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6. While pandas.Timedelta does not provide these attributes directly, it indeed provide a method called total_seconds, based on which days, hours, and minutes can be easily derived: import pandas as pd. td = pd.Timedelta("2 days 12:30:00") minutes = td.total_seconds()/60. hours = minutes/60. days = hours/ 24.
Jan 7, 2013 · def convert_timedelta(duration): days, seconds = duration.days, duration.seconds hours = seconds // 3600 minutes = (seconds % 3600) // 60 seconds = (seconds % 60) return days, hours, minutes, seconds If you want to convert this to a single value to store in a database, then convert that single value back to format it, do this:
Aug 28, 2009 · Using datetime example >>> from datetime import datetime >>> then = datetime(2012, 3, 5, 23, 8, 15) # Random date in the past >>> now = datetime.now() # Now ...
Jan 1, 2010 · How do I calculate the difference in time in minutes for the following timestamp in Python? 2010-01-01 17:31:22 2010-01-03 17:31:22
7. You screwed up a bit over here: // Number of minutes in a year. int year = minutes / 525600; int day = minutes / 1440; int remainingMinutes = day % 525600; You took the total number of minutes and divided by 1440, so the number of days you got was wrong. You should have taken the remainder and then divided by 1440.
Aug 17, 2011 · You can substract dates in Oracle. This will give you the difference in days. Multiply by 24 to get hours, and so on. SQL> select oldest - creation from my_table; If your date is stored as character data, you have to convert it to a date type first. SQL> select 24 * (to_date('2009-07-07 22:00', 'YYYY-MM-DD hh24:mi')
Jul 30, 2013 · long duration = endDate.getTime() - startDate.getTime(); //now we calculate the differences in different time units. //this long value will be the total time difference in each unit. //i.e; total difference in seconds, total difference in minutes etc... long diffInSeconds = TimeUnit.MILLISECONDS.toSeconds(duration);
Sep 29, 2008 · (There are 153 days in every 5 months starting from 1 March.) Step E. Divide R₁ by 2 to get a quotient Q₂ and ignore the remainder. Add R₁ × 31 - Q₂ to D. (Within each group of 5 months, there are 61 days in every 2 months, and within that the first of each pair of months is 31 days.
Jan 22, 2010 · 1) The function DATEDIFF () is responsible to calculate differences between two dates, the result could be " year quarter month dayofyear day week hour minute second millisecond microsecond nanosecond ", specified on the first parameter (datepart): select datediff(day,'1997-10-07','2011-09-11') 2) You can use the function GETDATE () to get the ...
Dec 17, 2012 · somehow if you add 29 days to current date - then the diff is 28 days, 23 hours 59 minutes 59 seconds, but if you add 30 days, then the diff is 30 days, 0 hours 59 minutes 59 seconds. I.e. one extra hour is added. Something is wrong with this and the other answers..
