准确获取跨越不同年份的两个日期之间的月数的最佳方法

Best way to accurately get the number of months between two dates spanning different years

我需要能够准确地找到两个日期之间的月数,即使它们跨越不同的年份。我从这个 Whosebug thread 中获得了以下解决方案,但在日期跨度不同的年份时它失败了。

跨越一年有效

from datetime import datetime
from dateutil.relativedelta import relativedelta

date1 = datetime.strptime('2010-01-01', '%Y-%m-%d')
date2 = datetime.strptime('2010-12-01', '%Y-%m-%d')
diff = relativedelta(date2, date1)
number_of_months = diff.months
print(number_of_months) # 11

跨越多年时失败

from datetime import datetime
from dateutil.relativedelta import relativedelta

date1 = datetime.strptime('2010-01-01', '%Y-%m-%d')
date2 = datetime.strptime('2012-02-01', '%Y-%m-%d')
diff = relativedelta(date2, date1)
number_of_months = diff.months
print(number_of_months) # 1

鉴于此,我考虑了以下解决方案:

from datetime import datetime
from dateutil.relativedelta import relativedelta


def month_diff(date1, date2):
    if date1.year == date2.year:
        diff = relativedelta(date2, date1)
        number_of_months = diff.months
    else:
        num_of_days = (date2 - date1).days
        number_of_months = num_of_days//30
    print(number_of_months)


date1 = datetime.strptime('2010-01-01', '%Y-%m-%d')
date2 = datetime.strptime('2010-12-01', '%Y-%m-%d')

date3 = datetime.strptime('2010-01-01', '%Y-%m-%d')
date4 = datetime.strptime('2012-02-01', '%Y-%m-%d')

month_diff(date1, date2) # 11
month_diff(date3, date4)  # 25

因此,我的问题是,是否有比我目前拥有的更好的解决方案?

怎么样

from datetime import datetime
from dateutil.relativedelta import relativedelta

date1 = datetime.strptime('2010-01-01', '%Y-%m-%d')
date2 = datetime.strptime('2012-02-01', '%Y-%m-%d')
diff = relativedelta(date2, date1)
number_of_months = 12 * diff.years + diff.months