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Question

How would I determine zodiac / astrological star sign from a birthday in Python?

Answer 1

+4 A:

You could give them some more information about position of the planets and the stars.

import ephem 
>>> u = ephem.Uranus()
>>> u.compute('1871/3/13')
>>> print u.ra, u.dec, u.mag
7:38:06.27 22:04:47.4 5.46
>>> print ephem.constellation(u)
('Gem', 'Gemini')

DiggyF 2010-07-18 06:58:19

In a perfect world, you'd be able to use the `ephem` library to find out the Zodiac sign by simply replacing Uranus with Sun in your example. However, it appears that the signs and the constellations [don't match up exactly](http://en.wikipedia.org/wiki/Zodiac#Table_of_dates). There is also an astrology-specific Python library called [oroboros](http://pypi.python.org/pypi/oroboros/20080712), but it's probably overkill.

Aram Dulyan 2010-07-18 07:26:43

`oroboros` uses `ephem`.

fmark 2010-07-18 15:39:11

Answer 2

+9 A:

I've done this before. The simplest solution that I ended up with was an array of the following key/values:

120:Sag, 218:Cap, 320:Aqu, 420:Pis, 521:Ari,
621:Tau, 722:Gem, 823:Can, 923:Leo, 1023:Vir
1122:Lib, 1222:Scorp, 1231: Sag

Then you write the birth date in the mdd format, ie, month number (starting with 1 for January) and two digit day number (01-31). Iterate through the array, and if the date is less than an item in the array, you have your star sign.

bluesmoon 2010-07-18 07:07:48

+1 - this looks like the best solution that can be done without having to install a new module.

R Samuel Klatchko 2010-07-18 07:19:54

An `array` or a `dict`?

Johnsyweb 2010-07-18 07:58:41

I'd use a dict for that.

Daenyth 2010-07-18 14:37:00

for the lookup you could use `bisect`, see the example in http://docs.python.org/library/bisect.html#bisect.bisect .

DiggyF 2010-07-18 14:58:25

Using bisect for 13 keys in function that is used not very often is little overoptimised, don't you think?Moreover log_2(13) is 3.7 so you have only 13/3.7 = 3.5 times less operations. I would be concern that one operation in bisect may take more than 3.5 times than simple loop, so you even have worse performance then.To sum up, bisect is very good suggestion form most searching cased, but this is probably not the one.

Tomasz Wysocki 2010-07-18 15:45:21

@Johnsyweb: dict, yes.

bluesmoon 2010-07-18 22:50:57

I would also use a dict, but like this: `d[bisect(d.keys(),121)]`, resulting in "Sag". Not so much for speed, I think it just looks better than writing a loop for that.

DiggyF 2010-07-19 17:23:58

Answer 3

+2 A:

Using bisect is more efficient than iterating until you find a match, but a lookup table for each day of the year is faster still and really not that big.

from bisect import bisect
signs = [(1,20,"Cap"), (2,18,"Aqu"), (3,20,"Pis"), (4,20,"Ari"),
         (5,21,"Tau"), (6,21,"Gem"), (7,22,"Can"), (8,23,"Leo"),
         (9,23,"Vir"), (10,23,"Lib"), (11,22,"Sco"), (12,22,"Sag"),
         (12,31,"Cap")]
def zodiac_sign(m,d):
    return signs[bisect(signs,(m,d))][2]

assert zodiac_sign(3,9) == "Pis"
assert zodiac_sign(6,30) == "Can"

gnibbler 2010-07-18 14:28:41

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How would I determine zodiac / astrological star sign from a birthday in Python?

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