Regular expression for finding a sub-string

Question

I am trying to find all occurances of a sub-string using regular expression. The sub-string is composed of three parts, starts with one or more 'A', followed by one or more 'N' and ended with one or more 'A'. Let a string 'AAANAANABNA' and if I parse the string I should get two sub-strings 'AAANAA' and 'AANA' as the output. So, I have tried the below code.

import regex as re
reg_a='A+N+A+'
s='AAANAANABNA'
sub_str=re.findall(reg_a,s,overlapped=True)
print(sub_str)

And, I am getting the below output,

['AAANAA', 'AANAA', 'ANAA', 'AANA', 'ANA']

But, I want the output as,

['AAANAA', 'AANA']

That is, the trailing A's of the first match should be the leading A's of the next match. How can I get that, any idea?

score 4 · Accepted Answer · answered Jun 16 '20 at 19:57

Make sure there are no A on the left:

>>> reg_a='(?<!A)A+N+A+'
>>> print( re.findall(reg_a,s,overlapped=True) )
['AAANAA', 'AANA']

The (?<!A)A+N+A+ matches

(?<!A) - a negative lookbehind that matches a location that is not immediately preceded with A
A+ - one or more As
N+ - one or more Ns
A+ - one or more As

Note you may use re to get the matches, too:

>>> import re
>>> re_a = r'(?=(?<!A)(A+N+A+))'
>>> print( re.findall(re_a, s) )
['AAANAA', 'AANA']

I never used the `overlapped=True`, learned something new. ++ — The fourth bird, Jun 16 '20 at 20:21

The fourth bird · Answer 2 · 2020-06-16T20:26:46.170

4

One option is to use a negative lookbehind with a capturing group inside a positive lookahead without the overlapped. re.findall will return the value of the capturing group.

(?=(?<!A)(A+N+A+))

(?= Positive lookahead, assert what is on the right is
- (?<!A) Negative lookahead, assert what is on the left is not A
- (A+N+A+) Match your pattern
) Close lookahead

Regex demo | Python demo

import re
s = "AAANAANABNA"
pattern = r"(?=(?<!A)(A+N+A+))"
print(re.findall(pattern, s))

Output

['AAANAA', 'AANA']

Or as suggested by @anubhava, separate the lookbehind from the lookahead:

(?<!A)(?=(A+N+A+))

Regex demo | Python demo

edited Jun 16 '20 at 20:26

answered Jun 16 '20 at 19:59

The fourth bird

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I think this has a slight edge over the 3rd party regex? Upvoted both, I quite like the question too =) – JvdV Jun 16 '20 at 20:12
I can't really tell, did you have a pattern in mind using the regex pypi module? – The fourth bird Jun 16 '20 at 20:18
1

`(?<!A)(?=(A+N+A+))` might be a bit faster with same matches – anubhava Jun 16 '20 at 20:23
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@Thefourthbird, thank you for sharing nice code and demo cheers – RavinderSingh13 Jun 17 '20 at 03:12

anubhava · Answer 3 · 2020-06-16T20:13:17.273

4

Here is a simpler way of achieving this with re module. We just need a lookahead for 1+ trailing As and have to use use 2 capture groups:

>>> import re
>>> s = 'AAANAANABNA'
>>> [''.join(x) for x in re.findall(r'(A+N+)(?=(A+))', s)]
['AAANAA', 'AANA']

RegEx Demo

edited Jun 16 '20 at 20:13

answered Jun 16 '20 at 20:05

anubhava

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Nice find joining the groups ++ – The fourth bird Jun 16 '20 at 20:19

Regular expression for finding a sub-string

3 Answers3