edtime 1.0.1

Extended Decimal Time

Imagine decimal representation of days of UNIX seconds, call it “Epoch Days”.

Then, imagine grouping the digits of such days representation as follows:

dYear ° dMonth ″ dWeek ′ day T dHour : dMinute : dSecond, such that:

• 1 dYear = 1000 days, 1 dMonth = 100 days, 1 dWeek = 10 days,

• 1 dHour = 0.1 days, 1 dMinute = 0.001 days, 1 dSecond = 0.00001 days.

Yes, you just divide your UNIX seconds / 86400 and group digits!

Exmaple

For example, 2021-09-30T11:25:45Z would be represented as 1633001147 UNIX seconds, therefore: 1633001147/86400 = 18900.476238425927 = 18°9″0′0T4:76:23.8425927, meaning it is 18th dYear, 9th dMonth, 0th dWeek, 0th day, 4th dHour, 76th dMinute, 23.8425927th dSecond.

Usage

pip install edtime

>>> from edtime import edtime
>>> from datetime import datetime
>>> t0 = edtime(datetime.utcnow()) # OR edtime.utcnow()
>>> t1 = edtime.datetime(2021, 12, 20, 4, 52, 47, 954297)
>>> str(t1)
'18°9″8′1T2:03:33.2804365'
>>> float(t1)
18981.203332804365
>>> t2 = edtime(18981.203332804365); t2
edtime.edtime(dyear=18, dmonth=9, dweek=8, dday=1, dhour=2, dminute=3, dsecond=33.2804365)
>>> t1 == t2
True
>>> t3 = edtime(16, 4, 5, 3, 4, 65, 40.5092593); t3 # 2015-01-18T11:10:11
edtime.edtime(dyear=16, dmonth=4, dweek=5, dday=3, dhour=4, dminute=65, dsecond=40.5092593)
>>> str(t1 + t2 + t3)
'54°4″1′5T8:72:07.070133'
>>> (t1 + t2 + t3).to_datetime().isoformat()
'2118-12-26T20:55:46.908595'

For astronomers:

ED = JD - 2440587.5

>>> from edtime import edtime
>>> d = edtime.datetime(2000, 1, 1, 12)
>>> d.to_jd()
>>> 2451545.0
>>> float(edtime.from_jd(2451545))
10957.5

Notice

1. => 1 dSecond is:

   • 0.864 standard SI seconds long.

2. => 1 dMinute is:

   • 86.4 standard SI seconds long.

3. => 1 dHour is:

   • 8640 standard SI seconds long.

4. => 1 dWeek is:

   • 10 days long.

5. => 1 dMonth is:

   • 100 days long.

6. => 1 dYear is:

   • 1000 days long.