Regex simplification using python re

Question

Have the following date string which I need to match:

release_date1 = 'July 29, 2016'
release_date2 = 'August 2016'

res = re.match(r'(\w+)\s(\d+),\s(\d+)$', release_date1)
if res:
 datestr = res.group(1) + '-' + res.group(2) + '-' + res.group(3)

res = re.match(r'(\w+)\s(\d+)$', release_date2)
if res:
 datestr = res.group(1) + '-01-' + res.group(2)

Currently I have 2 separate regex, I'm looking to simplify my two regex expressions into a single one. Any suggestions?

Skycc · Accepted Answer · 2016-10-25 02:15:41Z

1

alternative solution

release_date1 = 'July 29, 2016'
release_date2 = 'August 2016'

def grab_date(date1):
    # (\d+(?=,\s))? match day follow by (,space) if there is any
    # (?:,\s)? non capturing match (,space) if there day in date
    res = re.match('(\w+)\s(\d+(?=,\s))?(?:,\s)?(\d+)$', date1).groups()
    datestr = res[0] + '-' + str(res[1]).replace('None','01') + '-' + res[2]
    return datestr

print grab_date(release_date1) # July-29-2016
print grab_date(release_date2) # August-01-2016

answered Oct 25, 2016 at 2:15

Skycc

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Comments

midori · Accepted Answer · 2016-10-25 01:45:16Z

1

Dirty solution but will match both:

'\w+\s[\d,\s]*[\d]{4}'

You might also consider using or operator | which will give a little bit more precise result:

'(\w+\s[\d]{2},\s|\w+\s)*[\d]{4}'

edited Oct 25, 2016 at 1:45

answered Oct 25, 2016 at 1:34

midori

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