Match string using regular expression except specific string combinations python

Question

In a list I need to match specific instances, except for a specific combination of strings:

let's say I have a list of strings like the following:

l = [
'PSSTFRPPLYO',
'BNTETNTT',
'DE52 5055 0020 0005 9287 29',
'210-0601001-41',
'BSABESBBXXX',
'COMMERZBANK'
]

I need to match all the words that points to a swift / bic code, this code has the following form: 6 letters followed by 2 letters/digits followed by 3 optional letters / digits

hence I have written the following regex to match such specific pattern

import re
regex = re.compile(r'(?<!\w)[a-zA-Z]{6}[a-zA-Z0-9]{2}([a-zA-Z0-9]{3})?(?!\w)')
for item in l:
    match = regex.search(item)
    if match:
        print('found a match, the matched string {} the match {}'.format( item, item[match.start() : match.end()]
    else:
        print('found no match in {}'.format(item)

I need the following cases to be macthed:

result = ['PSSTFRPPLYO', 'BNTETNTT', 'BSABESBBXXX' ]

rather I get

result = ['PSSTFRPPLYO', 'BNTETNTT', 'BSABESBBXXX', 'COMMERZBANK' ]

so what I need is to match only the strings that don't contain the word 'bank'

to do so I have refined my regex to :

regex = re.compile((?<!bank/i)(?<!\w)[a-zA-Z]{6}[a-zA-Z0-9]{2}([a-zA-Z0-9]{3})?(?!\w)(?!bank/i))

simply I have used negative look behind and ahead for more information about theses two concepts refer to link

My regex doesn't do the filtration intended to do, what did I miss?

(?!.*bank.*)^[a-z]{6}(?:[a-z0-9]{2})(?:[a-z0-9]{3})?$ with i modifier? — ctwheels
– ctwheels, Commented Oct 24, 2017 at 13:39
My previous regex can actually be shortened to (?!.*bank.*)^[a-z]{6}(?:[a-z0-9]{2}|[a-z0-9]{5})$ (4 less characters and 2 less steps) or (?![a-z0-9]*bank[a-z0-9]*)^[a-z]{6}(?:[a-z0-9]{2}|[a-z0-9]{5})$ (more characters, but almost 400 less steps) — ctwheels
– ctwheels, Commented Oct 24, 2017 at 13:57

Ajax1234 · Accepted Answer · 2017-10-24 13:49:55Z

2

You can try this:

import re
final_vals = [i for i in l if re.findall('^[a-zA-Z]{6}\w{2}|(^[a-zA-Z]{6}\w{2}\w{3})', i) and not re.findall('BANK', i, re.IGNORECASE)]

Output:

['PSSTFRPPLYO', 'BNTETNTT', 'BSABESBBXXX']

edited Oct 24, 2017 at 13:49

answered Oct 24, 2017 at 13:43

Ajax1234

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1 Comment

rachid Over a year ago

Is there no way to include the condition in the regex? instead of traversion the string twice, don't get me wrong your solution does the job, thanks a lot.

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