How to parse string dates with 2-digit year?

问题:

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 docs, Python's default pivot year is 1969 which won't work for me.

Is there an easy way to override the pivot year, or can you suggest some other solution? I am using Python 2.7. Thanks!

回答1:

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)

回答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.

回答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' )

回答4:

You can also perform the following:

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

回答5:

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