Como calcular o número de dias entre duas datas determinadas?

Question

Se eu tiver duas datas (ex. '8/18/2008' e '9/26/2008'), qual é a melhor maneira de obter o número de dias entre as duas datas?

Answer 1

Se você tem dois objetos de data, você pode apenas subtrair-los.

from datetime import date

d0 = date(2008, 8, 18)
d1 = date(2008, 9, 26)
delta = d1 - d0
print delta.days

A seção relevante dos docs: https://docs.python.org/library/datetime.html .

Answer 2

Usando o poder de data e hora:

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

Answer 3

Dias até o Natal:

>>> import datetime
>>> today = datetime.date.today()
>>> someday = datetime.date(2008, 12, 25)
>>> diff = someday - today
>>> diff.days
86

Mais aritmética aqui .

Answer 4

Você quer que o módulo datetime.

>>> from datetime import datetime, timedelta 
>>> datetime(2008,08,18) - datetime(2008,09,26) 
datetime.timedelta(4) 

Ou outro exemplo:

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 

Como apontado aqui

Answer 5

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)

Answer 6

Ele também pode ser feito facilmente com arrow:

import arrow

a = arrow.get('2017-05-09')
b = arrow.get('2017-05-11')

delta = (b-a)
print delta.days

Para referência: http://arrow.readthedocs.io/en/latest/

Answer 7

sem usar Lib código apenas puro:

#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()

Answer 8

Para datas e horários cálculo existem várias opções, mas eu vou escrever a maneira simples:

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()

Hope isso ajuda

Answer 9

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')

Isso pressupõe, naturalmente, que você já tenha verificado que as datas estão no formato r'\d+/\d+/\d+'.

Answer 10

Aqui estão três maneiras de ir com este problema:

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

Answer 11

Há também um método datetime.toordinal() que ainda não foi mencionado:

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 ()

Voltar a ordinal gregoriano proleptic da data, onde 01 de janeiro do ano 1 tem ordinal 1. Para qualquer objeto date d , date.fromordinal(d.toordinal()) == d.

Parece bem adequado para calcular dias diferença, embora não tão legível como timedelta.days.