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
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。
其他回答
这个怎么样?没有使用其他库(dateutil)或timedelta? 基于vartec的回答,我这样做了,我相信它是有效的:
import datetime
today = datetime.date.today()
six_months_from_today = datetime.date(today.year + (today.month + 6)/12, (today.month + 6) % 12, today.day)
我尝试使用timedelta,但因为它是计算天数的,365/2或6*356/12并不总是转换为6个月,而是182天。如。
day = datetime.date(2015, 3, 10)
print day
>>> 2015-03-10
print (day + datetime.timedelta(6*365/12))
>>> 2015-09-08
我相信我们通常会假设某一天的6个月将在6个月后的同一天登陆(即2015-03-10—> 2015-09-10,而不是2015-09-08)
我希望这对你有帮助。
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))
user417751之前回答的重做。也许不是那么python的方式,但它照顾到不同的月份长度和闰年。在本例中,2012年1月31日+ 1个月= 2012年2月29日。
import datetime
import calendar
def add_mths(d, x):
newday = d.day
newmonth = (((d.month - 1) + x) % 12) + 1
newyear = d.year + (((d.month - 1) + x) // 12)
if newday > calendar.mdays[newmonth]:
newday = calendar.mdays[newmonth]
if newyear % 4 == 0 and newmonth == 2:
newday += 1
return datetime.date(newyear, newmonth, newday)
假设你的datetime变量叫做date:
date=datetime.datetime(year=date.year+int((date.month+6)/12),
month=(date.month+6)%13 + (1 if (date.month +
months>12) else 0), day=date.day)
我经常需要一个月的最后一天来保持上个月的最后一天。为了解决这个问题,我在计算前加一天,然后在返回前再减去它。
from datetime import date, timedelta
# it's a lot faster with a constant day
DAY = timedelta(1)
def add_month(a_date, months):
"Add months to date and retain last day in month."
next_day = a_date + DAY
# calculate new year and month
m_sum = next_day.month + months - 1
y = next_day.year + m_sum // 12
m = m_sum % 12 + 1
try:
return date(y, m, next_day.day) - DAY
except ValueError:
# on fail return last day in month
# can't fail on december so I don't bother changing the year
return date(y, m + 1, 1) - DAY