Finding last Sunday and going 4 weeks backward every week in SAS - date

I have a SAS job that runs every Thursday, but sometime it need to run on Wednesday, and maybe Tuesday evening. The job collects some data in 4 week intervals up until the closest Sunday. For example, today we have 19Mar2015, and I need data until 15Mar2015.
data get_some_data;
set all_the_data;
where date >= '16Feb2015' and date <= '15Mar2015';
run;
Next week I have to manually change the date parameters too
data get_some_data;
set all_the_data;
where date >= '23Feb2015' and date <= '22Mar2015';
run;
Anyway I can automate this?

I'll expand on the suggestion from #user667489 as it could take you a while to work it out. The key is to use the week time interval, which by default starts on a Sunday (you can change this with a shift index, read this for further details)
So your query just needs to be :
where intnx('week',today(),-4)<date<=intnx('week',today(),0);

Use the INTNX function to regress the date back to last Sunday:
data get_some_data;
set all_the_data;
lastsun=intnx('week',today(),0);
/*where date >= '23Feb2015' and date <= '22Mar2015';*/
where date between lastsun-27 and lastsun;
run;

You can try getting the last sunday date using weekday function and then using INTNX get the 4 week back date from that sunday date. Check the below ref code :
data mydata;
input input_date YYMMDD10.;
/* Loop to get the last sunday date, do the processing
and get out of loop */
do i =0 to 7 until(last_sunday_date>0);
/* Weekday Sunday=1 */
if weekday(sum(input_date,-i))=1 then do;
last_sunday_date=sum(input_date,-i);
/* INTNX to get the 4 week back date */
my_4_week_start=intnx('week',last_sunday_date,-4);
end;
end;
format input_date last_sunday_date my_4_week_start yymmdd10.;
datalines4;
2015-03-01
2015-03-07
2015-03-14
2015-03-21
2015-03-28
2015-04-05
2015-04-13
2015-04-20
;;;;
run;
proc print data=mydata;run;
let me know if this helps!

Related

How do you find the first date of the week in SAS from a date?

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

intck() giving negative value

I am new to SAS and I am having trouble with finding the difference between 2 dates.
I have 2 columns: checkin_date and checkout_date
the dates are in mmddyy10. format (mm/dd/yyyy).
I have used the following code:
stay_days= intck('day', checkin_day, checkout_day);
I am getting the right values for dates in the same month but wrong values for days that are across 2 months. For example, the difference between 02/06/2014 and 02/11/2014 is 5. But the difference between 1/31/2014 and 2/13/2014 is -18 which is incorrect.
I have also simply tried to subtract them both:
stay_day = checkout_day - checkin_day;
I am getting the same result for that too.
My entire code:
data hotel;
infile "XXXX\Hotel.dat";
input room_no num_guests checkin_month checkin_day checkin_year checkout_month checkout_day checkout_year internet_used $ days_used room_type $16. room_rate;
checkin_date = mdy(checkin_month,checkin_day,checkin_year);
informat checkin_date mmddyy.;
format checkin_date mmddyy10.;
checkout_date = mdy(checkout_month,checkout_day,checkout_year);
informat checkout_date mmddyy.;
format checkout_date mmddyy10.;
stay_day= intck('day', checkin_day, checkout_day);
Your problem is a typo - using wrong variables in intck() function. You are using variables "xxx_DAY" which is the DAY of month instead of the full DATE. Change to stay_day= intck('day', checkin_date, checkout_date);
Your data probably has the date values in the wrong variables. When using subtraction the order should be ENDDATE - STARTDATE. When using INTNX() function the order should be from STARTDATE to ENDDATE. In either case if the value in the STARTDATE variable is AFTER the value in the ENDDATE variable then the difference will be a negative number.
Perhaps you need to clean the data?
The only way to get -18 comparing 2014-01-31 and 2014-02-13 would be if you extracted the day of the month and subtracted them.
diff3 = day(end) - day(start);
which would be the same as subtracting 31 from 13.
Example using your dates:
data check;
input start end ;
informat start end mmddyy.;
format start end yymmdd10.;
diff1=intck('day',start,end);
diff2=end-start;
cards;
02/06/2014 02/11/2014
1/31/2014 2/13/2014
;
Results:
Obs start end diff1 diff2
1 2014-02-06 2014-02-11 5 5
2 2014-01-31 2014-02-13 13 13

Subtracting 1 ISO 8601 year from a date in BigQuery

I'm trying to manipulate a date value to go back in time exactly 1 ISO-8601 year.
The following does not work, but best describes what I want to accomplish:
date_add(date '2018-01-03', interval -1 isoyear)
I tried string conversion as an intermediate step, but that doesn't work either:
select parse_date('%G%V%u',safe_cast(safe_cast(format_date('%G%V%u',date '2018-01-03') as int64)-1000 as string))
The error provided for the last one is "Failed to parse input string "2017013"". I don't understand why, this should always resolve to a unique date value.
Is there another way in which I can subtract an ISO year from a date?
This gives the corresponding day of the previous ISO year by subtracting the appropriate number of weeks from the date. I based the calculation on the description of weeks per year from the Wikipedia page:
CREATE TEMP FUNCTION IsLongYear(d DATE) AS (
-- Year starting on Thursday
EXTRACT(DAYOFWEEK FROM DATE_TRUNC(d, YEAR)) = 5 OR
-- Leap year starting on Wednesday
(EXTRACT(DAY FROM DATE_ADD(DATE(EXTRACT(YEAR FROM d), 2, 28), INTERVAL 1 DAY)) = 29
AND EXTRACT(DAYOFWEEK FROM DATE_TRUNC(d, YEAR)) = 4)
);
CREATE TEMP FUNCTION PreviousIsoYear(d DATE) AS (
DATE_SUB(d, INTERVAL IF(IsLongYear(d), 53, 52) WEEK)
);
SELECT PreviousIsoYear('2018-01-03');
This returns 2017-01-04, which is the third day of the 2017 ISO year. 2018-01-03 is the third day of the 2018 ISO year.

Microsoft Access DateDiff + Difference in time if end time is next day

I have a table with records, where each record has a date column, then a start time column and end time column.
I am trying to do a datediff to get the duration in hours from start to end date with DateDiff('s',[Start Date[,[End Date])/3600.
This works perfectly for End dates that are on same day as date column, but sometimes the end date would be the next day like 12:45 AM. The date diff will give me a large negative number, how do I let it know its next day?
I dont own the data, so not much I can do with the table
Thanks!
Try something like this:
DateDiff('s',[Start Date],DateAdd('d',IIF([End date]<[Start Date],1,0),[End Date]))/3600
It can be done with pure math:
TotalHours = TimeValue(CDate([End Date] - [Start Date] + 1)) * 24

SAS date swap year and day

I am working with a dataset containing a date variable with the format MMDDYY10..
The problem is, that the day, month and the year have been swapped.
The data set as it looks now:
Obs Date
1 11/01/1931
2 11/06/1930
3 12/02/2003
4 12/07/2018
What I would like is a date variable with the format DDMMYY10., or a similar:
Obs Date
1 31/01/2011
2 30/06/2011
3 03/02/2012
4 18/07/2012
Observation 1 is hence written as the 1st of November 1931, but really it is the 31st of January 2011.
Does anyone know how I can change this?
Looks like you read the original raw data using the wrong INFORMAT. Most likely you had data in YYMMDD format and you read it as MMDDYY. You can use the PUT() and INPUT() functions to attempt to reverse it.
data have ;
input date mmddyy10.;
newdate = input(put(date,mmddyy6.),yymmdd6.);
format date newdate yymmdd10. ;
put (date newdate) (=);
cards;
11/01/1931
11/06/1930
12/02/2003
12/07/2018
;;;;
Results:
date=1931-11-01 newdate=2011-01-31
date=1930-11-06 newdate=2011-06-30
date=2003-12-02 newdate=2012-02-03
date=2018-12-07 newdate=2012-07-18