Add one year in current date PYTHON

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))

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).

It seems from your question that you would like to simply increment the year of your given date rather than worry about leap year implications. You can use the date class to do this by accessing its member year.

from datetime import date
startDate = date(2012, 12, 21)

# reconstruct date fully
endDate = date(startDate.year + 1, startDate.month, startDate.day)
# replace year only
endDate = startDate.replace(startDate.year + 1)

If you're having problems creating one given your format, let us know.

Another way would be to use pandas "DateOffset" class

link:-https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.tseries.offsets.DateOffset.html

Using ASGM's code(above in the answers):

from datetime import datetime
import pandas as pd

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 + pd.DateOffset(years=1)

new_date.date()

It will return the datetime object with the added year.

Something like this:-

datetime.date(2013, 4, 1)

This is what I do when I need to add months or years and don't want to import more libraries. Just create a datetime.date() object, call add_month(date) to add a month and add_year(date) to add a year.

import datetime
__author__ = 'Daniel Margarido'


# Check if the int given year is a leap year
# return true if leap year or false otherwise
def is_leap_year(year):
    if (year % 4) == 0:
        if (year % 100) == 0:
            if (year % 400) == 0:
                return True
            else:
                return False
        else:
            return True
    else:
        return False


THIRTY_DAYS_MONTHS = [4, 6, 9, 11]
THIRTYONE_DAYS_MONTHS = [1, 3, 5, 7, 8, 10, 12]

# Inputs -> month, year Booth integers
# Return the number of days of the given month
def get_month_days(month, year):
    if month in THIRTY_DAYS_MONTHS:   # April, June, September, November
        return 30
    elif month in THIRTYONE_DAYS_MONTHS:   # January, March, May, July, August, October, December
        return 31
    else:   # February
        if is_leap_year(year):
            return 29
        else:
            return 28

# Checks the month of the given date
# Selects the number of days it needs to add one month
# return the date with one month added
def add_month(date):
    current_month_days = get_month_days(date.month, date.year)
    next_month_days = get_month_days(date.month + 1, date.year)

    delta = datetime.timedelta(days=current_month_days)
    if date.day > next_month_days:
        delta = delta - datetime.timedelta(days=(date.day - next_month_days) - 1)

    return date + delta


def add_year(date):
    if is_leap_year(date.year):
        delta = datetime.timedelta(days=366)
    else:
        delta = datetime.timedelta(days=365)

    return date + delta


# Validates if the expected_value is equal to the given value
def test_equal(expected_value, value):
    if expected_value == value:
        print "Test Passed"
        return True

    print "Test Failed : " + str(expected_value) + " is not equal to " str(value)
    return False

# Test leap year
print "---------- Test leap year ----------"
test_equal(True, is_leap_year(2012))
test_equal(True, is_leap_year(2000))
test_equal(False, is_leap_year(1900))
test_equal(False, is_leap_year(2002))
test_equal(False, is_leap_year(2100))
test_equal(True, is_leap_year(2400))
test_equal(True, is_leap_year(2016))

# Test add month
print "---------- Test add month ----------"
test_equal(datetime.date(2016, 2, 1), add_month(datetime.date(2016, 1, 1)))
test_equal(datetime.date(2016, 6, 16), add_month(datetime.date(2016, 5, 16)))
test_equal(datetime.date(2016, 3, 15), add_month(datetime.date(2016, 2, 15)))
test_equal(datetime.date(2017, 1, 12), add_month(datetime.date(2016, 12, 12)))
test_equal(datetime.date(2016, 3, 1), add_month(datetime.date(2016, 1, 31)))
test_equal(datetime.date(2015, 3, 1), add_month(datetime.date(2015, 1, 31)))
test_equal(datetime.date(2016, 3, 1), add_month(datetime.date(2016, 1, 30)))
test_equal(datetime.date(2016, 4, 30), add_month(datetime.date(2016, 3, 30)))
test_equal(datetime.date(2016, 5, 1), add_month(datetime.date(2016, 3, 31)))

# Test add year
print "---------- Test add year ----------"
test_equal(datetime.date(2016, 2, 2), add_year(datetime.date(2015, 2, 2)))
test_equal(datetime.date(2001, 2, 2), add_year(datetime.date(2000, 2, 2)))
test_equal(datetime.date(2100, 2, 2), add_year(datetime.date(2099, 2, 2)))
test_equal(datetime.date(2101, 2, 2), add_year(datetime.date(2100, 2, 2)))
test_equal(datetime.date(2401, 2, 2), add_year(datetime.date(2400, 2, 2)))