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
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()
其他回答
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
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)
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
获取指定的开始日期和结束日期之间的日期范围(优化的时间和空间复杂性):
import datetime
start = datetime.datetime.strptime("21-06-2014", "%d-%m-%Y")
end = datetime.datetime.strptime("07-07-2014", "%d-%m-%Y")
date_generated = [start + datetime.timedelta(days=x) for x in range(0, (end-start).days)]
for date in date_generated:
print date.strftime("%d-%m-%Y")
你可以写一个生成器函数,返回从今天开始的日期对象:
import datetime
def date_generator():
from_date = datetime.datetime.today()
while True:
yield from_date
from_date = from_date - datetime.timedelta(days=1)
这个生成器返回从今天开始的日期,一次返回一天。以下是前3次约会的方法:
>>> import itertools
>>> dates = itertools.islice(date_generator(), 3)
>>> list(dates)
[datetime.datetime(2009, 6, 14, 19, 12, 21, 703890), datetime.datetime(2009, 6, 13, 19, 12, 21, 703890), datetime.datetime(2009, 6, 12, 19, 12, 21, 703890)]
与循环或列表推导相比,这种方法的优点是可以返回任意多次。
Edit
使用生成器表达式代替函数的更紧凑的版本:
date_generator = (datetime.datetime.today() - datetime.timedelta(days=i) for i in itertools.count())
用法:
>>> dates = itertools.islice(date_generator, 3)
>>> list(dates)
[datetime.datetime(2009, 6, 15, 1, 32, 37, 286765), datetime.datetime(2009, 6, 14, 1, 32, 37, 286836), datetime.datetime(2009, 6, 13, 1, 32, 37, 286859)]