How to convert Julian date to standard date?

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'

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