Dear all,
Today i came with another stupid question, That i am having problem with column representation and the dataset given bellow
Now I need to create another dataset with all these variables and an extra variable which is calculated from T1, T2, T3, T4, Lower. Consider T1 as 1, T2 as 2, T3 as 3, T4 as 4. nSuppose ” lower” contains value 0.53 which is positioned at “T2” so the new variable will contains 2 and “lower “ is 1 then new variable contains 3 ……
For the above dataset the calculated result will be given in dataset bellow. Resulted variable spoted in green color.
Thanks in advance.
The more succinct code is :
data have;
input t1-t4 low;
cards;
0 0.51 1 2 1
0 0.53 1 2.02 0.53
;
run;
data want;
set have;
array x{*} t:;
first=whichn(low,of x{*});
run;
This is simple enough, I was trying to find a simpler solution using scan but didn't get there. I'm hoping someone comes up with something more eligant in case this is a situation with a large dataset.
data have;
infile cards dsd dlm= ' ';
input c1 c2 t1 t2 t3 t4 lower;
cards;
.7514 10.215 0 .51 1 2 1
1.08674 9.123 0 .53 1 2.02 .53
1.2134 10.324 0 .5 1 2 2
.23421 12.1234 0 .5 1.03 2 1.03
;
run;
data want;
set have;
if lower = t1 then first = 1;
else if lower = t2 then first = 2;
else if lower = t3 then first = 3;
else if lower = t4 then first = 4;
run;
data have;
input t1-t4 low;
cards;
0 0.51 1 2 1
0 0.53 1 2.02 0.53
;
run;
data want;
set have;
array x{*} t1-t4;
do i=1 to dim(x);
if x{i}=low then do;first=i;leave;end;
end;
drop i;
run;
Xia,
I think I've seen this but do not remember the solution, if you do not know how many t values there are is there a way to say array x{*} t1-t 'nth; ?
Thanks,
The more succinct code is :
data have;
input t1-t4 low;
cards;
0 0.51 1 2 1
0 0.53 1 2.02 0.53
;
run;
data want;
set have;
array x{*} t:;
first=whichn(low,of x{*});
run;
Hi,
data want;
set have;
array t{4};
do i=1 to 4;
if t{i} <= lower and first=. then first=i;
end;
run;
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