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

一个带有datetime和dateutil的月日期范围生成器。简单易懂的:

import datetime as dt
from dateutil.relativedelta import relativedelta

def month_range(start_date, n_months):
        for m in range(n_months):
            yield start_date + relativedelta(months=+m)

其他回答

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。

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))
        )
import datetime    
def date_generator():
    cur = base = datetime.date.today()
    end  = base + datetime.timedelta(days=100)
    delta = datetime.timedelta(days=1)
    while(end>base):
        base = base+delta
        print base

date_generator()

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”。

如果有两个日期,你需要范围试试

from dateutil import rrule, parser
date1 = '1995-01-01'
date2 = '1995-02-28'
datesx = list(rrule.rrule(rrule.DAILY, dtstart=parser.parse(date1), until=parser.parse(date2)))