Add one year in current date PYTHON

Question

I have fetched a date from database with the following variable

{{ i.operation_date }} 

with which I got a value like

April 1, 2013 

I need to add one year to the above, so that I can get

April 1, 2014 

Please suggest, how can I do this?

Answer 1

AGSM's answer shows a convenient way of solving this problem using the python-dateutil package. But what if you don't want to install that package? You could solve the problem in vanilla Python like this:

from datetime import date  def add_years(d, years):     """Return a date that's `years` years after the date (or datetime)     object `d`. Return the same calendar date (month and day) in the     destination year, if it exists, otherwise use the following day     (thus changing February 29 to March 1).      """     try:         return d.replace(year = d.year + years)     except ValueError:         return d + (date(d.year + years, 1, 1) - date(d.year, 1, 1)) 

If you want the other possibility (changing February 29 to February 28) then the last line should be changed to:

        return d + (date(d.year + years, 3, 1) - date(d.year, 3, 1)) 

Answer 2

You can use Python-dateutil's relativedelta to increment a datetime object while remaining sensitive to things like leap years and month lengths. Python-dateutil comes packaged with matplotlib if you already have that. You can do the following:

from dateutil.relativedelta import relativedelta  new_date = old_date + relativedelta(years=1) 

(This answer was given by @Max to a similar question).

But if your date is a string (i.e. not already a datetime object) you can convert it using datetime:

from datetime import datetime from dateutil.relativedelta import relativedelta  your_date_string = "April 1, 2012" format_string = "%B %d, %Y"  datetime_object = datetime.strptime(your_date_string, format_string).date() new_date = datetime_object + relativedelta(years=1) new_date_string = datetime.strftime(new_date, format_string).replace(' 0', ' ') 

new_date_string will contain "April 1, 2013".

NB: Unfortunately, datetime only outputs day values as "decimal numbers" - i.e. with leading zeros if they're single digit numbers. The .replace() at the end is a workaround to deal with this issue copied from @Alex Martelli (see this question for his and other approaches to this problem).