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")
打印日期->正在打印'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
使用时间增量你可以做:
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
使用时间增量你可以做:
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
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)
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:利用熊猫的图书馆
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