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
按月开始计算:
from datetime import timedelta
from dateutil.relativedelta import relativedelta
end_date = start_date + relativedelta(months=delta_period) + timedelta(days=-delta_period)
其他回答
按月开始计算:
from datetime import timedelta
from dateutil.relativedelta import relativedelta
end_date = start_date + relativedelta(months=delta_period) + timedelta(days=-delta_period)
import datetime
'''
Created on 2011-03-09
@author: tonydiep
'''
def add_business_months(start_date, months_to_add):
"""
Add months in the way business people think of months.
Jan 31, 2011 + 1 month = Feb 28, 2011 to business people
Method: Add the number of months, roll back the date until it becomes a valid date
"""
# determine year
years_change = months_to_add / 12
# determine if there is carryover from adding months
if (start_date.month + (months_to_add % 12) > 12 ):
years_change = years_change + 1
new_year = start_date.year + years_change
# determine month
work = months_to_add % 12
if 0 == work:
new_month = start_date.month
else:
new_month = (start_date.month + (work % 12)) % 12
if 0 == new_month:
new_month = 12
# determine day of the month
new_day = start_date.day
if(new_day in [31, 30, 29, 28]):
#user means end of the month
new_day = 31
new_date = None
while (None == new_date and 27 < new_day):
try:
new_date = start_date.replace(year=new_year, month=new_month, day=new_day)
except:
new_day = new_day - 1 #wind down until we get to a valid date
return new_date
if __name__ == '__main__':
#tests
dates = [datetime.date(2011, 1, 31),
datetime.date(2011, 2, 28),
datetime.date(2011, 3, 28),
datetime.date(2011, 4, 28),
datetime.date(2011, 5, 28),
datetime.date(2011, 6, 28),
datetime.date(2011, 7, 28),
datetime.date(2011, 8, 28),
datetime.date(2011, 9, 28),
datetime.date(2011, 10, 28),
datetime.date(2011, 11, 28),
datetime.date(2011, 12, 28),
]
months = range(1, 24)
for start_date in dates:
for m in months:
end_date = add_business_months(start_date, m)
print("%s\t%s\t%s" %(start_date, end_date, m))
在这个函数中,n可以是正数,也可以是负数。
def addmonth(d, n):
n += 1
dd = datetime.date(d.year + n/12, d.month + n%12, 1)-datetime.timedelta(1)
return datetime.date(dd.year, dd.month, min(d.day, dd.day))
我使用replace()方法并编写了这个递归函数。dt是一个日期时间。datetime对象:
def month_timedelta(dt, m):
y = m // 12
dm = m % 12
if y == 0:
if dt.month + m <= 12:
return dt.replace(month = dt.month + m)
else:
dy = (dt.month + m) // 12
ndt = dt.replace(year=dt.year + dy)
return ndt.replace(month=(ndt.month + m) % 12)
else:
return month_timedelta(dt.replace(year=dt.year + y),dm)
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。