这里有一种更加高效的方法来实现下面的操作吗？我想要计算两个日期之间的年份差作为一个单一的标量。欢迎任何建议。
from datetime import datetime
start_date = datetime(2010,4,28,12,33)
end_date = datetime(2010,5,5,23,14)
difference = end_date - start_date
difference_in_years = (difference.days + difference.seconds/86400)/365.2425
from dateutil.relativedelta import relativedelta
difference_in_years = relativedelta(end_date, start_date).years
这是针对完整年份的计算（例如一个人的年龄）。如果您想要小数年份，则需要加上月份、天数、小时等，以达到所需的精度。
我使用以下方法来计算一个人的年龄：
import datetime
dob = datetime.date(1980, 10, 10)
def age():
today = datetime.date.today()
years = today.year - dob.year
if today.month < dob.month or (today.month == dob.month and today.day < dob.day):
years -= 1
return years
def age2():
today = datetime.date.today()
this_year_birthday = datetime.date(today.year, dob.month, dob.day)
if this_year_birthday < today:
years = today.year - dob.year
else:
years = today.year - dob.year - 1
return years
只需这样做：
from dateutil.relativedelta import relativedelta
myBirthday = datetime.datetime(1983,5,20,0,0,0,0)
now = datetime.datetime.now()
difference = relativedelta(now, myBirthday)
print("My years: "+str(difference.years))
from datetime import datetime
from calendar import isleap
start_date = datetime(2005,4,28,12,33)
end_date = datetime(2010,5,5,23,14)
diffyears = end_date.year - start_date.year
difference = end_date - start_date.replace(end_date.year)
days_in_year = isleap(end_date.year) and 366 or 365
difference_in_years = diffyears + (difference.days + difference.seconds/86400.0)/days_in_year
from datetime import datetime
from datetime import timedelta
from calendar import isleap
size_of_day = 1. / 366.
size_of_second = size_of_day / (24. * 60. * 60.)
def date_as_float(dt):
days_from_jan1 = dt - datetime(dt.year, 1, 1)
if not isleap(dt.year) and days_from_jan1.days >= 31+28:
days_from_jan1 += timedelta(1)
return dt.year + days_from_jan1.days * size_of_day + days_from_jan1.seconds * size_of_second
start_date = datetime(2010,4,28,12,33)
end_date = datetime(2010,5,5,23,14)
difference_in_years = date_as_float(end_time) - date_as_float(start_time)
我并不是在说这是唯一的解决方案，因为我认为完美的解决方案是不存在的。但它具有一些可取之处：
这是基于Kostyantyn在他的“age2”函数中发布的内容的一个衍生版本。它略微更短/更简洁，同时也使用了“年龄”或年份差异的传统/口语含义：
def ageInYears( d ):
today = datetime.date.today()
currentYrAnniversary = datetime.date( today.year, d.month, d.day )
return (today.year - d.year) - (1 if today < currentYrAnniversary else 0)
from dateutil import parser
from dateutil.relativedelta import relativedelta
rip = [
["Tim Bergling\t\t", " 8 Sep 1989", "20 Apr 2018"], # Avicii Swedish musician
["Stephen Hillenburg\t", "21 Aug 1961", "26 Nov 2018"], # Creator of Spongebob
["Stephen Hawking\t\t", " 8 Jan 1942", "14 Mar 2018"], # Theoretical physicist
["Stan Lee\t\t", "28 Dec 1922", "12 Nov 2018"], # American comic book writer
["Stefán Karl Stefánsson\t", "10 Jul 1975", "21 Aug 2018"] # Robbie Rotten from LazyTown
]
for name,born,died in rip:
print("%s %s\t %s\t died at %i"%(name,born,died,relativedelta(parser.parse(died),parser.parse(born)).years))
输出
Tim Bergling 8 Sep 1989 20 Apr 2018 died at 28
Stephen Hillenburg 21 Aug 1961 26 Nov 2018 died at 57
Stephen Hawking 8 Jan 1942 14 Mar 2018 died at 76
Stan Lee 28 Dec 1922 12 Nov 2018 died at 95
Stefán Karl Stefánsson 10 Jul 1975 21 Aug 2018 died at 43
我想您正在寻找的是：
difference_in_years = difference.dt.days / 365.25
from datetime import datetime as dt
def get_age(dob_str):
now_str = dt.strftime(dt.utcnow(), '%Y-%m-%d')
return int(now_str[:4]) - int(dob_str[:4]) - int(dob_str[5:] > now_str[5:])
或者，如果您想编写一些单元测试，请将now_str设置为命名参数：
from datetime import datetime as dt
def get_age(dob_str, now_str=dt.strftime(dt.utcnow(), '%Y-%m-%d')):
return int(now_str[:4]) - int(dob_str[:4]) - int(dob_str[5:] > now_str[5:])
这是我想出来的，没有使用外部依赖：
def year_diff(d1, d2):
"""Returns the number of years between the dates as a positive integer."""
later = max(d1, d2)
earlier = min(d1, d2)
result = later.year - earlier.year
if later.month < earlier.month or (later.month == earlier.month and later.day < earlier.day):
result -= 1
return result
difference_in_years = (difference.days + difference.seconds/86400.0)/365.2425，以在 Python 2.X 中运行时得到预期的答案。 - John Machinhttps://pumas.nasa.gov/files/04_21_97_1.pdf- Marichyasana