python 计算两个日期相差多少个月实例代码

站长资源 2024/11/25 佚名

2 0 1

近期，由于业务需要计算两个日期之前相差多少个月。我在网上找了很久，结果发现万能的python，居然没有一个模块计算两个日期的月数，像Java、C#之类的高级语言，都会有(date1-date2).months的现成方法，觉得不可思议。说句实在的，一直觉得python 的日期处理模块真心不好用。

哦，对了，别跟我说 datetime, calendar, dateutil 这些模块，因为我都试过了，都没用。有个竟然算出来还有错。datetime.timedelta只能计算出日时分秒。对年月却不支持。网上一搜，一大堆的。

dateutil.rrule这个例子是我好不容易找到的，请看我的测试结果：

import datetime

from dateutil import rrule

d1 = datetime.date(2016, 2, 29)
d2 = datetime.date(2019, 3, 31)

months = rrule.rrule(rrule.MONTHLY, dtstart=d1, until=d2).count()

print(f"months={months}")

看到这样的结果，我只能呵呵了。

废话不多少，献上自己写的代码：

该代码返回，（月,小数月）

import datetime

import calendar as c



d1 = datetime.date(2016, 2, 29)

d2 = datetime.date(2019, 3, 31)



def calmonths(startdate, enddate):

# 计算两个日期相隔月差

samemonthdate = None

try:

samemonthdate = datetime.date(enddate.year, enddate.month,

startdate.day)

except Exception as e:

print(e)

samemonthdate = datetime.date(enddate.year, enddate.month,

c.monthrange(enddate.year,

enddate.month)[1])


holdmonths = 0

decimalmonth = 0.0

if samemonthdate > enddate:

premanthdate = None

try:

premanthdate = datetime.date(enddate.year, enddate.month - 1,

startdate.day)

except Exception as e:

print(e)

premanthdate = datetime.date(enddate.year, enddate.month - 1,

c.monthrange(

enddate.year,

enddate.month - 1)[1])

currmonthdays = (samemonthdate - premanthdate).days

holdmonths = (premanthdate.year - startdate.year

) * 12 + premanthdate.month - startdate.month

decimalmonth = (enddate - premanthdate).days / currmonthdays


elif samemonthdate < enddate:

nextmonthdate = None

try:

nextmonthdate = datetime.date(enddate.year, enddate.month + 1,

startdate.day)

except Exception as e:

nextmonthdate = datetime.date(enddate.year, enddate.month + 1,

c.monthrange(

enddate.year,

enddate.month + 1)[1])

currmonthdays = (nextmonthdate - samemonthdate).days

holdmonths = (samemonthdate.year - startdate.year

) * 12 + samemonthdate.month - startdate.month

decimalmonth = (enddate - samemonthdate).days / currmonthdays


else:

holdmonths = (enddate.year - startdate.year

) * 12 + enddate.month - startdate.month


return holdmonths, decimalmonth


months = calmonths(d1, d2)


print(f"months={months}")