查找两个日期之间的月份的最佳方法

Question

我需要能够准确地找到 python 中两个日期之间的月份。我有一个可行的解决方案，但它不是很好（如优雅）或快速。

dateRange = [datetime.strptime(dateRanges[0], "%Y-%m-%d"), datetime.strptime(dateRanges[1], "%Y-%m-%d")]
months = [] 

tmpTime = dateRange[0]
oneWeek = timedelta(weeks=1)
tmpTime = tmpTime.replace(day=1)
dateRange[0] = tmpTime
dateRange[1] = dateRange[1].replace(day=1)
lastMonth = tmpTime.month
months.append(tmpTime)
while tmpTime < dateRange[1]:
    if lastMonth != 12:
        while tmpTime.month <= lastMonth:
            tmpTime += oneWeek
        tmpTime = tmpTime.replace(day=1)
        months.append(tmpTime)
        lastMonth = tmpTime.month

    else:
        while tmpTime.month >= lastMonth:
            tmpTime += oneWeek
        tmpTime = tmpTime.replace(day=1)
        months.append(tmpTime)
        lastMonth = tmpTime.month

所以只是为了解释一下，我在这里所做的是获取两个日期并将它们从 iso 格式转换为 python 日期时间对象。 然后我循环将一周添加到开始日期时间 object 并检查月份的数值是否更大（除非月份是十二月然后它检查日期是否更小），如果值更大我 append 它到列表几个月并不断循环直到我到达我的结束日期。

它工作得很好，只是看起来不是一个好方法......

Answer 1

首先定义一些测试用例，然后你会看到函数非常简单，不需要循环

from datetime import datetime

def diff_month(d1, d2):
    return (d1.year - d2.year) * 12 + d1.month - d2.month

assert diff_month(datetime(2010,10,1), datetime(2010,9,1)) == 1
assert diff_month(datetime(2010,10,1), datetime(2009,10,1)) == 12
assert diff_month(datetime(2010,10,1), datetime(2009,11,1)) == 11
assert diff_month(datetime(2010,10,1), datetime(2009,8,1)) == 14

您应该在您的问题中添加一些测试用例，因为有很多潜在的极端案例需要涵盖 - 定义两个日期之间的月数的方法不止一种。

Answer 2

一个班轮查找两个日期之间的日期时间列表，按月递增。

import datetime
from dateutil.rrule import rrule, MONTHLY

strt_dt = datetime.date(2001,1,1)
end_dt = datetime.date(2005,6,1)

dates = [dt for dt in rrule(MONTHLY, dtstart=strt_dt, until=end_dt)]

Answer 3

这对我有用 -

from datetime import datetime
from dateutil import relativedelta
date1 = datetime.strptime('2011-08-15 12:00:00', '%Y-%m-%d %H:%M:%S')
date2 = datetime.strptime('2012-02-15', '%Y-%m-%d')
r = relativedelta.relativedelta(date2, date1)
r.months + (12*r.years)

Answer 4

您可以使用dateutil模块中的 rrule 轻松计算：

from dateutil import rrule
from datetime import date

print(list(rrule.rrule(rrule.MONTHLY, dtstart=date(2013, 11, 1), until=date(2014, 2, 1))))

会给你：

 [datetime.datetime(2013, 11, 1, 0, 0),
 datetime.datetime(2013, 12, 1, 0, 0),
 datetime.datetime(2014, 1, 1, 0, 0),
 datetime.datetime(2014, 2, 1, 0, 0)]

Answer 5

from dateutil import relativedelta

r = relativedelta.relativedelta(date1, date2)

months_difference = (r.years * 12) + r.months

Answer 6

获取结束月份（相对于开始月份的年份和月份，例如：如果您的开始日期从 2010 年 10 月开始，则 2011 年 1 月 = 13），然后生成开始月份和结束月份的日期时间，如下所示：

dt1, dt2 = dateRange
start_month=dt1.month
end_months=(dt2.year-dt1.year)*12 + dt2.month+1
dates=[datetime.datetime(year=yr, month=mn, day=1) for (yr, mn) in (
          ((m - 1) / 12 + dt1.year, (m - 1) % 12 + 1) for m in range(start_month, end_months)
      )]

如果两个日期都在同一年，也可以简单地写成：

dates=[datetime.datetime(year=dt1.year, month=mn, day=1) for mn in range(dt1.month, dt2.month + 1)]

Answer 7

这篇文章一针见血！ 使用dateutil.relativedelta 。

from datetime import datetime
from dateutil import relativedelta
date1 = datetime.strptime(str('2011-08-15 12:00:00'), '%Y-%m-%d %H:%M:%S')
date2 = datetime.strptime(str('2012-02-15'), '%Y-%m-%d')
r = relativedelta.relativedelta(date2, date1)
r.months

Answer 8

我的简单解决方案：

import datetime

def months(d1, d2):
    return d1.month - d2.month + 12*(d1.year - d2.year)

d1 = datetime.datetime(2009, 9, 26)  
d2 = datetime.datetime(2019, 9, 26) 

print(months(d1, d2))

Answer 9

2018-04-20更新：似乎 OP @Joshkunz 要求查找两个日期之间的月份，而不是两个日期之间的“多少个月”。 所以我不确定为什么@JohnLaRooy 被投票超过 100 次。 @Joshkunz 在原始问题下的评论中指出，他想要实际日期 [或月份]，而不是找到总月数。

因此，在2018-04-11年 4 月 11 日到2018-06-01年 6 月 1 日这两个日期之间，出现了想要的问题

Apr 2018, May 2018, June 2018 

如果它在2014-04-11到2018-06-01之间怎么办？ 那么答案将是

Apr 2014, May 2014, ..., Dec 2014, Jan 2015, ..., Jan 2018, ..., June 2018

所以这就是我多年前有以下伪代码的原因。 它只是建议使用两个月作为终点并循环它们，一次增加一个月。 @Joshkunz 提到他想要“月份”，他还提到他想要“日期”，在不确切知道的情况下，很难编写确切的代码，但想法是使用一个简单的循环来遍历端点，并且一个月递增一次。

8年前的2010年的答案：

如果增加一周，那么它的工作量大约是所需工作量的 4.35 倍。 为什么不只是：

1. get start date in array of integer, set it to i: [2008, 3, 12], 
       and change it to [2008, 3, 1]
2. get end date in array: [2010, 10, 26]
3. add the date to your result by parsing i
       increment the month in i
       if month is >= 13, then set it to 1, and increment the year by 1
   until either the year in i is > year in end_date, 
           or (year in i == year in end_date and month in i > month in end_date)

现在只是伪代码，尚未测试，但我认为同一行的想法会奏效。

Answer 10

将“月”定义为¹ / ₁₂年，然后执行以下操作：

def month_diff(d1, d2): 
    """Return the number of months between d1 and d2, 
    such that d2 + month_diff(d1, d2) == d1
    """
    diff = (12 * d1.year + d1.month) - (12 * d2.year + d2.month)
    return diff

您可以尝试将一个月定义为“29、28、30 或 31 天（取决于年份）的时间段”。 但是你这样做了，你还有一个额外的问题要解决。

虽然通常很清楚 6 月 15^日+ 1 个月应该是 7 月 15^日，但通常不清楚 1 月 30^日+ 1 个月是在 2 月还是 3 月。 在后一种情况下，您可能不得不将日期计算为 2 月 30^日，然后将其“更正”为 3 月 2^日。 但是当你这样做时，你会发现 March 2 ^nd - 1 月显然是 February 2 ^nd 。 因此，reductio ad absurdum（这个操作没有很好的定义）。

Answer 11

以下是如何使用 Pandas FWIW 执行此操作：

import pandas as pd
pd.date_range("1990/04/03", "2014/12/31", freq="MS")

DatetimeIndex(['1990-05-01', '1990-06-01', '1990-07-01', '1990-08-01',
               '1990-09-01', '1990-10-01', '1990-11-01', '1990-12-01',
               '1991-01-01', '1991-02-01',
               ...
               '2014-03-01', '2014-04-01', '2014-05-01', '2014-06-01',
               '2014-07-01', '2014-08-01', '2014-09-01', '2014-10-01',
               '2014-11-01', '2014-12-01'],
              dtype='datetime64[ns]', length=296, freq='MS')

请注意，它从给定开始日期后的月份开始。

Answer 12

许多人已经给了你很好的答案来解决这个问题，但我还没有阅读任何使用列表理解的内容，所以我给你我在类似用例中使用的内容：

def compute_months(first_date, second_date):
    year1, month1, year2, month2 = map(
        int, 
        (first_date[:4], first_date[5:7], second_date[:4], second_date[5:7])
    )

    return [
        '{:0>4}-{:0>2}'.format(year, month)
        for year in range(year1, year2 + 1)
        for month in range(month1 if year == year1 else 1, month2 + 1 if year == year2 else 13)
    ]

>>> first_date = "2016-05"
>>> second_date = "2017-11"
>>> compute_months(first_date, second_date)
['2016-05',
 '2016-06',
 '2016-07',
 '2016-08',
 '2016-09',
 '2016-10',
 '2016-11',
 '2016-12',
 '2017-01',
 '2017-02',
 '2017-03',
 '2017-04',
 '2017-05',
 '2017-06',
 '2017-07',
 '2017-08',
 '2017-09',
 '2017-10',
 '2017-11']

Answer 13

@Vin-G 有点美化的解决方案。

import datetime

def monthrange(start, finish):
  months = (finish.year - start.year) * 12 + finish.month + 1 
  for i in xrange(start.month, months):
    year  = (i - 1) / 12 + start.year 
    month = (i - 1) % 12 + 1
    yield datetime.date(year, month, 1)

Answer 14

您还可以使用箭头库。 这是一个简单的例子：

from datetime import datetime
import arrow

start = datetime(2014, 1, 17)
end = datetime(2014, 6, 20)

for d in arrow.Arrow.range('month', start, end):
    print d.month, d.format('MMMM')

这将打印：

1 January
2 February
3 March
4 April
5 May
6 June

希望这可以帮助！

Answer 15

有一个基于 360 天年的简单解决方案，其中所有月份都有 30 天。 它适用于大多数用例，在这些用例中，给定两个日期，您需要计算整月数加上剩余天数。

from datetime import datetime, timedelta

def months_between(start_date, end_date):
    #Add 1 day to end date to solve different last days of month 
    s1, e1 = start_date , end_date  + timedelta(days=1)
    #Convert to 360 days
    s360 = (s1.year * 12 + s1.month) * 30 + s1.day
    e360 = (e1.year * 12 + e1.month) * 30 + e1.day
    #Count days between the two 360 dates and return tuple (months, days)
    return divmod(e360 - s360, 30)

print "Counting full and half months"
print months_between( datetime(2012, 01, 1), datetime(2012, 03, 31)) #3m
print months_between( datetime(2012, 01, 1), datetime(2012, 03, 15)) #2m 15d
print months_between( datetime(2012, 01, 16), datetime(2012, 03, 31)) #2m 15d
print months_between( datetime(2012, 01, 16), datetime(2012, 03, 15)) #2m
print "Adding +1d and -1d to 31 day month"
print months_between( datetime(2011, 12, 01), datetime(2011, 12, 31)) #1m 0d
print months_between( datetime(2011, 12, 02), datetime(2011, 12, 31)) #-1d => 29d
print months_between( datetime(2011, 12, 01), datetime(2011, 12, 30)) #30d => 1m
print "Adding +1d and -1d to 29 day month"
print months_between( datetime(2012, 02, 01), datetime(2012, 02, 29)) #1m 0d
print months_between( datetime(2012, 02, 02), datetime(2012, 02, 29)) #-1d => 29d
print months_between( datetime(2012, 02, 01), datetime(2012, 02, 28)) #28d
print "Every month has 30 days - 26/M to 5/M+1 always counts 10 days"
print months_between( datetime(2011, 02, 26), datetime(2011, 03, 05))
print months_between( datetime(2012, 02, 26), datetime(2012, 03, 05))
print months_between( datetime(2012, 03, 26), datetime(2012, 04, 05))

Answer 16

尝试这样的事情。 如果两个日期恰好在同一个月份，则它目前包括月份。

from datetime import datetime,timedelta

def months_between(start,end):
    months = []
    cursor = start

    while cursor <= end:
        if cursor.month not in months:
            months.append(cursor.month)
        cursor += timedelta(weeks=1)

    return months

输出如下所示：

>>> start = datetime.now() - timedelta(days=120)
>>> end = datetime.now()
>>> months_between(start,end)
[6, 7, 8, 9, 10]

Answer 17

获取两个日期之间的天数、月数和年数的差异。

import datetime    
from dateutil.relativedelta import relativedelta


iphead_proc_dt = datetime.datetime.now()
new_date = iphead_proc_dt + relativedelta(months=+25, days=+23)

# Get Number of Days difference bewtween two dates
print((new_date - iphead_proc_dt).days)

difference = relativedelta(new_date, iphead_proc_dt)

# Get Number of Months difference bewtween two dates
print(difference.months + 12 * difference.years)

# Get Number of Years difference bewtween two dates
print(difference.years)

Answer 18

您可以使用python-dateutil 。 请参阅Python：2 个日期时间的月差

Answer 19

就像range函数一样，当月份为13时，转到下一年

def year_month_range(start_date, end_date):
    '''
    start_date: datetime.date(2015, 9, 1) or datetime.datetime
    end_date: datetime.date(2016, 3, 1) or datetime.datetime
    return: datetime.date list of 201509, 201510, 201511, 201512, 201601, 201602
    '''
    start, end = start_date.strftime('%Y%m'), end_date.strftime('%Y%m')
    assert len(start) == 6 and len(end) == 6
    start, end = int(start), int(end)

    year_month_list = []
    while start < end:
        year, month = divmod(start, 100)
        if month == 13:
            start += 88  # 201513 + 88 = 201601
            continue
        year_month_list.append(datetime.date(year, month, 1))

        start += 1
    return year_month_list

python shell中的示例

>>> import datetime
>>> s = datetime.date(2015,9,1)
>>> e = datetime.date(2016, 3, 1)
>>> year_month_set_range(s, e)
[datetime.date(2015, 11, 1), datetime.date(2015, 9, 1), datetime.date(2016, 1, 1), datetime.date(2016, 2, 1),
 datetime.date(2015, 12, 1), datetime.date(2015, 10, 1)]

Answer 20

可以使用datetime.timedelta来完成，其中跳到下个月的天数可以通过calender.monthrange获取。 monthrange 返回给定年份和月份的工作日 (0-6 ~ Mon-Sun) 和天数 (28-31)。
例如：monthrange(2017, 1) 返回 (6,31)。

这是使用此逻辑在两个月之间迭代的脚本。

from datetime import timedelta
import datetime as dt
from calendar import monthrange

def month_iterator(start_month, end_month):
    start_month = dt.datetime.strptime(start_month,
                                   '%Y-%m-%d').date().replace(day=1)
    end_month = dt.datetime.strptime(end_month,
                                 '%Y-%m-%d').date().replace(day=1)
    while start_month <= end_month:
        yield start_month
        start_month = start_month + timedelta(days=monthrange(start_month.year, 
                                                         start_month.month)[1])

`

Answer 21

from datetime import datetime
from dateutil import relativedelta

def get_months(d1, d2):
    date1 = datetime.strptime(str(d1), '%Y-%m-%d')
    date2 = datetime.strptime(str(d2), '%Y-%m-%d')
    print (date2, date1)
    r = relativedelta.relativedelta(date2, date1)
    months = r.months +  12 * r.years
    if r.days > 0:
        months += 1
    print (months)
    return  months


assert  get_months('2018-08-13','2019-06-19') == 11
assert  get_months('2018-01-01','2019-06-19') == 18
assert  get_months('2018-07-20','2019-06-19') == 11
assert  get_months('2018-07-18','2019-06-19') == 12
assert  get_months('2019-03-01','2019-06-19') == 4
assert  get_months('2019-03-20','2019-06-19') == 3
assert  get_months('2019-01-01','2019-06-19') == 6
assert  get_months('2018-09-09','2019-06-19') == 10

Answer 22

#This definition gives an array of months between two dates.
import datetime
def MonthsBetweenDates(BeginDate, EndDate):
    firstyearmonths = [mn for mn in range(BeginDate.month, 13)]<p>
    lastyearmonths = [mn for mn in range(1, EndDate.month+1)]<p>
    months = [mn for mn in range(1, 13)]<p>
    numberofyearsbetween = EndDate.year - BeginDate.year - 1<p>
    return firstyearmonths + months * numberofyearsbetween + lastyearmonths<p>

#example
BD = datetime.datetime.strptime("2000-35", '%Y-%j')
ED = datetime.datetime.strptime("2004-200", '%Y-%j')
MonthsBetweenDates(BD, ED)

Answer 23

通常 90 天并不是字面上的 3 个月，只是一个参考。

因此，最后，您需要检查天数是否大于 15 才能将 +1 添加到月份计数器。 或者更好的是，添加另一个带有半个月计数器的 elif。

从this other stackoverflow answer我终于结束了：

#/usr/bin/env python
# -*- coding: utf8 -*-

import datetime
from datetime import timedelta
from dateutil.relativedelta import relativedelta
import calendar

start_date = datetime.date.today()
end_date = start_date + timedelta(days=111)
start_month = calendar.month_abbr[int(start_date.strftime("%m"))]

print str(start_date) + " to " + str(end_date)

months = relativedelta(end_date, start_date).months
days = relativedelta(end_date, start_date).days

print months, "months", days, "days"

if days > 16:
    months += 1

print "around " + str(months) + " months", "(",

for i in range(0, months):
    print calendar.month_abbr[int(start_date.strftime("%m"))],
    start_date = start_date + relativedelta(months=1)

print ")"

输出：

2016-02-29 2016-06-14
3 months 16 days
around 4 months ( Feb Mar Apr May )

我注意到，如果您添加的天数超过当年剩余的天数，这将不起作用，这是出乎意料的。

Answer 24

似乎答案不能令人满意，我已经使用了我自己的代码，这更容易理解

from datetime import datetime
from dateutil import relativedelta

date1 = datetime.strptime(str('2017-01-01'), '%Y-%m-%d')
date2 = datetime.strptime(str('2019-03-19'), '%Y-%m-%d')

difference = relativedelta.relativedelta(date2, date1)
months = difference.months
years = difference.years
# add in the number of months (12) for difference in years
months += 12 * difference.years
months

Answer 25

这是我的解决方案：

def calc_age_months(from_date, to_date):
    from_date = time.strptime(from_date, "%Y-%m-%d")
    to_date = time.strptime(to_date, "%Y-%m-%d")

    age_in_months = (to_date.tm_year - from_date.tm_year)*12 + (to_date.tm_mon - from_date.tm_mon)

    if to_date.tm_mday < from_date.tm_mday:
        return age_in_months -1
    else
        return age_in_months

这也将处理一些极端情况，其中 2018 年 12 月 31 日和 2019 年 1 月 1 日之间的月份差异为零（因为差异只有一天）。

Answer 26

我现在实际上需要做一些非常相似的事情

最后写了一个函数，它返回一个元组列表，指示两组日期之间每个月的start和end ，这样我就可以在它的后面写一些 SQL 查询来获取每月的销售总额等。

我敢肯定，知道自己在做什么的人可以改进它，但希望对您有所帮助...

返回值如下（以今天生成 - 到今天为止的 365 天为例）

[   (datetime.date(2013, 5, 1), datetime.date(2013, 5, 31)),
    (datetime.date(2013, 6, 1), datetime.date(2013, 6, 30)),
    (datetime.date(2013, 7, 1), datetime.date(2013, 7, 31)),
    (datetime.date(2013, 8, 1), datetime.date(2013, 8, 31)),
    (datetime.date(2013, 9, 1), datetime.date(2013, 9, 30)),
    (datetime.date(2013, 10, 1), datetime.date(2013, 10, 31)),
    (datetime.date(2013, 11, 1), datetime.date(2013, 11, 30)),
    (datetime.date(2013, 12, 1), datetime.date(2013, 12, 31)),
    (datetime.date(2014, 1, 1), datetime.date(2014, 1, 31)),
    (datetime.date(2014, 2, 1), datetime.date(2014, 2, 28)),
    (datetime.date(2014, 3, 1), datetime.date(2014, 3, 31)),
    (datetime.date(2014, 4, 1), datetime.date(2014, 4, 30)),
    (datetime.date(2014, 5, 1), datetime.date(2014, 5, 31))]

代码如下（有一些可以删除的调试内容）：

#! /usr/env/python
import datetime

def gen_month_ranges(start_date=None, end_date=None, debug=False):
    today = datetime.date.today()
    if not start_date: start_date = datetime.datetime.strptime(
        "{0}/01/01".format(today.year),"%Y/%m/%d").date()  # start of this year
    if not end_date: end_date = today
    if debug: print("Start: {0} | End {1}".format(start_date, end_date))

    # sense-check
    if end_date < start_date:
        print("Error. Start Date of {0} is greater than End Date of {1}?!".format(start_date, end_date))
        return None

    date_ranges = []  # list of tuples (month_start, month_end)

    current_year = start_date.year
    current_month = start_date.month

    while current_year <= end_date.year:
        next_month = current_month + 1
        next_year = current_year
        if next_month > 12:
            next_month = 1
            next_year = current_year + 1

        month_start = datetime.datetime.strptime(
            "{0}/{1}/01".format(current_year,
                                current_month),"%Y/%m/%d").date()  # start of month
        month_end = datetime.datetime.strptime(
            "{0}/{1}/01".format(next_year,
                                next_month),"%Y/%m/%d").date()  # start of next month
        month_end  = month_end+datetime.timedelta(days=-1)  # start of next month less one day

        range_tuple = (month_start, month_end)
        if debug: print("Month runs from {0} --> {1}".format(
            range_tuple[0], range_tuple[1]))
        date_ranges.append(range_tuple)

        if current_month == 12:
            current_month = 1
            current_year += 1
            if debug: print("End of year encountered, resetting months")
        else:
            current_month += 1
            if debug: print("Next iteration for {0}-{1}".format(
                current_year, current_month))

        if current_year == end_date.year and current_month > end_date.month:
            if debug: print("Final month encountered. Terminating loop")
            break

    return date_ranges


if __name__ == '__main__':
    print("Running in standalone mode. Debug set to True")
    from pprint import pprint
    pprint(gen_month_ranges(debug=True), indent=4)
    pprint(gen_month_ranges(start_date=datetime.date.today()+datetime.timedelta(days=-365),
                            debug=True), indent=4)

Answer 27

假设您想知道日期所在月份的“分数”，我就是这样做的，那么您需要做更多的工作。

from datetime import datetime, date
import calendar

def monthdiff(start_period, end_period, decimal_places = 2):
    if start_period > end_period:
        raise Exception('Start is after end')
    if start_period.year == end_period.year and start_period.month == end_period.month:
        days_in_month = calendar.monthrange(start_period.year, start_period.month)[1]
        days_to_charge = end_period.day - start_period.day+1
        diff = round(float(days_to_charge)/float(days_in_month), decimal_places)
        return diff
    months = 0
    # we have a start date within one month and not at the start, and an end date that is not
    # in the same month as the start date
    if start_period.day > 1:
        last_day_in_start_month = calendar.monthrange(start_period.year, start_period.month)[1]
        days_to_charge = last_day_in_start_month - start_period.day +1
        months = months + round(float(days_to_charge)/float(last_day_in_start_month), decimal_places)
        start_period = datetime(start_period.year, start_period.month+1, 1)

    last_day_in_last_month = calendar.monthrange(end_period.year, end_period.month)[1]
    if end_period.day != last_day_in_last_month:
        # we have lest days in the last month
        months = months + round(float(end_period.day) / float(last_day_in_last_month), decimal_places)
        last_day_in_previous_month = calendar.monthrange(end_period.year, end_period.month - 1)[1]
        end_period = datetime(end_period.year, end_period.month - 1, last_day_in_previous_month)

    #whatever happens, we now have a period of whole months to calculate the difference between

    if start_period != end_period:
        months = months + (end_period.year - start_period.year) * 12 + (end_period.month - start_period.month) + 1

    # just counter for any final decimal place manipulation
    diff = round(months, decimal_places)
    return diff

assert monthdiff(datetime(2015,1,1), datetime(2015,1,31)) == 1
assert monthdiff(datetime(2015,1,1), datetime(2015,02,01)) == 1.04
assert monthdiff(datetime(2014,1,1), datetime(2014,12,31)) == 12
assert monthdiff(datetime(2014,7,1), datetime(2015,06,30)) == 12
assert monthdiff(datetime(2015,1,10), datetime(2015,01,20)) == 0.35
assert monthdiff(datetime(2015,1,10), datetime(2015,02,20)) == 0.71 + 0.71
assert monthdiff(datetime(2015,1,31), datetime(2015,02,01)) == round(1.0/31.0,2) + round(1.0/28.0,2)
assert monthdiff(datetime(2013,1,31), datetime(2015,02,01)) == 12*2 + round(1.0/31.0,2) + round(1.0/28.0,2)

提供了一个示例，计算两个日期之间的月数，包括日期所在的每个月的分数。这意味着您可以计算出 2015-01-20 和 2015-02-14 之间的月数，其中一月份日期的分数由一月份的天数确定； 或者同样考虑到二月份的天数可以逐年变化。

供我参考，此代码也在 github 上 - https://gist.github.com/andrewyager/6b9284a4f1cdb1779b10

Answer 28

这有效...

from datetime import datetime as dt
from dateutil.relativedelta import relativedelta
def number_of_months(d1, d2):
    months = 0
    r = relativedelta(d1,d2)
    if r.years==0:
        months = r.months
    if r.years>=1:
        months = 12*r.years+r.months
    return months
#example 
number_of_months(dt(2017,9,1),dt(2016,8,1))

Answer 29

假设 upperDate 总是晚于 lowerDate 并且两者都是 datetime.date 对象：

if lowerDate.year == upperDate.year:
    monthsInBetween = range( lowerDate.month + 1, upperDate.month )
elif upperDate.year > lowerDate.year:
    monthsInBetween = range( lowerDate.month + 1, 12 )
    for year in range( lowerDate.year + 1, upperDate.year ):
        monthsInBetween.extend( range(1,13) )
    monthsInBetween.extend( range( 1, upperDate.month ) )

我还没有彻底测试过，但看起来它应该可以解决问题。

Answer 30

尝试这个：

 dateRange = [datetime.strptime(dateRanges[0], "%Y-%m-%d"),
             datetime.strptime(dateRanges[1], "%Y-%m-%d")]
delta_time = max(dateRange) - min(dateRange)
#Need to use min(dateRange).month to account for different length month
#Note that timedelta returns a number of days
delta_datetime = (datetime(1, min(dateRange).month, 1) + delta_time -
                           timedelta(days=1)) #min y/m/d are 1
months = ((delta_datetime.year - 1) * 12 + delta_datetime.month -
          min(dateRange).month)
print months

输入日期的顺序无关紧要，它会考虑月份长度的差异。

Answer 31

这是一个方法：

def months_between(start_dt, stop_dt):
    month_list = []
    total_months = 12*(stop_dt.year-start_dt.year)+(stop_dt.month-start_d.month)+1
    if total_months > 0:
        month_list=[ datetime.date(start_dt.year+int((start_dt+i-1)/12), 
                                   ((start_dt-1+i)%12)+1,
                                   1) for i in xrange(0,total_months) ]
    return month_list

这是首先计算两个日期之间的总月数，包括两个日期。 然后它使用第一个日期作为基础创建一个列表，并执行模运算来创建日期对象。

Answer 32

from datetime import datetime

def diff_month(start_date,end_date):
    qty_month = ((end_date.year - start_date.year) * 12) + (end_date.month - start_date.month)

    d_days = end_date.day - start_date.day

    if d_days >= 0:
        adjust = 0
    else:
        adjust = -1
    qty_month += adjust

    return qty_month

diff_month(datetime.date.today(),datetime(2019,08,24))


#Examples:
#diff_month(datetime(2018,02,12),datetime(2019,08,24)) = 18
#diff_month(datetime(2018,02,12),datetime(2018,08,10)) = 5

Answer 33

这是我这样做的方法：

Start_date = "2000-06-01"
End_date   = "2001-05-01"

month_num  = len(pd.date_range(start = Start_date[:7], end = End_date[:7] ,freq='M'))+1

我只是使用月份来创建日期范围并计算长度。

Answer 34

要获取两个日期之间的整月数：

import datetime

def difference_in_months(start, end):
    if start.year == end.year:
        months = end.month - start.month
    else:
        months = (12 - start.month) + (end.month)

    if start.day > end.day:
        months = months - 1

    return months

Answer 35

您可以使用以下代码获取两个日期之间的月份：

OrderedDict(((start_date + timedelta(_)).strftime(date_format), None) for _ in xrange((end_date - start_date).days)).keys()

其中start_date和end_date必须是正确的日期，而 date_format 是您希望日期结果的格式。

在您的情况下， date_format将为%b %Y 。

Answer 36

问题，实际上是询问两个日期之间的总月数，而不是它的差异

因此，具有一些额外功能的重新审视的答案，

from datetime import date, datetime
from dateutil.rrule import rrule, MONTHLY

def month_get_list(dt_to, dt_from, return_datetime=False, as_month=True):
    INDEX_MONTH_MAPPING = {1: 'january', 2: 'february', 3: 'march', 4: 'april', 5: 'may', 6: 'june', 7: 'july',
                           8: 'august',
                           9: 'september', 10: 'october', 11: 'november', 12: 'december'}
    if return_datetime:
        return [dt for dt in rrule(MONTHLY, dtstart=dt_from, until=dt_to)]
    if as_month:
        return [INDEX_MONTH_MAPPING[dt.month] for dt in rrule(MONTHLY, dtstart=dt_from, until=dt_to)]

    return [dt.month for dt in rrule(MONTHLY, dtstart=dt_from, until=dt_to)]

month_list = month_get_list(date(2021, 12, 31), date(2021, 1, 1))
total_months = len(month_list)

结果

month_list = ['january', 'february', 'march', 'april', 'may', 'june', 'july', 'august', 'september', 'october', 'november', 'december']
total_months = 12

as_month设置为 False

month_list = month_get_list(日期(2021, 12, 31), 日期(2021, 1, 1),

as_month=False)
# month_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

而是返回日期时间

month_list = month_get_list(date(2021, 12, 31), date(2021, 1, 1), return_datetime=True)
month_list = [datetime.datetime(2021, 1, 1, 0, 0), datetime.datetime(2021, 2, 1, 0, 0), datetime.datetime(2021, 3, 1, 0, 0), datetime.datetime(2021, 4, 1, 0, 0), datetime.datetime(2021, 5, 1, 0, 0), datetime.datetime(2021, 6, 1, 0, 0), datetime.datetime(2021, 7, 1, 0, 0), datetime.datetime(2021, 8, 1, 0, 0), datetime.datetime(2021, 9, 1, 0, 0), datetime.datetime(2021, 10, 1, 0, 0), datetime.datetime(2021, 11, 1, 0, 0), datetime.datetime(2021, 12, 1, 0, 0)]

Answer 37

如果月份的小数部分对您很重要，请利用下个月的第一天来计算每个月可能有的天数，无论是否在一个步骤年：

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

def month_fraction(this_date):
        this_date_df=this_date.strftime("%Y-%m-%d")
        day1_same_month_as_this_date_df=this_date_df[:8]+'01'
        day1_same_month_as_this_date=datetime.strptime(day1_same_month_as_this_date_df, '%Y-%m-%d').date()
        next_mon    th_as_this_date=this_date+relativedelta(months=1)
        next_month_as_this_date_df=next_month_as_this_date.strftime("%Y-%m-%d")
        day1_next_month_this_date_df=next_month_as_this_date_df[:8]+'01'
        day1_next_month_this_date=datetime.strptime(day1_next_month_this_date_df, '%Y-%m-%d').date()
        last_day_same_month_this_date=day1_next_month_this_date-timedelta(days=1)
        delta_days_from_month_beginning=(this_date-day1_same_month_as_this_date).days
        delta_days_whole_month=(last_day_same_month_this_date-day1_same_month_as_this_date).days    
        fraction_beginning_of_month=round(delta_days_from_month_beginning/delta_days_whole_month,4)
        return fraction_beginning_of_month

def delta_months_JLR(second_date,first_date):
        return (second_date.year - first_date.year) * 12 + second_date.month - first_date.month

def delta_months_float(first_date,second_date):
        outgoing_fraction_first_date  =  month_fraction(first_date)
        incoming_fraction_second_date  =  month_fraction(second_date)
        delta_months=delta_months_JLR(second_date,first_date) #as on John La Rooy’s response
        months_float=round(delta_months-outgoing_fraction_first_date+incoming_fraction_second_date,4)
        return months_float


first_date_df='2021-12-28'
first_date=datetime.strptime(first_date, '%Y-%m-%d').date()
second_date_df='2022-01-02'
second_date=datetime.strptime(second_date, '%Y-%m-%d').date()

print (delta_months_float(first_date,second_date))
0.1333

Answer 38

请给出以下功能的开始和结束日期，所有月份的开始和结束日期都可以。

$months = $this->getMonthsInRange('2022-03-15', '2022-07-12');
    
    
    public function getMonthsInRange($startDate, $endDate)
            {
                $months = array();
                while (strtotime($startDate) <= strtotime($endDate)) {
                    $start_date = date('Y-m-d', strtotime($startDate));
                    $end_date = date("Y-m-t", strtotime($start_date));
                    if(strtotime($end_date) >= strtotime($endDate)) {
                        $end_date = $endDate;
                    }
                    $months[] = array(
                        'start_date' => $start_date,
                        'end_date' => $end_date
                    );
                    $startDate = date('01 M Y', strtotime($startDate . '+ 1 month'));
                }
                return $months;
            }

Answer 39

这适用于 Python 3

import datetime

start = datetime.datetime.today()
end = start + datetime.timedelta(days=365)
month_range = [start.date()]
diff = end - start

months = round(diff.days / 30)
temp = start
for i in range(months):
    temp = temp + datetime.timedelta(days=30)
    if temp <= end:
        month_range.append(temp.date())

print(month_range)

Answer 40

你可以使用类似的东西：

import datetime
days_in_month = 365.25 / 12  # represent the average of days in a month by year
month_diff = lambda end_date, start_date, precision=0: round((end_date - start_date).days / days_in_month, precision)
start_date = datetime.date(1978, 12, 15)
end_date = datetime.date(2012, 7, 9)
month_diff(end_date, start_date)  # should show 403.0 months

Answer 41

import datetime
from calendar import monthrange

def date_dif(from_date,to_date):   # Изчислява разлика между две дати
    dd=(to_date-from_date).days
    if dd>=0:
        fromDM=from_date.year*12+from_date.month-1
        toDM=to_date.year*12+to_date.month-1
        mlen=monthrange(int((toDM)/12),(toDM)%12+1)[1]
        d=to_date.day-from_date.day
        dm=toDM-fromDM
        m=(dm-int(d<0))%12
        y=int((dm-int(d<0))/12)
        d+=int(d<0)*mlen
        # diference in Y,M,D, diference months,diference  days, days in to_date month
        return[y,m,d,dm,dd,mlen]
    else:
        return[0,0,0,0,dd,0]