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+'。

如果您有两个date对象，您可以将它们相减，这将计算一个timedelta对象。

from datetime import date

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

文档的相关部分: https://docs.python.org/library/datetime.html。

请看另一个例子。

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

以下是解决这个问题的三种方法:

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

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

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/