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up vote 82 down vote favorite
29

Lets say I have a string 'gfgfdAAA1234ZZZuijjk' I want to extract just the '1234' part.

I only know what will be the few characters directly before (AAA), and after (ZZZ) the part I am interested in (1234).

With sed it is possible to do something like this with a string:

echo "$STRING" | sed -e "s|.*AAA\(.*\)ZZZ.*|\1|"

And this will give me 1234 as a result.

How to do the same thing in Python?

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up vote 171 down vote accepted

Using regular expressions - documentation for further reference

import re

text = 'gfgfdAAA1234ZZZuijjk'

m = re.search('AAA(.+?)ZZZ', text)
if m:
    found = m.group(1)

# found: 1234

or:

import re

text = 'gfgfdAAA1234ZZZuijjk'

try:
    found = re.search('AAA(.+?)ZZZ', text).group(1)
except AttributeError:
    # AAA, ZZZ not found in the original string
    found = '' # apply your error handling

# found: 1234
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1  
The second solution is better, if the pattern matches most of the time, because its Easier to ask for forgiveness than permission.. – Bengt Jan 14 '13 at 16:11
3  
Doesn't the indexing start at 0? So you would need to use group(0) instead of group(1)? – Alexander Nov 8 '15 at 22:16
8  
@Alexander, no, group(0) will return full matched string: AAA1234ZZZ, and group(1) will return only characters matched by first group: 1234 – Yuri K Nov 12 '15 at 13:46
    
@Bengt: Why is that? The first solution looks quite simple to me, and it has fewer lines of code. – HelloGoodbye Jul 7 at 13:21
    
Why use .+? and not .*? Aren't they rather equivalent? – HelloGoodbye Jul 7 at 13:27
up vote 57 down vote 
>>> s = 'gfgfdAAA1234ZZZuijjk'
>>> start = s.find('AAA') + 3
>>> end = s.find('ZZZ', start)
>>> s[start:end]
'1234'

Then you can use regexps with the re module as well, if you want, but that's not necessary in your case.

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16  
+1 for without re. – user225312 Jan 12 '11 at 9:44
4  
The question seems to imply that the input text will always contain both "AAA" and "ZZZ". If this is not the case, your answer fails horribly (by that I mean it returns something completely wrong instead of an empty string or throwing an exception; think "hello there" as input string). – tzot Feb 6 '11 at 23:46
    
@user225312 Is the re method not faster though? – confused00 Jul 21 at 9:25
up vote 12 down vote 
import re
print re.search('AAA(.*?)ZZZ', 'gfgfdAAA1234ZZZuijjk').group(1)
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AttributeError: 'NoneType' object has no attribute 'groups' - if there is no AAA, ZZZ in the string... – eumiro Jan 12 '11 at 9:20
up vote 9 down vote

regular expression

import re

re.search(r"(?<=AAA).*?(?=ZZZ)", your_text).group(0)

The above as-is will fail with an AttributeError if there are no "AAA" and "ZZZ" in your_text

string methods

your_text.partition("AAA")[2].partition("ZZZ")[0]

The above will return an empty string if either "AAA" or "ZZZ" don't exist in your_text.

PS Python Challenge?

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1  
This answer probably deserves more up votes. The string method is the most robust way. It does not need a try/except. – ChaimG Dec 3 '15 at 2:59
    
... nice, though limited. partition is not regex based, so it only works in this instance because the search string was bounded by fixed literals – GreenAsJade Feb 29 at 2:07
up vote 5 down vote

You can use re module for that:

>>> import re
>>> re.compile(".*AAA(.*)ZZZ.*").match("gfgfdAAA1234ZZZuijjk").groups()
('1234,)
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up vote 4 down vote

With sed it is possible to do something like this with a string:

echo "$STRING" | sed -e "s|.*AAA\(.*\)ZZZ.*|\1|"

And this will give me 1234 as a result.

You could do the same with re.sub function using the same regex.

>>> re.sub(r'.*AAA(.*)ZZZ.*', r'\1', 'gfgfdAAA1234ZZZuijjk')
'1234'

In basic sed, capturing group are represented by \(..\), but in python it was represented by (..).

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up vote 1 down vote

Just in case somebody will have to do the same thing that I did. I had to extract everything inside parenthesis in a line. For example, if I have a line like 'US president (Barack Obama) met with ...' and I want to get only 'Barack Obama' this is solution:

regex = '.*\((.*?)\).*'
matches = re.search(regex, line)
line = matches.group(1) + '\n'

I.e. you need to block parenthesis with slash \ sign. Though it is a problem about more regular expressions that Python.

Also, in some cases you may see 'r' symbols before regex definition. If there is no r prefix, you need to use escape characters like in C. Here is more discussion on that.

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up vote 1 down vote 
>>> s = '/tmp/10508.constantstring'
>>> s.split('/tmp/')[1].split('constantstring')[0].strip('.')
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