Python使用正则表达式来查找子字符串的起始位置

Question

I need to find the position of a substring within a string.

The substring is the characters ",0*" followed by tow characters that are [0-9] or [AF] ie

kdjrnnj,0*B3;,w0l44
       ^^^^^
qui8ecc),0*21qxxcd4))
        ^^^^^

The substring is always exactly 5 characters in length. There are always some number (inknown) of chars before the substring. There may or may not be chars after the substring.

I'd like to use re.something to find the starting position of my substring within the string. My regex knowledge is quite poor - if someone could tell me how to do this you'd save me hours of hacking around.

Thanks

Answer 1

Use match object 's start() method:

>>> r = re.compile(r',0\*[0-9A-F]{2}')
>>> m = r.search("kdjrnnj,0*B3;,w0l44")
>>> if m : print m.start()
7
>>> m = r.search("qui8ecc),0*21qxxcd4))")
>>> if m : print m.start()
8

Next step is to remove everything after the substring

You don't need index for that, that can be done with regex too:

>>> strs = "qui8ecc),0*21qxxcd4))"
>>> re.search(r'.*?,0\*[0-9A-F]{2}', strs).group()
'qui8ecc),0*21'

>>> m = r.search("kdjrnnj,0*B3;,w0l44")
>>> if m : print m.group()
kdjrnnj,0*B3

re.search is faster than re.sub here:

>>> strs = 'kdjrnnj,0*B3;,w0l44'
>>> %timeit r.search(strs).group()
100000 loops, best of 3: 1.42 us per loop
>>> %timeit pattern.sub('', strs)
100000 loops, best of 3: 2.79 us per loop

>>> strs = 'kdjrnnj,0*B3;,w0l44'*1000
>>> %timeit r.search(strs).group()
100000 loops, best of 3: 1.43 us per loop
>>> %timeit pattern.sub('', strs)
10000 loops, best of 3: 59.9 us per loop

>>> strs = 'kdjrnnj'*1000 + ',0*B3;,w0l44'
>>> %timeit r.search(strs).group()
1000 loops, best of 3: 260 us per loop
>>> %timeit pattern.sub('', strs)
1000 loops, best of 3: 410 us per loop

Answer 2

Python re.search() returns a MatchObject() instance when a match is made, it includes a .start() method to give you the matched position:

import re

pattern = re.compile(r',0\*[0-9A-F]{2}')

match = pattern.search(inputstring)
if match:
    print match.start()

Note the \\* though; an asterisk ( * ) is a regular expression metacharacter, so it needs to be escaped with a slash to match a literal * .

The [0-9A-F] defines a character class that matches any character in the two named ranges, and the {2} following the class limits it to matching exactly two characters.

Demo:

>>> import re
>>> pattern = re.compile(r',0\*[0-9A-F]{2}')
>>> match = pattern.search('kdjrnnj,0*B3;,w0l44')
>>> match.start()
7
>>> match.group()
',0*B3'
>>> match = pattern.search('qui8ecc),0*21qxxcd4))')
>>> match.start()
8
>>> match.group()
',0*21'

If you need to remove everything after this string, use re.sub() instead:

pattern = re.compile(r'(?<=,0\*[0-9A-F]{2}).*')

newstring = pattern.sub('', oldstring)

This uses a look-behind assertion; it looks for your pattern, then matches everything that follows , and the re.sub() call then removes what was matched from the inputstring.

Demo:

>>> pattern = re.compile(r'(?<=,0\*[0-9A-F]{2}).*')
>>> pattern.sub('', 'kdjrnnj,0*B3;,w0l44')
'kdjrnnj,0*B3'
>>> pattern.sub('', 'qui8ecc),0*21qxxcd4))')
'qui8ecc),0*21'

Note how everything after ,0*B3 and ,0*21 is gone now.

Answer 3

这个的正则表达式应该非常简单： .*,0\\*[0-9A-F]{2} 。

Answer 4

使用re.search() ：

re.search(r',0*[0-9A-F][0-9A-F]', your_string).start()