有没有一种简单的方法可以在 Python 中将 datetime 对象增加一个月？

因此，我试图找到一种方法来增加一个日期时间对象一个月。然而，根据 这个问题的说法，这似乎并不那么简单。

我希望是这样的:

import datetime as dt


now = dt.datetime.now()
later = now + dt.timedelta(months=1)

但这不管用。我也希望能够去同一天(或最接近的选择)在下个月，如果可能的话。例如，设置在1月1日的日期时间对象会增加到2月1日，而设置在2月28日的日期时间对象会增加到3月31日，而不是3月28日或其他时间。

需要说明的是，2月28日(通常)会映射到3月31日，因为这是一个月的最后一天，因此它应该到下个月的最后一天。否则它将是一个直接链接: 增量应该到下一个月的具有相同编号的日期的那一天。

在当前的 Python 版本中有没有一种简单的方法来实现这一点？

165164 次浏览
小开 
>>> now
datetime.datetime(2016, 1, 28, 18, 26, 12, 980861)
>>> later = now.replace(month=now.month+1)
>>> later
datetime.datetime(2016, 2, 28, 18, 26, 12, 980861)

EDIT: Fails on

y = datetime.date(2016, 1, 31); y.replace(month=2) results in ValueError: day is out of range for month

Ther is no simple way to do it, but you can use your own function like answered below.

小开

Note: This answer shows how to achieve this using only the datetime and calendar standard library (stdlib) modules - which is what was explicitly asked for. The accepted answer shows how to better achieve this with one of the many dedicated non-stdlib libraries. If you can use non-stdlib libraries, by all means do so for these kinds of date/time manipulations!

How about this?

def add_one_month(orig_date):
# advance year and month by one month
new_year = orig_date.year
new_month = orig_date.month + 1
# note: in datetime.date, months go from 1 to 12
if new_month > 12:
new_year += 1
new_month -= 12


new_day = orig_date.day
# while day is out of range for month, reduce by one
while True:
try:
new_date = datetime.date(new_year, new_month, new_day)
except ValueError as e:
new_day -= 1
else:
break


return new_date

EDIT:

Improved version which:

  1. keeps the time information if given a datetime.datetime object
  2. doesn't use try/catch, instead using calendar.monthrange from the calendar module in the stdlib:
import datetime
import calendar


def add_one_month(orig_date):
# advance year and month by one month
new_year = orig_date.year
new_month = orig_date.month + 1
# note: in datetime.date, months go from 1 to 12
if new_month > 12:
new_year += 1
new_month -= 12


last_day_of_month = calendar.monthrange(new_year, new_month)[1]
new_day = min(orig_date.day, last_day_of_month)


return orig_date.replace(year=new_year, month=new_month, day=new_day)
小开

Question: Is there a simple way to do this in the current release of Python?

Answer: There is no simple (direct) way to do this in the current release of Python.

Reference: Please refer to docs.python.org/2/library/datetime.html, section 8.1.2. timedelta Objects. As we may understand from that, we cannot increment month directly since it is not a uniform time unit.

Plus: If you want first day -> first day and last day -> last day mapping you should handle that separately for different months.

小开
最佳答案

Check out from dateutil.relativedelta import * for adding a specific amount of time to a date, you can continue to use timedelta for the simple stuff i.e.

import datetime
from dateutil.relativedelta import *
use_date = datetime.datetime.now()


use_date = use_date + datetime.timedelta(minutes=+10)
use_date = use_date + datetime.timedelta(hours=+1)
use_date = use_date + datetime.timedelta(days=+1)
use_date = use_date + datetime.timedelta(weeks=+1)

or you can start using relativedelta

use_date = use_date+relativedelta(months=+1)


use_date = use_date+relativedelta(years=+1)

for the last day of next month:

use_date = use_date+relativedelta(months=+1)
use_date = use_date+relativedelta(day=31)

Right now this will provide 29/02/2016

for the penultimate day of next month:

use_date = use_date+relativedelta(months=+1)
use_date = use_date+relativedelta(day=31)
use_date = use_date+relativedelta(days=-1)

last Friday of the next month:

use_date = use_date+relativedelta(months=+1, day=31, weekday=FR(-1))

2nd Tuesday of next month:

new_date = use_date+relativedelta(months=+1, day=1, weekday=TU(2))

As @mrroot5 points out dateutil's rrule functions can be applied, giving you an extra bang for your buck, if you require date occurences.
for example:
Calculating the last day of the month for 9 months from the last day of last month.
Then, calculate the 2nd Tuesday for each of those months.

from dateutil.relativedelta import *
from dateutil.rrule import *
from datetime import datetime
use_date = datetime(2020,11,21)


#Calculate the last day of last month
use_date = use_date+relativedelta(months=-1)
use_date = use_date+relativedelta(day=31)


#Generate a list of the last day for 9 months from the calculated date
x = list(rrule(freq=MONTHLY, count=9, dtstart=use_date, bymonthday=(-1,)))
print("Last day")
for ld in x:
print(ld)


#Generate a list of the 2nd Tuesday in each of the next 9 months from the calculated date
print("\n2nd Tuesday")
x = list(rrule(freq=MONTHLY, count=9, dtstart=use_date, byweekday=TU(2)))
for tuesday in x:
print(tuesday)


Last day
2020-10-31 00:00:00
2020-11-30 00:00:00
2020-12-31 00:00:00
2021-01-31 00:00:00
2021-02-28 00:00:00
2021-03-31 00:00:00
2021-04-30 00:00:00
2021-05-31 00:00:00
2021-06-30 00:00:00


2nd Tuesday
2020-11-10 00:00:00
2020-12-08 00:00:00
2021-01-12 00:00:00
2021-02-09 00:00:00
2021-03-09 00:00:00
2021-04-13 00:00:00
2021-05-11 00:00:00
2021-06-08 00:00:00
2021-07-13 00:00:00

rrule could be used to find the next date occurring on a particular day.
e.g. the next 1st of January occurring on a Monday (Given today is the 4th November 2021)

from dateutil.relativedelta import *
from dateutil.rrule import *
from datetime import *
year = rrule(YEARLY,dtstart=datetime.now(),bymonth=1,bymonthday=1,byweekday=MO)[0].year
year
2024

or the next 5 x 1st of January's occurring on a Monday

years = rrule(YEARLY,dtstart=datetime.now(),bymonth=1,bymonthday=1,byweekday=MO)[0:5]
for i in years:print(i.year)
...
2024
2029
2035
2046
2052

The first Month next Year that starts on a Monday:

>>> month = rrule(YEARLY,dtstart=datetime.date(2023, 1, 1),bymonthday=1,byweekday=MO)[0]
>>> month.strftime('%Y-%m-%d : %B')
'2023-05-01 : May'

If you need the months that start on a Monday between 2 dates:

months = rrule(YEARLY,dtstart=datetime.date(2025, 1, 1),until=datetime.date(2030, 1, 1),bymonthday=1,byweekday=MO)
>>> for m in months:
...     print(m.strftime('%Y-%m-%d : %B'))
...
2025-09-01 : September
2025-12-01 : December
2026-06-01 : June
2027-02-01 : February
2027-03-01 : March
2027-11-01 : November
2028-05-01 : May
2029-01-01 : January
2029-10-01 : October

This is by no means an exhaustive list of what is available. Documentation is available here: https://dateutil.readthedocs.org/en/latest/


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