一个比@DannyWAdairs稍微优雅一点的解决方案可能是使用.timetuple()方法[Python-doc]:

from datetime import date

def calculate_age(born):
    today = date.today()
    return today.year - born.year - (today.timetuple()[1:3] < born.timetuple()[1:3])

你可以很容易地使用这个来进一步推广它，将其粒度增加到秒，这样，如果它大于或等于当天的秒数，年龄就会增加，例如born是一个datetime对象:

from datetime import datetime

def calculate_age_with_seconds(born):
    today = datetime.now()
    return today.year - born.year - (today.timetuple()[1:6] < born.timetuple()[1:6])

这对于date或datetime对象都适用。

不幸的是，您不能只使用时间数据，因为它使用的最大单位是日，闰年将使您的计算无效。因此，让我们找到年数，然后如果最后一年没有满，就按1调整:

from datetime import date
birth_date = date(1980, 5, 26)
years = date.today().year - birth_date.year
if (datetime.now() - birth_date.replace(year=datetime.now().year)).days >= 0:
    age = years
else:
    age = years - 1

Upd:

这个解决方案在2月29日开始时确实会导致一个异常。以下是正确的检查:

from datetime import date
birth_date = date(1980, 5, 26)
today = date.today()
years = today.year - birth_date.year
if all((x >= y) for x,y in zip(today.timetuple(), birth_date.timetuple()):
   age = years
else:
   age = years - 1

Upd2:

将多次调用now()称为性能损失是荒谬的，除非在极端特殊的情况下，否则这无关紧要。使用变量的真正原因是数据不一致的风险。

考虑到int(True)是1,int(False)是0，元组比较从左到右，这可以做得更简单:

from datetime import date

def calculate_age(born):
    today = date.today()
    return today.year - born.year - ((today.month, today.day) < (born.month, born.day))

延伸到丹尼·w·阿代尔回答，得到月也

def calculate_age(b):
    t = date.today()
    c = ((t.month, t.day) < (b.month, b.day))
    c2 = (t.day< b.day)
    return t.year - b.year - c,c*12+t.month-b.month-c2

最简单的方法是使用python-dateutil

import datetime

import dateutil

def birthday(date):
    # Get the current date
    now = datetime.datetime.utcnow()
    now = now.date()

    # Get the difference between the current date and the birthday
    age = dateutil.relativedelta.relativedelta(now, date)
    age = age.years

    return age

from datetime import date

def age(birth_date):
    today = date.today()
    y = today.year - birth_date.year
    if today.month < birth_date.month or today.month == birth_date.month and today.day < birth_date.day:
        y -= 1
    return y