我想创建一个日期列表,从今天开始,追溯到任意天数,例如,在我的示例中是100天。还有比这更好的办法吗?

import datetime

a = datetime.datetime.today()
numdays = 100
dateList = []
for x in range (0, numdays):
    dateList.append(a - datetime.timedelta(days = x))
print dateList

当前回答

from datetime import datetime , timedelta, timezone


start_date = '2022_01_25'
end_date = '2022_01_30'

start = datetime.strptime(start_date, "%Y_%m_%d")
print(type(start))
end =  datetime.strptime(end_date, "%Y_%m_%d")
##pDate = str(pDate).replace('-', '_')
number_of_days = (end - start).days

print("number_of_days: ", number_of_days)

##
date_list = []
for day in range(number_of_days):
    a_date = (start + timedelta(days = day)).astimezone(timezone.utc)
    a_date = a_date.strftime('%Y-%m-%d')
    date_list.append(a_date)

print(date_list)

其他回答

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)]
from datetime import datetime, timedelta
from dateutil import parser
def getDateRange(begin, end):
    """  """
    beginDate = parser.parse(begin)
    endDate =  parser.parse(end)
    delta = endDate-beginDate
    numdays = delta.days + 1
    dayList = [datetime.strftime(beginDate + timedelta(days=x), '%Y%m%d') for x in range(0, numdays)]
    return dayList

我想用一个简单(不完整)的日期范围实现来发表我的意见:

from datetime import date, timedelta, datetime

class DateRange:
    def __init__(self, start, end, step=timedelta(1)):
        self.start = start
        self.end = end
        self.step = step

    def __iter__(self):
        start = self.start
        step = self.step
        end = self.end

        n = int((end - start) / step)
        d = start

        for _ in range(n):
            yield d
            d += step

    def __contains__(self, value):
        return (
            (self.start <= value < self.end) and 
            ((value - self.start) % self.step == timedelta(0))
        )

我知道这个问题已经有人回答了,但为了历史的目的,我还是把我的答案写下来,因为我认为这是直截了当的。

import numpy as np
import datetime as dt
listOfDates=[date for date in np.arange(firstDate,lastDate,dt.timedelta(days=x))]

当然,它不会像代码高尔夫那样赢得任何东西,但我认为它很优雅。

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”。