将timedelta转换为年? [英] Convert timedelta to years?

问题描述

我需要检查自某个日期以来是否已有数年的时间.目前，我从 datetime 模块获得了 timedelta ，而且我不知道如何将其转换为年份.

I need to check if some number of years have been since some date. Currently I've got timedelta from datetime module and I don't know how to convert it to years.

推荐答案

您需要的不仅仅是一个 timedelta 来告诉已经过去了多少年.您还需要知道开始(或结束)日期.(这是year年.)

You need more than a timedelta to tell how many years have passed; you also need to know the beginning (or ending) date. (It's a leap year thing.)

您最好的选择是使用 dateutil.relativedelta 对象，但是这是一个第三方模块.如果您想知道距某个日期 n 年的 datetime (默认为现在)，则可以执行以下操作:

Your best bet is to use the dateutil.relativedelta object, but that's a 3rd party module. If you want to know the datetime that was n years from some date (defaulting to right now), you can do the following::

from dateutil.relativedelta import relativedelta

def yearsago(years, from_date=None):
    if from_date is None:
        from_date = datetime.now()
    return from_date - relativedelta(years=years)

如果您愿意使用标准库，答案会稍微复杂一些::

If you'd rather stick with the standard library, the answer is a little more complex::

from datetime import datetime
def yearsago(years, from_date=None):
    if from_date is None:
        from_date = datetime.now()
    try:
        return from_date.replace(year=from_date.year - years)
    except ValueError:
        # Must be 2/29!
        assert from_date.month == 2 and from_date.day == 29 # can be removed
        return from_date.replace(month=2, day=28,
                                 year=from_date.year-years)

如果是2/29，而18年前还没有2/29，则此函数将返回2/28.如果您希望返回3/1，只需将最后一个 return 语句更改为::

If it's 2/29, and 18 years ago there was no 2/29, this function will return 2/28. If you'd rather return 3/1, just change the last return statement to read::

    return from_date.replace(month=3, day=1,
                             year=from_date.year-years)

您的问题最初是说您想知道自某个日期以来已有多少年了.假设您想要整数年，则可以根据每年365.25天进行猜测，然后使用上面定义的 yearsago 函数之一进行检查:

Your question originally said you wanted to know how many years it's been since some date. Assuming you want an integer number of years, you can guess based on 365.25 days per year and then check using either of the yearsago functions defined above::

def num_years(begin, end=None):
    if end is None:
        end = datetime.now()
    num_years = int((end - begin).days / 365.25)
    if begin > yearsago(num_years, end):
        return num_years - 1
    else:
        return num_years

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