How to convert Julian date to standard date?

Question

I have a string as Julian date like \"16152\" meaning 152\'nd day of 2016 or \"15234\" meaning 234\'th day of 2015.

How can I convert these

Answer 1

Easy way

• Convert from regular date to Julian date

print datetime.datetime.now().strftime("%y%j")

• Convert from Julian date to regular date

print datetime.datetime.strptime('19155', '%y%j').strftime("%d-%m-%Y")

Answer 2

I used this for changing a Juian date to xml xsd:datetime

def julianDate2ISO8601(d, offset='+00:00'): """ return ISO8601 formated datetime from julian date optional offset [+|-]hh:mm """ d = str(d) # make sure it is a string # replace leading number with correct century centuryArray = ['19','20','21'] d = centuryArray[int(d[:1])] + d[1:] # format string to iso 8601 datetime return datetime.datetime.strptime(d, '%Y%j').date().strftime( '%Y-%m-%dT00:00:00') + offset

Answer 3

Well, first, create a datetime object (from the module datetime)

from datetime import datetime
from datetime import timedelta
julian = ... # Your julian datetime
date = datetime.strptime("1/1/" + jul[:2], "%m/%d/%y") 
# Just initializing the start date, which will be January 1st in the year of the Julian date (2 first chars)

Now add the days from the start date:

daysToAdd = int(julian[2:]) # Taking the days and converting to int
date += timedelta(days = daysToAdd - 1)

Now, you can just print it as is:

print(str(date))

Or you can use strftime() function.

print(date.strftime("%d/%m/%y"))

Read more about strftime format string here

Answer 4

The .strptime() method supports the day of year format:

>>> import datetime
>>>
>>> datetime.datetime.strptime('16234', '%y%j').date()
datetime.date(2016, 8, 21)

And then you can use strftime() to reformat the date

>>> date = datetime.date(2016, 8, 21)
>>> date.strftime('%d/%m/%Y')
'21/08/2016'