Parse a date from a complex string using python

Question

I have a number of strings that have different date formats in them. I would like to be able to extract the date from the string. For example:

Today is August 2012. Tomorrow isn't
Another day 12 August, another time
12/08 is another format
have another ? 08/12/12 could be
finally august 12 would be

What I would expect to get from each of these results is 2012-08-01 00:00:00, 2013-08-12 00:00:00, 2013-08-12 00:00:00, 2012-08-12 00:00:00, 2013-08-12 00:00:00.

I currently have this code:

from dateutil import parser
print parser.parse("Today is August 2012. Tomorrow isn't",fuzzy=True)

You will see from this that the date prints as 2012-08-27 00:00:00 (because today is the 27th of the month). What I would want in this example is 2012-08-01 00:00:00.

How do I force it to always put the first of the month if a day is not given? (For example if I give August 2012 it should return 2012-08-01, if I give it 12 August 2012 it should return 2012-08-12.)

Hack: set date to first of month, call parser, set date back... — Floris
– Floris, Commented Jan 27, 2013 at 14:26

root · Accepted Answer · 2013-01-27 15:14:55Z

4

Use the default argument to set the default date. This should handle all the cases except the third one, which is somewhat ambiguous and probably needs some parser tweaking or a mindreader:

In [15]: from datetime import datetime

In [16]: from dateutil import parser

In [17]: DEFAULT_DATE = datetime(2013,1,1)

In [18]: dates=["Today is August 2012. Tomorrow isn't",
    ...:        "Another day 12 August, another time",
    ...:        "12/08 is another format",
    ...:        "have another ? 08/12/12 could be", 
    ...:        "finally august 12 would be"]


In [19]: for date in dates:
    ...:     print parser.parse(date,fuzzy=True, default=DEFAULT_DATE)
    ...:     
2012-08-01 00:00:00
2013-08-12 00:00:00
2013-12-08 00:00:00  # wrong
2012-08-12 00:00:00
2013-08-12 00:00:00

edited Jan 27, 2013 at 15:14

answered Jan 27, 2013 at 14:27

root

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1 Comment

user1464409 Over a year ago

A superb answer root, thank-you very much for your time in editing your answer to make it even more useful.

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