topic Output every 30 records in SAS Procedures

Output every 30 records

Solph — Wed, 16 Nov 2011 16:58:52 GMT

Hi,

I've a datadata consisting of daily data from 0 day to 7300 day (20 years of data); so far there is no day gap between records, so the days are consecutive. I'd like to output the data for every 30 days (30th, 60th) and ideally after 360 days, I'd like to jump to 366 and start again for every 30 days until the next 360th days and jump to 731, and so on. How do I do it? If it is too complicate, I can do every 30 days (plus the first record). The data would look something like the following:

data aa; input

day amount;

datalines;

0 12 -> output

1 23

2 50

30 34 -> output

31 34

60 41 -> output

....

360 3 -> output

365 3

366 1

367 4

....

;

Thanks in advance for your help.

Re: Output every 30 records

FriedEgg — Wed, 16 Nov 2011 17:26:34 GMT

data _null_;

input day amount;

if mod(day,30)=0 then put (day amount) (/=);

cards;

0 12

1 23

2 50

30 34

31 34

60 41

360 3

365 3

366 1

367 4

;

run;

day=0

amount=12

day=30

amount=34

day=60

amount=41

day=360

amount=3

Output every 30 records

Solph — Wed, 16 Nov 2011 17:52:20 GMT

Thanks a lot. It is a really neat way to do it. I thought about something like that (use division day/30) but didn't know how to proceed after that. I also just found the following way and it seemed to work (although it doesn't address the 365 days problem.)

data aa2 set aa;

do i = 0 to 7300 by 30;

if day=i then output;

end;

drop i;

run;

Output every 30 records

art297 — Wed, 16 Nov 2011 18:20:04 GMT

Do you have dates in the file? Otherwise, the 365 seems arbitray/biased, as it doesn't account for leap years.

Re: Output every 30 records

Linlin — Wed, 16 Nov 2011 18:39:45 GMT

Hi Solph,

Is this what you want?

data have;

do day=0 to 7300;

output;

end;

run;

data want(drop=i j);

set have;

if day=0 then output;

do j=0 to 19;

do i=j*365 to j*365+360 by 30;

if day=i and mod(day,365)ne 0 then output;

end;

run;

Output every 30 records

Solph — Wed, 16 Nov 2011 18:45:36 GMT

Not in this file, although I could create a pseudo date variable if it helps. Unfortunately we'd have to to output for every 30 days (has to be integer), although ultimately we are reporting the data by year. It is going to be difficult to reconcile the days differences one way or another. Another way is program by month of course, if so I guess a date variable would be useful, right?

Output every 30 records

Solph — Wed, 16 Nov 2011 22:36:10 GMT

Thanks Linlin. I just saw your reply and tested it. Yes it is the alternative approach I was thinking about. I'll just need to understand the logic of the code. Thanks so much.

Output every 30 records

Reeza — Wed, 16 Nov 2011 22:50:01 GMT

Do you need equal intervals or can you take the last day of every month?

Output every 30 records

Ksharp — Thu, 17 Nov 2011 05:07:22 GMT

data have;
do day=0 to 7300;
  output;
  end;
run;
data want;
 set have;
 if _n_ eq 1 then do;output;return;end;
 count+1; 
 if mod(group,12)=0 and count=31 then do;count=-4;end;
 mod=mod(count,30);
 if mod=1 then do;group+1; count=1;end;
 if mod=0 and count ne 0 then output;
run;

Ksharp

Output every 30 records

ieva — Thu, 17 Nov 2011 06:47:50 GMT

I think you need to know the date. Otherwise you don’t know if ‘366’ is last day of a leap year or first day of next year. Other solution could be to have these day numbers starting from 1 every new year:

363

364

365

Re: Output every 30 records

Howles — Fri, 18 Nov 2011 03:10:41 GMT

Even without the presence of an explicit date variable, it seems possible to use the INTNX function to avoid drift caused by leap years.

data have ;

do Day= 0 to 7300 ;

output ;

end ;

run;

data want (keep = day) ;

counter = 0 ;

do outer = 0 to 19 ;

anniversary = intnx('year', 0, outer, 'same') ;

do inner = 0 to ifn(outer LT 19, 330, 360) by 30 ;

next = anniversary + inner ;

do counter = counter to next ;

set have ;

end ;

output ;

end ;

run ;

art297 wrote:

Do you have dates in the file? Otherwise, the 365 seems arbitray/biased, as it doesn't account for leap years.

Output every 30 records

Peter_C — Sat, 19 Nov 2011 16:33:39 GMT

sometimes refered to as a DOW loop, because it loops around a SET statement (but iirc named after presenters who first brought that idea to prominence Paul Dorfman and Ian Whitlock),

this step might do what was wanted = re-start counting after 365, 730,

I tested on the first 1200 rows of sashelp.prdsale >>>>

data thirtys ;

do row = 1 to 365 ;

set sashelp.prdsale( obs=1200) ;

read+1 ;

if not mod( row, 30 ) then output ;

end ;

put read= ;

run