不使用Lib只是纯代码:

#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()

还有一个尚未提到的datetime.toordinal()方法:

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 () 返回日期的预期格里高利历序数，其中第一年的1月1日序数为1。对于任意日期对象d， Date.fromordinal (d.toordinal()) == d。

似乎很适合计算日差，但可读性不如timedelta.days。

它也可以很容易地用箭头完成:

import arrow

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

delta = (b-a)
print delta.days

参考:http://arrow.readthedocs.io/en/latest/

对于计算日期和时间，有几个选项，但我将写简单的方式:

from datetime import timedelta, datetime, date
import dateutil.relativedelta

# current time
date_and_time = datetime.now()
date_only = date.today()
time_only = datetime.now().time()

# calculate date and time
result = date_and_time - 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)

# week
results = date_only - dateutil.relativedelta.relativedelta(weeks=1)

# days
result = date_only - dateutil.relativedelta.relativedelta(days=10)

# calculate time 
result = date_and_time - timedelta(hours=26, minutes=25, seconds=10)
result.time()

希望能有所帮助

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')

当然，这假设您已经验证了日期的格式为r'\d+/\d+/\d+'。

每个人都用日期回答得很好， 让我用熊猫来回答这个问题

dt = pd.to_datetime('2008/08/18', format='%Y/%m/%d')
dt1 = pd.to_datetime('2008/09/26', format='%Y/%m/%d')

(dt1-dt).days

这就会给出答案。 如果其中一个输入是dataframe列。简单地使用dt。日子换成了日子

(dt1-dt).dt.days