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
稍微更好……
base = datetime.datetime.today()
date_list = [base - datetime.timedelta(days=x) for x in range(numdays)]
其他回答
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)
一个泛型方法,允许在参数化窗口大小(天,分钟,小时,秒)上创建日期范围:
from datetime import datetime, timedelta
def create_date_ranges(start, end, **interval):
start_ = start
while start_ < end:
end_ = start_ + timedelta(**interval)
yield (start_, min(end_, end))
start_ = end_
测试:
def main():
tests = [
('2021-11-15:00:00:00', '2021-11-17:13:00:00', {'days': 1}),
('2021-11-15:00:00:00', '2021-11-16:13:00:00', {'hours': 12}),
('2021-11-15:00:00:00', '2021-11-15:01:45:00', {'minutes': 30}),
('2021-11-15:00:00:00', '2021-11-15:00:01:12', {'seconds': 30})
]
for t in tests:
print("\nInterval: %s, range(%s to %s)" % (t[2], t[0], t[1]))
start = datetime.strptime(t[0], '%Y-%m-%d:%H:%M:%S')
end = datetime.strptime(t[1], '%Y-%m-%d:%H:%M:%S')
ranges = list(create_date_ranges(start, end, **t[2]))
x = list(map(
lambda x: (x[0].strftime('%Y-%m-%d:%H:%M:%S'), x[1].strftime('%Y-%m-%d:%H:%M:%S')),
ranges
))
print(x)
main()
测试输出:
Interval: {'days': 1}, range(2021-11-15:00:00:00 to 2021-11-17:13:00:00)
[('2021-11-15:00:00:00', '2021-11-16:00:00:00'), ('2021-11-16:00:00:00', '2021-11-17:00:00:00'), ('2021-11-17:00:00:00', '2021-11-17:13:00:00')]
Interval: {'hours': 12}, range(2021-11-15:00:00:00 to 2021-11-16:13:00:00)
[('2021-11-15:00:00:00', '2021-11-15:12:00:00'), ('2021-11-15:12:00:00', '2021-11-16:00:00:00'), ('2021-11-16:00:00:00', '2021-11-16:12:00:00'), ('2021-11-16:12:00:00', '2021-11-16:13:00:00')]
Interval: {'minutes': 30}, range(2021-11-15:00:00:00 to 2021-11-15:01:45:00)
[('2021-11-15:00:00:00', '2021-11-15:00:30:00'), ('2021-11-15:00:30:00', '2021-11-15:01:00:00'), ('2021-11-15:01:00:00', '2021-11-15:01:30:00'), ('2021-11-15:01:30:00', '2021-11-15:01:45:00')]
Interval: {'seconds': 30}, range(2021-11-15:00:00:00 to 2021-11-15:00:01:12)
[('2021-11-15:00:00:00', '2021-11-15:00:00:30'), ('2021-11-15:00:00:30', '2021-11-15:00:01:00'), ('2021-11-15:00:01:00', '2021-11-15:00:01:12')]
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))
)
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
以下是我从自己的代码中创建的要点,这可能会有所帮助。(我知道这个问题太老了,但其他人可以用)
https://gist.github.com/2287345
(下同)
import datetime
from time import mktime
def convert_date_to_datetime(date_object):
date_tuple = date_object.timetuple()
date_timestamp = mktime(date_tuple)
return datetime.datetime.fromtimestamp(date_timestamp)
def date_range(how_many=7):
for x in range(0, how_many):
some_date = datetime.datetime.today() - datetime.timedelta(days=x)
some_datetime = convert_date_to_datetime(some_date.date())
yield some_datetime
def pick_two_dates(how_many=7):
a = b = convert_date_to_datetime(datetime.datetime.now().date())
for each_date in date_range(how_many):
b = a
a = each_date
if a == b:
continue
yield b, a