python - 计算两个日期之间的年数，但以标准方式四舍五入

Question

我想获得两个日期之间的整数年数。 relativedelta 的 years 属性几乎可以做到这一点，但因为它服务于不同的用例，所以它总是向下舍入而不是使用标准舍入规则:

from datetime import datetime, date
from dateutil.relativedelta import relativedelta

almost_one_year_ago = date.today() - relativedelta(years=1) + relativedelta(days=1)
print(relativedelta(date.today(), almost_one_year_ago).years)

这输出 0，我想要 1。

有没有一种好的方法可以按照标准舍入规则将年数作为整数？如:如果低于 .5 年则向下舍入，如果等于或高于 .5 则向上舍入？

Answer 1

我找到了一个与闰年兼容的解决方案。它仍然没有我想要的那么干净，但它完成了工作:

from datetime import datetime, date
from dateutil.relativedelta import relativedelta

def how_many_years_ago(input_date):
    difference = relativedelta(date.today(), input_date)
    result = difference.years

    # Round up if closer to the next year than the previous
    if difference.months >= 6:
        result += 1

    return result

# Exactly 1 year ago, should output 1
print(how_many_years_ago(date.today() - relativedelta(years=1)))

# Almost 1 year ago, should round up to 1
print(how_many_years_ago(date.today() - relativedelta(years=1) + relativedelta(days=1)))

# Too far removed from 35 years, should round down to 34
print(how_many_years_ago(date.today() - relativedelta(years=35) + relativedelta(months=8)))

# Almost 38 years ago, should round up to 38
print(how_many_years_ago(date.today() - relativedelta(years=38) + relativedelta(months=4)))

如果我没有忽略某些注意事项(例如闰年)，此实现可以解决我的问题。