在Python中为日期添加天数

我有一个日期"10/10/11(m-d-y)"，我想用Python脚本添加5天。请考虑一个在月底也适用的通用解决方案。

我使用以下代码:

import re
from datetime import datetime


StartDate = "10/10/11"


Date = datetime.strptime(StartDate, "%m/%d/%y")

print Date ->打印'2011-10-10 00:00:00'

现在我想在这个日期上加5天。我使用了以下代码:

EndDate = Date.today()+timedelta(days=10)

返回以下错误:

name 'timedelta' is not defined
792653 次浏览
小开

首先导入timedelta和date。

from datetime import timedelta, date

date.today()将返回今天的日期时间，可能是你想要的

EndDate = date.today() + timedelta(days=10)
小开

我猜你错过了一些类似的东西:

from datetime import timedelta
小开
最佳答案

前面的答案是正确的，但通常更好的做法是:

import datetime

然后你会有，使用datetime.timedelta:

date_1 = datetime.datetime.strptime(start_date, "%m/%d/%y")


end_date = date_1 + datetime.timedelta(days=10)
小开

如果你碰巧已经在使用熊猫，你可以通过不指定格式来节省一点空间:

import pandas as pd
startdate = "10/10/2011"
enddate = pd.to_datetime(startdate) + pd.DateOffset(days=5)
小开

这里是一个从现在开始+指定天数的函数

import datetime


def get_date(dateFormat="%d-%m-%Y", addDays=0):


timeNow = datetime.datetime.now()
if (addDays!=0):
anotherTime = timeNow + datetime.timedelta(days=addDays)
else:
anotherTime = timeNow


return anotherTime.strftime(dateFormat)

用法:

addDays = 3 #days
output_format = '%d-%m-%Y'
output = get_date(output_format, addDays)
print output
小开

下面是另一个使用dateutil的relativedelta添加日期的方法。

from datetime import datetime
from dateutil.relativedelta import relativedelta


print 'Today: ',datetime.now().strftime('%d/%m/%Y %H:%M:%S')
date_after_month = datetime.now()+ relativedelta(days=5)
print 'After 5 Days:', date_after_month.strftime('%d/%m/%Y %H:%M:%S')

输出:

今天:25/06/2015 15:56:09

5天后:30/06/2015 15:56:09

小开

为了有一个更少的详细代码，避免名称冲突 在datetime和datetime.datetime之间，你应该重命名类的名称为CamelCase。

from datetime import datetime as DateTime, timedelta as TimeDelta

你可以这样做，我认为这样更清楚。

date_1 = DateTime.today()
end_date = date_1 + TimeDelta(days=10)

此外，如果您稍后想要import datetime，则会有无名称冲突。

小开

如果现在想要添加日期，可以使用这段代码

from datetime import datetime
from datetime import timedelta




date_now_more_5_days = (datetime.now() + timedelta(days=5) ).strftime('%Y-%m-%d')
小开

使用timedeltas你可以做到:

import datetime
today=datetime.date.today()




time=datetime.time()
print("today :",today)


# One day different .
five_day=datetime.timedelta(days=5)
print("one day :",five_day)
#output - 1 day , 00:00:00




# five day extend .
fitfthday=today+five_day
print("fitfthday",fitfthday)




# five day extend .
fitfthday=today+five_day
print("fitfthday",fitfthday)
#output -
today : 2019-05-29
one day : 5 days, 0:00:00
fitfthday 2019-06-03
小开

一般来说，你现在已经有了一个答案，但也许我创建的类也会有帮助。对我来说，它解决了我在Pyhon项目中遇到的所有需求。

class GetDate:
def __init__(self, date, format="%Y-%m-%d"):
self.tz = pytz.timezone("Europe/Warsaw")


if isinstance(date, str):
date = datetime.strptime(date, format)


self.date = date.astimezone(self.tz)


def time_delta_days(self, days):
return self.date + timedelta(days=days)


def time_delta_hours(self, hours):
return self.date + timedelta(hours=hours)


def time_delta_seconds(self, seconds):
return self.date + timedelta(seconds=seconds)


def get_minimum_time(self):
return datetime.combine(self.date, time.min).astimezone(self.tz)


def get_maximum_time(self):
return datetime.combine(self.date, time.max).astimezone(self.tz)


def get_month_first_day(self):
return datetime(self.date.year, self.date.month, 1).astimezone(self.tz)


def current(self):
return self.date


def get_month_last_day(self):
lastDay = calendar.monthrange(self.date.year, self.date.month)[1]
date = datetime(self.date.year, self.date.month, lastDay)
return datetime.combine(date, time.max).astimezone(self.tz)

如何使用

  1. self.tz = pytz.timezone("Europe/Warsaw") -在这里你可以定义你想在项目中使用的时区
  2. GetDate("2019-08-08").current() -将你的字符串日期转换为你在pt 1中定义的时区的时间感知对象。默认字符串格式是format="%Y-%m-%d"，但可以随意更改。(如。# EYZ2)
  3. GetDate("2019-08-08").get_month_first_day()返回给定日期(字符串或对象)月的第一天
  4. GetDate("2019-08-08").get_month_last_day()返回给定日期上个月的最后一天
  5. GetDate("2019-08-08").minimum_time()返回给定日期日期开始
  6. GetDate("2019-08-08").maximum_time()返回给定日期
  7. GetDate("2019-08-08").time_delta_days({number_of_days})返回给定日期+添加{天数}(你也可以调用:GetDate(timezone.now()).time_delta_days(-1)昨天)
  8. GetDate("2019-08-08").time_delta_haours({number_of_hours})类似于pt 7，但按小时工作
  9. GetDate("2019-08-08").time_delta_seconds({number_of_seconds})类似于pt 7，但工作在秒
小开

有时我们需要使用from date &到目前为止。如果我们使用date__range，那么我们需要为to_date添加1天，否则queryset将为空。

例子:

from datetime import timedelta


from_date  = parse_date(request.POST['from_date'])


to_date    = parse_date(request.POST['to_date']) + timedelta(days=1)


attendance_list = models.DailyAttendance.objects.filter(attdate__range = [from_date, to_date])
小开

这可能会有帮助:

from datetime import date, timedelta
date1 = date(2011, 10, 10)
date2 = date1 + timedelta(days=5)
print (date2)
小开

我已经看到了一个熊猫的例子，但这里有一个转折，你可以直接导入Day类

from pandas.tseries.offsets import Day


date1 = datetime(2011, 10, 10)
date2 = date1 + 5 * Day()
小开

试试这个:

在当前日期上增加5天。

from datetime import datetime, timedelta


current_date = datetime.now()
end_date = current_date + timedelta(days=5) # Adding 5 days.
end_date_formatted = end_date.strftime('%Y-%m-%d')
print(end_date_formatted)

从当前日期减去5天。

from datetime import datetime, timedelta


current_date = datetime.now()
end_date = current_date + timedelta(days=-5) # Subtracting 5 days.
end_date_formatted = end_date.strftime('%Y-%m-%d')
print(end_date_formatted)
小开 
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)
小开

我刚看到一条旧帖子:

我查过了，但大部分答案都是一样的。我喜欢其中的两个答案，所以我想检查一下这两种方法的效率。

第一种方法:使用DateTime模块 第二种方法:使用熊猫的库

所以我运行了大约10k次测试，熊猫库方法要慢得多。所以我建议使用内置的DateTime模块。

from datetime import date, timedelta
import pandas as pd
import timeit


def using_datetime():
pre_date = date(2013, 10, 10)
day_date = pre_date + timedelta(days=5)
return day_date


def using_pd():
start_date = "10/10/2022"
pd_date = pd.to_datetime(start_date)
end_date = pd_date + pd.DateOffset(days=5)
return end_date
    



for func in [using_datetime, using_pd]:
print(f"{func.__name__} Time Took: ",  timeit.timeit(stmt=func, number=10000))
    

# Output
# using_datetime Time Took:  0.009390000021085143
# using_pd Time Took:  2.1051381999859586

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