如何使用datetime Python模块计算当前日期六个月的日期？-Java 学习之路

261

我正在使用datetime Python模块 . 我想计算从当前日期起6个月的日期 . 有人可以给我一点帮助吗？

我想要从当前日期起6个月生成日期的原因是生成审核日期 . 如果用户将数据输入系统，则其审核日期为自输入数据之日起6个月 .

30 回答

import datetime
print (datetime.date.today() + datetime.timedelta(6*365/12)).isoformat()

回复于 2024-04-18T09:38:11+08:00

我发现这个解决方案很好 . （这使用python-dateutil extension）

from datetime import date
from dateutil.relativedelta import relativedelta

six_months = date.today() + relativedelta(months=+6)

这种方法的优点是它可以处理28,30,31天等问题 . 这在处理业务规则和方案（比如发票生成等）时非常有用 .

$ date(2010,12,31)+relativedelta(months=+1)
  datetime.date(2011, 1, 31)

$ date(2010,12,31)+relativedelta(months=+2)
  datetime.date(2011, 2, 28)

回复于 2024-04-18T09:38:11+08:00

2
那么，这取决于您从当前日期起6个月的含义 .
- 使用自然月份：
```
(day, month, year) = (day, (month+6)%12, year+(month+6)/12)
```
- 使用银行家的定义，6 * 30：
```
date += datetime.timedelta(6*30)
```
回复于 2024-04-18T09:38:11+08:00

对于月初计算的开始：

from datetime import timedelta
from dateutil.relativedelta import relativedelta

end_date = start_date + relativedelta(months=delta_period) + timedelta(days=-delta_period)

回复于 2024-04-18T09:38:11+08:00

你是什么意思'6个月' . 是2009-02-13 6个月== 2009-08-13还是2009-02-13 6 * 30天？

import mx.DateTime as dt

#6 Months
dt.now()+dt.RelativeDateTime(months=6)
#result is '2009-08-13 16:28:00.84'

#6*30 days
dt.now()+dt.RelativeDateTime(days=30*6)
#result is '2009-08-12 16:30:03.35'

有关更多信息mx.DateTime

回复于 2024-04-18T09:38:11+08:00

此解决方案适用于12月，此页面上的大多数答案都没有 . 在使用模数（％）或整数除法（//）之前，您需要首先将月份从基数1（即Jan = 1）转移到基数0（即Jan = 0），否则11月（11）加1个月给出12 ，当找到余数（12％12）给出0 .

（并且不建议“（月％12）1”或10月1日= 12月！）

def AddMonths(d,x):
    newmonth = ((( d.month - 1) + x ) % 12 ) + 1
    newyear  = d.year + ((( d.month - 1) + x ) / 12 ) 
    return datetime.date( newyear, newmonth, d.day)

但是......这并不像1月31日那样解决问题 . 所以我们回到OP - 添加一个月是什么意思？一个解决方案是回溯直到你到达有效的一天，因为大多数人会认为jan的最后一天，加上一个月，等于2月的最后一天 . 这也将在负数月份工作 . 证明：

>>> import datetime
>>> AddMonths(datetime.datetime(2010,8,25),1)
datetime.date(2010, 9, 25)
>>> AddMonths(datetime.datetime(2010,8,25),4)
datetime.date(2010, 12, 25)
>>> AddMonths(datetime.datetime(2010,8,25),5)
datetime.date(2011, 1, 25)
>>> AddMonths(datetime.datetime(2010,8,25),13)
datetime.date(2011, 9, 25)
>>> AddMonths(datetime.datetime(2010,8,25),24)
datetime.date(2012, 8, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-1)
datetime.date(2010, 7, 25)
>>> AddMonths(datetime.datetime(2010,8,25),0)
datetime.date(2010, 8, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-12)
datetime.date(2009, 8, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-8)
datetime.date(2009, 12, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-7)
datetime.date(2010, 1, 25)>>>

回复于 2024-04-18T09:38:11+08:00

2
我知道这已经有6个月了，但是如果你要添加一个月，谷歌的答案显示“在python中增加几个月”：
```
import calendar

date = datetime.date.today()    //Or your date

datetime.timedelta(days=calendar.monthrange(date.year,date.month)[1])
```
这将计算当前月份的天数，并将它们添加到当前日期，使用365/12将一年中的1/12可能会导致短期/长期的问题，如果您的迭代日期 .
回复于 2024-04-18T09:38:11+08:00
6

使用Python的日期时间没有直接的方法 .

查看python-dateutil处的relativedelta类型 . 它允许您指定以月为单位的时间增量 .

回复于 2024-04-18T09:38:11+08:00

所以，这里有一个 dateutil.relativedelta 的例子，我发现它可以用于迭代过去的一年，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，你必须在“6个月后”找出你的真正含义 . 如果你的意思是“未来六个月的今天这个月”，那么这样做：

datetime.datetime.now() + relativedelta(months=6)

回复于 2024-04-18T09:38:11+08:00

0

Dateutil package已实现此类功能 . 但请注意，这将是天真的，正如其他人已经指出的那样 .

回复于 2024-04-18T09:38:11+08:00
0
只需使用timetuple方法提取月份，添加月份并构建新的日期对象 . 如果有一个已经存在的方法，我不知道 .
```
import datetime

def in_the_future(months=1):
    year, month, day = datetime.date.today().timetuple()[:3]
    new_month = month + months
    return datetime.date(year + (new_month / 12), new_month % 12, day)
```
API有点笨拙，但作为一个例子 . 显然也不会像2008-01-31 1月那样在角落案件上工作 . :)
回复于 2024-04-18T09:38:11+08:00

使用Python标准库，即没有 dateutil 或其他，并解决'February 31st'问题：

import datetime
import calendar

def add_months(date, months):
    months_count = date.month + months

    # Calculate the year
    year = date.year + int(months_count / 12)

    # Calculate the month
    month = (months_count % 12)
    if month == 0:
        month = 12

    # Calculate the day
    day = date.day
    last_day_of_month = calendar.monthrange(year, month)[1]
    if day > last_day_of_month:
        day = last_day_of_month

    new_date = datetime.date(year, month, day)
    return new_date

测试：

>>>date = datetime.date(2018, 11, 30)

>>>print(date, add_months(date, 3))
(datetime.date(2018, 11, 30), datetime.date(2019, 2, 28))

>>>print(date, add_months(date, 14))
(datetime.date(2018, 12, 31), datetime.date(2020, 2, 29))

回复于 2024-04-18T09:38:11+08:00

使用Python 3.x，您可以这样做：

from datetime import datetime, timedelta
from dateutil.relativedelta import *

date = datetime.now()
print(date)
# 2018-09-24 13:24:04.007620

date = date + relativedelta(months=+6)
print(date)
# 2019-03-24 13:24:04.007620

但是你需要安装 python-dateutil 模块：

pip install python-dateutil

回复于 2024-04-18T09:38:11+08:00

我有更好的方法来解决'2月31日'的问题：

def add_months(start_date, months):
    import calendar

    year = start_date.year + (months / 12)
    month = start_date.month + (months % 12)
    day = start_date.day

    if month > 12:
        month = month % 12
        year = year + 1

    days_next = calendar.monthrange(year, month)[1]
    if day > days_next:
        day = days_next

    return start_date.replace(year, month, day)

我认为它也适用于负数（减去几个月），但我没有对此进行过多次测试 .

回复于 2024-04-18T09:38:11+08:00

PyQt4的QDate类具有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)

回复于 2024-04-18T09:38:11+08:00

修改AddMonths（）以在Zope中使用并处理无效的日期数：

def AddMonths(d,x):
    days_of_month = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    newmonth = ((( d.month() - 1) + x ) % 12 ) + 1
    newyear  = d.year() + ((( d.month() - 1) + x ) // 12 ) 
    if d.day() > days_of_month[newmonth-1]:
      newday = days_of_month[newmonth-1]
    else:
      newday = d.day() 
    return DateTime( newyear, newmonth, newday)

回复于 2024-04-18T09:38:11+08:00

import time

def add_month(start_time, months):  

        ret = time.strptime(start_time, '%Y-%m-%d')
        t = list(ret)

        t[1] += months

        if t[1] > 12:
            t[0] += 1 + int(months / 12)

            t[1] %= 12

        return int(time.mktime(tuple(t)))

回复于 2024-04-18T09:38:11+08:00

1
这个怎么样？不使用其他库（ dateutil ）或 timedelta ？ Build 在vartec的答案上，我这样做了，我相信它有效：
```
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 ，但因为它计算的是日期， 365/2 或 6*356/12 并不总是转换为6个月，而是182天 . 例如
```
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从某一天开始的6个月将在该月的同一天降落但是在6个月之后（即 2015-03-10 - > 2015-09-10 ，不是 2015-09-08 ）

我希望你觉得这有帮助 .
回复于 2024-04-18T09:38:11+08:00

765

我这样解决了这个问题：

import calendar
from datetime import datetime
moths2add = 6
now = datetime.now()
current_year = now.year
current_month = now.month
#count days in months you want to add using calendar module
days = sum(
  [calendar.monthrange(current_year, elem)[1] for elem in range(current_month, current_month + moths)]
    )
print now + days

回复于 2024-04-18T09:38:11+08:00

在1new_month = 121的案例中修改了Johannes Wei的答案 . 这对我来说非常有效 . 这几个月可能是积极的或消极的 .

def addMonth(d,months=1):
    year, month, day = d.timetuple()[:3]
    new_month = month + months
    return datetime.date(year + ((new_month-1) / 12), (new_month-1) % 12 +1, day)

回复于 2024-04-18T09:38:11+08:00

还有另一种解决方案 - 希望有人会喜欢它：

def add_months(d, months):
    return d.replace(year=d.year+months//12).replace(month=(d.month+months)%12)

对于所有情况，此解决方案在第29,30,31天都不起作用，因此需要更强大的解决方案（这不再那么好了:)）：

def add_months(d, months):
    for i in range(4):
        day = d.day - i
        try:
            return d.replace(day=day).replace(year=d.year+int(months)//12).replace(month=(d.month+int(months))%12)
        except:
            pass
    raise Exception("should not happen")

回复于 2024-04-18T09:38:11+08:00

从this answer，请参阅parsedatetime . 代码示例如下 . 更多细节：unit test with many natural-language -> YYYY-MM-DD conversion examples，明显parsedatetime conversion challenges/bugs .

#!/usr/bin/env python
# -*- coding: utf-8 -*-
import time, calendar
from datetime import date

# from https://github.com/bear/parsedatetime
import parsedatetime as pdt

def print_todays_date():
    todays_day_of_week = calendar.day_name[date.today().weekday()]
    print "today's date = " + todays_day_of_week + ', ' + \
                              time.strftime('%Y-%m-%d')

def convert_date(natural_language_date):
    cal = pdt.Calendar()
    (struct_time_date, success) = cal.parse(natural_language_date)
    if success:
        formal_date = time.strftime('%Y-%m-%d', struct_time_date)
    else:
        formal_date = '(conversion failed)'
    print '{0:12s} -> {1:10s}'.format(natural_language_date, formal_date)

print_todays_date()
convert_date('6 months')

该上面的代码从MacOSX机器生成以下内容：

$ ./parsedatetime_simple.py 
today's date = Wednesday, 2015-05-13
6 months     -> 2015-11-13
$

回复于 2024-04-18T09:38:11+08:00

这是一个示例，允许用户决定如何返回日期大于该月份天数的日期 .

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)

回复于 2024-04-18T09:38:11+08:00

假设您的datetime变量名为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)

回复于 2024-04-18T09:38:11+08:00

在x个月之后/之前获得下一个日期的一般功能 .

from datetime import date

def after_month(given_date, month):
    yyyy = int(((given_date.year * 12 + given_date.month) + month)/12)
    mm = int(((given_date.year * 12 + given_date.month) + month)%12)

    if mm == 0:
        yyyy -= 1
        mm = 12
    return given_date.replace(year=yyyy, month=mm)


if __name__ == "__main__":
    today = date.today()
    print(today)

    for mm in [-12, -1, 0, 1, 2, 12, 20 ]:
        next_date = after_month(today, mm)
        print(next_date)

回复于 2024-04-18T09:38:11+08:00

0

使用python datetime模块向datetime.today（）添加六个月的timedelta .

http://docs.python.org/library/datetime.html

你当然要解决JohannesWeiß提出的问题 - 你的意思是6个月？

回复于 2024-04-18T09:38:11+08:00

这就是我提出的 . 它移动了正确的月数和年数，但忽略了几天（这是我在我的情况下所需要的） .

import datetime

month_dt = 4
today = datetime.date.today()
y,m = today.year, today.month
m += month_dt-1
year_dt = m//12
new_month = m%12
new_date = datetime.date(y+year_dt, new_month+1, 1)

回复于 2024-04-18T09:38:11+08:00

我用这个函数来改变年份和月份但是保持一天：

def replace_month_year(date1, year2, month2):
    try:
        date2 = date1.replace(month = month2, year = year2)
    except:
        date2 = datetime.date(year2, month2 + 1, 1) - datetime.timedelta(days=1)
    return date2

你应该写：

new_year = my_date.year + (my_date.month + 6) / 12
new_month = (my_date.month + 6) % 12
new_date = replace_month_year(my_date, new_year, new_month)

回复于 2024-04-18T09:38:11+08:00

我认为做这样的事情而不是手动添加天数会更安全：

import datetime
today = datetime.date.today()

def addMonths(dt, months = 0):
    new_month = months + dt.month
    year_inc = 0
    if new_month>12:
        year_inc +=1
        new_month -=12
    return dt.replace(month = new_month, year = dt.year+year_inc)

newdate = addMonths(today, 6)

回复于 2024-04-18T09:38:11+08:00

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))

回复于 2024-04-18T09:38:11+08:00

如何使用datetime Python模块计算当前日期六个月的日期？

30 回答

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