如何获得每月的最后一天?

问题:如何获得每月的最后一天?

是否可以使用Python的标准库轻松确定(即调用一个函数)给定月份的最后一天?

如果标准库不支持该功能,dateutil包是否支持此功能?

Is there a way using Python’s standard library to easily determine (i.e. one function call) the last day of a given month?

If the standard library doesn’t support that, does the dateutil package support this?


回答 0

查看该calendar模块文档时,我没有注意到这一点,但是称为的方法monthrange提供了以下信息:

monthrange(year,month)
    返回指定年份和月份的月份的第一天的工作日以及月份中的天数。

>>> import calendar
>>> calendar.monthrange(2002,1)
(1, 31)
>>> calendar.monthrange(2008,2)
(4, 29)
>>> calendar.monthrange(2100,2)
(0, 28)

所以:

calendar.monthrange(year, month)[1]

似乎是最简单的方法。

明确一点,也monthrange支持supports年:

>>> from calendar import monthrange
>>> monthrange(2012, 2)
(2, 29)

我以前的答案仍然有效,但显然不是最佳选择。

I didn’t notice this earlier when I was looking at the documentation for the calendar module, but a method called monthrange provides this information:

monthrange(year, month)
    Returns weekday of first day of the month and number of days in month, for the specified year and month.

>>> import calendar
>>> calendar.monthrange(2002,1)
(1, 31)
>>> calendar.monthrange(2008,2)
(4, 29)
>>> calendar.monthrange(2100,2)
(0, 28)

so:

calendar.monthrange(year, month)[1]

seems like the simplest way to go.

Just to be clear, monthrange supports leap years as well:

>>> from calendar import monthrange
>>> monthrange(2012, 2)
(2, 29)

My previous answer still works, but is clearly suboptimal.


回答 1

如果您不想导入calendar模块,那么一个简单的两步函数也可以是:

import datetime

def last_day_of_month(any_day):
    next_month = any_day.replace(day=28) + datetime.timedelta(days=4)  # this will never fail
    return next_month - datetime.timedelta(days=next_month.day)

输出:

>>> for month in range(1, 13):
...     print last_day_of_month(datetime.date(2012, month, 1))
...
2012-01-31
2012-02-29
2012-03-31
2012-04-30
2012-05-31
2012-06-30
2012-07-31
2012-08-31
2012-09-30
2012-10-31
2012-11-30
2012-12-31

If you don’t want to import the calendar module, a simple two-step function can also be:

import datetime

def last_day_of_month(any_day):
    next_month = any_day.replace(day=28) + datetime.timedelta(days=4)  # this will never fail
    return next_month - datetime.timedelta(days=next_month.day)

Outputs:

>>> for month in range(1, 13):
...     print last_day_of_month(datetime.date(2012, month, 1))
...
2012-01-31
2012-02-29
2012-03-31
2012-04-30
2012-05-31
2012-06-30
2012-07-31
2012-08-31
2012-09-30
2012-10-31
2012-11-30
2012-12-31

回答 2

编辑:请参阅@Blair Conrad的答案以获得更清洁的解决方案


>>> import datetime
>>> datetime.date(2000, 2, 1) - datetime.timedelta(days=1)
datetime.date(2000, 1, 31)

EDIT: See @Blair Conrad’s answer for a cleaner solution


>>> import datetime
>>> datetime.date(2000, 2, 1) - datetime.timedelta(days=1)
datetime.date(2000, 1, 31)

回答 3

dateutil.relativedelta(使用pip软件包python-datetutil)实际上很容易。day=31始终会返回该月的最后一天。

例:

from datetime import datetime
from dateutil.relativedelta import relativedelta

date_in_feb = datetime.datetime(2013, 2, 21)
print datetime.datetime(2013, 2, 21) + relativedelta(day=31)  # End-of-month
>>> datetime.datetime(2013, 2, 28, 0, 0)

This is actually pretty easy with dateutil.relativedelta (package python-datetutil for pip). day=31 will always always return the last day of the month.

Example:

from datetime import datetime
from dateutil.relativedelta import relativedelta

date_in_feb = datetime.datetime(2013, 2, 21)
print datetime.datetime(2013, 2, 21) + relativedelta(day=31)  # End-of-month
>>> datetime.datetime(2013, 2, 28, 0, 0)

回答 4

编辑:看到我的其他答案。它的实现比该方法更好,我留在这里是为了防止有人感兴趣地看一看如何“滚动自己的”计算器。

@John Millikin提供了一个很好的答案,增加了计算下个月第一天的复杂性。

以下内容并不是特别优雅,但是要弄清楚任何给定日期所在的月份的最后一天,您可以尝试:

def last_day_of_month(date):
    if date.month == 12:
        return date.replace(day=31)
    return date.replace(month=date.month+1, day=1) - datetime.timedelta(days=1)

>>> last_day_of_month(datetime.date(2002, 1, 17))
datetime.date(2002, 1, 31)
>>> last_day_of_month(datetime.date(2002, 12, 9))
datetime.date(2002, 12, 31)
>>> last_day_of_month(datetime.date(2008, 2, 14))
datetime.date(2008, 2, 29)

EDIT: see my other answer. It has a better implementation than this one, which I leave here just in case someone’s interested in seeing how one might “roll your own” calculator.

@John Millikin gives a good answer, with the added complication of calculating the first day of the next month.

The following isn’t particularly elegant, but to figure out the last day of the month that any given date lives in, you could try:

def last_day_of_month(date):
    if date.month == 12:
        return date.replace(day=31)
    return date.replace(month=date.month+1, day=1) - datetime.timedelta(days=1)

>>> last_day_of_month(datetime.date(2002, 1, 17))
datetime.date(2002, 1, 31)
>>> last_day_of_month(datetime.date(2002, 12, 9))
datetime.date(2002, 12, 31)
>>> last_day_of_month(datetime.date(2008, 2, 14))
datetime.date(2008, 2, 29)

回答 5

使用dateutil.relativedelta您会得到这样的月份的最后日期:

from dateutil.relativedelta import relativedelta
last_date_of_month = datetime(mydate.year, mydate.month, 1) + relativedelta(months=1, days=-1)

这个想法是获取月份的第一天,并relativedelta习惯于提前一个月再返回一天,这样您就可以获得想要的月份的最后一天。

Using dateutil.relativedelta you would get last date of month like this:

from dateutil.relativedelta import relativedelta
last_date_of_month = datetime(mydate.year, mydate.month, 1) + relativedelta(months=1, days=-1)

The idea is to get the first day of the month and use relativedelta to go 1 month ahead and 1 day back so you would get the last day of the month you wanted.


回答 6

另一个解决方案是做这样的事情:

from datetime import datetime

def last_day_of_month(year, month):
    """ Work out the last day of the month """
    last_days = [31, 30, 29, 28, 27]
    for i in last_days:
        try:
            end = datetime(year, month, i)
        except ValueError:
            continue
        else:
            return end.date()
    return None

并使用如下功能:

>>> 
>>> last_day_of_month(2008, 2)
datetime.date(2008, 2, 29)
>>> last_day_of_month(2009, 2)
datetime.date(2009, 2, 28)
>>> last_day_of_month(2008, 11)
datetime.date(2008, 11, 30)
>>> last_day_of_month(2008, 12)
datetime.date(2008, 12, 31)

Another solution would be to do something like this:

from datetime import datetime

def last_day_of_month(year, month):
    """ Work out the last day of the month """
    last_days = [31, 30, 29, 28, 27]
    for i in last_days:
        try:
            end = datetime(year, month, i)
        except ValueError:
            continue
        else:
            return end.date()
    return None

And use the function like this:

>>> 
>>> last_day_of_month(2008, 2)
datetime.date(2008, 2, 29)
>>> last_day_of_month(2009, 2)
datetime.date(2009, 2, 28)
>>> last_day_of_month(2008, 11)
datetime.date(2008, 11, 30)
>>> last_day_of_month(2008, 12)
datetime.date(2008, 12, 31)

回答 7

from datetime import timedelta
(any_day.replace(day=1) + timedelta(days=32)).replace(day=1) - timedelta(days=1)
from datetime import timedelta
(any_day.replace(day=1) + timedelta(days=32)).replace(day=1) - timedelta(days=1)

回答 8

>>> import datetime
>>> import calendar
>>> date  = datetime.datetime.now()

>>> print date
2015-03-06 01:25:14.939574

>>> print date.replace(day = 1)
2015-03-01 01:25:14.939574

>>> print date.replace(day = calendar.monthrange(date.year, date.month)[1])
2015-03-31 01:25:14.939574
>>> import datetime
>>> import calendar
>>> date  = datetime.datetime.now()

>>> print date
2015-03-06 01:25:14.939574

>>> print date.replace(day = 1)
2015-03-01 01:25:14.939574

>>> print date.replace(day = calendar.monthrange(date.year, date.month)[1])
2015-03-31 01:25:14.939574

回答 9

如果您愿意使用外部库,请访问http://crsmithdev.com/arrow/

然后,您可以使用以下命令获取月份的最后一天:

import arrow
arrow.utcnow().ceil('month').date()

这将返回一个日期对象,您可以随后对其进行操作。

if you are willing to use an external library, check out http://crsmithdev.com/arrow/

U can then get the last day of the month with:

import arrow
arrow.utcnow().ceil('month').date()

This returns a date object which you can then do your manipulation.


回答 10

要获取该月的最后日期,我们需要执行以下操作:

from datetime import date, timedelta
import calendar
last_day = date.today().replace(day=calendar.monthrange(date.today().year, date.today().month)[1])

现在解释一下我们在这里做什么,我们将其分为两部分:

首先是获取当月的天数,我们使用Blair Conrad已经提到他的解决方案的monthrange

calendar.monthrange(date.today().year, date.today().month)[1]

二是让我们的帮助下做的最后日期本身代替

>>> date.today()
datetime.date(2017, 1, 3)
>>> date.today().replace(day=31)
datetime.date(2017, 1, 31)

当我们按顶部所述将它们组合在一起时,便得到了动态解决方案。

To get the last date of the month we do something like this:

from datetime import date, timedelta
import calendar
last_day = date.today().replace(day=calendar.monthrange(date.today().year, date.today().month)[1])

Now to explain what we are doing here we will break it into two parts:

first is getting the number of days of the current month for which we use monthrange which Blair Conrad has already mentioned his solution:

calendar.monthrange(date.today().year, date.today().month)[1]

second is getting the last date itself which we do with the help of replace e.g

>>> date.today()
datetime.date(2017, 1, 3)
>>> date.today().replace(day=31)
datetime.date(2017, 1, 31)

and when we combine them as mentioned on the top we get a dynamic solution.


回答 11

在Python 3.7中,有未记录的calendar.monthlen(year, month)函数

>>> calendar.monthlen(2002, 1)
31
>>> calendar.monthlen(2008, 2)
29
>>> calendar.monthlen(2100, 2)
28

它等效于记录的calendar.monthrange(year, month)[1]呼叫

In Python 3.7 there is the undocumented calendar.monthlen(year, month) function:

>>> calendar.monthlen(2002, 1)
31
>>> calendar.monthlen(2008, 2)
29
>>> calendar.monthlen(2100, 2)
28

It is equivalent to the documented calendar.monthrange(year, month)[1] call.


回答 12

import datetime

now = datetime.datetime.now()
start_month = datetime.datetime(now.year, now.month, 1)
date_on_next_month = start_month + datetime.timedelta(35)
start_next_month = datetime.datetime(date_on_next_month.year, date_on_next_month.month, 1)
last_day_month = start_next_month - datetime.timedelta(1)
import datetime

now = datetime.datetime.now()
start_month = datetime.datetime(now.year, now.month, 1)
date_on_next_month = start_month + datetime.timedelta(35)
start_next_month = datetime.datetime(date_on_next_month.year, date_on_next_month.month, 1)
last_day_month = start_next_month - datetime.timedelta(1)

回答 13

使用熊猫!

def isMonthEnd(date):
    return date + pd.offsets.MonthEnd(0) == date

isMonthEnd(datetime(1999, 12, 31))
True
isMonthEnd(pd.Timestamp('1999-12-31'))
True
isMonthEnd(pd.Timestamp(1965, 1, 10))
False

Use pandas!

def isMonthEnd(date):
    return date + pd.offsets.MonthEnd(0) == date

isMonthEnd(datetime(1999, 12, 31))
True
isMonthEnd(pd.Timestamp('1999-12-31'))
True
isMonthEnd(pd.Timestamp(1965, 1, 10))
False

回答 14

对我来说,这是最简单的方法:

selected_date = date(some_year, some_month, some_day)

if selected_date.month == 12: # December
     last_day_selected_month = date(selected_date.year, selected_date.month, 31)
else:
     last_day_selected_month = date(selected_date.year, selected_date.month + 1, 1) - timedelta(days=1)

For me it’s the simplest way:

selected_date = date(some_year, some_month, some_day)

if selected_date.month == 12: # December
     last_day_selected_month = date(selected_date.year, selected_date.month, 31)
else:
     last_day_selected_month = date(selected_date.year, selected_date.month + 1, 1) - timedelta(days=1)

回答 15

最简单的方法(无需导入日历)是获取下个月的第一天,然后从中减去一天。

import datetime as dt
from dateutil.relativedelta import relativedelta

thisDate = dt.datetime(2017, 11, 17)

last_day_of_the_month = dt.datetime(thisDate.year, (thisDate + relativedelta(months=1)).month, 1) - dt.timedelta(days=1)
print last_day_of_the_month

输出:

datetime.datetime(2017, 11, 30, 0, 0)

PS:与该import calendar方法相比,此代码运行速度更快;见下文:

import datetime as dt
import calendar
from dateutil.relativedelta import relativedelta

someDates = [dt.datetime.today() - dt.timedelta(days=x) for x in range(0, 10000)]

start1 = dt.datetime.now()
for thisDate in someDates:
    lastDay = dt.datetime(thisDate.year, (thisDate + relativedelta(months=1)).month, 1) - dt.timedelta(days=1)

print ('Time Spent= ', dt.datetime.now() - start1)


start2 = dt.datetime.now()
for thisDate in someDates:
    lastDay = dt.datetime(thisDate.year, 
                          thisDate.month, 
                          calendar.monthrange(thisDate.year, thisDate.month)[1])

print ('Time Spent= ', dt.datetime.now() - start2)

输出:

Time Spent=  0:00:00.097814
Time Spent=  0:00:00.109791

此代码假定您需要该月最后一天的日期(即,不仅是DD部分,而且是整个YYYYMMDD日期)

The easiest way (without having to import calendar), is to get the first day of the next month, and then subtract a day from it.

import datetime as dt
from dateutil.relativedelta import relativedelta

thisDate = dt.datetime(2017, 11, 17)

last_day_of_the_month = dt.datetime(thisDate.year, (thisDate + relativedelta(months=1)).month, 1) - dt.timedelta(days=1)
print last_day_of_the_month

Output:

datetime.datetime(2017, 11, 30, 0, 0)

PS: This code runs faster as compared to the import calendarapproach; see below:

import datetime as dt
import calendar
from dateutil.relativedelta import relativedelta

someDates = [dt.datetime.today() - dt.timedelta(days=x) for x in range(0, 10000)]

start1 = dt.datetime.now()
for thisDate in someDates:
    lastDay = dt.datetime(thisDate.year, (thisDate + relativedelta(months=1)).month, 1) - dt.timedelta(days=1)

print ('Time Spent= ', dt.datetime.now() - start1)


start2 = dt.datetime.now()
for thisDate in someDates:
    lastDay = dt.datetime(thisDate.year, 
                          thisDate.month, 
                          calendar.monthrange(thisDate.year, thisDate.month)[1])

print ('Time Spent= ', dt.datetime.now() - start2)

OUTPUT:

Time Spent=  0:00:00.097814
Time Spent=  0:00:00.109791

This code assumes that you want the date of the last day of the month (i.e., not just the DD part, but the entire YYYYMMDD date)


回答 16

这是另一个答案。无需额外的程序包。

datetime.date(year + int(month/12), month%12+1, 1)-datetime.timedelta(days=1)

获取下个月的第一天并从中减去一天。

Here is another answer. No extra packages required.

datetime.date(year + int(month/12), month%12+1, 1)-datetime.timedelta(days=1)

Get the first day of the next month and subtract a day from it.


回答 17

您可以自己计算结束日期。简单的逻辑是从下个月的start_date减去一天。:)

因此,编写一个自定义方法,

import datetime

def end_date_of_a_month(date):


    start_date_of_this_month = date.replace(day=1)

    month = start_date_of_this_month.month
    year = start_date_of_this_month.year
    if month == 12:
        month = 1
        year += 1
    else:
        month += 1
    next_month_start_date = start_date_of_this_month.replace(month=month, year=year)

    this_month_end_date = next_month_start_date - datetime.timedelta(days=1)
    return this_month_end_date

打电话

end_date_of_a_month(datetime.datetime.now().date())

它将返回本月的结束日期。将任何日期传递给此功能。返回该月的结束日期。

You can calculate the end date yourself. the simple logic is to subtract a day from the start_date of next month. :)

So write a custom method,

import datetime

def end_date_of_a_month(date):


    start_date_of_this_month = date.replace(day=1)

    month = start_date_of_this_month.month
    year = start_date_of_this_month.year
    if month == 12:
        month = 1
        year += 1
    else:
        month += 1
    next_month_start_date = start_date_of_this_month.replace(month=month, year=year)

    this_month_end_date = next_month_start_date - datetime.timedelta(days=1)
    return this_month_end_date

Calling,

end_date_of_a_month(datetime.datetime.now().date())

It will return the end date of this month. Pass any date to this function. returns you the end date of that month.


回答 18

您可以使用relativedelta https://dateutil.readthedocs.io/en/stable/relativedelta.html month_end = <your datetime value within the month> + relativedelta(day=31) 来给您最后的一天。

you can use relativedelta https://dateutil.readthedocs.io/en/stable/relativedelta.html month_end = <your datetime value within the month> + relativedelta(day=31) that will give you the last day.


回答 19

仅使用标准日期时间库,这对我来说是最简单的解决方案:

import datetime

def get_month_end(dt):
    first_of_month = datetime.datetime(dt.year, dt.month, 1)
    next_month_date = first_of_month + datetime.timedelta(days=32)
    new_dt = datetime.datetime(next_month_date.year, next_month_date.month, 1)
    return new_dt - datetime.timedelta(days=1)

This is the simplest solution for me using just the standard datetime library:

import datetime

def get_month_end(dt):
    first_of_month = datetime.datetime(dt.year, dt.month, 1)
    next_month_date = first_of_month + datetime.timedelta(days=32)
    new_dt = datetime.datetime(next_month_date.year, next_month_date.month, 1)
    return new_dt - datetime.timedelta(days=1)

回答 20

这没有解决主要问题,但是使用一个月中的最后一个工作日的一个不错的技巧是使用calendar.monthcalendar,该方法返回一个日期矩阵,将星期一作为第一列,将星期日作为最后一个列。

# Some random date.
some_date = datetime.date(2012, 5, 23)

# Get last weekday
last_weekday = np.asarray(calendar.monthcalendar(some_date.year, some_date.month))[:,0:-2].ravel().max()

print last_weekday
31

整个过程[0:-2]就是刮掉周末专栏文章,然后将它们扔掉。月份以外的日期用0表示,因此最大值实际上会忽略它们。

使用的numpy.ravel是不是绝对必要的,但我不喜欢依靠单纯的惯例numpy.ndarray.max将压平的数组,如果没有被告知哪个轴计算过。

This does not address the main question, but one nice trick to get the last weekday in a month is to use calendar.monthcalendar, which returns a matrix of dates, organized with Monday as the first column through Sunday as the last.

# Some random date.
some_date = datetime.date(2012, 5, 23)

# Get last weekday
last_weekday = np.asarray(calendar.monthcalendar(some_date.year, some_date.month))[:,0:-2].ravel().max()

print last_weekday
31

The whole [0:-2] thing is to shave off the weekend columns and throw them out. Dates that fall outside of the month are indicated by 0, so the max effectively ignores them.

The use of numpy.ravel is not strictly necessary, but I hate relying on the mere convention that numpy.ndarray.max will flatten the array if not told which axis to calculate over.


回答 21

import calendar
from time import gmtime, strftime
calendar.monthrange(int(strftime("%Y", gmtime())), int(strftime("%m", gmtime())))[1]

输出:

31



这将打印当前月份的最后一天。在此示例中,日期是2016年5月15日。因此您的输出可能会有所不同,但是输出将是当月的几天。如果您想通过运行每日Cron作业来检查每月的最后一天,那就太好了。

所以:

import calendar
from time import gmtime, strftime
lastDay = calendar.monthrange(int(strftime("%Y", gmtime())), int(strftime("%m", gmtime())))[1]
today = strftime("%d", gmtime())
lastDay == today

输出:

False

除非是每月的最后一天。

import calendar
from time import gmtime, strftime
calendar.monthrange(int(strftime("%Y", gmtime())), int(strftime("%m", gmtime())))[1]

Output:

31



This will print the last day of whatever the current month is. In this example it was 15th May, 2016. So your output may be different, however the output will be as many days that the current month is. Great if you want to check the last day of the month by running a daily cron job.

So:

import calendar
from time import gmtime, strftime
lastDay = calendar.monthrange(int(strftime("%Y", gmtime())), int(strftime("%m", gmtime())))[1]
today = strftime("%d", gmtime())
lastDay == today

Output:

False

Unless it IS the last day of the month.


回答 22

我喜欢这样

import datetime
import calendar

date=datetime.datetime.now()
month_end_date=datetime.datetime(date.year,date.month,1) + datetime.timedelta(days=calendar.monthrange(date.year,date.month)[1] - 1)

I prefer this way

import datetime
import calendar

date=datetime.datetime.now()
month_end_date=datetime.datetime(date.year,date.month,1) + datetime.timedelta(days=calendar.monthrange(date.year,date.month)[1] - 1)

回答 23

如果要创建自己的小函数,这是一个很好的起点:

def eomday(year, month):
    """returns the number of days in a given month"""
    days_per_month = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    d = days_per_month[month - 1]
    if month == 2 and (year % 4 == 0 and year % 100 != 0 or year % 400 == 0):
        d = 29
    return d

为此,您必须了解the年的规则:

  • 每四年
  • 每100年除外
  • 但是每400年一次

If you want to make your own small function, this is a good starting point:

def eomday(year, month):
    """returns the number of days in a given month"""
    days_per_month = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    d = days_per_month[month - 1]
    if month == 2 and (year % 4 == 0 and year % 100 != 0 or year % 400 == 0):
        d = 29
    return d

For this you have to know the rules for the leap years:

  • every fourth year
  • with the exception of every 100 year
  • but again every 400 years

回答 24

如果输入日期范围,则可以使用以下方法:

def last_day_of_month(any_days):
    res = []
    for any_day in any_days:
        nday = any_day.days_in_month -any_day.day
        res.append(any_day + timedelta(days=nday))
    return res

If you pass in a date range, you can use this:

def last_day_of_month(any_days):
    res = []
    for any_day in any_days:
        nday = any_day.days_in_month -any_day.day
        res.append(any_day + timedelta(days=nday))
    return res

回答 25

“ get_last_day_of_month(dt)”下面的代码中,将为您提供此日期,日期格式为“ YYYY-MM-DD”。

import datetime

def DateTime( d ):
    return datetime.datetime.strptime( d, '%Y-%m-%d').date()

def RelativeDate( start, num_days ):
    d = DateTime( start )
    return str( d + datetime.timedelta( days = num_days ) )

def get_first_day_of_month( dt ):
    return dt[:-2] + '01'

def get_last_day_of_month( dt ):
    fd = get_first_day_of_month( dt )
    fd_next_month = get_first_day_of_month( RelativeDate( fd, 31 ) )
    return RelativeDate( fd_next_month, -1 )

In the code below ‘get_last_day_of_month(dt)’ will give you this, with date in string format like ‘YYYY-MM-DD’.

import datetime

def DateTime( d ):
    return datetime.datetime.strptime( d, '%Y-%m-%d').date()

def RelativeDate( start, num_days ):
    d = DateTime( start )
    return str( d + datetime.timedelta( days = num_days ) )

def get_first_day_of_month( dt ):
    return dt[:-2] + '01'

def get_last_day_of_month( dt ):
    fd = get_first_day_of_month( dt )
    fd_next_month = get_first_day_of_month( RelativeDate( fd, 31 ) )
    return RelativeDate( fd_next_month, -1 )

回答 26

最简单的方法是使用datetime和一些日期数学,例如从下个月的第一天减去一天:

import datetime

def last_day_of_month(d: datetime.date) -> datetime.date:
    return (
        datetime.date(d.year + d.month//12, d.month % 12 + 1, 1) -
        datetime.timedelta(days=1)
    )

或者,您可以calendar.monthrange()用来获取一个月中的天数(考虑leap年)并相应地更新日期:

import calendar, datetime

def last_day_of_month(d: datetime.date) -> datetime.date:
    return d.replace(day=calendar.monthrange(d.year, d.month)[1])

快速基准测试表明,第一个版本的速度明显更快:

In [14]: today = datetime.date.today()

In [15]: %timeit last_day_of_month_dt(today)
918 ns ± 3.54 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [16]: %timeit last_day_of_month_calendar(today)
1.4 µs ± 17.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

The simplest way is to use datetime and some date math, e.g. subtract a day from the first day of the next month:

import datetime

def last_day_of_month(d: datetime.date) -> datetime.date:
    return (
        datetime.date(d.year + d.month//12, d.month % 12 + 1, 1) -
        datetime.timedelta(days=1)
    )

Alternatively, you could use calendar.monthrange() to get the number of days in a month (taking leap years into account) and update the date accordingly:

import calendar, datetime

def last_day_of_month(d: datetime.date) -> datetime.date:
    return d.replace(day=calendar.monthrange(d.year, d.month)[1])

A quick benchmark shows that the first version is noticeably faster:

In [14]: today = datetime.date.today()

In [15]: %timeit last_day_of_month_dt(today)
918 ns ± 3.54 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [16]: %timeit last_day_of_month_calendar(today)
1.4 µs ± 17.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

回答 27

这是一个很长的版本(易于理解),但是照顾了leap年。

干杯,JK

def last_day_month(year, month):
    leap_year_flag = 0
    end_dates = {
        1: 31,
        2: 28,
        3: 31,
        4: 30,
        5: 31,
        6: 30,
        7: 31,
        8: 31,
        9: 30,
        10: 31,
        11: 30,
        12: 31
    }

    # Checking for regular leap year    
    if year % 4 == 0:
        leap_year_flag = 1
    else:
        leap_year_flag = 0

    # Checking for century leap year    
    if year % 100 == 0:
        if year % 400 == 0:
            leap_year_flag = 1
        else:
            leap_year_flag = 0
    else:
        pass

    # return end date of the year-month
    if leap_year_flag == 1 and month == 2:
        return 29
    elif leap_year_flag == 1 and month != 2:
        return end_dates[month]
    else:
        return end_dates[month]

Here is a long (easy to understand) version but takes care of leap years.

cheers, JK

def last_day_month(year, month):
    leap_year_flag = 0
    end_dates = {
        1: 31,
        2: 28,
        3: 31,
        4: 30,
        5: 31,
        6: 30,
        7: 31,
        8: 31,
        9: 30,
        10: 31,
        11: 30,
        12: 31
    }

    # Checking for regular leap year    
    if year % 4 == 0:
        leap_year_flag = 1
    else:
        leap_year_flag = 0

    # Checking for century leap year    
    if year % 100 == 0:
        if year % 400 == 0:
            leap_year_flag = 1
        else:
            leap_year_flag = 0
    else:
        pass

    # return end date of the year-month
    if leap_year_flag == 1 and month == 2:
        return 29
    elif leap_year_flag == 1 and month != 2:
        return end_dates[month]
    else:
        return end_dates[month]

回答 28

这是一个基于解决方案的python lambdas:

next_month = lambda y, m, d: (y, m + 1, 1) if m + 1 < 13 else ( y+1 , 1, 1)
month_end  = lambda dte: date( *next_month( *dte.timetuple()[:3] ) ) - timedelta(days=1)

next_month拉姆达发现到明年下个月的第一天的元组表示,和卷。该month_end拉姆达转换日期(dte)的元组,应用next_month和创造新的日期。然后,“月底”就是下个月的第一天减去timedelta(days=1)

Here is a solution based python lambdas:

next_month = lambda y, m, d: (y, m + 1, 1) if m + 1 < 13 else ( y+1 , 1, 1)
month_end  = lambda dte: date( *next_month( *dte.timetuple()[:3] ) ) - timedelta(days=1)

The next_month lambda finds the tuple representation of the first day of the next month, and rolls over to the next year. The month_end lambda transforms a date (dte) to a tuple, applies next_month and creates a new date. Then the “month’s end” is just the next month’s first day minus timedelta(days=1).