I need to plus some days for the date. When I use this code, I miss a leap year. How not to miss it?
import java.time.LocalDate
LocalDate dob = LocalDate.of(1900, 1, 1).plusDays(40176)
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I'm scraping a website with requests. The URL requires dynamic dates so i'm generating the dates to use with the following variables:
stmDt = (pd.to_datetime('today').date()+ pd.offsets.MonthBegin(-1)).strftime('%d-%b-%Y')
todDt = (pd.to_datetime('today').date()).strftime('%d-%b-%Y')
snmDt = (pd.to_datetime('today').date()+ pd.offsets.MonthBegin(1)).strftime('%d-%b-%Y')
enmDt = (pd.to_datetime('today').date()+ pd.offsets.MonthEnd(2)).strftime('%d-%b-%Y')
For this example, i'm running this script on 11/30/2022 (last day of the month).
stmDt = start this month date (first day of month on which we run script - 11/1/2022)
todDt = today (11/30/2022)
snmDt = start next month (12/1/2022)
enmDt = end next month (12/31/2022)
The date variables are correct most of the time, but it appears (for november run dates):
On the first day of the month: stmDt shows the first day of the
previous month (10/1/22)
On the last day of the month: enmDt shows the last day two months
from now (1/31/23)
How can I tweak these so they always give me the correct dates? Happy to use other packages to accomplish etc
Thanks
You can do this using dateutil.relativedelta like this:
from datetime import datetime
from dateutil.relativedelta import relativedelta
today = datetime.now()
stmDt = today.replace(day=1).strftime('%d-%b-%Y')
todDt = today.strftime('%d-%b-%Y')
snmDt = (today + relativedelta(months=1)).replace(day=1).strftime('%d-%b-%Y')
enmDt = ((today + relativedelta(months=2)).replace(day=1) - relativedelta(days=1)).strftime('%d-%b-%Y')
I am writing a webscraper and the URL is based on the dates:
checkin_month=2&checkin_monthday=21&checkin_year=2020&checkout_month=2&checkout_monthday=22
There are 4 variables: checkin day and month, and checkout day and month. Checkout day will be always be = checkin day + 1. But when the newmonth is ending, checkout day is 1 and checkout month = checkin month +1.
Is there any function or library I can use to somehow implement it or do I have to write my own code to solve this?
You can use:
from datetime import timedelta
then :
yourDate += timedelta(days=1)
This previous code increments your date by one day. But, to increment the date, your date must be of type datetime.
Below, the documentation of thhis class.
class datetime.timedelta(days=0, seconds=0, microseconds=0, milliseconds=0, minutes=0, hours=0, weeks=0)
You can see a tutorial to manipulate date here : https://www.dataquest.io/blog/python-datetime-tutorial/
I need to get the difference between the days in months and days (eg. 3months 20days).
from datetime import datetime
from dateutil import relativedelta
date1 = datetime.strptime('2019-06-23', "%Y-%m-%d")
date2 = datetime.strptime('2018-04-17', '%Y-%m-%d')
r = relativedelta.relativedelta(date1, date2)
print(r)
This gives me result like relativedelta(years=+1, months=+2, days=+6) whereas I need result like 14 months 6 days
Thank you
Incorporate a minor modification to get the answer.
print(r.years, 'years,', r.months,'months and', r.days, 'days')
What I want to do is figure out what X of the month this is, and returning the relative delta constant for it (Su Mo Tu...). I have found many examples of jumping to a specific day of the month (1). For instance today is the 3rd Tuesday of December and I can get to it by doing this: + relativedelta(month=12, day=1, weekday=TU(3))) but what I want to do is the opposite:
Put in today's date and subtract the first of the month and get something like TU(3) or if it were the 4th wednesday to get: WE(4)
My ultimate goal is to then be able to transfer this constant to a different month or timedelta object and find the equivalent 3rd Tuesday, or 4th Wednesday, etc...
This is a solution that I have come up with, maybe you'll find it less complicated.
It also seems to be about 4 times faster, which if you process a lot of dates can make a difference.
from datetime import *
from dateutil.relativedelta import *
def weekDayOfTheMonth(xdate):
daylist = [MO,TU,WE,TH,FR,SA,SU]
weekday = xdate.weekday()
firstDayOfTheMonth = datetime(xdate.year, xdate.month, 1)
interval = (weekday + 7 - firstDayOfTheMonth.weekday() ) % 7
firstOfThisWeekDay = datetime(xdate.year, xdate.month, 1 + interval)
n = ((xdate.day - firstOfThisWeekDay.day) / 7) + 1
return daylist[weekday](n)
print(weekDayOfTheMonth(datetime.today()))
print(weekDayOfTheMonth(datetime(2018,11,24)))
Basically what happens is that:
I find what day of the week is the first day of given month.
Based on that information I can easily calculate first day of any given weekday in given month.
Then I can even more easily calculate that for example 18th of December 2018 is third Tuesday of this month.
Ok I found a way using rrule to create a list of days in the month that share the current weekday up until today, then length of this list becomes the Nth. Than I use a list as a lookup table for the weekday constants. Not tested to see if this will work for every day of the month but this is a start.
from datetime import *; from dateutil.relativedelta import *
from dateutil.rrule import rrule, WEEKLY
import calendar
def dayofthemonth(xdate):
daylist = [MO,TU,WE,TH,FR,SA,SU]
thisweekday = daylist[xdate.weekday()]
thisdaylist = list(rrule(freq=WEEKLY, dtstart=xdate+relativedelta(day=1), until=xdate, byweekday=xdate.weekday()))
return thisweekday(len(thisdaylist))
print(dayofthemonth(datetime.today())) #correctly returns TU(+3) for 2018, 12, 18
I want to know the total number of days from current date.
if given date is 2 Feb 2018. and i want to calculate number of days after 2 weeks/3 months/1 year. keeping in mind of leap year or not.
Language - python3
Is there any library in python which give me this.
The dateutil module has a way to do this very conveniently. Sample code:
import datetime
import dateutil
today = datetime.date.today()
delta = dateutil.relativedelta.relativedelta(years=1, months=3, weeks=2)
desired_date = today + delta
desired_date
Output:
datetime.date(2019, 5, 16)