How do I parse an ISO 8601-formatted date?

Question

I need to parse RFC 3339 strings like "2008-09-03T20:56:35.450686Z" into Python's datetime type.

I have found strptime in the Python standard library, but it is not very convenient.

What is the best way to do this?

Answer 1

The python-dateutil package can parse not only RFC 3339 datetime strings like the one in the question, but also other ISO 8601 date and time strings that don't comply with RFC 3339 (such as ones with no UTC offset, or ones that represent only a date).

>>> import dateutil.parser
>>> dateutil.parser.isoparse('2008-09-03T20:56:35.450686Z') # RFC 3339 format
datetime.datetime(2008, 9, 3, 20, 56, 35, 450686, tzinfo=tzutc())
>>> dateutil.parser.isoparse('2008-09-03T20:56:35.450686') # ISO 8601 extended format
datetime.datetime(2008, 9, 3, 20, 56, 35, 450686)
>>> dateutil.parser.isoparse('20080903T205635.450686') # ISO 8601 basic format
datetime.datetime(2008, 9, 3, 20, 56, 35, 450686)
>>> dateutil.parser.isoparse('20080903') # ISO 8601 basic format, date only
datetime.datetime(2008, 9, 3, 0, 0)

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Note that dateutil.parser.isoparse is presumably stricter than the more hacky dateutil.parser.parse, but both of them are quite forgiving and will attempt to interpret the string that you pass in. If you want to eliminate the possibility of any misreads, you need to use something stricter than either of these functions.

The Pypi name is python-dateutil, not dateutil (thanks code3monk3y):

pip install python-dateutil

If you're using Python 3.7, have a look at this answer about datetime.datetime.fromisoformat.