import time
import datetime

# use mktime to step by one day
# end - the last day, numdays - count of days to step back
def gen_dates_list(end, numdays):
  start = end - datetime.timedelta(days=numdays+1)
  end   = int(time.mktime(end.timetuple()))
  start = int(time.mktime(start.timetuple()))
  # 86400 s = 1 day
  return xrange(start, end, 86400)

# if you need reverse the list of dates
for dt in reversed(gen_dates_list(datetime.datetime.today(), 100)):
    print datetime.datetime.fromtimestamp(dt).date()

使用datetime.datetime.date()方法:

datetime.datetime.now().date()

显然，上面的表达式可以(恕我直言)写成:

datetime.date.today()

您可以使用datetime对象的date()方法将datetime对象转换为日期，如下所示:

<datetime_object>.date()

如果你正在使用熊猫，那么这可以解决你的问题:

假设你在你的数据帧中有一个名为start_time的datetime64类型的变量，然后你可以像这样得到日期部分:

df.start_time.dt.date

答案已更新到Python 3.7及更高版本

下面介绍如何转动日期和时间对象

(又名datetime。Datetime对象，即存储在模型中的对象。django模型字段

变成一个日期对象(又名datetime。约会对象):

from datetime import datetime

#your date-and-time object
# let's supposed it is defined as
datetime_element = datetime(2020, 7, 10, 12, 56, 54, 324893)

# where
# datetime_element = datetime(year, month, day, hour, minute, second, milliseconds)

# WHAT YOU WANT: your date-only object
date_element = datetime_element.date()

为了更清楚，如果你打印这些元素，这是输出:

print(datetime_element)

2020-07-10 12:56:54.324893

print(date_element)

2020-07-10

使用date()方法:

datetime.datetime.now().date()