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 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 aritmetic operation on the original datetime value with the information in the relativedelta. weekday: One of the weekday instances (MO, TU, etc). 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. 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.
Here is the behavior of operations with relativedelta:
 Calculate the absolute year, using the ‘year’ argument, or the original datetime year, if the argument is not present.
 Add the relative ‘years’ argument to the absolute year.
 Do steps 1 and 2 for month/months.
 Calculate the absolute day, using the ‘day’ argument, or the original datetime day, if the argument is not present. Then, subtract from the day until it fits in the year and month found after their operations.
 Add the relative ‘days’ argument to the absolute day. Notice that the ‘weeks’ argument is multiplied by 7 and added to ‘days’.
 Do steps 1 and 2 for hour/hours, minute/minutes, second/seconds, microsecond/microseconds.
 If the ‘weekday’ argument is present, calculate the weekday, with the given (wday, nth) tuple. wday is the index of the weekday (06, 0=Mon), and nth is the number of weeks to add forward or backward, depending on its signal. Notice that if the calculated date is already Monday, for example, using (0, 1) or (0, 1) won’t change the day.

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 [http://www.cl.cam.ac.uk/~mgk25/isotime.html 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)