Parsing a date in python without using a default

Question

I\'m using python\'s dateutil.parser tool to parse some dates I\'m getting from a third party feed. It allows specifying a default date, which itself defaults

Answer 1

simple-date does this for you (it does try multiple formats, internally, but not as many as you might think, because the patterns it uses extend python's date patterns with optional parts, like regexps).

see https://github.com/andrewcooke/simple-date - but only python 3.2 and up (sorry).

it's more lenient than what you want by default:

>>> for date in ('2011-10-12', '2011-10', '2011', '10-12', '2011-10-12T11:45:30', '10-12 11:45', ''):
...   print(date)
...   try: print(SimpleDate(date).naive.datetime)
...   except: print('nope')
... 
2011-10-12
2011-10-12 00:00:00
2011-10
2011-10-01 00:00:00
2011
2011-01-01 00:00:00
10-12
nope
2011-10-12T11:45:30
2011-10-12 11:45:30
10-12 11:45
nope

nope

but you could specify your own format. for example:

>>> from simpledate import SimpleDateParser, invert
>>> parser = SimpleDateParser(invert('Y-m-d(%T| )?(H:M(:S)?)?'))
>>> for date in ('2011-10-12', '2011-10', '2011', '10-12', '2011-10-12T11:45:30', '10-12 11:45', ''):
...   print(date)
...   try: print(SimpleDate(date, date_parser=parser).naive.datetime)
...   except: print('nope')
... 
2011-10-12
2011-10-12 00:00:00
2011-10
nope
2011
nope
10-12
nope
2011-10-12T11:45:30
2011-10-12 11:45:30
10-12 11:45
nope

nope

ps the invert() just switches the presence of % which otherwise become a real mess when specifying complex date patterns. so here only the literal T character needs a % prefix (in standard python date formatting it would be the only alpha-numeric character without a prefix)