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Jenc
Calcite | Level 5

Hello,
I have two data sets (i.e., Temp1 and Temp2). Temp1 has one variable x with 20000 observations, and Temp2 contains four variables y1, y2, z1 and z2 with 60000 observations.
I used the following codes to obtain variable mu, but it took a lot of time to get the result with the large sample sizes. Is there a way that can have SAS run faster? Thanks!

 

proc sql noprint; select x into :x1 - :x20000 from temp1;quit;

data temp3;
set temp2;
%do i= 1 %to 20000;
T&i.= (constant('E')**(y1*(&&x&i.-z1)) - constant('E')**(y2*(&&x&i.-z2)))**2
%end;
mu = mean (of T1-T20000)); 
run;
1 ACCEPTED SOLUTION

Accepted Solutions
Astounding
PROC Star

Then I would get rid of the macro language and the abundant variables. Given that you have no missing values, start with:

data want;
array xval [20000] _temporary_;
if _n_=1 then do k=1 to 20000;
set temp1;
Xval[k]=x;
end;

That makes all 20,000 values of x available. Next, accumulate the numerator:

set temp2;

mu = 0;
do k=1 to 20000;
mu + same formula, using xval[k] not &&x&I;
end;
mu = mu/ 20000;

drop x k;

run;

View solution in original post

4 REPLIES 4
Astounding
PROC Star
When all is said and done, do you need T1-T20000, or do you just need mu?
Jenc
Calcite | Level 5

Thanks for your response, I just need mu. 

Astounding
PROC Star

Then I would get rid of the macro language and the abundant variables. Given that you have no missing values, start with:

data want;
array xval [20000] _temporary_;
if _n_=1 then do k=1 to 20000;
set temp1;
Xval[k]=x;
end;

That makes all 20,000 values of x available. Next, accumulate the numerator:

set temp2;

mu = 0;
do k=1 to 20000;
mu + same formula, using xval[k] not &&x&I;
end;
mu = mu/ 20000;

drop x k;

run;

Jenc
Calcite | Level 5

It did get much faster. Thanks so much for your help!!!

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