The scenario is I need to get last week date in format yyyy-mm-dd HH:mm:ss from last Sunday till Saturday(date will increment by 1 from last week of Sunday till Saturday) and like wise one more is to get date from past 20 weeks(till last week).
How may I get it?
Your requirement is not very clear so I can provide only a generic solution:
Getting the "last Sunday"
def calendar = Calendar.instance
def delta = Calendar.SUNDAY - calendar.get(Calendar.DAY_OF_WEEK)
calendar.add(Calendar.DAY_OF_WEEK, delta)
log.info('Last Sunday: ' + calendar.time.format("yyyy-MM-dd HH:mm:ss"))
Getting all previous "dates" for last 20 weeks:
def now = new Date()
use(groovy.time.TimeCategory) {
def twentyWeeksAgo = now - 30.weeks
def duration = now - twentyWeeksAgo
1.upto(duration.days, {
twentyWeeksAgo = twentyWeeksAgo + 1.days
log.info(twentyWeeksAgo.format('yyyy-MM-dd HH:mm:ss'))
})
}
More information:
Groovy TimeCategory
Creating and Testing Dates in JMeter - Learn How
Related
I am trying to create a list of the last days of each month for the past n months from the current date but not including current month
I tried different approaches:
def last_n_month_end(n_months):
"""
Returns a list of the last n month end dates
"""
return [datetime.date.today().replace(day=1) - datetime.timedelta(days=1) - datetime.timedelta(days=30*i) for i in range(n_months)]
somehow this partly works if each every month only has 30 days and also not work in databricks pyspark. It returns AttributeError: 'method_descriptor' object has no attribute 'today'
I also tried the approach mentioned in Generate a sequence of the last days of all previous N months with a given month
def previous_month_ends(date, months):
year, month, day = [int(x) for x in date.split('-')]
d = datetime.date(year, month, day)
t = datetime.timedelta(1)
s = datetime.date(year, month, 1)
return [(x - t).strftime('%Y-%m-%d')
for m in range(months - 1, -1, -1)
for x in (datetime.date(s.year, s.month - m, s.day) if s.month > m else \
datetime.date(s.year - 1, s.month - (m - 12), s.day),)]
but I am not getting it correctly.
I also tried:
df = spark.createDataFrame([(1,)],['id'])
days = df.withColumn('last_dates', explode(expr('sequence(last_day(add_months(current_date(),-3)), last_day(add_months(current_date(), -1)), interval 1 month)')))
I got the last three months (Sep, oct, nov), but all of them are the 30th but Oct has Oct 31st. However, it gives me the correct last days when I put more than 3.
What I am trying to get is this:
(last days of the last 4 months not including last_day of current_date)
daterange = ['2022-08-31','2022-09-30','2022-10-31','2022-11-30']
Not sure if this is the best or optimal way to do it, but this does it...
Requires the following package since datetime does not seem to have anyway to subtract months as far as I know without hardcoding the number of days or weeks. Not sure, so don't quote me on this....
Package Installation:
pip install python-dateutil
Edit: There was a misunderstanding from my end. I had assumed that all dates were required and not just the month ends. Anyways hope the updated code might help. Still not the most optimal, but easy to understand I guess..
# import datetime package
from datetime import date, timedelta
from dateutil.relativedelta import relativedelta
def previous_month_ends(months_to_subtract):
# get first day of current month
first_day_of_current_month = date.today().replace(day=1)
print(f"First Day of Current Month: {first_day_of_current_month}")
# Calculate and previous month's Last date
date_range_list = [first_day_of_current_month - relativedelta(days=1)]
cur_iter = 1
while cur_iter < months_to_subtract:
# Calculate First Day of previous months relative to first day of current month
cur_iter_fdom = first_day_of_current_month - relativedelta(months=cur_iter)
# Subtract one day to get the last day of previous month
cur_iter_ldom = cur_iter_fdom - relativedelta(days=1)
# Append to the list
date_range_list.append(cur_iter_ldom)
# Increment Counter
cur_iter+=1
return date_range_list
print(previous_month_ends(3))
Function to calculate date list between 2 dates:
Calculate the first of current month.
Calculate start and end dates and then loop through them to get the list of dates.
I have ignored the date argument, since I have assumed that it will be for current date. alternatively it can be added following your own code which should work perfectly.
# import datetime package
from datetime import date, timedelta
from dateutil.relativedelta import relativedelta
def gen_date_list(months_to_subtract):
# get first day of current month
first_day_of_current_month = date.today().replace(day=1)
print(f"First Day of Current Month: {first_day_of_current_month}")
start_date = first_day_of_current_month - relativedelta(months=months_to_subtract)
end_date = first_day_of_current_month - relativedelta(days=1)
print(f"Start Date: {start_date}")
print(f"End Date: {end_date}")
date_range_list = [start_date]
cur_iter_date = start_date
while cur_iter_date < end_date:
cur_iter_date += timedelta(days=1)
date_range_list.append(cur_iter_date)
# print(date_range_list)
return date_range_list
print(gen_date_list(3))
Hope it helps...Edits/Comments are welcome - I am learning myself...
I just thought a work around I can use since my last codes work:
df = spark.createDataFrame([(1,)],['id'])
days = df.withColumn('last_dates', explode(expr('sequence(last_day(add_months(current_date(),-3)), last_day(add_months(current_date(), -1)), interval 1 month)')))
is to enter -4 and just remove the last_date that I do not need days.pop(0) that should give me the list of needed last_dates.
from datetime import datetime, timedelta
def get_last_dates(n_months):
'''
generates a list of lastdates for each month for the past n months
Param:
n_months = number of months back
'''
last_dates = [] # initiate an empty list
for i in range(n_months):
last_dates.append((datetime.today() - timedelta(days=i*30)).replace(day=1) - timedelta(days=1))
return last_dates
This should give you a more accurate last_days
I have a variable in a SAS dataset that has a number of dates (e.g. 01APR21). What I'm looking to do is create a new variable that shows the date of the first Monday of that week. So using the above example of 01APR21, the output would be 29/03/2021 as that what was when the Monday in that week was. I'm assuming it's using intnx, but I can't get my head around it.
data test;
format date date8.;
format first_day date10.;
date = '01APR21'd;
first_day = ?;
run;
INTNX Parameters:
Interval : WEEK
Increment: 0 (same week)
Alignment: Beginning
(Sunday)
Then add 1 to get to Monday instead of Sunday. You could probably play with the SHIFT INDEX parameter as well.
Monday = intnx('week', dateVariable, 0, 'B') + 1
I'm looking to do a difference between 2 date in scala. These 2 date are should not saturday not sunday.
I did a scala function to test if the day if Saturday or Sunday:
I edited my question, this is my code to test if a day is saturday or sunday.
I should use it using tow dates : start_date and finish_date, because after this operation I'll do the difference between these tow dates.
My function jourouvree took one parameter, not a dates.
How can I modify my code to pass the tow dates.
Check if Day is Saturday or Sunday:
import java.time.{LocalDate, DayOfWeek}
def isWeekend(day: LocalDate) =
day.getDayOfWeek == DayOfWeek.SATURDAY ||
day.getDayOfWeek == DayOfWeek.SUNDAY
Using Java 8 datetime api:
import java.time._
import java.time.format.DateTimeFormatter
val formatter = DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss")
//Assume d2,d2 themselves are not weekend days
def jourOuvree(sd1:String, sd2:String): Unit = {
val d1 = LocalDateTime.parse(sd1, formatter)
val d2 = LocalDateTime.parse(sd2, formatter)
//return if d1 is not earlier than d2. TODO: handle what to be done
if(d2.isAfter(d1) == false) return
var (days, dayCtr) = (1, 1)
while(d1.plusDays(dayCtr).isBefore(d2)){
dayCtr+=1
val dow = d1.plusDays(dayCtr).getDayOfWeek()
if(!dow.equals(DayOfWeek.SATURDAY) && !dow.equals(DayOfWeek.SUNDAY))
days+=1
}
println(days)
}
Invoke as below:
jourOuvree("2011-03-31 07:55:00", "2011-04-06 15:41:00")
You get 5 printed.
NOTE: The code doesn't handle exceptions in parsing.
Also, there may be other requirement fine points, for which you are the best judge to make required changes.
I have a spreadsheet that asks people to enter in a day of the month when we need to send out a bill. What I want to do is create a calendar event based on that. So, essentially what I need is an event that starts at the current month, day from the spreadsheet, and continues to a specified point in time.
var monthlyDate = row[6]; // Seventh column, monthly date of payment
var curDate = new Date();
var curMonth = curDate.getMonth();
var curYear = curDate.getYear();
curDate.setDate(curMonth, monthlyDate, curYear);
Logger.log("Day of month: %s", monthlyDate);
Logger.log("Current Date: %s", curDate);
Logger.log("Current Date: %s", Date());
What I'm seeing is that the monthly date is coming in as a float "6.0" for example, and no matter what I enter in for monthlyDate in the setDate line, it keeps setting the date to 10/9/15 (Today is 10/15/15). I've hard-coded that value to many different numbers, but for some reason it's just not working.
How can I create a date (in any format) that follows the scheme "Current Month / Day from Speadsheet / Current Year" ?
The getMonth() method returns a "zero-indexed" number. So, it returns the number 9 for the 10th month. setDate() doesn't set the date, it sets the "Day of the Month". The name of that method is misleading.
Documentation - setDate()
So, the last two parameters that you are using in setDate() are doing nothing. You are setting the day of the month to 9.
If you want to set multiple date parameters at the same time, you need to use the new Date() method:
var d = new Date(year, month, day, hours, minutes, seconds, milliseconds);
The month parameter accept values from 0 to 11, 0 is Jan and 11 is Dec
Date Reference
I need to make my Access query always return the Monday of the current week. I have seen a few solutions on Google/StackOverflow but they are written in SQL and I am a beginner in creating Access queries (I am using the Design view to make them).
Goal: The week should be considered as M T W T F S S. Then, the query should always return the Monday of the current week. Therefore, if it is Sunday, it should still return the Monday before, NOT the next week's Monday. Can anyone explain how to do this using the Design View in Access 2010?
Keep in mind that in this context we are working with dates, so if we do Date() - 1, we will get 1 day prior to today.
Date() ~ Today's date
DatePart(
"w" - Weekday
Date() - Today's date
2 - vBMonday (Access assumes Sunday is the first day of the week, which is why this is necessary.)
1 - vbFirstJan1 - This gets into using the first week of the year. We could have omitted this, as 1 is the default.
)
-1 - Subtract 1 from the DatePart value.
Values
Date() = 4/27/2015 (at time of this writing)
DatePart("w",Date(),2,1) = 1
DatePart("w",Date(),2,1)-1 = 0
So we have Date()-0... Okay, what's so great about that? Well, let's look at a more useful scenario where today's date is a day other than Monday.
Let's act like today is 4/28/2015 (Tuesday)
Date() = 4/28/2015
DatePart("w",Date(),2,1) = 2
DatePart("w",Date(),2,1)-1 = 1
So, from the outside, in; give me the current weekday value. (1 = Monday, 2 = Tuesday, etc.), and subtract 1 from that -> that's how many days we need to subtract from the current date to get back to the weekday value of 1 (Monday).
Here's a function that will do this:
Public Function DatePrevWeekday( _
ByVal datDate As Date, _
Optional ByVal bytWeekday As VbDayOfWeek = vbMonday) _
As Date
' Returns the date of the previous weekday, as spelled in vbXxxxday, prior to datDate.
' 2000-09-06. Cactus Data ApS.
' No special error handling.
On Error Resume Next
DatePrevWeekday = DateAdd("d", 1 - Weekday(datDate, bytWeekday), datDate)
End Function
As vbMonday is 2 and your date is today, you can use the core expression in a query:
PreviousMonday: DateAdd("d",1-Weekday(Date(),2),Date())