import datetime

今天的日期

td=datetime.datetime.now().date() 

你的出生年月日

bd=datetime.date(1989,3,15)

你的年龄

age_years=int((td-bd).days /365.25)

import datetime

今天的日期

td=datetime.datetime.now().date() 

你的出生年月日

bd=datetime.date(1989,3,15)

你的年龄

age_years=int((td-bd).days /365.25)

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。

from datetime import date

days_in_year = 365.2425    
age = int((date.today() - birth_date).days / days_in_year)

在Python 3中，你可以对datetime.timedelta执行除法:

from datetime import date, timedelta

age = (date.today() - birth_date) // timedelta(days=365.2425)

扩展了Danny的解决方案，但有各种各样的方法来报告年轻人的年龄(注意，今天是datetime.date(2015,7,17)):

def calculate_age(born):
    '''
        Converts a date of birth (dob) datetime object to years, always rounding down.
        When the age is 80 years or more, just report that the age is 80 years or more.
        When the age is less than 12 years, rounds down to the nearest half year.
        When the age is less than 2 years, reports age in months, rounded down.
        When the age is less than 6 months, reports the age in weeks, rounded down.
        When the age is less than 2 weeks, reports the age in days.
    '''
    today = datetime.date.today()
    age_in_years = today.year - born.year - ((today.month, today.day) < (born.month, born.day))
    months = (today.month - born.month - (today.day < born.day)) %12
    age = today - born
    age_in_days = age.days
    if age_in_years >= 80:
        return 80, 'years or older'
    if age_in_years >= 12:
        return age_in_years, 'years'
    elif age_in_years >= 2:
        half = 'and a half ' if months > 6 else ''
        return age_in_years, '%syears'%half
    elif months >= 6:
        return months, 'months'
    elif age_in_days >= 14:
        return age_in_days/7, 'weeks'
    else:
        return age_in_days, 'days'

示例代码:

print '%d %s' %calculate_age(datetime.date(1933,6,12)) # >=80 years
print '%d %s' %calculate_age(datetime.date(1963,6,12)) # >=12 years
print '%d %s' %calculate_age(datetime.date(2010,6,19)) # >=2 years
print '%d %s' %calculate_age(datetime.date(2010,11,19)) # >=2 years with half
print '%d %s' %calculate_age(datetime.date(2014,11,19)) # >=6 months
print '%d %s' %calculate_age(datetime.date(2015,6,4)) # >=2 weeks
print '%d %s' %calculate_age(datetime.date(2015,7,11)) # days old

80 years or older
52 years
5 years
4 and a half years
7 months
6 weeks
7 days

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