Date regex python

Question

I am trying to match dates in a string where the date is formatted as (month dd, yyyy). I am confused by what I see when I use my regex pattern below. It only matches strings that begin with a date. What am I missing?

 >>> p = re.compile('[A-z]{3}\s{1,}\d{1,2}[,]\s{1,}\d{4}')
 >>> s = "xyz Dec 31, 2013 - Jan 4, 2014"
 >>> print p.match(s).start()
 Traceback (most recent call last):
 File "<stdin>", line 1, in <module>
 AttributeError: 'NoneType' object has no attribute 'start'

 >>> s = "Dec 31, 2013 - Jan 4, 2014"
 >>> print p.match(s).start()
 0 #Correct

Iron Fist · Accepted Answer · 2015-06-20 20:39:06Z

2

Use re.findall rather than re.match, it will return to you list of all matches:

>>> s = "Dec 31, 2013 - Jan 4, 2014"
>>> r = re.findall(r'[A-z]{3}\s{1,}\d{1,2}[,]\s{1,}\d{4}',s)
>>> r
['Dec 31, 2013', 'Jan 4, 2014']
>>>
>>> s = 'xyz Dec 31, 2013 - Jan 4, 2014'
>>> r = re.findall(r'[A-z]{3}\s{1,}\d{1,2}[,]\s{1,}\d{4}',s)
>>> r
['Dec 31, 2013', 'Jan 4, 2014']

From Python docs:

re.match(pattern, string, flags=0) If zero or more characters at the beginning of string match the regular expression pattern, return a corresponding MatchObject instance

In the other hand:

findall() matches all occurrences of a pattern, not just the first one as search() does.

answered Jun 20, 2015 at 20:39

Iron Fist

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SBH · Accepted Answer · 2015-06-20 19:50:19Z

1

Use the search method instead of match. Match compares the whole string but search finds the matching part.

answered Jun 20, 2015 at 19:50

SBH

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vks · Accepted Answer · 2015-06-20 19:49:35Z

1

p = re.compile(r'.*?[A-Za-z]{3}\s{1,}\d{1,2}[,]\s{1,}\d{4}')

match matches a string from start.if start does is not same it will fail.In the first example xyz will be consumed by [A-Za-z]{3} but rest of the string will not match.

You can directly use your regex with re.findall and get the result without caring about the location of the match.

answered Jun 20, 2015 at 19:49

vks

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