Timestamp to Decimal day of year

I have a dataset with a Timestamp column in the format (yyyy-mm-dd hour:minute:second.millisecond) (example: 2003-02-20 01:00:00.100) that I need to convert into decimal day of year, could I please get some guidance on how to do this?

thanks

Answer 1

[Walter Roberson]

d = datetime('2003-02-20 01:00:00.100')

d = datetime
   20-Feb-2003 01:00:00

floor(days(d - dateshift(d, 'start', 'year'))) + 1

ans = 51

The +1 is because the start of year is day 1 of the year, and the subtraction gives the difference in days -- so Jan 2 would be 1 day difference, Jan 3 would be 2 days difference, and so on. Therefore to get day of year with Jan 1 being day 1, then you need to add 1.

Answer 2

[Steven Lord]

dt = datetime('now', 'TimeZone', 'EST')

Warning: 'EST' specifies a time zone with a fixed offset from UTC, -05:00. This zone does not follow daylight saving time, and so may give unexpected results. See the datetime.TimeZone property for details about specifying time zones.

dt = datetime
   30-Nov-2020 21:49:15

startOfYear = dateshift(dt, 'start', 'year')

startOfYear = datetime
   01-Jan-2020

fractionOfYear = years(dt - startOfYear)

fractionOfYear = 0.9170

11/12

ans = 0.9167

Given that as I type this it's the evening of November 30th (roughly) 11/12th of the year has elapsed.

[Walter Roberson]

However, this would be decimal year, not decimal day. And Jan 1 would be 0.

Also this uses fixed-length years, not leap-years.

dt = datetime('2020-12-31 23:59')

dt = datetime
   31-Dec-2020 23:59:00

startOfYear = datetime('2020-1-1 00:00')

startOfYear = datetime
   01-Jan-2020

fractionOfYear = years(dt - startOfYear)

fractionOfYear = 1.0021

sy2 = datetime('2019-1-1 00:00')

sy2 = datetime
   01-Jan-2019

eoy2 = datetime('2019-12-31 23:59')

eoy2 = datetime
   31-Dec-2019 23:59:00

foy2 = years(eoy2 - sy2)

foy2 = 0.9993

Notice that the fractionOfYear for December 31 23:59 2020 is more than 1: that is because years() is fixed-length years but 2020 is a leap year

[Noelle Cielito Soriano]

Thank you for this, this makes more sense! I also want to ask how this would work if I have a whole column of several timestamps that I want to find the decimal day of?

Would I just edit the dt datetime array to

dt = datetime(myTimestamp)?

In that case, I don't think that fractionOfYear = years(dt - startOfYear) would work since they are of different lengths?

Thanks!

[Walter Roberson]

What output do you want? Date as fraction of a year since the beginning of the yea? Day number of the year with January 1 being day 1 ?

If you want fraction of a year, do you want the fraction to use fixed length years, or actual length of the year in question? e.g., years(0.75) + datetime('2019-01-01') being October 1 2019, but years(0.75) + datetime('2020-01-01') being September 30, 2020 ?

[Noelle Cielito Soriano]

Hi,

I apologize for the lack of clarity, but I am looking to convert my timestamps to decimal days. I wouldn't need to worry about leap years in this case. Since the previous examples were looking at decimal years, I went along with the example code.

Thank you again for all the help!

[Walter Roberson]

The Answer that I posted computes decimal day number of the year. If the d variable were a vector of values, then the code I posted with floor() would compute the appropriate day number for each element of the array, basing each calculation upon the year appropriate to it. For example, the code would be perfectly happy to be fed dec 2 2020 and jan 2 2021 and would compute the day of the year relative to 2020 for the first and the day of the year relative to 2021 for the second.

[Noelle Cielito Soriano]

Hello Steven and Walter, my apologies for replying so late. I wanted to let you know that I really appreciate the help you gave me with this question! Noelle