How to parse string dates with 2-digit year?

Question

I need to parse strings representing 6-digit dates in the format yymmdd where yy ranges from 59 to 05 (1959 to 2005). According to the time module

Answer 1

Recently had a similar case, ended up with this basic calculation and logic:

pivotyear = 1969
century = int(str(pivotyear)[:2]) * 100

def year_2to4_digit(year):
    return century + year if century + year > pivotyear else (century + 100) + year

Answer 2

Prepend the century to your date using your own pivot:

  year = int(date[0:2])
  if 59 <= year <= 99:
      date = '19' + date
  else
      date = '20' + date

and then use strptime with the %Y directive instead of %y.

Answer 3

import datetime
date = '20-Apr-53'
dt = datetime.datetime.strptime( date, '%d-%b-%y' )
if dt.year > 2000:
    dt = dt.replace( year=dt.year-100 )
                     ^2053   ^1953
print dt.strftime( '%Y-%m-%d' )

Answer 4

If you are dealing with very recent dates as well as very old dates and want to use the current date as a pivot (not just the current year), try this code:

import datetime
def parse_date(date_str):
    parsed = datetime.datetime.strptime(date_str,'%y%m%d')
    current_date = datetime.datetime.now()
    if parsed > current_date:
        parsed = parsed.replace(year=parsed.year - 100)
    return parsed

Answer 5

I'd use datetime and parse it out normally. Then I'd use datetime.datetime.replace on the object if it is past your ceiling date -- Adjusting it back 100 yrs.:

import datetime
dd = datetime.datetime.strptime(date,'%y%m%d')
if dd.year > 2005:
   dd = dd.replace(year=dd.year-100)

Answer 6

You can also perform the following:

today=datetime.datetime.today().strftime("%m/%d/%Y")
today=today[:-4]+today[-2:]