我正在使用datetime Python模块。我希望从当前日期计算6个月的日期。有人能帮我一下吗?
我想从当前日期生成一个6个月后的日期的原因是为了生成一个回顾日期。如果用户在系统中输入数据,系统将有从输入数据之日起6个月的审查日期。
我正在使用datetime Python模块。我希望从当前日期计算6个月的日期。有人能帮我一下吗?
我想从当前日期生成一个6个月后的日期的原因是为了生成一个回顾日期。如果用户在系统中输入数据,系统将有从输入数据之日起6个月的审查日期。
当前回答
Dateutil包实现了这样的功能。但要知道,这将是天真的,因为其他人已经指出。
其他回答
我找不到这个问题的确切解决方案,所以我将发布我的解决方案,以防使用标准日历和datetime库可能有任何帮助。这适用于添加和减去月份,并考虑月末滚动和最后一个月比第一个月天数少的情况。如果你正在寻找更复杂的操作,我还有一个更通用的解决方案,它添加了定期间隔(天,月,年,季度,学期等),如:“1m”,“-9m”,“-1.5y”,“-3q”,“1s”等。
from datetime import datetime
from calendar import monthrange
def date_bump_months(start_date, months):
"""
bumps months back and forth.
--> if initial date is end-of-month, i will move to corresponding month-end
--> ir inital date.day is greater than end of month of final date, it casts it to momth-end
"""
signbit = -1 if months < 0 else 1
d_year, d_month = divmod(abs(months),12)
end_year = start_date.year + d_year*signbit
end_month = 0
if signbit ==-1:
if d_month < start_date.month:
end_month = start_date.month - d_month
else:
end_year -=1
end_month = 12 - (d_month - start_date.month)
else:
end_month +=start_date.month
if end_month > 12:
end_year +=1
end_month -=12
# check if we are running end-of-month dates
eom_run = monthrange(start_date.year, start_date.month)[1]==start_date.day
eom_month = monthrange((end_year), (end_month))[1]
if eom_run:
end_day = eom_month
else:
end_day = min(start_date.day, eom_month )
return date(end_year, end_month, end_day)
PyQt4的QDate类有一个addmonths函数。
>>>from PyQt4.QtCore import QDate
>>>dt = QDate(2009,12,31)
>>>required = dt.addMonths(6)
>>>required
PyQt4.QtCore.QDate(2010, 6, 30)
>>>required.toPyDate()
datetime.date(2010, 6, 30)
我对Tony Diep的答案的修改,可能稍微更优雅(当然,Python 2,匹配问题和原始答案的日期,对于Python 3,必要时修改,至少包括/ to //):
def add_months(date, months):
month = date.month + months - 1
year = date.year + (month / 12)
month = (month % 12) + 1
day = date.day
while (day > 0):
try:
new_date = date.replace(year=year, month=month, day=day)
break
except:
day = day - 1
return new_date
根据“业务需求”解释添加月份,即日期映射到月底之后,应该映射到月底,而不是下一个月。
这里有一个dateutil的例子。relativedelta,我发现它对于迭代过去一年很有用,每次跳过一个月到现在的日期:
>>> import datetime
>>> from dateutil.relativedelta import relativedelta
>>> today = datetime.datetime.today()
>>> month_count = 0
>>> while month_count < 12:
... day = today - relativedelta(months=month_count)
... print day
... month_count += 1
...
2010-07-07 10:51:45.187968
2010-06-07 10:51:45.187968
2010-05-07 10:51:45.187968
2010-04-07 10:51:45.187968
2010-03-07 10:51:45.187968
2010-02-07 10:51:45.187968
2010-01-07 10:51:45.187968
2009-12-07 10:51:45.187968
2009-11-07 10:51:45.187968
2009-10-07 10:51:45.187968
2009-09-07 10:51:45.187968
2009-08-07 10:51:45.187968
和其他答案一样,你必须弄清楚你说的“6个月后”到底是什么意思。如果你的意思是“六年后某个月的今天”,那么这个是:
datetime.datetime.now() + relativedelta(months=6)
下面是一个示例,它允许用户决定如何返回一个日期,其中一天大于一个月中的天数。
def add_months(date, months, endOfMonthBehaviour='RoundUp'):
assert endOfMonthBehaviour in ['RoundDown', 'RoundIn', 'RoundOut', 'RoundUp'], \
'Unknown end of month behaviour'
year = date.year + (date.month + months - 1) / 12
month = (date.month + months - 1) % 12 + 1
day = date.day
last = monthrange(year, month)[1]
if day > last:
if endOfMonthBehaviour == 'RoundDown' or \
endOfMonthBehaviour == 'RoundOut' and months < 0 or \
endOfMonthBehaviour == 'RoundIn' and months > 0:
day = last
elif endOfMonthBehaviour == 'RoundUp' or \
endOfMonthBehaviour == 'RoundOut' and months > 0 or \
endOfMonthBehaviour == 'RoundIn' and months < 0:
# we don't need to worry about incrementing the year
# because there will never be a day in December > 31
month += 1
day = 1
return datetime.date(year, month, day)
>>> from calendar import monthrange
>>> import datetime
>>> add_months(datetime.datetime(2016, 1, 31), 1)
datetime.date(2016, 3, 1)
>>> add_months(datetime.datetime(2016, 1, 31), -2)
datetime.date(2015, 12, 1)
>>> add_months(datetime.datetime(2016, 1, 31), -2, 'RoundDown')
datetime.date(2015, 11, 30)