是的，重新发明轮子.... 只要搜索论坛，你就会得到这样的东西:

from dateutil import rrule
from datetime import datetime

list(rrule.rrule(rrule.DAILY,count=100,dtstart=datetime.now()))

根据我自己的回答:

import datetime;
print [(datetime.date.today() - datetime.timedelta(days=x)).strftime('%Y-%m-%d') for x in range(-5, 0)]

输出:

['2017-12-11', '2017-12-10', '2017-12-09', '2017-12-08', '2017-12-07']

区别在于我得到的是'date'对象，而不是'datetime '对象。datetime”。

获取指定的开始日期和结束日期之间的日期范围(优化的时间和空间复杂性):

import datetime

start = datetime.datetime.strptime("21-06-2014", "%d-%m-%Y")
end = datetime.datetime.strptime("07-07-2014", "%d-%m-%Y")
date_generated = [start + datetime.timedelta(days=x) for x in range(0, (end-start).days)]

for date in date_generated:
    print date.strftime("%d-%m-%Y")

下面是一个稍微不同的答案，基于S.Lott的答案，给出了两个日期开始和结束之间的日期列表。在下面的例子中，从2017年初到今天。

start = datetime.datetime(2017,1,1)
end = datetime.datetime.today()
daterange = [start + datetime.timedelta(days=x) for x in range(0, (end-start).days)]

稍微更好……

base = datetime.datetime.today()
date_list = [base - datetime.timedelta(days=x) for x in range(numdays)]

另一个向前或向后计数的例子，从桑迪普的回答开始。

from datetime import date, datetime, timedelta
from typing import Sequence
def range_of_dates(start_of_range: date, end_of_range: date) -> Sequence[date]:

    if start_of_range <= end_of_range:
        return [
            start_of_range + timedelta(days=x)
            for x in range(0, (end_of_range - start_of_range).days + 1)
        ]
    return [
        start_of_range - timedelta(days=x)
        for x in range(0, (start_of_range - end_of_range).days + 1)
    ]

start_of_range = datetime.today().date()
end_of_range = start_of_range + timedelta(days=3)
date_range = range_of_dates(start_of_range, end_of_range)
print(date_range)

给了

[datetime.date(2019, 12, 20), datetime.date(2019, 12, 21), datetime.date(2019, 12, 22), datetime.date(2019, 12, 23)]

and

start_of_range = datetime.today().date()
end_of_range = start_of_range - timedelta(days=3)
date_range = range_of_dates(start_of_range, end_of_range)
print(date_range)

给了

[datetime.date(2019, 12, 20), datetime.date(2019, 12, 19), datetime.date(2019, 12, 18), datetime.date(2019, 12, 17)]

请注意，开始日期包含在返回中，因此如果需要四个总日期，请使用timedelta(days=3)