Complex regex in Python

Question

I am trying to write a generic pattern using regex so that it fetches only particular things from the string. Let's say we have strings like GigabitEthernet0/0/0/0 or FastEthernet0/4 or Ethernet0/0.222. The regex should fetch the first 2 characters and all the numerals. Therefore, the fetched result should be something like Gi0000 or Fa04 or Et00222 depending on the above cases.

x = 'GigabitEthernet0/0/0/2
m = re.search('([\w+]{2}?)[\\\.(\d+)]{0,}',x)

I am not able to understand how shall I write the regular expression. The values can be fetched in the form of a list also. I write few more patterns but it isn't helping.

If you're going to process a lot of strings you should compile your regex - it's a lot more efficient than creating the same regex over & over in a loop. — PM 2Ring
– PM 2Ring, Commented Jun 4, 2015 at 11:30

Avinash Raj · Accepted Answer · 2015-06-04 11:28:04Z

5

In regex, you may use re.findall function.

>>> import re
>>> s = 'GigabitEthernet0/0/0/0 '
>>> s[:2]+''.join(re.findall(r'\d', s))
'Gi0000'

OR

>>> ''.join(re.findall(r'^..|\d', s))
'Gi0000'
>>> ''.join(re.findall(r'^..|\d', 'Ethernet0/0.222'))
'Et00222'

OR

>>> s = 'GigabitEthernet0/0/0/0 '
>>> s[:2]+''.join([i for i in s if i.isdigit()])
'Gi0000'

edited Jun 4, 2015 at 11:28

answered Jun 4, 2015 at 11:19

Avinash Raj

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3 Comments

Kasravnd Over a year ago

you can use s[:2] and a list comprehension within join for more performance

PM 2Ring Over a year ago

Now you just need to turn that last gen exp into a list comp. :) As Kasra said, you get better performance that way. That's because join has to scan its iterable arg twice - the 1st scan is to determine the size of the string it needs to allocate. So if you pass it a generator it has to build that into a list before it can join it.

PM 2Ring Over a year ago

Lots of people don't know that. :) So I'll leave that comment for the benefit of future readers, even though you've fixed it now.

vks · Accepted Answer · 2015-06-04 11:35:38Z

0

z="Ethernet0/0.222."
print z[:2]+"".join(re.findall(r"(\d+)(?=[\d\W]*$)",z))

You can try this.This will make sure only digits from end come into play .

edited Jun 4, 2015 at 11:35

answered Jun 4, 2015 at 11:30

vks

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3 Comments

PM 2Ring Over a year ago

But from the OP it looks like Vipul wants all the digits, eg Ethernet0/0.222 -> Et00222

vks Over a year ago

@PM2Ring Gig9abitEthernet0/0/0/2 i dont think OP would want 9 from this ex..He should clarify though

vks Over a year ago

@Vipul what is the expected output for Gig9abitEthernet0/0/0/2 do you want 9 here?

Dejoskii · Accepted Answer · 2020-05-05 13:06:39Z

0

Here is another option:

s = 'Ethernet0/0.222'

"".join(re.findall('^\w{2}|[\d]+', s))

edited May 5, 2020 at 13:06

answered May 5, 2020 at 7:33

Dejoskii

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