)

but It didn't work for me .. how can I get my result list as '

result = ['germany2006','edition=4395','edition=1013' , ...]

Answer 1

If you know for sure that the pattern will match always, you can use this

import re
print [re.search("archive/(.*?)/index.html", l).group(1) for l in l1]

Or you can simply split like this

print [l.rsplit("/", 2)[-2] for l in l1]

Answer 2

Look-arounds is what you need. You need to use it like this:

>>> [re.search(r"(?<=archive/).*?(?=/index.html)", s).group() for s in l1]
[u'southafrica2010', u'germany2006', u'edition=4395', u'edition=1013', u'edition=84', u'edition=76', u'edition=68', u'edition=59', u'edition=50', u'edition=39', u'edition=32', u'edition=26', u'edition=21', u'edition=15', u'edition=9', u'edition=7', u'edition=5', u'edition=3', u'edition=1']

Answer 3

The regular expression

m = re.search(r'(?<=archive\/).+(?=\/index.html)', s)

can solve this, suppose that s is a string from your list.

Answer 4

You can take help from below code .It will solve your problem.

>>> import re
>>> p = '/worldcup/archive/southafrica2010/index.html'
>>> r = re.compile('archive/(.*?)/index.html')
>>> m = r.search(p)
>>> m.group(1)
'southafrica2010'