如何从今天的日期和一个人的生日在 python 中找到年龄？生日来自 Django 模型中的 DateField。
【问题讨论】：
datetime模块不够用时，可以试试：labix.org/python-dateutil
dateutil.relativedelta.relativedelta
标签： python
考虑到 int(True) 为 1 且 int(False) 为 0，这可以更简单地完成：
from datetime import date
def calculate_age(born):
today = date.today()
return today.year - born.year - ((today.month, today.day) < (born.month, born.day))
【讨论】：
date.today() 返回可能与出生地不同的本地时区日期。您可能需要use timezones explicitly
int(True) == 1和int(False) == 0，事实证明Python允许你对整数进行加减布尔运算，所以你甚至不需要用@987654326包装布尔值@！直到。 1 - True == 0 在 Python 2 和 3 中。
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's solution，这样更好
【讨论】：
except 块应该只捕获可能引发的一个特定异常。
ValueError。已更新。
def calculate_age(dob) 有什么问题？ 我认为这很简单。
datetime.date(2014, 1, 1) 给出 -1，它应该给出 0。您的 today > dob 正在检查 DOB 是否在过去，而不是同年早些时候。 datetime.date.today() 包含年份信息，这就是我在解决方案中将其替换为当前年份的原因。
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 datetime.timedelta 执行除法：
from datetime import date, timedelta
age = (date.today() - birth_date) // timedelta(days=365.2425)
【讨论】：
365.25) 和公历年 (365.2425) 之间的差异不到一天。
(date(2017, 3, 1) - date(2004, 3, 1)) / timedelta(days=365.2425) 应该返回 13，但它返回 12。未下限，结果为12.999582469181433。
正如@[Tomasz Zielinski] 和@Williams python-dateutil 所建议的，只需 5 行即可。
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)`
【讨论】：
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
【讨论】：
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
【讨论】：
不幸的是，您不能只使用 timedelata，因为它使用的最大单位是天，闰年会使您的计算无效。因此，让我们找出年数，如果最后一年未满，则调整一：
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
更新2：
多次调用now() 性能受到影响是荒谬的，除了极端特殊的情况外，这无关紧要。使用变量的真正原因是数据不一致的风险。
【讨论】：
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 }}
【讨论】：
|timesince 来计算几年的时间增量，因为它没有考虑闰年，因此会产生不准确的结果。有关这方面的更多信息，请参阅 Django 票证 #19210。
这个场景中的经典问题是如何处理 2 月 29 日出生的人。示例：您需要年满 18 岁才能投票、开车、购买酒精等……如果您出生于 2004-02-29，您被允许做这些事情的第一天是什么时候：2022-02 -28，还是 2022-03-01？ AFAICT，主要是第一个，但少数人可能会说后者。
以下代码适用于当天出生的 0.068%（大约）的人口：
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
【讨论】：
扩展至Danny's Solution，但有各种方式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
【讨论】：
import datetime
今天的日期
td=datetime.datetime.now().date()
你的生日
bd=datetime.date(1989,3,15)
您的年龄
age_years=int((td-bd).days /365.25)
【讨论】：
这是一种以年、月或日为单位查找人年龄的解决方案。
假设一个人的出生日期是 2012-01-17T00:00:00 因此，他在 2013-01-16T00:00:00 的年龄将是 11 个月
或者如果他出生于 2012-12-17T00:00:00， 他在 2013-01-12T00:00:00 的年龄将是 26 天
或者如果他出生于 2000-02-29T00:00:00， 他在 2012-02-29T00:00:00 的年龄将是 12 岁
您需要导入日期时间。
代码如下：
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
以上代码中用到的一些额外功能是：
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，我们必须向 get_date_format() 提供像 2000-02-29T00:00:00
这样的字符串它将其转换为日期类型对象，该对象将被提供给 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年龄。
【讨论】：
由于我没有看到正确的实现，我以这种方式重新编码...
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 2 月 28 日出生于 29 日是“18”是错误的。 可以省略交换边界......这只是我的代码的个人方便:)
【讨论】：
扩展到Danny W. Adair Answer，也可以得到月份
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
【讨论】：
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
在你的情况下：
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
【讨论】：
对Danny's solution稍作修改，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
【讨论】：
serializers.py
age = serializers.SerializerMethodField('get_age')
class Meta:
model = YourModel
fields = [..,'','birthdate','age',..]
import datetime
def get_age(self, instance):
return datetime.datetime.now().year - instance.birthdate.year
【讨论】：
您可以使用 Python 3 来完成这一切。只需运行以下代码即可查看。
# Creating a variables:
greeting = "Hello, "
name = input("what is your name?")
birth_year = input("Which year you were born?")
response = "Your age is "
# Converting string variable to int:
calculation = 2020 - int(birth_year)
# Printing:
print(f'{greeting}{name}. {response}{calculation}')
【讨论】：