Python时间增量(以年为单位)

timedelta

dateutil.relativedelta object ,但那是第三方模块。如果你想知道 datetime 那是 n

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)

如果你宁愿坚持使用标准库,答案会稍微复杂一些:

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)

return

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

你的问题最初说你想知道从某个日期到现在已经有多少年了。假设你想要一个整数年,你可以根据每年365.2425天进行猜测,然后使用以下任一方法进行检查 yearsago

def num_years(begin, end=None):
    if end is None:
        end = datetime.now()
    num_years = int((end - begin).days / 365.2425)
    if begin > yearsago(num_years, end):
        return num_years - 1
    else:
        return num_years

如果你想检查某人是否年满18岁,请使用 timedelta 由于闰年的原因,在某些边缘情况下无法正常工作。例如,2000年1月1日出生的人将在2018年1月3日(包括5个闰年)后恰好6575天满18岁,但2001年1月2日出生的个人将在2019年1月4日(包括4个闰年在内)后正好6574天满18周岁。因此,如果一个人恰好是6574天大,在不知道更多关于他们出生日期的信息的情况下,你无法确定他们是17岁还是18岁。

delta_in_days / (365.25)
delta_in_seconds / (365.25*24*60*60)

…或者别的什么。远离月份,因为它们比年份更不明确。

这是一个更新的DOB函数,它以与人类相同的方式计算生日:

import datetime
import locale


# Source: https://en.wikipedia.org/wiki/February_29
PRE = [
    'US',
    'TW',
]
POST = [
    'GB',
    'HK',
]


def get_country():
    code, _ = locale.getlocale()
    try:
        return code.split('_')[1]
    except IndexError:
        raise Exception('Country cannot be ascertained from locale.')


def get_leap_birthday(year):
    country = get_country()
    if country in PRE:
        return datetime.date(year, 2, 28)
    elif country in POST:
        return datetime.date(year, 3, 1)
    else:
        raise Exception('It is unknown whether your country treats leap year '
                      + 'birthdays as being on the 28th of February or '
                      + 'the 1st of March. Please consult your country\'s '
                      + 'legal code for in order to ascertain an answer.')
def age(dob):
    today = datetime.date.today()
    years = today.year - dob.year

    try:
        birthday = datetime.date(today.year, dob.month, dob.day)
    except ValueError as e:
        if dob.month == 2 and dob.day == 29:
            birthday = get_leap_birthday(today.year)
        else:
            raise e

    if today < birthday:
        years -= 1
    return years

print(age(datetime.date(1988, 2, 29)))

你需要它有多精确? td.days / 365.25 如果你担心闰年,它会让你非常接近。

def age(dob):
    import datetime
    today = datetime.date.today()

    if today.month < dob.month or \
      (today.month == dob.month and today.day < dob.day):
        return today.year - dob.year - 1
    else:
        return today.year - dob.year

>>> import datetime
>>> datetime.date.today()
datetime.date(2009, 12, 1)
>>> age(datetime.date(2008, 11, 30))
1
>>> age(datetime.date(2008, 12, 1))
1
>>> age(datetime.date(2008, 12, 2))
0

` 简单的解决方案!

import datetime as dt
from dateutil.relativedelta import relativedelta

dt1 = dt.datetime(1990,2,1)
dt2 = dt.datetime(2021,5,16)
out = relativedelta(dt2, dt1)

print(f'Complete: {out}')
print(f'years:{out.years}, months:{out.months}, days:{out.days}') `

年:31,月:3,日:15

b = birthday
today = datetime.datetime.today()
age = today.year - b.year + (today.month - b.month > 0 or 
                             (today.month == b.month > 0 and 
                              today.day - b.day > 0))

def calculate_age(birthday_date, fmt="%Y-%m-%d"):
    birthday = datetime.datetime.strptime(birthday_date, fmt)
    age = datetime.datetime.now() - birthday
    # first datime valid is (1, 1, 1), I use (1, 12, 31) => (2, 0, 0) to hack the lib
    age = (datetime.datetime(1, 12, 31) + age)
    return age.year - 2

这个解决方案很奇怪,所以我和你分享了。这可能不是最优雅的功能。

now : 2019-09-21

print(calculate_age("2019-09-21")) => 2 (Done) 

-> age + (1,12,31) = 0004-01-01 23:42:17.767031

print(calculate_age("2020-09-21")) => 0 (Undone)

-> age + (1,12,31) = 0002-12-31 23:46:39.144091

要纠正这种不希望的行为,您需要使用生日年份来添加结果。

这里没有提到的另一个第三方库是mxDateTime(这两个python的前身) datetime timeutil )可用于此任务。

前述的 yearsago 将是:

from mx.DateTime import now, RelativeDateTime

def years_ago(years, from_date=None):
    if from_date == None:
        from_date = now()
    return from_date-RelativeDateTime(years=years)

第一个参数应该是 DateTime 例子

日期时间 到 日期时间 您可以将其用于1秒精度):

def DT_from_dt_s(t):
    return DT.DateTimeFromTicks(time.mktime(t.timetuple()))

或者,对于1微秒的精度:

def DT_from_dt_u(t):
    return DT.DateTime(t.year, t.month, t.day, t.hour,
  t.minute, t.second + t.microsecond * 1e-6)

是的,与使用timeutil(由Rick Copeland建议)相比,为这个有问题的单一任务添加依赖关系肯定是一种过度的做法。

时间增量/(365*4)+1)/4=时间增量*4/(365*4+1)

这是我想出的解决方案,希望能有所帮助;-)

def menor_edad_legal(birthday):
    """ returns true if aged<18 in days """ 
    try:

        today = time.localtime()                        

        fa_divuit_anys=date(year=today.tm_year-18, month=today.tm_mon, day=today.tm_mday)

        if birthday>fa_divuit_anys:
            return True
        else:
            return False            

    except Exception, ex_edad:
        logging.error('Error menor de edad: %s' % ex_edad)
        return True

为了我的目的,从减两开始 datetime timedelta 只有a days

>>> admit = datetime.strptime('20130530', '%Y%m%d')
>>> birth = datetime.strptime('20010621', '%Y%m%d')
>>> age = (admit - birth).days/365.2425
>>> age
11.940012457476882

def validatedate(date):
    parts = date.strip().split('-')

    if len(parts) == 3 and False not in [x.isdigit() for x in parts]: 
        birth = datetime.date(int(parts[2]), int(parts[1]), int(parts[0]))
        today = datetime.date.today()

        b = (birth.year * 10000) + (birth.month * 100) + (birth.day)
        t = (today.year * 10000) + (today.month * 100) + (today.day)

        if (t - 18 * 10000) >= b:
            return True

    return False

此函数返回两个日期之间的年差(以ISO格式作为字符串,但可以很容易地修改为任何格式)

import time
def years(earlydateiso,  laterdateiso):
    """difference in years between two dates in ISO format"""

    ed =  time.strptime(earlydateiso, "%Y-%m-%d")
    ld =  time.strptime(laterdateiso, "%Y-%m-%d")
    #switch dates if needed
    if  ld < ed:
        ld,  ed = ed,  ld            

    res = ld[0] - ed [0]
    if res > 0:
        if ld[1]< ed[1]:
            res -= 1
        elif  ld[1] == ed[1]:
            if ld[2]< ed[2]:
                res -= 1
    return res

Pyfdate

语言及其工作特性 它们应该是用户友好的 pyfdate的目的是弥补这一点 使用以下日期和时间 Python的其余部分。

tutorial

import datetime

def check_if_old_enough(years_needed, old_date):

    limit_date = datetime.date(old_date.year + years_needed,  old_date.month, old_date.day)

    today = datetime.datetime.now().date()

    old_enough = False

    if limit_date <= today:
        old_enough = True

    return old_enough



def test_ages():

    years_needed = 30

    born_date_Logan = datetime.datetime(1988, 3, 5)

    if check_if_old_enough(years_needed, born_date_Logan):
        print("Logan is old enough")
    else:
        print("Logan is not old enough")


    born_date_Jessica = datetime.datetime(1997, 3, 6)

    if check_if_old_enough(years_needed, born_date_Jessica):
        print("Jessica is old enough")
    else:
        print("Jessica is not old enough")


test_ages()

这是卡鲁塞尔操作员在洛根的《奔跑》电影中运行的代码;)

https://en.wikipedia.org/wiki/Logan%27s_Run_(film)

我遇到了这个问题,发现亚当斯的回答最有帮助 https://stackoverflow.com/a/765862/2964689

但他的方法没有python示例,但这是我最终使用的方法。

def age(birthday):
    birthday = birthday.date()
    today = date.today()

    years = today.year - birthday.year

    if (today.month < birthday.month or
       (today.month == birthday.month and today.day < birthday.day)):

        years = years - 1

    return years

def age(dob):
    import datetime
    today = datetime.date.today()
    age = today.year - dob.year
    if ( today.month == dob.month == 2 and
         today.day == 28 and dob.day == 29 ):
         pass
    elif today.month < dob.month or \
      (today.month == dob.month and today.day < dob.day):
        age -= 1
    return age

我会用 datetime.date 数据类型,因为检查已经过去了多少年、几个月和几天更简单:

now = date.today()
birthday = date(1993, 4, 4)
print("you are", now.year - birthday.year, "years,", now.month - birthday.month, "months and",
  now.day - birthday.day, "days old")

输出:

you are 27 years, 7 months and 11 days old

我用 timedelta

age = now - birthday
print("addition of days to a date: ", birthday + timedelta(days=age.days))

输出:

addition of days to a date:  2020-11-15