Django template filter that returns appropriate zodiac for a given date (in Turkish)

zodiac_tag.py

register = template.Library()
@register.filter(name='zodiac')
def zodiac(value):
    """ Return appropriate zodiac for a given date
    Usage in templates: {{ your_object.date_of_birth|zodiac }} """
    signs = (
    (u"Oğlak", range(356, 365)),
    (u"Oğlak", range(1, 20)),
    (u"Kova", range(21, 52)),
    (u"Balık", range(52, 79)),
    (u"Koç", range(79, 111)),
    (u"Boğa", range(111, 141)),
    (u"İkizler", range(141, 173)),
    (u"Yengeç", range(173, 204)),
    (u"Aslan", range(204, 236)),
    (u"Başak", range(236, 267)),
    (u"Terazi", range(267, 297)),
    (u"Akrep", range(297, 327)),
    (u"Yay", range(327, 355))
    )

    birthday_day_of_year = value.timetuple()[7]

    for sign, days in signs:
        if birthday_day_of_year in days:
            return sign