最好的方法来find两个date之间的月份

我有需要能够准确地find两个date之间的python的几个月。我有一个解决scheme，但不是很好（如在优雅）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，我在这里做的是把两个date，并从iso格式转换成pythondate时间对象。然后我循环通过增加一个星期到开始date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date是否较less），如果值更大，我将它追加到列表几个月，并保持循环，直到我到我的结束date。

它完美的工作，似乎不是一个好办法…

如果增加一个星期，那么它将近似地按照需要工作4.35倍的工作。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pseduo的代码，还没有testing过，但我认为沿同一条线的想法将工作。

首先定义一些testing用例，然后你会看到这个函数非常简单，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

您应该为您的问题添加一些testing案例，因为有很多潜在的angular落案例可以覆盖 – 有两种以上的方法来定义两个date之间的月份数。

一个class轮在两个date之间查找date时间列表，递增月份。

 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)]

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

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，如果开始date从2010年10月开始），然后生成开始月份和结束月份的date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date都在同一年，那么也可以简单地写成：

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

您可以使用从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)]

这篇文章钉了它！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

有点美化解决scheme@ Vin-G。

 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)

尝试这样的事情。如果两个date恰好在同一个月份，它现在包括月份。

 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]

你可以使用python-dateutil 。请参阅Python：2个date时间的差异

您也可以使用箭头库。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！

有一个基于360天的简单解决scheme，所有的月份都有30天。它适合大多数使用情况，在给定两个date的情况下，您需要计算完整月份的数量加上剩下的date。

 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))

将“月”定义为¹ / ₁₂年，然后执行此操作：

 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）”。但是你这样做，你还有一个额外的问题需要解决。

通常情况下，6月15 ^日 + 1月应该是7月15 ^日，通常情况下并不清楚1月30 ^日 + 1月是2月还是3月。在后一种情况下，你可能被迫计算date为2月30 ^日，然后“纠正”到3月2日。但是当你这样做的时候，你会发现3月2日 – 1月显然是2月2日。人类，还原和荒诞（这个操作没有很好的定义）。

 #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)

就像rangefunction一样，当月份是13时 ，进入下一年

 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

在python shell中的例子

 >>> 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)]

以下是如何用pandasFWIW做到这一点：

 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')

注意它从给定开始date之后的月份开始。

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 ) )

我没有彻底地testing过，但看起来它应该做的伎俩。

这里是一个方法：

 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

首先计算两个date之间的总月数。然后它使用第一个date作为基础创build一个列表并执行modula算术来创builddate对象。

我其实刚刚需要做一些非常类似的事情

最后写了一个函数，它返回一组元组列表，指出两个date之间的每个月的start和end ，所以我可以写一些SQL查询后面的月销售总额等。

我相信可以通过一个知道自己在做什么的人来改善它，但是希望这有助于…

返回的值如下所示（今天生成 – 直到今天365天为止）

 [ (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debugging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)

假设你想知道date所在月份的“分数”，那么我需要做更多的工作。

 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)

提供一个计算两个date之间的月份数的示例，包括date所在的每个月份的小数部分。这意味着您可以计算2015-01-20和2015-02-14之间的月份数其中1月份的date部分由1月份的天数决定; 或同样考虑到二月份的天数可能每年都会改变。

对于我的参考，这个代码也在github上 – https://gist.github.com/andrewyager/6b9284a4f1cdb1779b10

通常90天不是字面上的3个月，只是一个参考。

所以，最后，你需要检查天数是否大于15，为月份计数器添加+1。或者更好，再加上一个半月柜台elif。

从这个其他的stackoverflow答案我终于结束了：

 #/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，如果你在当年剩余的时间多于这个时间，这是行不通的，那是意想不到的。

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

不pipe你inputdate的顺序是什么，它都考虑到月份长度的差异。

这工作…

 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))

你可以使用像这样的东西：

 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