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
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)
其他回答
稍微更好……
base = datetime.datetime.today()
date_list = [base - datetime.timedelta(days=x) for x in range(numdays)]
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
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”。
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()))
你也可以使用日期序数来简化:
def date_range(start_date, end_date):
for ordinal in range(start_date.toordinal(), end_date.toordinal()):
yield datetime.date.fromordinal(ordinal)
或者像评论中建议的那样,你可以创建一个这样的列表:
date_range = [
datetime.date.fromordinal(ordinal)
for ordinal in range(
start_date.toordinal(),
end_date.toordinal(),
)
]