如何从日期中减去一天?
Posted
技术标签:
【中文标题】如何从日期中减去一天?【英文标题】:How to subtract a day from a date? 【发布时间】:2010-10-01 06:24:18 【问题描述】:我有一个 Python datetime.datetime
对象。减去一天的最佳方法是什么?
【问题讨论】:
相关:Find if 24 hrs have passed between datetimes - Python 这个线程直接用于将一年中的某一天转换为日期。 【参考方案1】:您可以使用timedelta
对象:
from datetime import datetime, timedelta
d = datetime.today() - timedelta(days=days_to_subtract)
【讨论】:
如果你不忽略时区,那么the answer is more complex。 另外,您如何将其与特定日期联系起来。请参阅我的问题:***.com/questions/43092508/… 它也可以与其他单位一起使用,例如,我已将其与timedelta(minutes=12)
一起使用。
文档说这将返回“表示两个日期、时间或日期时间实例之间差异的持续时间......”。你如何获得实际日期,比如,5 天前还是 5 天后?
@jfs 我认为这个答案在时区方面很好。如果减去一天,那可能实际上会减去 23 或 25 小时,但时间部分将保持不变。这就是我所期望的正常行为。夏令时 (DST) 使得 1 天并不总是 24 小时。【参考方案2】:
减去datetime.timedelta(days=1)
【讨论】:
【参考方案3】:只是为了详细说明一种替代方法和一个有用的用例:
从当前日期时间减去 1 天:在案例中有用,如果您想从当前日期时间增加 5 天并减少 5 小时。即从现在起 5 天后的日期时间是多少,但少了 5 小时?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
【讨论】:
【参考方案4】: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(或由于其他原因导致 UTC 偏移量发生变化)的错误:
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
一般来说,如果本地时区的 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/夏令时于 2014 年 11 月 2 日星期日凌晨 02:00:00 结束。在 America/Los_Angeles 时区,因此如果:
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
那么day_ago
和yesterday
不同:
day_ago
正好是 24 小时前(相对于 now
),但在上午 11 点,而不是在上午 10 点,因为 now
yesterday
是昨天上午 10 点,但它是 25 小时前(相对于 now
),而不是 24 小时。
pendulum
module 自动处理:
>>> 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]>
【讨论】:
【参考方案5】:也是我想计算时喜欢使用的另一个不错的函数,即上个月的第一天/最后一天或其他相对时间增量等...
来自dateutil 函数的 relativedelta 函数(对 datetime 库的强大扩展)
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
【讨论】:
【参考方案6】:通用箭头模块存在
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
【讨论】:
【参考方案7】: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
【讨论】:
以上是关于如何从日期中减去一天?的主要内容,如果未能解决你的问题,请参考以下文章
MySQL - 如何自动化视图查询,从最近的日期减去前一天的指标和日期戳最近的数据