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，我在这里做的是将两个日期转换为iso格式转换为python datetime对象。然后我循环通过在开始日期时间对象中添加一周，并检查月份的数值是否更大（除非月份是12月，然后检查日期是否更少），如果值更大，我将其附加到列表几个月，并一直循环，直到我到达我的结束日期。
它完美无缺，它似乎不是一种好方法......
答案 0 :(得分：159)
首先定义一些测试用例，然后你会发现该函数非常简单，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
您应该在问题中添加一些测试用例，因为有很多潜在的角落案例要涵盖 - 有两种方法可以定义两个日期之间的月数。
答案 1 :(得分：33)
一个班轮，用于查找两个日期之间按月递增的日期时间列表。
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)]
答案 2 :(得分：24)
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 * (r.years+1)
答案 3 :(得分：10)
获取结束月份（相对于开始月份的年份和月份，例如：2011年1月= 13，如果您的开始日期从2010年10月开始），然后生成从开始月份和结束月份开始的日期时间，如下所示： / p>
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，它也可以简单地写成：
dates=[datetime.datetime(year=dt1.year, month=mn, day=1) for mn in range(dt1.month, dt2.month + 1)]
答案 4 :(得分：9)
您可以使用dateutil模块中的rrule轻松计算出来：
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)]
答案 5 :(得分：5)
This post指甲！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
答案 6 :(得分：4)
您还可以使用arrow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
希望这有帮助！
答案 7 :(得分：4)
@ Vin-G稍微有点美化解决方案。
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)
答案 8 :(得分：3)
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))
答案 9 :(得分：3)
有一个基于360天的简单解决方案，所有月份都有30天。 它适用于大多数用例，在给定两个日期的情况下，您需要计算完整月数和剩余天数。
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))
答案 10 :(得分：3)
答案 11 :(得分：3)
尝试这样的事情。如果两个日期恰好在同一个月，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：
>>> start = datetime.now() - timedelta(days=120)
>>> end = datetime.now()
>>> months_between(start,end)
[6, 7, 8, 9, 10]
答案 12 :(得分：2)
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
您可能会尝试将某月定义为“29天，28天，30天或31天（取决于年份）”。但是，如果你这样做，你还有一个问题需要解决。
虽然通常很明显6月15日 th + 1个月应该是7月15日 th ，但通常不清楚1月30日 th + 1个月是2月或3月。在后一种情况下，您可能被迫将日期计算为2月30日 th ，然后将其“更正”为3月2日 nd 。但是当你这样做时，你会发现3月2日 nd - 1个月显然是2月2日 nd 。 Ergo，reductio ad absurdum（此操作没有明确定义）。
答案 13 :(得分：1)
#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)
答案 14 :(得分：1)
许多人已经给您很好的答案来解决此问题，但是我还没有阅读任何使用列表理解的文章，因此我向您提供了我用于类似用例的内容：
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']
答案 15 :(得分：1)
要获取两个日期之间的完整月份数：
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
答案 16 :(得分：1)
就像range函数一样，当月份 13 时，请转到明年
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)]
答案 17 :(得分：1)
通常90天不是字面上的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，才能为月份计数器添加+1。或者更好的是，添加另一个半月计数器的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 )
我注意到如果你在当年添加的天数超过了几天就行不通了，这是出乎意料的。
答案 18 :(得分：1)
可以使用datetime.timedelta完成，其中可以通过calender.monthrange获得跳至下个月的天数。 monthrange返回给定年份和月份的工作日（0-6〜周一至周日）和天数（28-31）。
例如：monthrange（2017，1）返回（6,31）。
以下是使用此逻辑在两个月之间进行迭代的脚本。
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])
`
答案 19 :(得分：1)
以下是如何使用Pandas FWIW：
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，它以之后给定的开始日期开始。
答案 20 :(得分：0)
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
答案 21 :(得分：0)
这是我的方法：
Get property() { ret Date.now()}
我只是使用月份创建日期范围并计算长度。
答案 22 :(得分：0)
这是我的解决方案：
DateFormatter这还将处理一些边缘情况，这些情况在2018年12月31日到2019年1月1日之间的月份差异为零（因为差异仅为一天）。
答案 23 :(得分：0)
答案似乎不尽人意，此后我一直使用自己的代码，这更易于理解
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
答案 24 :(得分：0)
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)
答案 25 :(得分：0)
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
答案 26 :(得分：0)
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))
答案 27 :(得分：0)
假设您想知道日期所在的月份的“分数”，我做了，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，包括日期所在的每月的分数。这意味着您可以计算出2015-01-20和2015之间的月数 - 02-14，1月份的日期分数由1月份的天数决定;或者同样考虑到2月份的天数每年都会发生变化。
供我参考，此代码也在github上 - https://gist.github.com/andrewyager/6b9284a4f1cdb1779b10
答案 28 :(得分：0)
我实际上需要做一些非常相似的事情
结束编写一个函数，该函数返回一个元组列表，指示两组日期之间每个月的start和end，这样我就可以在其后面写一些SQL查询以获得每月总计销售等。
我相信一个知道自己在做什么的人可以改善它，但希望它有所帮助...
返回的值如下所示（今天生成 - 直到今天为例365天）
[ (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)
答案 29 :(得分：0)
这是一种方法：
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（包括两者）。然后它使用第一个日期作为基础创建一个列表，并执行模块运算来创建日期对象。
答案 30 :(得分：0)
试试这个：
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
您输入日期的顺序无关紧要，并且考虑到月份长度的差异。
答案 31 :(得分：0)
假设upperDate总是晚于lowerDate，并且都是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 ) )
我没有对此进行过彻底的测试，但看起来应该可以解决这个问题。
答案 32 :(得分：-1)
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]
答案 33 :(得分：-1)
您可以使用以下内容：
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
答案 34 :(得分：-3)
更新2018-04-20：似乎OP @Joshkunz要求找到哪个月介于两个日期之间，而不是“多少个月”之间两个约会。所以我不确定为什么@JohnLaRooy被投票超过100次。 @Joshkunz在原始问题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到2018-06-01
Apr 2018, May 2018, June 2018
如果是2014-04-11到2018-06-01之间怎么办？那么答案就是
Apr 2014, May 2014, ..., Dec 2014, Jan 2015, ..., Jan 2018, ..., June 2018
这就是我多年前拥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通过它们，一次增加一个月。 @Joshkunz提到他想要“月”，他还提到他想要“约会”，但不知道确切地说，编写确切的代码很困难，但想法是使用一个简单的循环来遍历终点，并且一次递增一个月。
8年前的答案2010年：
如果按一周添加，那么它将大约按需要工作4.35倍。为什么不呢：
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pseduo代码，尚未经过测试，但我认为沿着同一条线的想法是可行的。