Define end and beginning of regular expression

Question

Say I have a list of procedure codes like the following:

  procedure_codes = ['038911.B', '32431.1', 'fdsfd.B', '13635.1B', '78935.1b']

In python I now want to extract all codes:

• without a leading zero
• where the first characters are digits
• that includes a '.' somewhere
• that have digits following the '.'
• that do not end in either a 'b' or 'B'

To me the most logical approach seemed using a regular expression, so I tried

• ([^0]+) to not select strings with a leading zero
• (\d+) to only select strings where the first characters are digits
• (.) to only select strings that includes a '.' somewhere
• (\d?) to only select strings that include a digit after the '.'
• ([^bB]?) to not select strings that end in either a 'b' or 'B'

combined (for each element in the list):

  for i in procedure_codes:
      if re.search('[^0]\d+\.\d[^bB]?', i):
          print(i)

I would suspect python to only return the code: '32431.1' but instead the return is :

• 32431.1
• 13635.1B
• 78935.1b

And thus it seems like the final negation is just completely ignored

Since I'm completely new to using regular expressions I do not know if I am simply making a syntax error or if I have misunderstood regular expressions completely.

Answer 1

Just wrap your regex with ^ and $ to match beginning of line and end of line respectively:

^[^0]\d+\.\d[^bB]?$

Because without them, it finds 13635.1 group in 13635.1B and the same with 78935.1b.