Python 从日期中减去日期或天数得到年、月、日

Question

我试图从今天的日期中减去两个日期或天数。我想在几年、几个月、几天内得到结果。该怎么做？

期待1 year, 2 months, 5 days、20 days、3 months, 2 days 等之前，而不是仅仅几天。

import datetime
import pytz
tz='US/Pacific'
birthday = datetime.datetime(2020, 2, 19, 12, 0, 0)

>>> import datetime
>>> import pytz
>>> tz='US/Pacific'
>>> birthday = datetime.datetime(2020, 2, 19, 12, 0, 0)
>>> diff = datetime.datetime.now() - birthday
>>>
>>> diff
datetime.timedelta(days=326, seconds=39130, microseconds=319509)
>>>
>>> birthday = datetime.datetime(2015, 2, 19, 12, 0, 0)
>>> diff = datetime.datetime.now() - birthday
>>> diff
datetime.timedelta(days=2152, seconds=39151, microseconds=823846)
>>>
>>> diff.days
2152

Answer 1

使用dateutil 包中的dateutil.relativedelta：

import datetime
from dateutil.relativedelta import relativedelta

>>> relativedelta(datetime.datetime.now(), datetime.datetime(2020, 2, 19, 12, 0, 0))
relativedelta(months=+10, days=+23, hours=+4, minutes=+8, seconds=+42, microseconds=+204978)

>>> relativedelta(datetime.datetime.now(), datetime.datetime(2015, 2, 19, 12, 0, 0))
relativedelta(years=+5, months=+10, days=+23, hours=+4, minutes=+9, seconds=+10, microseconds=+624971)

您可以提取年份、月份等，例如relativedelta().years、relativedelta().months 等。

Answer 2

这个answer 可能会对你有所帮助。但是您也可以使用timetuple() 进行检查

from datetime import datetime
import datetime
import pytz
tz = 'US/Pacific'
birthday = datetime.datetime(2020, 2, 19, 12, 0, 0)


b = birthday.timetuple()
n = datetime.datetime.now().timetuple()
print(f'Year diff: {n.tm_year-b.tm_year}')
print(f'Month diff: {n.tm_mon-b.tm_mon}')
print(f'Day diff: {n.tm_mday-b.tm_mday}')