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/

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)

圣诞节前几天:

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

这里有更多的算术。

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。

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+'。

使用datetime函数:

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