Comment calculer le nombre de jours entre deux dates données?
Si j'ai deux dates (ex. '8/18/2008' et '9/26/2008'), quel est le meilleur moyen d'obtenir le nombre de jours entre ces deux dates?
Si vous avez deux objets de date, vous pouvez simplement les soustraire.
from datetime import date
d0 = date(2008, 8, 18)
d1 = date(2008, 9, 26)
delta = d1 - d0
print delta.days
La section pertinente de la documentation: https://docs.python.org/library/datetime.html .
Utiliser le pouvoir de datetime:
from datetime import datetime
date_format = "%m/%d/%Y"
a = datetime.strptime('8/18/2008', date_format)
b = datetime.strptime('9/26/2008', date_format)
delta = b - a
print delta.days # that's it
Jours jusqu'à Noël:
>>> import datetime
>>> today = datetime.date.today()
>>> someday = datetime.date(2008, 12, 25)
>>> diff = someday - today
>>> diff.days
86
Plus d'arithmétique ici .
Vous voulez le module datetime.
>>> from datetime import datetime, timedelta
>>> datetime(2008,08,18) - datetime(2008,09,26)
datetime.timedelta(4)
Ou autre exemple:
Python 2.5.2 (r252:60911, Feb 22 2008, 07:57:53)
[GCC 4.0.1 (Apple Computer, Inc. build 5363)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import datetime
>>> today = datetime.date.today()
>>> print today
2008-09-01
>>> last_year = datetime.date(2007, 9, 1)
>>> print today - last_year
366 days, 0:00:00
Comme souligné ici
from datetime import datetime
start_date = datetime.strptime('8/18/2008', "%m/%d/%Y")
end_date = datetime.strptime('9/26/2008', "%m/%d/%Y")
print abs((end_date-start_date).days)
Cela peut aussi être facilement fait avec arrow:
import arrow
a = arrow.get('2017-05-09')
b = arrow.get('2017-05-11')
delta = (b-a)
print delta.days
Pour référence: http://arrow.readthedocs.io/en/latest/
sans utiliser Lib simplement du code pur:
#Calculate the Days between Two Date
daysOfMonths = [ 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
def isLeapYear(year):
# Pseudo code for this algorithm is found at
# http://en.wikipedia.org/wiki/Leap_year#Algorithm
## if (year is not divisible by 4) then (it is a common Year)
#else if (year is not divisable by 100) then (ut us a leap year)
#else if (year is not disible by 400) then (it is a common year)
#else(it is aleap year)
return (year % 4 == 0 and year % 100 != 0) or year % 400 == 0
def Count_Days(year1, month1, day1):
if month1 ==2:
if isLeapYear(year1):
if day1 < daysOfMonths[month1-1]+1:
return year1, month1, day1+1
else:
if month1 ==12:
return year1+1,1,1
else:
return year1, month1 +1 , 1
else:
if day1 < daysOfMonths[month1-1]:
return year1, month1, day1+1
else:
if month1 ==12:
return year1+1,1,1
else:
return year1, month1 +1 , 1
else:
if day1 < daysOfMonths[month1-1]:
return year1, month1, day1+1
else:
if month1 ==12:
return year1+1,1,1
else:
return year1, month1 +1 , 1
def daysBetweenDates(y1, m1, d1, y2, m2, d2,end_day):
if y1 > y2:
m1,m2 = m2,m1
y1,y2 = y2,y1
d1,d2 = d2,d1
days=0
while(not(m1==m2 and y1==y2 and d1==d2)):
y1,m1,d1 = Count_Days(y1,m1,d1)
days+=1
if end_day:
days+=1
return days
# Test Case
def test():
test_cases = [((2012,1,1,2012,2,28,False), 58),
((2012,1,1,2012,3,1,False), 60),
((2011,6,30,2012,6,30,False), 366),
((2011,1,1,2012,8,8,False), 585 ),
((1994,5,15,2019,8,31,False), 9239),
((1999,3,24,2018,2,4,False), 6892),
((1999,6,24,2018,8,4,False),6981),
((1995,5,24,2018,12,15,False),8606),
((1994,8,24,2019,12,15,True),9245),
((2019,12,15,1994,8,24,True),9245),
((2019,5,15,1994,10,24,True),8970),
((1994,11,24,2019,8,15,True),9031)]
for (args, answer) in test_cases:
result = daysBetweenDates(*args)
if result != answer:
print "Test with data:", args, "failed"
else:
print "Test case passed!"
test()
Pour calculer les dates et les heures, il y a plusieurs options, mais j'écrirai simplement:
import datetime
import dateutil.relativedelta
# current time
date_and_time = datetime.datetime.now()
date_only = date.today()
time_only = datetime.datetime.now().time()
# calculate date and time
result = date_and_time - datetime.timedelta(hours=26, minutes=25, seconds=10)
# calculate dates: years (-/+)
result = date_only - dateutil.relativedelta.relativedelta(years=10)
# months
result = date_only - dateutil.relativedelta.relativedelta(months=10)
# days
result = date_only - dateutil.relativedelta.relativedelta(days=10)
# calculate time
result = date_and_time - datetime.timedelta(hours=26, minutes=25, seconds=10)
result.time()
J'espère que ça aide
from datetime import date
def d(s):
[month, day, year] = map(int, s.split('/'))
return date(year, month, day)
def days(start, end):
return (d(end) - d(start)).days
print days('8/18/2008', '9/26/2008')
Cela suppose, bien entendu, que vous ayez déjà vérifié que vos dates sont au format r'\d+/\d+/\d+'.
Voici trois solutions pour résoudre ce problème:
from datetime import datetime
Now = datetime.now()
StartDate = datetime.strptime(str(Now.year) +'-01-01', '%Y-%m-%d')
NumberOfDays = (Now - StartDate)
print(NumberOfDays.days) # Starts at 0
print(datetime.now().timetuple().tm_yday) # Starts at 1
print(Now.strftime('%j')) # Starts at 1
Il y a aussi une méthode datetime.toordinal() qui n'a pas encore été mentionnée:
import datetime
print(datetime.date(2008,9,26).toordinal() - datetime.date(2008,8,18).toordinal()) # 39
https://docs.python.org/3/library/datetime.html#datetime.date.toordinal
date.toordinal ()Renvoie l'ordinal proleptique grégorien de la date, où le 1 er janvier de l'année 1 a l'ordinal 1. Pour tout
dateobjet d ,date.fromordinal(d.toordinal()) == d.
Semble bien convenir au calcul de la différence de jours, bien que moins lisible que timedelta.days.