Correcting for Drift between Two Dates
steven at REMOVE.THIS.cybersource.com.au
Tue Sep 9 09:41:41 CEST 2008
On Mon, 08 Sep 2008 21:53:18 -0700, W. eWatson wrote:
> I have two dates, ts1, ts2 as below in the sample program. I know the
> clock drift in seconds per day. I would like to calculate the actual
> date of ts2. See my question at the end of the program.
When faced with a complicated task, break it down into simpler subtasks.
Functions are your friends. Here you go:
from __future__ import division
from datetime import datetime as DT
from datetime import timedelta
SITE_DRIFT = 4.23 # drift in seconds per day
# negative drift means the clock falls slow
SEC_PER_DAY = 60*60*24 # number of seconds per day
def calc_drift(when, base, drift=SITE_DRIFT):
"""Return the amount of drift at date when since date base."""
x = when - base
days = x.days + x.seconds/SEC_PER_DAY
def fix_date(when, base, drift=SITE_DRIFT):
"""Return date when adjusted to the correct time."""
d = calc_drift(when, base, drift)
delta = timedelta(seconds=-d)
return when + delta
And here it is in action:
>>> fix_date(DT(2008,9,9), DT(2008,9,8))
datetime.datetime(2008, 9, 8, 23, 59, 55, 770000)
I leave it to you to convert date/time strings into datetime objects.
More information about the Python-list