2015-01-11, 11:47
Die Oster-Formel der Wikipedia kann man recht einfach in einem Python-Algorithmus nutzen:
import datetime
def calcEaster(jahr):
a = jahr % 19
b = jahr % 4
c = jahr % 7
k = jahr / 100
p = (8 * k + 13) / 25
q = k / 4
M = (15 + k - p - q) % 30
d = (19 * a + M) % 30
N = (4 + k - q) % 7
e = (2 * b + 4 * c + 6 * d + N) % 7
Ostermontag = (22 + d + e) - 1
start = datetime.datetime.strptime("01.03." + str(jahr), "%d.%m.%Y")
easter = start + datetime.timedelta(days=round(Ostermontag))
return easter.strftime('%d.%m.%Y')
for i in range(2014,2050):
print(calcEaster(i)) |
import datetime
def calcEaster(jahr):
a = jahr % 19
b = jahr % 4
c = jahr % 7
k = jahr / 100
p = (8 * k + 13) / 25
q = k / 4
M = (15 + k - p - q) % 30
d = (19 * a + M) % 30
N = (4 + k - q) % 7
e = (2 * b + 4 * c + 6 * d + N) % 7
Ostermontag = (22 + d + e) - 1
start = datetime.datetime.strptime("01.03." + str(jahr), "%d.%m.%Y")
easter = start + datetime.timedelta(days=round(Ostermontag))
return easter.strftime('%d.%m.%Y')
for i in range(2014,2050):
print(calcEaster(i))
Ergänzung: Das Osterparadoxon ist nicht berücksichtigt!
Uwe Ziegenhagen likes LaTeX and Python, sometimes even combined.
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