I have the following code to match the dates
import re
date_reg_exp2 = re.compile(r'\d{2}([-/.])(\d{2}|[a-zA-Z]{3})\1(\d{4}|\d{2})|\w{3}\s\d{2}[,.]\s\d{4}')
matches_list = date_reg_exp2.findall("23-SEP-2015 and 23-09-2015 and 23-09-15 and Sep 23, 2015")
print matches_list
The output I expect is
["23-SEP-2015","23-09-2015","23-09-15","Sep 23, 2015"]
What I am getting is:
[('-', 'SEP', '2015'), ('-', '09', '2015'), ('-', '09', '15'), ('', '', '')]
Please check the link for regex here.
The problem you have is that re.findall returns captured texts only excluding Group 0 (the whole match). Since you need the whole match (Group 0), you just need to use re.finditer and grab the group() value:
matches_list = [x.group() for x in date_reg_exp2.finditer("23-SEP-2015 and 23-09-2015 and 23-09-15 and Sep 23, 2015")]
See IDEONE demo
re.findall(pattern, string, flags=0)
Return all non-overlapping matches of pattern in string, as a list of strings... If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group.
re.finditer(pattern, string, flags=0)
Return an iterator yieldingMatchObjectinstances over all non-overlapping matches for the RE pattern in string.
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