relativedelta¶

class
dateutil.relativedelta.
relativedelta
(dt1=None, dt2=None, years=0, months=0, days=0, leapdays=0, weeks=0, hours=0, minutes=0, seconds=0, microseconds=0, year=None, month=None, day=None, weekday=None, yearday=None, nlyearday=None, hour=None, minute=None, second=None, microsecond=None)[source]¶ The relativedelta type is designed to be applied to an existing datetime and can replace specific components of that datetime, or represents an interval of time.
It is based on the specification of the excellent work done by M.A. Lemburg in his mx.DateTime extension. However, notice that this type does NOT implement the same algorithm as his work. Do NOT expect it to behave like mx.DateTime’s counterpart.
There are two different ways to build a relativedelta instance. The first one is passing it two date/datetime classes:
relativedelta(datetime1, datetime2)
The second one is passing it any number of the following keyword arguments:
relativedelta(arg1=x,arg2=y,arg3=z...) year, month, day, hour, minute, second, microsecond: Absolute information (argument is singular); adding or subtracting a relativedelta with absolute information does not perform an arithmetic operation, but rather REPLACES the corresponding value in the original datetime with the value(s) in relativedelta. years, months, weeks, days, hours, minutes, seconds, microseconds: Relative information, may be negative (argument is plural); adding or subtracting a relativedelta with relative information performs the corresponding arithmetic operation on the original datetime value with the information in the relativedelta. weekday: One of the weekday instances (MO, TU, etc) available in the relativedelta module. These instances may receive a parameter N, specifying the Nth weekday, which could be positive or negative (like MO(+1) or MO(2)). Not specifying it is the same as specifying +1. You can also use an integer, where 0=MO. This argument is always relative e.g. if the calculated date is already Monday, using MO(1) or MO(1) won't change the day. To effectively make it absolute, use it in combination with the day argument (e.g. day=1, MO(1) for first Monday of the month). leapdays: Will add given days to the date found, if year is a leap year, and the date found is post 28 of february. yearday, nlyearday: Set the yearday or the nonleap year day (jump leap days). These are converted to day/month/leapdays information.
There are relative and absolute forms of the keyword arguments. The plural is relative, and the singular is absolute. For each argument in the order below, the absolute form is applied first (by setting each attribute to that value) and then the relative form (by adding the value to the attribute).
The order of attributes considered when this relativedelta is added to a datetime is:
 Year
 Month
 Day
 Hours
 Minutes
 Seconds
 Microseconds
Finally, weekday is applied, using the rule described above.
For example
>>> from datetime import datetime >>> from dateutil.relativedelta import relativedelta, MO >>> dt = datetime(2018, 4, 9, 13, 37, 0) >>> delta = relativedelta(hours=25, day=1, weekday=MO(1)) >>> dt + delta datetime.datetime(2018, 4, 2, 14, 37)
First, the day is set to 1 (the first of the month), then 25 hours are added, to get to the 2nd day and 14th hour, finally the weekday is applied, but since the 2nd is already a Monday there is no effect.

normalized
()[source]¶ Return a version of this object represented entirely using integer values for the relative attributes.
>>> relativedelta(days=1.5, hours=2).normalized() relativedelta(days=+1, hours=+14)
Returns: Returns a dateutil.relativedelta.relativedelta
object.

weeks
¶
Examples¶
>>> from datetime import *; from dateutil.relativedelta import *
>>> import calendar
>>> NOW = datetime(2003, 9, 17, 20, 54, 47, 282310)
>>> TODAY = date(2003, 9, 17)
Let’s begin our trip:
>>> from datetime import *; from dateutil.relativedelta import *
>>> import calendar
Store some values:
>>> NOW = datetime.now()
>>> TODAY = date.today()
>>> NOW
datetime.datetime(2003, 9, 17, 20, 54, 47, 282310)
>>> TODAY
datetime.date(2003, 9, 17)
Next month
>>> NOW+relativedelta(months=+1)
datetime.datetime(2003, 10, 17, 20, 54, 47, 282310)
Next month, plus one week.
>>> NOW+relativedelta(months=+1, weeks=+1)
datetime.datetime(2003, 10, 24, 20, 54, 47, 282310)
Next month, plus one week, at 10am.
>>> TODAY+relativedelta(months=+1, weeks=+1, hour=10)
datetime.datetime(2003, 10, 24, 10, 0)
Here is another example using an absolute relativedelta. Notice the use of year and month (both singular) which causes the values to be replaced in the original datetime rather than performing an arithmetic operation on them.
>>> NOW+relativedelta(year=1, month=1)
datetime.datetime(1, 1, 17, 20, 54, 47, 282310)
Let’s try the other way around. Notice that the hour setting we get in the relativedelta is relative, since it’s a difference, and the weeks parameter has gone.
>>> relativedelta(datetime(2003, 10, 24, 10, 0), TODAY)
relativedelta(months=+1, days=+7, hours=+10)
One month before one year.
>>> NOW+relativedelta(years=+1, months=1)
datetime.datetime(2004, 8, 17, 20, 54, 47, 282310)
How does it handle months with different numbers of days? Notice that adding one month will never cross the month boundary.
>>> date(2003,1,27)+relativedelta(months=+1)
datetime.date(2003, 2, 27)
>>> date(2003,1,31)+relativedelta(months=+1)
datetime.date(2003, 2, 28)
>>> date(2003,1,31)+relativedelta(months=+2)
datetime.date(2003, 3, 31)
The logic for years is the same, even on leap years.
>>> date(2000,2,28)+relativedelta(years=+1)
datetime.date(2001, 2, 28)
>>> date(2000,2,29)+relativedelta(years=+1)
datetime.date(2001, 2, 28)
>>> date(1999,2,28)+relativedelta(years=+1)
datetime.date(2000, 2, 28)
>>> date(1999,3,1)+relativedelta(years=+1)
datetime.date(2000, 3, 1)
>>> date(2001,2,28)+relativedelta(years=1)
datetime.date(2000, 2, 28)
>>> date(2001,3,1)+relativedelta(years=1)
datetime.date(2000, 3, 1)
Next friday
>>> TODAY+relativedelta(weekday=FR)
datetime.date(2003, 9, 19)
>>> TODAY+relativedelta(weekday=calendar.FRIDAY)
datetime.date(2003, 9, 19)
Last friday in this month.
>>> TODAY+relativedelta(day=31, weekday=FR(1))
datetime.date(2003, 9, 26)
Next wednesday (it’s today!).
>>> TODAY+relativedelta(weekday=WE(+1))
datetime.date(2003, 9, 17)
Next wednesday, but not today.
>>> TODAY+relativedelta(days=+1, weekday=WE(+1))
datetime.date(2003, 9, 24)
Following ISO year week number notation find the first day of the 15th week of 1997.
>>> datetime(1997,1,1)+relativedelta(day=4, weekday=MO(1), weeks=+14)
datetime.datetime(1997, 4, 7, 0, 0)
How long ago has the millennium changed?
>>> relativedelta(NOW, date(2001,1,1))
relativedelta(years=+2, months=+8, days=+16,
hours=+20, minutes=+54, seconds=+47, microseconds=+282310)
How old is John?
>>> johnbirthday = datetime(1978, 4, 5, 12, 0)
>>> relativedelta(NOW, johnbirthday)
relativedelta(years=+25, months=+5, days=+12,
hours=+8, minutes=+54, seconds=+47, microseconds=+282310)
It works with dates too.
>>> relativedelta(TODAY, johnbirthday)
relativedelta(years=+25, months=+5, days=+11, hours=+12)
Obtain today’s date using the yearday:
>>> date(2003, 1, 1)+relativedelta(yearday=260)
datetime.date(2003, 9, 17)
We can use today’s date, since yearday should be absolute in the given year:
>>> TODAY+relativedelta(yearday=260)
datetime.date(2003, 9, 17)
Last year it should be in the same day:
>>> date(2002, 1, 1)+relativedelta(yearday=260)
datetime.date(2002, 9, 17)
But not in a leap year:
>>> date(2000, 1, 1)+relativedelta(yearday=260)
datetime.date(2000, 9, 16)
We can use the nonleap year day to ignore this:
>>> date(2000, 1, 1)+relativedelta(nlyearday=260)
datetime.date(2000, 9, 17)