Python Regex: how to parse repeated groups in a string?

Question

I want to match a pattern with a string including pure numbers, such as '2324235235980980' with a pattern like as described below:

The pattern is '2-6-8-7-4', in which the pattern starts with 2, transit to 6, either self-loop at 6 or transit to 8, then it could go back and forth between 6 and 8, could self-loop at 8, or could transit to 7. And the same thing for 7. One more thing for 7 is 7-8-6-8-7 could happen. Finally, 7 could reach 4, once it reaches 4, the pattern is done. During the process, if it reaches out to other points, then it has to start with 2 again to be counted. I use

import re    
re.findall(r'(2((6+8+)+)7)', test_string)

the output includes '2666686888668887', but when I add 4, I don't know the syntax to compile this. Has anyone an idea? Thanks a lot!

When you say "then it could go back and forth between 6 and 8, could self-loop at 8, or could transit to 7" does that mean it could go 6-8-6-7 or does it need to go back to 8 before going to 7? — CAustin
– CAustin, Commented Nov 15, 2017 at 1:02
Also, when you say "the same thing for 7" what exactly are you referring to? 8's property of going back and forth between 6 and itself? It would be helpful if you provided a list of valid and invalid strings that outline as many of the edge cases as possible. — CAustin
– CAustin, Commented Nov 15, 2017 at 1:06
My bad for not clarifying the question. When you reach 8, you could transit to 6, or stay at 8, or transit to 7. "The same thing for 7" is when you reach 7, you could transit back to 8, or stay at 7, or transit forward to 4. Samples of valid strings are like '266868874', '268787866668887774', '268688788774'. Invalid strings are like '263874', '268734', '2688668778868714', etc... — Richard Riverlands
– Richard Riverlands, Commented Nov 15, 2017 at 3:29
Think 2-6-8-7-4 as a tunnel, once you enter from 2, in each time period, you move one step in the tunnel. You don't break out the tunnel in the middle, while you could stay at any intermediate point for any amount of time. You could also back and forth among the intermediate points. Once you reach the end point 4, you are out of the tunnel. — Richard Riverlands
– Richard Riverlands, Commented Nov 15, 2017 at 3:29

wp78de · Accepted Answer · 2017-11-15 21:57:27Z

1

I think this is easier achieved than initially expected:

26[68]+?[687]+?4

2-followed-by-6-followed-by-6|8-followed-by-6|8|7-followed-by-4.

The only not so obvious part is to make the pattern lazy.

Here is an even better pattern:

\b26?([^7]6|8|[^6]7)+?4\b

2-followed-by-(not7)6|8|(not6)7-followed-by-4.

edited Nov 15, 2017 at 21:57

answered Nov 15, 2017 at 4:49

wp78de

19.1k7 gold badges49 silver badges78 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Richard Riverlands Over a year ago

This is brilliant! Perfectly solve my problem! Thank you!

Richard Riverlands Over a year ago

Wait, there is a problem. When I test string' 26876874', it matches what it should not match. You CAN NOT transit from 7 to 6 in one step. Only '7-8-6-8-7-4' would happen, not '7-6-8-7-4'. Do you have an idea to fix this?

Tania Reyes · Accepted Answer · 2017-11-15 05:12:09Z

0

I don't know if I understand what you need, but maybe this can work for you:

string = "2666686888668887748926874"
index = [(m.start(0), m.end(0)) for m in re.finditer(r'2(6+8+)+7+\1?4', string)]
print(index)

Prints: [(0, 18), (20, 25)].

Is a list of tuples with the start and end index for every occurrence.

answered Nov 15, 2017 at 5:12

Tania Reyes

11 bronze badge

1 Comment

Richard Riverlands Over a year ago

This may be helpful to my further exploration. Thanks for your work!

Collectives™ on Stack Overflow

Python Regex: how to parse repeated groups in a string?

2 Answers 2

2 Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related