Parsing string in python3 using regex

Question

I want to parse string which has length & precision in it eg:'(10,2)' and I need to take out length & precision.

Need output as:

_len = 10, _pre = 2

I tried below but it's not working,

>>> import re
>>> my_str = 'numeric(10,2)'
>>> m = re.match(r'\d+,\d+', my_str)
>>> m
>>> m = re.match(r'(\d+,\d+)', my_str)
>>> m
>>> m = re.match('\((+d),(+d)\)', my_str)
>>> m = re.match('\((+d),(+d)\)', my_str)
Traceback (most recent call last):

Read this post for more info http://stackoverflow.com/questions/180986/what-is-the-difference-between-pythons-re-search-and-re-match — Mazdak, Apr 08 '16 at 10:02
And documentation https://docs.python.org/3/library/re.html#search-vs-match — Mazdak, Apr 08 '16 at 10:03

timgeb · Answer 1 · 2016-04-08T10:00:29.563

2

re.match starts searching from the beginning of the line, that's why you are not getting any match.

Use re.search instead:

>>> m = re.search(r'(\d+),(\d+)', my_str)
>>> if m:
...     _len, _pre = map(int, m.groups())
... 
>>> _len, _pre
(10, 2)

edited Apr 08 '16 at 10:00

answered Apr 08 '16 at 09:58

timgeb

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You could use capturing groups in the regex to avoid the need for the split. – Daniel Roseman Apr 08 '16 at 09:59
@DanielRoseman good suggestion – timgeb Apr 08 '16 at 10:00

Parsing string in python3 using regex

1 Answers1