How can I perform divison on a datetime.timedelta in python?

Question

I\'d like to be able to do the following:

num_intervals = (cur_date - previous_date) / interval_length

or

print (datetime.n         


        

Answer 1

Sure, just convert to a number of seconds (minutes, milliseconds, hours, take your pick of units) and do the division.

EDIT (again): so you can't assign to timedelta.__div__. Try this, then:

divtdi = datetime.timedelta.__div__
def divtd(td1, td2):
    if isinstance(td2, (int, long)):
        return divtdi(td1, td2)
    us1 = td1.microseconds + 1000000 * (td1.seconds + 86400 * td1.days)
    us2 = td2.microseconds + 1000000 * (td2.seconds + 86400 * td2.days)
    return us1 / us2 # this does integer division, use float(us1) / us2 for fp division

And to incorporate this into nadia's suggestion:

class MyTimeDelta:
    __div__ = divtd

Example usage:

>>> divtd(datetime.timedelta(hours = 12), datetime.timedelta(hours = 2))
6
>>> divtd(datetime.timedelta(hours = 12), 2)
datetime.timedelta(0, 21600)
>>> MyTimeDelta(hours = 12) / MyTimeDelta(hours = 2)
6

etc. Of course you could even name (or alias) your custom class timedelta so it gets used in place of the real timedelta, at least in your code.

Answer 2

You can override the division operator like this:

class MyTimeDelta(timedelta):
     def __div__(self, value):
          # Dome something about the object

Answer 3

Division and multiplication by integers seems to work out of the box:

>>> from datetime import timedelta
>>> timedelta(hours=6)
datetime.timedelta(0, 21600)
>>> timedelta(hours=6) / 2
datetime.timedelta(0, 10800)