You can (and should) use pd.PeriodIndex as a first step, then convert to timestamp using PeriodIndex.to_timestamp: qs = df[‘Quarter’].str.replace(r'(Q\d) (\d+)’, r’\2-\1′) qs 0 1996-Q3 1 1996-Q4 2 1997-Q1 Name: Quarter, dtype: object df[‘date’] = pd.PeriodIndex(qs, freq=’Q’).to_timestamp() df Quarter date 0 Q3 1996 1996-07-01 1 Q4 1996 1996-10-01 2 Q1 1997 1997-01-01 The initial replace step … Read more
In Java SE 8, it is the responsibility of the application to create a class linking Period and Duration if that is needed. Note that a Duration contains an amount of seconds, not separate amounts of seconds, minutes and hours. The amount of seconds can exceed 24 hours, thus a Duration can represent a “day”. … Read more
Use datetime.timedelta: from datetime import date, datetime, timedelta def perdelta(start, end, delta): curr = start while curr < end: yield curr curr += delta >>> for result in perdelta(date(2011, 10, 10), date(2011, 12, 12), timedelta(days=4)): … print result … 2011-10-10 2011-10-14 2011-10-18 2011-10-22 2011-10-26 2011-10-30 2011-11-03 2011-11-07 2011-11-11 2011-11-15 2011-11-19 2011-11-23 2011-11-27 2011-12-01 2011-12-05 2011-12-09 … Read more
You need to normalize the period because if you construct it with the total number of seconds, then that’s the only value it has. Normalizing it will break it down into the total number of days, minutes, seconds, etc. Edit by ripper234 – Adding a TL;DR version: PeriodFormat.getDefault().print(period) For example: public static void main(String[] args) … Read more