# [Tutor] time challange

paul brian paul1brian at gmail.com
Thu Sep 22 18:35:46 CEST 2005

```You are fairly close

>>> t1 = today()
>>> t1
<DateTime object for '2005-09-22 00:00:00.00' at 11656e0>
>>> t2 = today() + RelativeDateTime(hours=20)
>>> t2
<DateTime object for '2005-09-22 20:00:00.00' at 1165760>
>>> t3 = t2 - t1
>>> t3.hours
20.0
>>> slice = t3/20
>>> slice
<DateTimeDelta object for '01:00:00.00' at 116cc50>

t3 is a "Delta" - that is an abstract representation
of time - it is not the 20 hours since midnight, just 20 hours
at any time in the universe.
slice is just 1/20th of that same abstract time. But because of the munificence
of Marc we can add that abstract hour to a real fixed time (ignoring
Einstein of course)

>>> t1 + slice
<DateTime object for '2005-09-22 01:00:00.00' at 11655e0>

And so that is a "real" datetime 1/20th of the way forward from t1

so a simple loop will get you your 20 evenly spaced time periods,
which is what i think you were asking for.

cheers

On 9/22/05, nephish <nephish at xit.net> wrote:
> Hey there,
>
> i use mx.DateTime.RelativeDateTimeDiff to get the difference between
> date_x and date_y.
> what i need to do divide this amount of time into 20 different times
> that spaced out between the date_x and the date_y.
>
> so if the difference between date_x and date_y is 20 hours, i need 20
> DateTimes that are one hour apart from each other. If the difference is
> 40 minutes, i need the 20 DateTimes to be spaced out 2 minutes from each
> other..
>
> what would be a way to pull this off? i have looked at the docs for
> mxDateTime
> http://www.egenix.com/files/python/mxDateTime.html
> and there seems to be a divide operation, but i dont quite know what it
> with the deltas.
>
> anyone have a good start point?
>
> thanks
> shawn
> _______________________________________________
> Tutor maillist  -  Tutor at python.org
> http://mail.python.org/mailman/listinfo/tutor
>

--
--------------------------
Paul Brian
m. 07875 074 534
t. 0208 352 1741
```