How can I find an age in python from today's date and a persons birthdate? The birthdate is a from a DateField in a Django model.
19 Answers
That can be done much simpler considering that int(True) is 1 and int(False) is 0:
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
Update: Use Danny's solution, it's better
from datetime import date
days_in_year = 365.2425
age = int((date.today() - birth_date).days / days_in_year)
In Python 3, you could perform division on datetime.timedelta:
from datetime import date, timedelta
age = (date.today() - birth_date) // timedelta(days=365.2425)
Unfortunately, you cannot just use timedelata as the largest unit it uses is day and leap years will render you calculations invalid. Therefore, let's find number of years then adjust by one if the last year isn't full:
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:
This solution really causes an exception when Feb, 29 comes into play. Here's correct check:
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:
Calling multiple calls to now() a performance hit is ridiculous, it does not matter in all but extremely special cases. The real reason to use a variable is the risk of data incosistency.
The classic gotcha in this scenario is what to do with people born on the 29th day of February. Example: you need to be aged 18 to vote, drive a car, buy alcohol, etc ... if you are born on 2004-02-29, what is the first day that you are permitted to do such things: 2022-02-28, or 2022-03-01? AFAICT, mostly the first, but a few killjoys might say the latter.
Here's code that caters for the 0.068% (approx) of the population born on that day:
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
Here's the output:
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
import datetime
Todays date
td=datetime.datetime.now().date()
Your birthdate
bd=datetime.date(1989,3,15)
Your age
age_years=int((td-bd).days /365.25)
Extend to Danny W. Adair Answer, to get month also
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
A slightly more elegant solution than @DannyWAdairs solution might be to work with the .timetuple() method [Python-doc]:
from datetime import date
def calculate_age(born):
today = date.today()
return today.year - born.year - (today.timetuple()[1:3] < born.timetuple()[1:3])You can easily use this to generalize this further to increase its granularity to seconds, such that the age increments if it is greater than or equal to the number of seconds of that day, if born is for example a datetime object:
from datetime import datetime
def calculate_age_with_seconds(born):
today = datetime.now()
return today.year - born.year - (today.timetuple()[1:6] < born.timetuple()[1:6])This will work for born both being a date or a datetime object.
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
You can use Python 3 to do all this. Just run the following code and see.
# 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}')
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