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
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”。
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))
)
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)
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()
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)))