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
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”。