What is the correct date parser format string for the ISO date format?

Question

I am trying to parse dates in the ISO date format, but for reasons I don't understand, the pattern I am using isn't working.

From the shell:

>>> s
datetime.date(2014, 1, 3)
>>> str(s)
'2014-01-03'
>>> datetine.strptime(str(s),"%y-%m-%d").date()
Traceback (most recent call last):
  File "<interactive input>", line 1, in <module>
  File "C:\Python27\lib\_strptime.py", line 325, in _strptime
    (date_string, format))
ValueError: time data '2014-01-30' does not match format '%y-%m-$d'

But 2014-01-03 should be a match to %y-%m-%d, right? Why am I getting this error?

Capital `Y` for a 4 digit year - eg: `%Y-%m-%d`. What exactly are you trying to achieve anyway? It's odd to take the str rep of a date only to convert it to a datetime, then take the date back... — Jon Clements, Mar 09 '14 at 15:37
@GrijeshChauhan lower case `y` is 2 digit year... See [this site](http://strftime.org/) for info... Or the [official docs](http://docs.python.org/2/library/datetime.html#strftime-and-strptime-behavior) — Jon Clements, Mar 09 '14 at 15:40
@JonClements thanks, its almost same as I have been [used in C](http://pic.dhe.ibm.com/infocenter/iseries/v7r1m0/index.jsp?topic=/rtref/strpti.htm) also — Grijesh Chauhan, Mar 09 '14 at 15:44

score 4 · Accepted Answer · answered Mar 09 '14 at 15:45

From the docs (emphasis mine):

%y Year without century as a zero-padded decimal number. 00, 01, ..., 99
%Y Year with century as a decimal number. 1970, 1988, 2001, 2013

Therefore %y-%m-%d is expecting 14-01-03 while you have 2014-01-03 which requires %Y-%m-%d.

score 3 · Answer 2 · answered Mar 09 '14 at 15:42

Difference between upper and lower case:

>>> datetime.datetime.strptime('2014-01-03', '%Y-%m-%d').date()
datetime.date(2014, 1, 3)
>>> datetime.datetime.strptime('14-01-03', '%y-%m-%d').date()
datetime.date(2014, 1, 3)

Documented in strptime and strftime behavior.

What is the correct date parser format string for the ISO date format?

2 Answers2