Convert a timedelta to days, hours and minutes
I've got a timedelta. I want the days, hours and minutes from that - either as a tuple or a dictionary... I'm not fussed.
I must have done this a dozen times in a dozen languages over the years but Python usually has a simple answer to everything so I thought I'd ask here before busting out some nauseatingly simple (yet verbose) mathematics.
Mr Fooz raises a good point.
I'm dealing with "listings" (a bit like ebay listings) where each one has a duration. I'm trying to find the time left by doing when_added + duration - now
Am I right in saying that wouldn't account for DST? If not, what's the simplest way to add/subtract an hour?
Solution 1:
If you have a datetime.timedelta
value td
, td.days
already gives you the "days" you want. timedelta
values keep fraction-of-day as seconds (not directly hours or minutes) so you'll indeed have to perform "nauseatingly simple mathematics", e.g.:
def days_hours_minutes(td):
return td.days, td.seconds//3600, (td.seconds//60)%60
Solution 2:
This is a bit more compact, you get the hours, minutes and seconds in two lines.
days = td.days
hours, remainder = divmod(td.seconds, 3600)
minutes, seconds = divmod(remainder, 60)
# If you want to take into account fractions of a second
seconds += td.microseconds / 1e6
Solution 3:
days, hours, minutes = td.days, td.seconds // 3600, td.seconds // 60 % 60
As for DST, I think the best thing is to convert both datetime
objects to seconds. This way the system calculates DST for you.
>>> m13 = datetime(2010, 3, 13, 8, 0, 0) # 2010 March 13 8:00 AM
>>> m14 = datetime(2010, 3, 14, 8, 0, 0) # DST starts on this day, in my time zone
>>> mktime(m14.timetuple()) - mktime(m13.timetuple()) # difference in seconds
82800.0
>>> _/3600 # convert to hours
23.0
Solution 4:
I don't understand
days, hours, minutes = td.days, td.seconds // 3600, td.seconds // 60 % 60
how about this
days, hours, minutes = td.days, td.seconds // 3600, td.seconds % 3600 / 60.0
You get minutes and seconds of a minute as a float.