It is possible by: df[‘duration1’] = df[‘duration’].astype(str).str[-18:-10] But solution is not general, if input is 3 days 05:01:11 it remove 3 days too. So solution working only for timedeltas less as one day correctly. More general solution is create custom format: N = 10 np.random.seed(11230) rng = pd.date_range(‘2017-04-03 15:30:00′, periods=N, freq=’13.5H’) df = pd.DataFrame({‘duration’: np.abs(np.random.choice(rng, … Read more
Simple diff (without lib) /** * Get a diff between two dates * @param date1 the oldest date * @param date2 the newest date * @param timeUnit the unit in which you want the diff * @return the diff value, in the provided unit */ public static long getDateDiff(Date date1, Date date2, TimeUnit timeUnit) { … Read more
datetime.time objects do not support addition with datetime.timedeltas. There is one natural definition though, clock arithmetic. You could compute it like this: import datetime as dt now = dt.datetime.now() delta = dt.timedelta(hours = 12) t = now.time() print(t) # 12:39:11.039864 print((dt.datetime.combine(dt.date(1,1,1),t) + delta).time()) # 00:39:11.039864 dt.datetime.combine(…) lifts the datetime.time t to a datetime.datetime object, the … Read more
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.
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
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.
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
The previous answers are correct but it’s generally a better practice to do: import datetime Then you’ll have, using datetime.timedelta: date_1 = datetime.datetime.strptime(start_date, “%m/%d/%y”) end_date = date_1 + datetime.timedelta(days=10)
Format timedelta to string