topic Re: using array function in SAS Programming

using array function

asinusdk — Wed, 16 Jan 2019 22:11:10 GMT

year	id	censor
2001	1	1
2002	1	1
2003	1	1
2004	1	1
2005	1	1
2006	1	1
2002	2	0
2003	2	0
2004	2	0
2005	2	0
2006	2	0
2002	3	0
2003	3	0
2004	3	0
2005	3	0
2002	4	1
2003	4	1
2004	4	1

From this table, to make a table below,

year	id	censor	y01	y02	y03	y04	y05	y06	outcome
2001	1	1	1	0	0	0	0	0	0
2002	1	1	0	1	0	0	0	0	0
2003	1	1	0	0	1	0	0	0	0
2004	1	1	0	0	0	1	0	0	0
2005	1	1	0	0	0	0	1	0	0
2006	1	1	0	0	0	0	0	1	0
2002	2	0	0	1	0	0	0	0	0
2003	2	0	0	0	1	0	0	0	0
2004	2	0	0	0	0	1	0	0	0
2005	2	0	0	0	0	0	1	0	0
2006	2	0	0	0	0	0	0	1	1
2002	3	0	0	1	0	0	0	0	0
2003	3	0	0	0	1	0	0	0	0
2004	3	0	0	0	0	1	0	0	0
2005	3	0	0	0	0	0	1	0	1
2002	4	1	0	1	0	0	0	0	0
2003	4	1	0	0	1	0	0	0	0
2004	4	1	0	0	0	1	0	0	0

I tried to use array function, but it didn't work out well.

Could you please help me to make a code for this?

Thanks so much.

Re: using array function

Reeza — Wed, 16 Jan 2019 22:13:51 GMT

Please show what you've tried so far.

Re: using array function

PGStats — Wed, 16 Jan 2019 22:49:01 GMT

I would guess:

data have;
input year	id	censor;
datalines;
2001	1	1
2002	1	1
2003	1	1
2004	1	1
2005	1	1
2006	1	1
2002	2	0
2003	2	0
2004	2	0
2005	2	0
2006	2	0
2002	3	0
2003	3	0
2004	3	0
2005	3	0
2002	4	1
2003	4	1
2004	4	1
;

data want;
set have; by id;
array y y01-y06;
do i = 1 to dim(y);
    y{i} = i = mod(year, 100);
    end;
outcome = last.id and not censor;
drop i;
run;

proc print data=want noobs; run;

Re: using array function

Reeza — Wed, 16 Jan 2019 22:50:10 GMT

Index the array with the year?

Re: using array function

novinosrin — Wed, 16 Jan 2019 23:07:13 GMT

Same idea as PG, but easier than having to use mod


data want;
set have; 
by id;
array y(2001:2006) y01-y06;
do i = lbound(y) to hbound(y);
    y{i} = i = year;
end;
drop i;
run;