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

您需要datetime模块。

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

另一个例子:

>>> 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 

正如这里所指出的

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。

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

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

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。