Updated expression used in my previous code:
import re
recs = [
"New SFR w/Attached Garage, covered parch and patio B24501000 0", # B24501000 0
"B34567000 0 30 X 48 Pole Barn", # B34567000 0
"NEW 22,778 SQ FT B33561000 0 Addition to existing hop building", # B33561000 0
"Residential B143560000 Storage", # B14356000
"Single Family Res B34561000A0", # B34561000A0
"New SFR w/Attached Garage, covered parch and patio B24501000 0" # B2450100000
]
pattern = re.compile(r"([\d]+[A-Z][\d]*|[\d]+[\s][\d]*)"
for rec in recs:
print(pattern.findall(rec)[0].strip())
'''
Results:
B24501000 0
B34567000 0
B33561000 0
B143560000
B34561000A0
B24501000 0
'''
Then replace the space in the results with a zero or whatever.
To experiment with regular expressions, see: Regular Expressions 101. This is the explanation given by the website of the expression used in line 12.
"[\d]+[A-Z][\d]*|[\d]+[\s][\d]*"
gm
1st Alternative [\d]+[A-Z][\d]*
Match a single character present in the list below
B matches the character B literally (case sensitive)
Match a single character present in the list below [\d]+
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
\d matches a digit (equal to [0-9])
Match a single character present in the list below [A-Z]
A-Z a single character in the range between A (index 65) and Z (index 90) (case sensitive)
Match a single character present in the list below [\d]*
* Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
\d matches a digit (equal to [0-9])
2nd Alternative [\d]+[\s][\d]*
Match a single character present in the list below
B matches the character B literally (case sensitive)
Match a single character present in the list below [\d]+
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
\d matches a digit (equal to [0-9])
Match a single character present in the list below [\s]
\s matches any whitespace character (equal to [\r\n\t\f\v ])
Match a single character present in the list below [\d]*
* Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
\d matches a digit (equal to [0-9])
Global pattern flags
g modifier: global. All matches (don't return after first match)
m modifier: multi line. Causes ^ and $ to match the begin/end of each line (not only begin/end of string)
EDIT:
The following simpler expression also works for the examples in the code above:
pattern = re.compile(r"([\d][\d A-Z]{7,9})")
