Python中生日的年龄

如何从今天的日期和人的出生日期中找到python的年龄？生日是来自Django模型中的DateField的。

考虑到int（True）为1且int（False）为0，可以轻松得多：

from datetime import date

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

from datetime import date

def calculate_age(born):
    today = date.today()
    try: 
        birthday = born.replace(year=today.year)
    except ValueError: # raised when birth date is February 29 and the current year is not a leap year
        birthday = born.replace(year=today.year, month=born.month+1, day=1)
    if birthday > today:
        return today.year - born.year - 1
    else:
        return today.year - born.year

更新：使用Danny的解决方案更好

from datetime import date

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

在Python 3中，您可以对进行除法datetime.timedelta：

from datetime import date, timedelta

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

如@ [Tomasz Zielinski]和@Williams的建议，python-dateutil只能执行5行。

from dateutil.relativedelta import *
from datetime import date
today = date.today()
dob = date(1982, 7, 5)
age = relativedelta(today, dob)

>>relativedelta(years=+33, months=+11, days=+16)`

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

不幸的是，您不能仅仅使用timedelata，因为它使用的最大单位是天，leap年将使您的计算无效。因此，让我们找到年数，如果最后一年未满，则将其调整为一：

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

更新：

当2月29日生效时，此解决方案确实会导致异常。这是正确的检查：

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：

多次致电来now()提高绩效很可笑，这在所有情况下都没有关系，但在非常特殊的情况下。使用变量的真正原因是数据不一致的风险。

在这种情况下，典型的陷阱是如何处理2月29日出生的人。例如：您必须年满18岁才能投票，开车，买酒等...如果您出生于2004-02-29，那么您被允许做此类事情的第一天是2022-02 -28还是2022-03-01？AFAICT主要是第一个，但有些杀人狂可能会说后者。

以下代码可满足当日出生人口的0.068％（大约）的需求：

def age_in_years(from_date, to_date, leap_day_anniversary_Feb28=True):
    age = to_date.year - from_date.year
    try:
        anniversary = from_date.replace(year=to_date.year)
    except ValueError:
        assert from_date.day == 29 and from_date.month == 2
        if leap_day_anniversary_Feb28:
            anniversary = datetime.date(to_date.year, 2, 28)
        else:
            anniversary = datetime.date(to_date.year, 3, 1)
    if to_date < anniversary:
        age -= 1
    return age

if __name__ == "__main__":
    import datetime

    tests = """

    2004  2 28 2010  2 27  5 1
    2004  2 28 2010  2 28  6 1
    2004  2 28 2010  3  1  6 1

    2004  2 29 2010  2 27  5 1
    2004  2 29 2010  2 28  6 1
    2004  2 29 2010  3  1  6 1

    2004  2 29 2012  2 27  7 1
    2004  2 29 2012  2 28  7 1
    2004  2 29 2012  2 29  8 1
    2004  2 29 2012  3  1  8 1

    2004  2 28 2010  2 27  5 0
    2004  2 28 2010  2 28  6 0
    2004  2 28 2010  3  1  6 0

    2004  2 29 2010  2 27  5 0
    2004  2 29 2010  2 28  5 0
    2004  2 29 2010  3  1  6 0

    2004  2 29 2012  2 27  7 0
    2004  2 29 2012  2 28  7 0
    2004  2 29 2012  2 29  8 0
    2004  2 29 2012  3  1  8 0

    """

    for line in tests.splitlines():
        nums = [int(x) for x in line.split()]
        if not nums:
            print
            continue
        datea = datetime.date(*nums[0:3])
        dateb = datetime.date(*nums[3:6])
        expected, anniv = nums[6:8]
        age = age_in_years(datea, dateb, anniv)
        print datea, dateb, anniv, age, expected, age == expected

这是输出：

2004-02-28 2010-02-27 1 5 5 True
2004-02-28 2010-02-28 1 6 6 True
2004-02-28 2010-03-01 1 6 6 True

2004-02-29 2010-02-27 1 5 5 True
2004-02-29 2010-02-28 1 6 6 True
2004-02-29 2010-03-01 1 6 6 True

2004-02-29 2012-02-27 1 7 7 True
2004-02-29 2012-02-28 1 7 7 True
2004-02-29 2012-02-29 1 8 8 True
2004-02-29 2012-03-01 1 8 8 True

2004-02-28 2010-02-27 0 5 5 True
2004-02-28 2010-02-28 0 6 6 True
2004-02-28 2010-03-01 0 6 6 True

2004-02-29 2010-02-27 0 5 5 True
2004-02-29 2010-02-28 0 5 5 True
2004-02-29 2010-03-01 0 6 6 True

2004-02-29 2012-02-27 0 7 7 True
2004-02-29 2012-02-28 0 7 7 True
2004-02-29 2012-02-29 0 8 8 True
2004-02-29 2012-03-01 0 8 8 True

如果您希望使用django模板在页面中打印此内容，那么以下内容可能就足够了：

{{ birth_date|timesince }}

这是找到一个人的年龄（数月，数月或数天）的解决方案。

假设某人的出生日期是2012-01-17T00：00：00 因此，他在2013-01-16T00：00：00的年龄将是11个月

或如果他出生于2012-12-17T00：00：00，则他在2013-01-12T00：00：00的年龄为26天

或如果他出生于2000-02-29T00：00：00，则他在2012-02-29T00：00：00的年龄将为12岁

您将需要导入datetime。

这是代码：

def get_person_age(date_birth, date_today):

"""
At top level there are three possibilities : Age can be in days or months or years.
For age to be in years there are two cases: Year difference is one or Year difference is more than 1
For age to be in months there are two cases: Year difference is 0 or 1
For age to be in days there are 4 possibilities: Year difference is 1(20-dec-2012 - 2-jan-2013),
                                                 Year difference is 0, Months difference is 0 or 1
"""
years_diff = date_today.year - date_birth.year
months_diff = date_today.month - date_birth.month
days_diff = date_today.day - date_birth.day
age_in_days = (date_today - date_birth).days

age = years_diff
age_string = str(age) + " years"

# age can be in months or days.
if years_diff == 0:
    if months_diff == 0:
        age = age_in_days
        age_string = str(age) + " days"
    elif months_diff == 1:
        if days_diff < 0:
            age = age_in_days
            age_string = str(age) + " days"
        else:
            age = months_diff
            age_string = str(age) + " months"
    else:
        if days_diff < 0:
            age = months_diff - 1
        else:
            age = months_diff
        age_string = str(age) + " months"
# age can be in years, months or days.
elif years_diff == 1:
    if months_diff < 0:
        age = months_diff + 12
        age_string = str(age) + " months" 
        if age == 1:
            if days_diff < 0:
                age = age_in_days
                age_string = str(age) + " days" 
        elif days_diff < 0:
            age = age-1
            age_string = str(age) + " months"
    elif months_diff == 0:
        if days_diff < 0:
            age = 11
            age_string = str(age) + " months"
        else:
            age = 1
            age_string = str(age) + " years"
    else:
        age = 1
        age_string = str(age) + " years"
# The age is guaranteed to be in years.
else:
    if months_diff < 0:
        age = years_diff - 1
    elif months_diff == 0:
        if days_diff < 0:
            age = years_diff - 1
        else:
            age = years_diff
    else:
        age = years_diff
    age_string = str(age) + " years"

if age == 1:
    age_string = age_string.replace("years", "year").replace("months", "month").replace("days", "day")

return age_string

以上代码中使用的一些额外功能是：

def get_todays_date():
    """
    This function returns todays date in proper date object format
    """
    return datetime.now()

和

def get_date_format(str_date):
"""
This function converts string into date type object
"""
str_date = str_date.split("T")[0]
return datetime.strptime(str_date, "%Y-%m-%d")

现在，我们必须使用类似2000-02-29T00：00：00的字符串来填充get_date_format（）

它将转换为日期类型对象，该对象将被馈送到get_person_age（date_birth，date_today）。

函数get_person_age（date_birth，date_today）将以字符串格式返回age。

扩展了Danny的解决方案，但提供了多种报告年轻人群年龄的方法（请注意，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'

样例代码：

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

由于我没有看到正确的实现方式，因此以这种方式重新编码了我的代码...

    def age_in_years(from_date, to_date=datetime.date.today()):
  if (DEBUG):
    logger.debug("def age_in_years(from_date='%s', to_date='%s')" % (from_date, to_date))

  if (from_date>to_date): # swap when the lower bound is not the lower bound
    logger.debug('Swapping dates ...')
    tmp = from_date
    from_date = to_date
    to_date = tmp

  age_delta = to_date.year - from_date.year
  month_delta = to_date.month - from_date.month
  day_delta = to_date.day - from_date.day

  if (DEBUG):
    logger.debug("Delta's are : %i  / %i / %i " % (age_delta, month_delta, day_delta))

  if (month_delta>0  or (month_delta==0 and day_delta>=0)): 
    return age_delta 

  return (age_delta-1)

假设在2月28日出生时在29日成为“ 18”是错误的。可以忽略界限...这只是我的代码的个人方便:)

扩展到Danny W. Adair答案，还可以获得月份

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

import datetime

今天的日期

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

你的生日

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

你的年龄

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

导入日期时间

def age(date_of_birth):
    if date_of_birth > datetime.date.today().replace(year = date_of_birth.year):
        return datetime.date.today().year - date_of_birth.year - 1
    else:
        return datetime.date.today().year - date_of_birth.year

在您的情况下：

import datetime

# your model
def age(self):
    if self.birthdate > datetime.date.today().replace(year = self.birthdate.year):
        return datetime.date.today().year - self.birthdate.year - 1
    else:
        return datetime.date.today().year - self.birthdate.year

稍微修改了Danny的解决方案，以便于阅读和理解

    from datetime import date

    def calculate_age(birth_date):
        today = date.today()
        age = today.year - birth_date.year
        full_year_passed = (today.month, today.day) < (birth_date.month, birth_date.day)
        if not full_year_passed:
            age -= 1
        return age