Instead of using regex, it is generally better to parse the string as a datetime.datetime
object:
In [140]: datetime.datetime.strptime("11/12/98","%m/%d/%y")
Out[140]: datetime.datetime(1998, 11, 12, 0, 0)
In [141]: datetime.datetime.strptime("11/12/98","%d/%m/%y")
Out[141]: datetime.datetime(1998, 12, 11, 0, 0)
You could then access the day, month, and year (and hour, minutes, and seconds) as attributes of the datetime.datetime
object:
In [143]: date.year
Out[143]: 1998
In [144]: date.month
Out[144]: 11
In [145]: date.day
Out[145]: 12
To test if a sequence of digits separated by forward-slashes represents a valid date, you could use a try..except
block. Invalid dates will raise a ValueError
:
In [159]: try:
.....: datetime.datetime.strptime("99/99/99","%m/%d/%y")
.....: except ValueError as err:
.....: print(err)
.....:
.....:
time data '99/99/99' does not match format '%m/%d/%y'
If you need to search a longer string for a date, you could use regex to search for digits separated by forward-slashes:
In [146]: import re
In [152]: match = re.search(r'(\d+/\d+/\d+)','The date is 11/12/98')
In [153]: match.group(1)
Out[153]: '11/12/98'
Of course, invalid dates will also match:
In [154]: match = re.search(r'(\d+/\d+/\d+)','The date is 99/99/99')
In [155]: match.group(1)
Out[155]: '99/99/99'
To check that match.group(1)
returns a valid date string, you could then parsing it using datetime.datetime.strptime
as shown above.
Well, from my understanding, simply for matching this format in a given string, I prefer this regular expression:
pattern='[0-9|/]+'
to match the format in a more strict way, the following works:
pattern='(?:[0-9]{2}/){2}[0-9]{2}'
Personally, I cannot agree with unutbu's answer since sometimes we use regular expression for "finding" and "extract", not only "validating".
I use something like this
>>> import datetime
>>> regex = datetime.datetime.strptime
>>>
>>> # TEST
>>> assert regex('2020-08-03', '%Y-%m-%d')
>>>
>>> assert regex('2020-08', '%Y-%m-%d')
ValueError: time data '2020-08' does not match format '%Y-%m-%d'
>>> assert regex('08/03/20', '%m/%d/%y')
>>>
>>> assert regex('08-03-2020', '%m/%d/%y')
ValueError: time data '08-03-2020' does not match format '%m/%d/%y'
Sometimes we need to get the date from a string. One example with grouping:
record = '1518-09-06 00:57 some-alphanumeric-charecter'
pattern_date_time = ([0-9]{4}-[0-9]{2}-[0-9]{2} [0-9]{2}:[0-9]{2}) .+
match = re.match(pattern_date_time, record)
if match is not None:
group = match.group()
date = group[0]
print(date) // outputs 1518-09-06 00:57
Using this regular expression you can validate different kinds of Date/Time samples, just a little change is needed.
^\d\d\d\d/(0?[1-9]|1[0-2])/(0?[1-9]|[12][0-9]|3[01]) (00|[0-9]|1[0-9]|2[0-3]):([0-9]|[0-5][0-9]):([0-9]|[0-5][0-9])$
-->validate this: 2018/7/12 13:00:00
for your format you cad change it to:
^(0?[1-9]|[12][0-9]|3[01])/(0?[1-9]|1[0-2])/\d\d$
--> validates this: 11/12/98
I find the below RE working fine for Date in the following format;
It can accept year from 2000-2099
Please do not forget to add $ at the end,if not it accept 14-11-201 or 20177
date="13-11-2017"
x=re.search("^([1-9] |1[0-9]| 2[0-9]|3[0-1])(.|-)([1-9] |1[0-2])(.|-|)20[0-9][0-9]$",date)
x.group()
output = '13-11-2017'
I built my solution on top of @aditya Prakash appraoch:
print(re.search("^([1-9]|0[1-9]|1[0-9]|2[0-9]|3[0-1])(\.|-|/)([1-9]|0[1-9]|1[0-2])(\.|-|/)([0-9][0-9]|19[0-9][0-9]|20[0-9][0-9])$|^([0-9][0-9]|19[0-9][0-9]|20[0-9][0-9])(\.|-|/)([1-9]|0[1-9]|1[0-2])(\.|-|/)([1-9]|0[1-9]|1[0-9]|2[0-9]|3[0-1])$",'01/01/2018'))
The first part (^([1-9]|0[1-9]|1[0-9]|2[0-9]|3[0-1])(\.|-|/)([1-9]|0[1-9]|1[0-2])(\.|-|/)([0-9][0-9]|19[0-9][0-9]|20[0-9][0-9])$
) can handle the following formats:
The second part (^([0-9][0-9]|19[0-9][0-9]|20[0-9][0-9])(\.|-|/)([1-9]|0[1-9]|1[0-2])(\.|-|/)([1-9]|0[1-9]|1[0-9]|2[0-9]|3[0-1])$
) can basically do the same, but in inverse order, where the year comes first, followed by month, and then day.
As delimiters it allows ., /, -. As years it allows everything from 1900-2099, also giving only two numbers is fine.
If you have suggestions for improvement please let me know in the comments, so I can update the answer.
As the question title asks for a regex that finds many dates, I would like to propose a new solution, although there are many solutions already.
In order to find all dates of a string that are in this millennium (2000 - 2999), for me it worked the following:
dates = re.findall('([1-9]|1[0-9]|2[0-9]|3[0-1]|0[0-9])(.|-|\/)([1-9]|1[0-2]|0[0-9])(.|-|\/)(20[0-9][0-9])',dates_ele)
dates = [''.join(dates[i]) for i in range(len(dates))]
This regex is able to find multiple dates in the same string, like bla Bla 8.05/2020 \n BLAH bla15/05-2020 blaa. As one could observe, instead of / the date can have . or -, not necessary at the same time.
Some explaining
More specifically it can find dates of format day , moth year. Day is an one digit integer or a zero followed by one digit integer or 1 or 2 followed by an one digit integer or a 3 followed by 0 or 1. Month is an one digit integer or a zero followed by one digit integer or 1 followed by 0, 1, or 2. Year is the number 20 followed by any number between 00 and 99.
Useful notes
One can add more date splitting symbols by adding | symbol
at the end of both (.|-|\/)
. For example for adding -- one would do (.|-|\/|--)
To have years outside of this millennium one has to modify (20[0-9][0-9])
to ([0-9][0-9][0-9][0-9])
Source: Stackoverflow.com