我怎么能从今天的日期和一个人的出生日期找到一个python年龄?出生日期来自Django模型中的DateField。
当前回答
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
其他回答
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:使用丹尼的解决方案,效果更好
在这种情况下,最经典的问题是如何对待2月29日出生的人。例如:你必须年满18岁才能投票、开车、买酒等等。如果你出生在2004-02-29,你被允许做这些事情的第一天是哪一天:2022-02-28,或2022-03-01?AFAICT,大多数是前者,但一些扫兴的人可能会说是后者。
下面的代码迎合了那一天出生的0.068%(大约)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
输出如下:
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
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))
由于我没有看到正确的实现,我以这种方式重新编码了我的…
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)
如果你出生在2月29日,你就认为自己在2月28日是“18岁”,这是错误的。 交换边界可以省略…这只是我的代码的个人方便:)
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)
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