Hi All,
Following is the data set:
id marks1 marks2
1 10 20
1 30 30
1 20 40
1 20 40
2 20 10
2 100 90
2 30 20
2 20 40
Expected result : I would like to know how we could calculate the final variable using only LAG function?
id | marks1 | marks2 | final | |
1 | 10 | 20 | 10 | if first.id then MARKS1 |
1 | 30 | 30 | 40 | if not first.id then lag(final)+marks2 |
1 | 20 | 40 | 80 | |
1 | 20 | 40 | 120 | |
2 | 20 | 10 | 20 | |
2 | 100 | 90 | 110 | |
2 | 30 | 20 | 130 | |
2 | 20 | 40 | 170 |
Thanks and Regards,
Mushy
Hi @Mushy,
@Mushy wrote:
I would like to know how we could calculate the final variable using only LAG function?
Here's how:
data want(drop=lf);
set have;
by id;
if first.id then final=marks1;
else do;
link op_q; /* retrieve previous FINAL from the queue */
final=lf+marks2;
end;
op_q: lf=lag(final); /* operate the queue */
run;
As you can see, a computed value can be saved in a LAG queue for the next iteration of the DATA step, but it's easier to use a retained variable for this purpose (as Kurt_Bremser has suggested).
No LAG needed, just a retained variable:
data want;
set have;
by id;
if first.id
then final = marks1;
else final + marks2;
run;
Hello Kurt,
Thanks for the response. Indeed I know this solution. But I want to know how a LAG could be applied on the calculated value.
Thanks in advance ,
Mushy
Hi @Mushy,
@Mushy wrote:
I would like to know how we could calculate the final variable using only LAG function?
Here's how:
data want(drop=lf);
set have;
by id;
if first.id then final=marks1;
else do;
link op_q; /* retrieve previous FINAL from the queue */
final=lf+marks2;
end;
op_q: lf=lag(final); /* operate the queue */
run;
As you can see, a computed value can be saved in a LAG queue for the next iteration of the DATA step, but it's easier to use a retained variable for this purpose (as Kurt_Bremser has suggested).
@FreelanceReinh Thanks for the solution 🙂
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