topic SAS Roll_up Summary in SAS Procedures

SAS Roll_up Summary

sasuser00 — Fri, 20 Apr 2012 16:22:12 GMT

I'd like to know how to do 10 day roll up summary. Thanks.

date	count	10day sum
1/1/2012	2
1/2/2012	4
1/3/2012	3
1/4/2012	6
1/5/2012	7
1/6/2012	9
1/7/2012	65
1/8/2012	4
1/9/2012	8
1/10/2012	56	164
1/11/2012	4	166
1/12/2012	5	167
1/13/2012	4	168
1/14/2012	2	164
1/15/2012	2	159
1/16/2012	3	153
1/17/2012	4	92
1/18/2012	3	91
1/19/2012	2	85
1/20/2012	4	33
1/21/2012	5	34
1/22/2012	3	32
1/23/2012	5	33
1/24/2012	8	39
1/25/2012	9	46
1/26/2012	0	43
1/27/2012	7	46
1/28/2012	6	49
1/29/2012	6	53
1/30/2012	45	94

Re: SAS Roll_up Summary

FriedEgg — Fri, 20 Apr 2012 16:44:54 GMT

data foo;

input date mmddyy10. count;

cards;

1/1/2012 2

1/2/2012 4

1/3/2012 3

1/4/2012 6

1/5/2012 7

1/6/2012 9

1/7/2012 65

1/8/2012 4

1/9/2012 8

1/10/2012 56

1/11/2012 4

1/12/2012 5

1/13/2012 4

1/14/2012 2

1/15/2012 2

1/16/2012 3

1/17/2012 4

1/18/2012 3

1/19/2012 2

1/20/2012 4

1/21/2012 5

1/22/2012 3

1/23/2012 5

1/24/2012 8

1/25/2012 9

1/26/2012 0

1/27/2012 7

1/28/2012 6

1/29/2012 6

1/30/2012 45

;

run;

data bar;

if _n_=1 then

do i=1 to 9;

set foo;

call missing(sum);

output;

end;

sum=0;

set foo (firstobs=10);

do i=1 to 10;

p=(_n_-1)+i;

set foo point=p;

sum+count;

end;

output;

drop i;

run;

date	count	sum
18993	2	.
18994	4	.
18995	3	.
18996	6	.
18997	7	.
18998	9	.
18999	65	.
19000	4	.
19001	8	.
19002	56	164
19003	4	166
19004	5	167
19005	4	168
19006	2	164
19007	2	159
19008	3	153
19009	4	92
19010	3	91
19011	2	85
19012	4	33
19013	5	34
19014	3	32
19015	5	33
19016	8	39
19017	9	46
19018	0	43
19019	7	46
19020	6	49
19021	6	53
19022	45	94

Re: SAS Roll_up Summary

Haikuo — Fri, 20 Apr 2012 16:55:50 GMT

Here we are, again. without involving ETS, in addition to FriedEgg's solution, here is another approach (credit to Howard, Ksharp):

data have;

infile cards truncover;

input date :mmddyy10. count;

format date mmddyy10.;

cards;

1/1/2012 2

1/2/2012 4

1/3/2012 3

1/4/2012 6

1/5/2012 7

1/6/2012 9

1/7/2012 65

1/8/2012 4

1/9/2012 8

1/10/2012 56

1/11/2012 4

1/12/2012 5

1/13/2012 4

1/14/2012 2

1/15/2012 2

1/16/2012 3

1/17/2012 4

1/18/2012 3

1/19/2012 2

1/20/2012 4

1/21/2012 5

1/22/2012 3

1/23/2012 5

1/24/2012 8

1/25/2012 9

1/26/2012 0

1/27/2012 7

1/28/2012 6

1/29/2012 6

1/30/2012 45

;

data want;

array v(0:9) _temporary_;

set have;

v(mod(_n_,10))=count;

if _n_>=10 then day_sum=sum( of v(*));

run;

proc print;run;

Haikuo

Re: SAS Roll_up Summary

JasonDiVirgilio — Fri, 20 Apr 2012 18:08:33 GMT

Beat me to it...and that code is definitely slick and concise. If it is too slick:

data want;
    set have;
    array sumz {10};
    retain sumz rowcnt 0;

    rowcnt +1;
    sumz{rowcnt}=count;
    if _n_ >=10 then do;
      tendaycount=sum(of sumz1-sumz10);
      if _n_ in (10,20,30)then rowcnt=0;
    end;
    keep date1 count tendaycount;
run;

Re: SAS Roll_up Summary

FriedEgg — Fri, 20 Apr 2012 18:17:49 GMT

using macro and lag function:

%macro lag_sum(var,n);

%let lag_sum=lag(&var);

%do i=2 %to &n;

%let lag_sum=&lag_sum,lag&i(&var);

%end;

sum_&var&n=sum(&lag_sum);

if _n_ < &n then call missing(sum_&var&n);

%mend;

data bar;

set foo;

%lag_sum(count,10);

run;

Re: SAS Roll_up Summary

art297 — Fri, 20 Apr 2012 20:05:09 GMT

I like Haikuo's FIFO approach but, if you are concerned about 10 day sums and don't really have all dates, then you might need to expand the code a little. E.g.:

%let roll=10;

data want;

array v{0:%eval(&roll.-1)} _temporary_;

array d{0:%eval(&roll.-1)} _temporary_;

set have;

v{mod(_n_,&roll.)}=count;

d{mod(_n_,&roll.)}=date;

do i=0 to 9;

if not missing(d{i}) then do;

if date-d{i} gt %eval(&roll.-1) then do;

call missing(v{i});

call missing(d{i});

end;

if _n_>=&roll. then day_sum=sum( of v{*});

run;

Re: SAS Roll_up Summary

FloydNevseta — Fri, 20 Apr 2012 22:14:37 GMT

It doesn't get any easier than this:

data want (drop=_:);

set have;

retain _cum 0;

_cum + ( count - coalesce( lag10( count ),0 ));

if _n_ <= 10 then cum_sum = .;

else cum_sum = _cum;

run;

Each iteration of the data step adds to the cumulative sum the new value for count and subtracts the 10th lag of count.

Re: SAS Roll_up Summary

Linlin — Fri, 20 Apr 2012 23:22:54 GMT

Nice!

if _n_ <= 10 then cum_sum = .;

should be

if _n_ < 10 then cum_sum = .;

Re: SAS Roll_up Summary

FloydNevseta — Fri, 20 Apr 2012 23:43:55 GMT

Good catch. Thanks.

Re: SAS Roll_up Summary

MikeZdeb — Sat, 21 Apr 2012 03:28:22 GMT

hi ... here's a tweak of SAS_Bigot's nice solution ...

data want (drop=_cum);

set have;

_cum + sum(count, lag10(-count));

cum_sum = ifn(_n_ lt 10, . , _cum);

run;

Re: SAS Roll_up Summary

PGStats — Sat, 21 Apr 2012 04:08:16 GMT

I agree with your point Arthur. Solutions that depend on sorted values can be guarantied to work but if you rely on complete data, that's more difficult to enforce. So better be safe than sorry... Moreover, the overhead cost can be minimal (I removed one data line to show the effect) :

data have;
infile cards truncover;
input date :mmddyy10. count;
format date mmddyy10.;
cards;
1/1/2012 2
1/2/2012 4
1/3/2012 3
1/4/2012 6
1/5/2012 7
1/6/2012 9
1/7/2012 65
1/8/2012 4
1/9/2012 8
1/10/2012 56
1/11/2012 4
1/12/2012 5
1/13/2012 4
1/14/2012 2
1/16/2012 3
1/17/2012 4
1/18/2012 3
1/19/2012 2
1/20/2012 4
1/21/2012 5
1/22/2012 3
1/23/2012 5
1/24/2012 8
1/25/2012 9
1/26/2012 0
1/27/2012 7
1/28/2012 6
1/29/2012 6
1/30/2012 45
;

%let wnd=10;
data want;
array d{0:0} _temporary_;
array s{0:&wnd.} _temporary_;
set have;
d{0} = 1 + coalesce(d{0}, date-1);
do while (d{0} < date);
s{mod(d{0}, &wnd)} = .; /* creates a hole when sum is incomplete (preferred)*/
*s{mod(d{0}, &wnd)} = 0; /* accepts incomplete sums */
d{0} + 1;
end;
s{mod(date, &wnd)} = count;
if nmiss(of s{*}) = 1 then day_sum = sum(of s{*});
else day_sum = .;
run;

proc print data=want; run;

Re: SAS Roll_up Summary

Ksharp — Mon, 23 Apr 2012 04:09:25 GMT

I have a SQL solution instead of it.Not need to pre-sort dataset.

data foo;
input date mmddyy10. count;
format date mmddyy10.;
cards;
1/1/2012  2
1/2/2012  4
1/3/2012  3
1/4/2012  6
1/5/2012  7
1/6/2012  9
1/7/2012  65
1/8/2012  4
1/9/2012  8
1/10/2012 56
1/11/2012 4
1/12/2012 5
1/13/2012 4
1/14/2012 2
1/15/2012 2
1/16/2012 3
1/17/2012 4
1/18/2012 3
1/19/2012 2
1/20/2012 4
1/21/2012 5
1/22/2012 3
1/23/2012 5
1/24/2012 8
1/25/2012 9
1/26/2012 0
1/27/2012 7
1/28/2012 6
1/29/2012 6
1/30/2012 45
;
run;
proc sql;
create table want as
 select date,count,case when date lt min(date)+9 then . else (select sum(count) from foo as a where a.date between b.date-9 and b.date) end as sum
  from foo as b;
quit;

Ksharp

Re: SAS Roll_up Summary

Haikuo — Mon, 23 Apr 2012 14:09:12 GMT

Liked! SQL is indeed a powerful tool!

Haikuo

Re: SAS Roll_up Summary

Haikuo — Mon, 23 Apr 2012 15:49:47 GMT

Awesome! So far we have Array, Lag(), SQL, Macro, DOW, and let's see how far this great thread can lead us to!

Haikuo

Re: SAS Roll_up Summary

PGStats — Mon, 23 Apr 2012 17:52:31 GMT

Of course, as a rule, SQL is the only way to do this properly (duck :smileysilly:) as in the bullet proof :

proc sql;

create table want as

select A.date, A.count, ifn(count(distinct B.date)=10, sum(B.count), .) as ten_day_sum

from foo as A left join foo as B

on B.date between intnx("DAY", A.date, -9, "Beginning") and A.date

group by A.date, A.count;

select * from want;

quit;

Re: SAS Roll_up Summary

FriedEgg — Thu, 26 Apr 2012 20:55:40 GMT

Use double set with key lookup to account for missing dates, similar to my previous method using double set statements and point:

data foo(index=(date /nomiss /unique));
call streaminit(12345);
do date='01JAN1960'd to today();
count=int(rand('uniform')*10e2);
if rand('table',.95)=1 then output;
end;
format date date9.;
run;

data bar;
set foo(rename=(date=_date)) nobs=nobs;
sum=.;
miss=0;
do date=_date-10 by 1 while ( date < _date and _n_ > 10 /* and miss=0 */ );
set foo key=date/unique;
if _iorc_ then do;
   _iorc_=0;
   _error_=0;
   miss+1;
end;
else do;
   sum+count;
end;
end;

*if miss>1 then call missing(sum);
date=_date;
drop _date;
run;