如何从今天的日期和生日中找到蟒蛇的年龄？生日是来自Django模型中的DateField。
答案 0 :(得分：235)
考虑到int（True）为1且int（False）为0，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))
答案 1 :(得分：67)
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，更好
答案 2 :(得分：16)
from datetime import date
days_in_year = 365.2425
age = int((date.today() - birth_date).days / days_in_year)
在Python 3中，您可以在datetime.timedelta上执行除法：
from datetime import date, timedelta
age = (date.today() - birth_date) // timedelta(days=365.2425)
答案 3 :(得分：9)
正如@ [Tomasz Zielinski]和@Williams python-dateutil所建议的那样只能做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)`
答案 4 :(得分：8)
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
答案 5 :(得分：6)
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
答案 6 :(得分：5)
不幸的是，你不能只使用timedelata作为它使用的最大单位是白天和闰年将使你的计算无效。因此，如果去年未满，我们可以找到年数，然后调整一年：
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
<强> UPD：
当2月29日发挥作用时，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
<强> UPD2：
调用多次调用now()性能命中是荒谬的，除了非常特殊的情况外，它并不重要。使用变量的真正原因是数据不完整的风险。
答案 7 :(得分：4)
这个场景中的经典问题是如何处理2月29日出生的人。例如：你需要年满18岁才能投票，开车，买酒等等......如果你出生于2004-02-29，那么你被允许做这些事的第一天是什么时间：2022-02 -28，或2022-03-01？ AFAICT，大部分是第一个，但是一些杀戮者可能会说后者。
此代码适用于当天出生的0.068％（约）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
答案 8 :(得分：2)
以下是将人的年龄定为年，月或日的解决方案。
让我们说一个人的出生日期 2012-01-17T00：00：00 因此，他 2013-01-16T00：00：00 的年龄 11个月
或者如果他出生于 2012-12-17T00：00：00 ， 他 2013-01-12T00：00：00 的年龄 26天
或者如果他出生于 2000-02-29T00：00：00 ， 他 2012-02-29T00：00：00 的年龄 12年
您需要导入日期时间。
以下是代码：
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
上述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 2000-02-29T00：00：00
等字符串提供 get_date_format（）它会将其转换为日期类型对象，该对象将被输入 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。
答案 9 :(得分：2)
如果您希望使用django模板在页面中打印，那么以下内容可能就足够了：
{{ birth_date|timesince }}
答案 10 :(得分：2)
扩展Danny's Solution，但有各种方式报告年轻人的年龄（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'
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
答案 11 :(得分：1)
由于我没有看到正确的实施，我用这种方式重新编码......
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)
2月28日出生于29日的假设为“18”是错误的。 交换边界可以省略......这只是我的代码的个人便利：）
答案 12 :(得分：1)
延长至Danny W. Adair Answer，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
答案 13 :(得分：0)
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
答案 14 :(得分：0)
略微修改Danny's solution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
答案 15 :(得分：0)
import datetime
今天日期
td=datetime.datetime.now().date()
您的生日
bd=datetime.date(1989,3,15)
您的年龄
age_years=int((td-bd).days /365.25)
答案 16 :(得分：0)
您可以使用Python 3来完成所有这些工作。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}')