Because timedelta is defined like: class datetime.timedelta([days,] [seconds,] [microseconds,] [milliseconds,] [minutes,] [hours,] [weeks]) All arguments are optional and default to 0. You can easily say “Three days and four milliseconds” with optional arguments that way. >>> datetime.timedelta(days=3, milliseconds=4) datetime.timedelta(3, 0, 4000) >>> datetime.timedelta(3, 0, 0, 4) #no need for that. datetime.timedelta(3, 0, 4000) And for … Read more
If you need to be exact use the dateutil module to calculate relative dates from datetime import datetime from dateutil.relativedelta import relativedelta three_yrs_ago = datetime.now() – relativedelta(years=3)
dateutil is an extension package to the python standard datetime module. As you say, it provides extra functionality, such as timedeltas that are expressed in units larger than a day. This is useful if you have to ask questions such as how many months can I save before my girlfriend’s birthday comes up, or what’s … Read more
To me the most elegant solution, without having to resort to external libraries such as dateutil or manually parsing the input, is to use datetime’s powerful strptime string parsing method. from datetime import datetime, timedelta # we specify the input and the format… t = datetime.strptime(“05:20:25″,”%H:%M:%S”) # …and use datetime’s hour, min and sec properties … Read more
Try this: from datetime import datetime, date datetime.combine(date.today(), exit) – datetime.combine(date.today(), enter) combine builds a datetime, that can be subtracted.
You need more than a timedelta to tell how many years have passed; you also need to know the beginning (or ending) date. (It’s a leap year thing.) Your best bet is to use the dateutil.relativedelta object, but that’s a 3rd party module. If you want to know the datetime that was n years from … Read more
The Series class has a pandas.Series.dt accessor object with several useful datetime attributes, including dt.days. Access this attribute via: timedelta_series.dt.days You can also get the seconds and microseconds attributes in the same way.
from datetime import datetime, timedelta nine_hours_from_now = datetime.now() + timedelta(hours=9) #datetime.datetime(2012, 12, 3, 23, 24, 31, 774118) And then use string formatting to get the relevant pieces: >>> ‘{:%H:%M:%S}’.format(nine_hours_from_now) ’23:24:31′ If you’re only formatting the datetime then you can use: >>> format(nine_hours_from_now, ‘%H:%M:%S’) ’23:24:31′ Or, as @eumiro has pointed out in comments – strftime
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
You can just convert the timedelta to a string with str(). Here’s an example: import datetime start = datetime.datetime(2009,2,10,14,00) end = datetime.datetime(2009,2,10,16,00) delta = end-start print(str(delta)) # prints 2:00:00