Python, Regular Expression Postcode search

纵然是瞬间 提交于 2019-12-01 00:14:26

repeating your address 3 times with postcode PA23 6NH, PA2 6NH and PA2Q 6NH as test for you pattern and using the regex from wikipedia against yours, the code is..

import re

s="123 Some Road Name\nTown, City\nCounty\nPA23 6NH\n123 Some Road Name\nTown, City"\
    "County\nPA2 6NH\n123 Some Road Name\nTown, City\nCounty\nPA2Q 6NH"

#custom                                                                                                                                               
print re.findall(r'\b[A-Z]{1,2}[0-9][A-Z0-9]? [0-9][ABD-HJLNP-UW-Z]{2}\b', s)

#regex from #http://en.wikipedia.orgwikiUK_postcodes#Validation                                                                                            
print re.findall(r'[A-Z]{1,2}[0-9R][0-9A-Z]? [0-9][A-Z]{2}', s)

the result is

['PA23 6NH', 'PA2 6NH', 'PA2Q 6NH']
['PA23 6NH', 'PA2 6NH', 'PA2Q 6NH']

both the regex's give the same result.

Try

import re
re.findall("[A-Z]{1,2}[0-9][A-Z0-9]? [0-9][ABD-HJLNP-UW-Z]{2}", x)

You don't need the \b.

#!/usr/bin/env python

import re

ADDRESS="""123 Some Road Name
Town, City
County
PA23 6NH"""

reobj = re.compile(r'(\b[A-Z]{1,2}[0-9][A-Z0-9]? [0-9][ABD-HJLNP-UW-Z]{2}\b)')
matchobj = reobj.search(ADDRESS)
if matchobj:
    print matchobj.group(1)

Example output:

[user@host]$ python uk_postcode.py 
PA23 6NH
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