[英]Adding days to a date in Python
我有一个日期"10/10/11(mdy)"
,我想使用 Python 脚本添加 5 天。 请考虑一个适用于月底的通用解决方案。
我正在使用以下代码:
import re
from datetime import datetime
StartDate = "10/10/11"
Date = datetime.strptime(StartDate, "%m/%d/%y")
print Date
-> 正在打印'2011-10-10 00:00:00'
现在我想在这个日期上增加 5 天。 我使用了以下代码:
EndDate = Date.today()+timedelta(days=10)
返回此错误:
name 'timedelta' is not defined
以前的答案是正确的,但通常是更好的做法:
import datetime
然后你将拥有,使用datetime.timedelta
:
date_1 = datetime.datetime.strptime(start_date, "%m/%d/%y")
end_date = date_1 + datetime.timedelta(days=10)
首先导入timedelta
和date
。
from datetime import timedelta, date
而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)
如果您碰巧已经在使用pandas ,则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)
如果您想现在添加日期,您可以使用此代码
from datetime import datetime
from datetime import timedelta
date_now_more_5_days = (datetime.now() + timedelta(days=5) ).strftime('%Y-%m-%d')
这是使用dateutil 的 relativedelta在日期上添加天数的另一种方法。
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
我想你错过了这样的东西:
from datetime import timedelta
这是从现在开始+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
为了减少冗长的代码,并避免datetime 和 datetime.datetime 之间的名称冲突,您应该使用CamelCase名称重命名类。
from datetime import datetime as DateTime, timedelta as TimeDelta
所以你可以做下面的事情,我认为这更清楚。
date_1 = DateTime.today()
end_date = date_1 + TimeDelta(days=10)
此外,如果您想稍后import datetime
时间,也不会有名称冲突。
这可能有帮助:
from datetime import date, timedelta
date1 = date(2011, 10, 10)
date2 = date1 + timedelta(days=5)
print (date2)
使用timedelta
s,您可以执行以下操作:
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
一般来说,您现在已经有了答案,但也许我创建的课程也会有所帮助。 对我来说,它解决了我在 Pyhon 项目中遇到的所有要求。
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)
如何使用它
self.tz = pytz.timezone("Europe/Warsaw")
- 在这里定义要在项目中使用的时区GetDate("2019-08-08").current()
- 这会将您的字符串日期转换为您在 pt 1 中定义的时区的时间感知对象。默认字符串格式为format="%Y-%m-%d"
但随意改变它。 (例如。 GetDate("2019-08-08 08:45", format="%Y-%m-%d %H:%M").current()
)GetDate("2019-08-08").get_month_first_day()
返回给定日期(字符串或对象)月份第一天GetDate("2019-08-08").get_month_last_day()
返回给定日期月份最后一天GetDate("2019-08-08").minimum_time()
返回给定的日期开始日期GetDate("2019-08-08").maximum_time()
返回给定的日期日期GetDate("2019-08-08").time_delta_days({number_of_days})
返回给定日期 + 添加 {number of days}(您也可以调用: GetDate(timezone.now()).time_delta_days(-1)
为昨天)GetDate("2019-08-08").time_delta_haours({number_of_hours})
类似于 pt 7 但按小时工作GetDate("2019-08-08").time_delta_seconds({number_of_seconds})
类似于 pt 7 但在秒上工作有时我们需要使用 from date & to date 搜索。 如果我们使用date__range
那么我们需要增加1天至to_date
否则查询集将是空的。
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])
我已经看到了一个大熊猫的例子,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()
尝试这个:
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)
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)
我刚刚遇到了这个旧线程:
我查过了,但大多数答案都是一样的。 我喜欢所有这些中的两个答案,所以我想检查这两种方法的效率。
第一种方法:使用 DateTime 模块第二种方法:使用 panda 的库
所以我运行了大约 10k 次测试,而 pandas 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
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)
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