python - 查找两个日期之间月份的最佳方法

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，我在这里做的是获取两个日期并将它们从 iso 格式转换为 python datetime 对象。然后我循环将一周添加到开始日期时间对象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），如果值更大我将其附加到列表中几个月，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......

score 237 · Accepted Answer

Start by defining some test cases, then you will see that the function is very simple and needs no loops

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

You should add some test cases to your question, as there are lots of potential corner cases to cover - there is more than one way to define the number of months between two dates.

score 58 · Accepted Answer

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)]

score 45 · Accepted Answer

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)

score 14 · Accepted Answer

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)]

score 11 · Accepted Answer

from dateutil import relativedelta

r = relativedelta.relativedelta(date1, date2)

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

score 10 · Accepted Answer

获取结束月份（相对于开始月份的年份和月份，例如：如果开始日期从 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)]

score 8 · Accepted Answer

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

score 7 · Accepted Answer

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))

score 6 · Accepted Answer

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）。

score 5 · Accepted Answer

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，似乎需要在两个日期之间提出2018-04-11问题2018-06-01

Apr 2018, May 2018, June 2018

如果它介于2014-04-11to之间2018-06-01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。

score 4 · Accepted Answer

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))

score 4 · Accepted Answer

@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)

score 4 · Accepted Answer

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！

score 4 · Accepted Answer

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']

score 4 · Accepted Answer

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。

score 3 · Accepted Answer

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]

score 3 · Accepted Answer

3

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

于 2011-08-10T18:25:20.867 回答

score 2 · Accepted Answer

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

score 2 · Accepted Answer

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)

score 2 · Accepted Answer

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])

`

score 1 · Accepted Answer

#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)

score 1 · Accepted Answer

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

score 1 · Accepted Answer

就像range函数一样，当月份是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)]

score 1 · Accepted Answer

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。

score 1 · Accepted Answer

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 日之间的月差为零（因为差只有一天）。

score 0 · Accepted Answer

Here is a method:

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

This is first computing the total number of months between the two dates, inclusive. Then it creates a list using the first date as the base and performs modula arithmetic to create the date objects.

score 0 · Accepted Answer

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。

score 0 · Accepted Answer

我现在实际上需要做一些非常相似的事情

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

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)

score 0 · Accepted Answer

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。

score 0 · Accepted Answer

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()

where start_dateand end_datemust be proper date and date_format 是您希望日期结果的格式。

在你的情况下，date_format将是%b %Y.

score 0 · Accepted Answer

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))

score 0 · Accepted Answer

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

score 0 · Accepted Answer

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)]

score 0 · Accepted Answer

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。

score 0 · Accepted Answer

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

score 0 · Accepted Answer

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

score -1 · Accepted Answer

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

score -1 · Accepted Answer

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]

python - 查找两个日期之间月份的最​​佳方法

38 回答 38

获取两个日期之间的天数、月数和年数的差异。

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python - 查找两个日期之间月份的最佳方法