How to convert Python's .isoformat() string back into datetime object

Question

So in Python 3, you can generate an ISO 8601 date with .isoformat(), but you can't convert a string created by isoformat() back into a datetime object because Python's own datetime directives don't match properly. That is, %z = 0500 instead of 05:00 (which is produced by .isoformat()).

For example:

>>> strDate = d.isoformat()
>>> strDate
'2015-02-04T20:55:08.914461+00:00'

>>> objDate = datetime.strptime(strDate,"%Y-%m-%dT%H:%M:%S.%f%z")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\Python34\Lib\_strptime.py", line 500, in _strptime_datetime
    tt, fraction = _strptime(data_string, format)
  File "C:\Python34\Lib\_strptime.py", line 337, in _strptime
    (data_string, format))
ValueError: time data '2015-02-04T20:55:08.914461+00:00' does not match format '%Y-%m-%dT%H:%M:%S.%f%z'

From Python's strptime documentation: (https://docs.python.org/2/library/datetime.html#strftime-strptime-behavior)

%z UTC offset in the form +HHMM or -HHMM (empty string if the the object is naive). (empty), +0000, -0400, +1030

So, in short, Python does not even adhere to its own string formatting directives.

I know datetime is already terrible in Python, but this really goes beyond unreasonable into the land of plain stupidity.

Tell me this isn't true.

Answer 1

As it turns out, this is the current best "solution" to this question:

pip install python-dateutil

Then...

import datetime
import dateutil

def getDateTimeFromISO8601String(s):
    d = dateutil.parser.parse(s)
    return d