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)

稍微更好……

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

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”。

一个泛型方法，允许在参数化窗口大小(天，分钟，小时，秒)上创建日期范围:

from datetime import datetime, timedelta

def create_date_ranges(start, end, **interval):
    start_ = start
    while start_ < end:
        end_ = start_ + timedelta(**interval)
        yield (start_, min(end_, end))
        start_ = end_

测试:

def main():
    tests = [
        ('2021-11-15:00:00:00', '2021-11-17:13:00:00', {'days': 1}),
        ('2021-11-15:00:00:00', '2021-11-16:13:00:00', {'hours': 12}),
        ('2021-11-15:00:00:00', '2021-11-15:01:45:00', {'minutes': 30}),
        ('2021-11-15:00:00:00', '2021-11-15:00:01:12', {'seconds': 30})
    ]
    for t in tests:
        print("\nInterval: %s, range(%s to %s)" % (t[2], t[0], t[1]))
        start = datetime.strptime(t[0], '%Y-%m-%d:%H:%M:%S')
        end =  datetime.strptime(t[1], '%Y-%m-%d:%H:%M:%S')
        ranges = list(create_date_ranges(start, end, **t[2]))        
        x = list(map(
            lambda x: (x[0].strftime('%Y-%m-%d:%H:%M:%S'), x[1].strftime('%Y-%m-%d:%H:%M:%S')),
            ranges
        ))
        print(x)
main()

测试输出:

Interval: {'days': 1}, range(2021-11-15:00:00:00 to 2021-11-17:13:00:00)
[('2021-11-15:00:00:00', '2021-11-16:00:00:00'), ('2021-11-16:00:00:00', '2021-11-17:00:00:00'), ('2021-11-17:00:00:00', '2021-11-17:13:00:00')]

Interval: {'hours': 12}, range(2021-11-15:00:00:00 to 2021-11-16:13:00:00)
[('2021-11-15:00:00:00', '2021-11-15:12:00:00'), ('2021-11-15:12:00:00', '2021-11-16:00:00:00'), ('2021-11-16:00:00:00', '2021-11-16:12:00:00'), ('2021-11-16:12:00:00', '2021-11-16:13:00:00')]

Interval: {'minutes': 30}, range(2021-11-15:00:00:00 to 2021-11-15:01:45:00)
[('2021-11-15:00:00:00', '2021-11-15:00:30:00'), ('2021-11-15:00:30:00', '2021-11-15:01:00:00'), ('2021-11-15:01:00:00', '2021-11-15:01:30:00'), ('2021-11-15:01:30:00', '2021-11-15:01:45:00')]

Interval: {'seconds': 30}, range(2021-11-15:00:00:00 to 2021-11-15:00:01:12)
[('2021-11-15:00:00:00', '2021-11-15:00:00:30'), ('2021-11-15:00:00:30', '2021-11-15:00:01:00'), ('2021-11-15:00:01:00', '2021-11-15:00:01:12')]

你可以写一个生成器函数，返回从今天开始的日期对象:

import datetime

def date_generator():
  from_date = datetime.datetime.today()
  while True:
    yield from_date
    from_date = from_date - datetime.timedelta(days=1)

这个生成器返回从今天开始的日期，一次返回一天。以下是前3次约会的方法:

>>> import itertools
>>> dates = itertools.islice(date_generator(), 3)
>>> list(dates)
[datetime.datetime(2009, 6, 14, 19, 12, 21, 703890), datetime.datetime(2009, 6, 13, 19, 12, 21, 703890), datetime.datetime(2009, 6, 12, 19, 12, 21, 703890)]

与循环或列表推导相比，这种方法的优点是可以返回任意多次。

Edit

使用生成器表达式代替函数的更紧凑的版本:

date_generator = (datetime.datetime.today() - datetime.timedelta(days=i) for i in itertools.count())

用法:

>>> dates = itertools.islice(date_generator, 3)
>>> list(dates)
[datetime.datetime(2009, 6, 15, 1, 32, 37, 286765), datetime.datetime(2009, 6, 14, 1, 32, 37, 286836), datetime.datetime(2009, 6, 13, 1, 32, 37, 286859)]