我有一个Python datetime.datetime对象。减去一天的最佳方法是什么？
答案 0 :(得分：1039)
您可以使用timedelta对象：
from datetime import datetime, timedelta
d = datetime.today() - timedelta(days=days_to_subtract)
答案 1 :(得分：67)
减去datetime.timedelta(days=1)
答案 2 :(得分：52)
如果您的Python日期时间对象是时区感知的，那么您应该小心避免DST转换周围的错误（或出于其他原因的UTC偏移更改）：
from datetime import datetime, timedelta
from tzlocal import get_localzone # pip install tzlocal
DAY = timedelta(1)
local_tz = get_localzone() # get local timezone
now = datetime.now(local_tz) # get timezone-aware datetime object
day_ago = local_tz.normalize(now - DAY) # exactly 24 hours ago, time may differ
naive = now.replace(tzinfo=None) - DAY # same time
yesterday = local_tz.localize(naive, is_dst=None) # but elapsed hours may differ
一般情况下，如果当地时区的UTC偏移量在最后一天发生变化，day_ago和yesterday可能会有所不同。
例如，夏令时/夏令时将于2014年11月2日凌晨02:00:00结束。因此，在America / Los_Angeles时区：
import pytz # pip install pytz
local_tz = pytz.timezone('America/Los_Angeles')
now = local_tz.localize(datetime(2014, 11, 2, 10), is_dst=None)
# 2014-11-02 10:00:00 PST-0800
然后day_ago和yesterday不同：
day_ago正是24小时前（相对于now）但是在上午11点，而不是在上午10点now yesterday是昨天上午10点，但是是25小时前（相对于now），而不是24小时。 pendulum module会自动处理它：
>>> import pendulum # $ pip install pendulum
>>> now = pendulum.create(2014, 11, 2, 10, tz='America/Los_Angeles')
>>> day_ago = now.subtract(hours=24) # exactly 24 hours ago
>>> yesterday = now.subtract(days=1) # yesterday at 10 am but it is 25 hours ago
>>> (now - day_ago).in_hours()
24
>>> (now - yesterday).in_hours()
25
>>> now
<Pendulum [2014-11-02T10:00:00-08:00]>
>>> day_ago
<Pendulum [2014-11-01T11:00:00-07:00]>
>>> yesterday
<Pendulum [2014-11-01T10:00:00-07:00]>
答案 3 :(得分：31)
只是详细说明替代方法以及有用的用例：
from datetime import datetime, timedelta print datetime.now() + timedelta(days=-1) # Here, I am adding a negative timedelta
from datetime import datetime, timedelta print datetime.now() + timedelta(days=5, hours=-5)
它可以类似地与其他参数一起使用，例如秒，周等
答案 4 :(得分：8)
另外，当我想要计算上个月的第一天/最后一天或其他相对时间等时，我喜欢使用另一个不错的功能......
来自dateutil函数的relativedelta函数（对datetime lib的强大扩展）
import datetime as dt
from dateutil.relativedelta import relativedelta
#get first and last day of this and last month)
today = dt.date.today()
first_day_this_month = dt.date(day=1, month=today.month, year=today.year)
last_day_last_month = first_day_this_month - relativedelta(days=1)
print (first_day_this_month, last_day_last_month)
>2015-03-01 2015-02-28
答案 5 :(得分：8)
Genial arrow模块存在
import arrow
utc = arrow.utcnow()
utc_yesterday = utc.shift(days=-1)
print(utc, '\n', utc_yesterday)
输出：
2017-04-06T11:17:34.431397+00:00
2017-04-05T11:17:34.431397+00:00
答案 6 :(得分：1)
我推荐pendulum，这是一个3 rd 的聚会库，可以使处理日期时间更加人性化。
减法时间是使用subtract()方法完成的。 （注意：示例已从文档中进行了修改。）
import pendulum
dt = pendulum.datetime(2012, 1, 31)
dt.to_datetime_string()
# '2019-03-26 23:39:29'
dt.subtract(days=1)
# '2019-03-25 23:39:29'
# You can subtract different/multiple units of time.
dt.subtract(years=3, months=2, days=6, hours=12, minutes=31, seconds=43)
# '2012-01-28 00:00:00'
注意：使用
add_timedelta()和sub_timedelta()，可以添加或减去timedelta对象，如果您已经拥有 这些。# You can also add or remove a timedelta dt.add_timedelta(timedelta(hours=3, minutes=4, seconds=5)) # '2012-01-28 03:04:05' dt.sub_timedelta(timedelta(hours=3, minutes=4, seconds=5)) # '2012-01-28 00:00:00'
类似地，可以使用add()方法来增加时间。
dt.add(days=1)
# '2019-03-27 23:39:29'
dt.add(years=5)
'2017-01-31 00:00:00'
# You can add different/multiple units of time.
dt.add(years=3, months=2, days=6, hours=12, minutes=31, seconds=43)
# '2015-04-03 12:31:43'
答案 7 :(得分：0)
class myDate:
def __init__(self):
self.day = 0
self.month = 0
self.year = 0
## for checking valid days month and year
while (True):
d = int(input("Enter The day :- "))
if (d > 31):
print("Plz 1 To 30 value Enter ........")
else:
self.day = d
break
while (True):
m = int(input("Enter The Month :- "))
if (m > 13):
print("Plz 1 To 12 value Enter ........")
else:
self.month = m
break
while (True):
y = int(input("Enter The Year :- "))
if (y > 9999 and y < 0000):
print("Plz 0000 To 9999 value Enter ........")
else:
self.year = y
break
## method for aday ands cnttract days
def adayDays(self, n):
## aday days to date day
nd = self.day + n
print(nd)
## check days subtract from date
if nd == 0: ## check if days are 7 subtracted from 7 then,........
if(self.year % 4 == 0):
if(self.month == 3):
self.day = 29
self.month -= 1
self.year = self. year
else:
if(self.month == 3):
self.day = 28
self.month -= 1
self.year = self. year
if (self.month == 5) or (self.month == 7) or (self.month == 8) or (self.month == 10) or (self.month == 12):
self.day = 30
self.month -= 1
self.year = self. year
elif (self.month == 2) or (self.month == 4) or (self.month == 6) or (self.month == 9) or (self.month == 11):
self.day = 31
self.month -= 1
self.year = self. year
elif(self.month == 1):
self.month = 12
self.year -= 1
## nd == 0 if condition over
## after subtract days to day io goes into negative then
elif nd < 0 :
n = abs(n)## return positive if no is negative
for i in range (n,0,-1): ##
if self.day == 0:
if self.month == 1:
self.day = 30
self.month = 12
self.year -= 1
else:
self.month -= 1
if(self.month == 1) or (self.month == 3)or (self.month == 5) or (self.month == 7) or (self.month == 8) or (self.month == 10) or (self.month ==12):
self.day = 30
elif(self.month == 4)or (self.month == 6) or (self.month == 9) or (self.month == 11):
self.day = 29
elif(self.month == 2):
if(self.year % 4 == 0):
self.day == 28
else:
self.day == 27
else:
self.day -= 1
## enf of elif negative days
## adaying days to DATE
else:
cnt = 0
while (True):
if self.month == 2: # check leap year
if(self.year % 4 == 0):
if(nd > 29):
cnt = nd - 29
nd = cnt
self.month += 1
else:
self.day = nd
break
## if not leap year then
else:
if(nd > 28):
cnt = nd - 28
nd = cnt
self.month += 1
else:
self.day = nd
break
## checking month other than february month
elif(self.month == 1) or (self.month == 3) or (self.month == 5) or (self.month == 7) or (self.month == 8) or (self.month == 10) or (self.month == 12):
if(nd > 31):
cnt = nd - 31
nd = cnt
if(self.month == 12):
self.month = 1
self.year += 1
else:
self.month += 1
else:
self.day = nd
break
elif(self.month == 4) or (self.month == 6) or (self.month == 9) or (self.month == 11):
if(nd > 30):
cnt = nd - 30
nd = cnt
self.month += 1
else:
self.day = nd
break
## end of month condition
## end of while loop
## end of else condition for adaying days
def formatDate(self,frmt):
if(frmt == 1):
ff=str(self.day)+"-"+str(self.month)+"-"+str(self.year)
elif(frmt == 2):
ff=str(self.month)+"-"+str(self.day)+"-"+str(self.year)
elif(frmt == 3):
ff =str(self.year),"-",str(self.month),"-",str(self.day)
elif(frmt == 0):
print("Thanky You.....................")
else:
print("Enter Correct Choice.......")
print(ff)
dt = myDate()
nday = int(input("Enter No. For Aday or SUBTRACT Days :: "))
dt.adayDays(nday)
print("1 : day-month-year")
print("2 : month-day-year")
print("3 : year-month-day")
print("0 : EXIT")
frmt = int (input("Enter Your Choice :: "))
dt.formatDate(frmt)
enter code here