Julian Date in Python

Two different “Julian dates”

The ordinal form is a year plus day-of-year (1–365/366), like 2026202, common in manufacturing and IT. The astronomical form is a continuous count of days since noon on 1 January 4713 BC, like 2461242.5. Prefer a browser tool? Use the Julian date converter.

Day-of-year (ordinal) Julian dates

Everything here comes from the standard library.

from datetime import date, datetime

The raw day of year (1–366) is available on the time tuple:

>>> date(2026, 7, 21).timetuple().tm_yday
202

For the padded string codes, use strftime. The %j directive is the zero-padded three-digit day of year, %Y the four-digit year and %y the two-digit year:

>>> datetime(2026, 7, 21).strftime('%Y%j')   # 7-digit YYYYDDD
'2026202'
>>> datetime(2026, 7, 21).strftime('%y%j')   # 5-digit YYDDD
'26202'

Parsing a code back into a date uses the same directives with strptime:

>>> datetime.strptime('2026202', '%Y%j').date()
datetime.date(2026, 7, 21)
>>> datetime.strptime('26202', '%y%j').date()
datetime.date(2026, 7, 21)

Two-digit years are ambiguous

%y follows Python’s convention: values 00–68 map to 2000–2068 and 69–99 map to 1969–1999. If your data spans a different range, parse the century yourself rather than trusting %y.

Astronomical Julian Date (JD) and MJD

The datetime module has no built-in JD, so use the classic algorithm. It works for any calendar date and includes the fractional time of day:

def to_jd(dt):
    """Astronomical Julian Date for a datetime assumed to be UTC.
    Valid across the Gregorian and Julian calendars (Fliegel/Meeus)."""
    y, m = dt.year, dt.month
    # fractional day, including the time of day
    d = dt.day + (dt.hour + dt.minute / 60 + dt.second / 3600) / 24
    if m <= 2:
        y -= 1
        m += 12
    a = y // 100
    b = 2 - a + a // 4
    jd = int(365.25 * (y + 4716)) + int(30.6001 * (m + 1)) + d + b - 1524.5
    return jd

def to_mjd(dt):
    return to_jd(dt) - 2400000.5

>>> to_jd(datetime(2026, 7, 21, 0, 0, 0))
2461242.5
>>> to_jd(datetime(2026, 7, 21, 12, 0, 0))
2461243.0
>>> to_mjd(datetime(2026, 7, 21))
61242.0

Remember the JD day starts at noon, so midnight UTC lands on a .5. The MJD (JD − 2400000.5) instead starts at midnight and drops the leading millions.

Shortcut for date-only values

For a plain date at 00:00, toordinal() gives a proleptic Gregorian day count where date(1, 1, 1) is 1. Add the JD of that epoch to get the Julian Date:

>>> d = date(2026, 7, 21)
>>> d.toordinal() + 1721424.5      # JD at 00:00 UTC
2461242.5

# reverse, for the date-only case:
>>> date.fromordinal(round(2461242.5 - 1721424.5))
datetime.date(2026, 7, 21)

Proleptic Gregorian

toordinal() extends the Gregorian calendar backwards, so for dates before the 1582 reform it will differ from the Julian-calendar to_jd result above. For modern dates the two agree exactly.

With astropy

For serious astronomy, let astropy handle time scales, leap seconds and the reverse conversion:

from astropy.time import Time

>>> Time('2026-07-21 00:00:00', scale='utc').jd
2461242.5
>>> Time(2461242.5, format='jd', scale='utc').iso
'2026-07-21 00:00:00.000'
>>> Time(2461242.5, format='jd').mjd
61242.0

Lighter alternatives exist too: jdcal offers small gcal2jd / jd2gcal helpers, and Astropy.time.Time covers JD, MJD and dozens of other formats in one object.

Worked example: 21 July 2026

import datetime as dt

d = dt.datetime(2026, 7, 21)

d.strftime('%Y%j')          # -> '2026202'  (ordinal, 7-digit)
d.strftime('%y%j')          # -> '26202'    (ordinal, 5-digit)
d.timetuple().tm_yday       # -> 202        (day of year)
to_jd(d)                    # -> 2461242.5  (astronomical JD)
to_jd(d) - 2400000.5        # -> 61242.0    (MJD)

Next steps

Want the same conversions in the browser? See the JavaScript examples or the Julian day converter. Other languages: SQL and Excel.