Python中特定年份的时差

Posted 2023-02-22

【中文标题】Python中特定年份的时差

【英文标题】：Time difference in specific year in Python

【发布时间】：2022-01-23 12:11:35

【问题描述】：

我有一个这样的时间范围：

runtime_start = datetime.date(2021,1,1)
runtime_end = datetime.date(2022,3,1)
current_year = datetime.date.today().year

如何计算当前年份的月份数？

一些例子：

runtime_start = 2021,1,1 | runtime_end = 2022,3,1 | current_year = 2021 | output = 12

runtime_start = 2021,1,1 | runtime_end = 2021,6,1 | current_year = 2021 | output= 5

【问题讨论】：

这很棘手，因为这取决于您认为“月数”是多少。你总是只有一个月的第一天吗？在这种情况下，.year * 12 和 .month（在 date 对象上）会有所帮助。 EV。处理.day 的差异（例如，

你能解释一下你的输出吗？

【参考方案1】：

import datetime

runtime_start = datetime.date(2021,1,1)
runtime_end = datetime.date(2022,3,1)
current_year = datetime.date.today().year


def calculate_differentmonths(runtime_start, runtime_end, current_year):
     if current_year == runtime_end.year:
         run_months = runtime_end.month - runtime_start.month 
     else:    
         years_month = (current_year - runtime_start.year) * 12
         run_months =  datetime.date.today().month + years_month

     return run_months

检查结果：

print(calculate_differentmonths(runtime_start, runtime_end, current_year)) 

结果 12

print(calculate_differentmonths(datetime.date(2021,1,1), datetime.date(2021,6,1), current_year))

结果 5

【参考方案2】：

您可以通过 timedelta 的.days 估算月数：

import datetime

current_year = datetime.date.today().year
start_of_curr = datetime.date(current_year,1,1)
end_of_curr = datetime.date(current_year,12,31)

data = [(datetime.date(2021,1,1), datetime.date(2022,3,1),  12), 
        (datetime.date(2021,1,1), datetime.date(2021,6,1),  5)]

for runtime_start, runtime_end, months in data:

    # limit the used start/end dates
    frm = start_of_curr  if runtime_start < start_of_curr else runtime_start
    to = runtime_end if runtime_end <= end_of_curr else end_of_curr

    print(int(round((to-frm).days / ((end_of_curr-start_of_curr).days/12),0)), 
        "vs expected: ", months)

输出：

12 vs expected:  12
5 vs expected:  5