如何从日期中减去一天？-IT科技

摘要：问题描述：我有一个 Pythondatetime.datetime对象。减去一天的最佳方法是什么？解决方案 1：您可以使用一个timedelta对象：from datetime import datetime, timedelta d = datetime.today() - timedelta(d...

问题描述：

我有一个 Pythondatetime.datetime对象。减去一天的最佳方法是什么？

解决方案 1：

您可以使用一个timedelta对象：

from datetime import datetime, timedelta
    
d = datetime.today() - timedelta(days=days_to_subtract)

解决方案 2：

减去datetime.timedelta(days=1)

解决方案 3：

如果您的 Python datetime 对象具有时区感知能力，那么您应该小心避免在 DST 转换期间出现错误（或由于其他原因导致 UTC 偏移量发生变化）：

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

一般而言，如果当地时区的 UTC 偏移量在过去一天发生变化，day_ago则可能会有所不同。yesterday

例如，夏令时/夏令时于 2014 年 11 月 2 日星期日凌晨 02:00:00（美国/洛杉矶时区）结束，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

然后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，而不是上午 10 点，因为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模块自动处理：

>>> 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]>

解决方案 4：

仅详细说明一种有用的替代方法和用例：

从当前日期时间减去 1 天：

from datetime import datetime, timedelta
print datetime.now() + timedelta(days=-1)  # Here, I am adding a negative timedelta

在这种情况下很有用，如果您想在当前日期时间上增加 5 天并减去 5 小时。即，从现在起 5 天但少 5 小时的日期时间是什么？

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

解决方案 5：

另外，当我想计算上个月的第一天/最后一天或其他相对时间增量等时，我喜欢使用另一个很好的函数......

dateutil函数中的relativedelta函数（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

解决方案 6：

存在 Genial arrow 模块

import arrow
utc = arrow.utcnow()
utc_yesterday = utc.shift(days=-1)
print(utc, '
', utc_yesterday)

输出：

2017-04-06T11:17:34.431397+00:00 
 2017-04-05T11:17:34.431397+00:00

解决方案 7：

您还可以使用 pandas pd.Timedelta('1D')（它非常灵活，您甚至可以传递pd.Timedelta('1d 5h 9s')1 天 5 小时 9 秒之类的内容）。

pandas 的一个方便之处在于它的日期时间对象是建立在之上的datetime.datetime，因此任何涉及 Pythondatetime对象的操作都可以在 pandas 日期时间对象上正常工作，反之亦然。

import pandas as pd
import numpy as np
from datetime import datetime, date, timedelta

datetime.now() - pd.Timedelta('1d')           # datetime.datetime(2023, 2, 21, 15, 35, 23, 603832)
pd.Timestamp('now') - timedelta(days=1)       # Timestamp('2023-02-21 15:35:23.741866')
pd.Timestamp('now') - pd.Timedelta('1d')      # Timestamp('2023-02-21 15:35:23.882746')
pd.Timestamp('now') - np.timedelta64(1, 'D')  # Timestamp('2023-02-21 15:35:24.032356')

date(2022, 2, 22) - pd.Timedelta('10d')        # datetime.date(2022, 2, 12)

pandas 的优点是可以执行矢量化操作（即使 dtype 是 object）。您可以使用pd.Timedelta/ datetime.timedelta/中的任意一个np.timedelta64。

pd.Series([datetime(2023,2,22), datetime(2023,2,21), datetime(2023,2,20)]) - pd.Timedelta('1d')
pd.Series([date(2023,2,22), date(2023,2,21), date(2023,2,20)]) - timedelta(days=1)
pd.Series([date(2023,2,22), date(2023,2,21), date(2023,2,20)]) - np.timedelta64(1, 'D')

解决方案 8：

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