Shape Function in IML

[Sanscrit_2prov]

Can anyone tell me what the following line of code does:

X1=X;

do i=1 to 24;

 do k=1 to 4;

X[,indices[i,k]+1]=shape(0,nrow(X1),1);

end;

end;

X is a 200 row by 5 column matrix of 1's and 0's.

indices is a 24 row by 4 column matrix with the numbers 1 to 4 in different order for each row.

[PeterClemmensen]

Are you able to post your entire IML code?

Otherwise, see if the SHAPE Function Doc helps you.

[Sanscrit_2prov]

This is the entire code that someone wrote. It's just the shape function that doesn't make sense to me.

%macro averageAF (out, riskfactors, data);

proc logistic data=&data descending outdesign=_design_ outest=_model_ noprint;
model &out = &riskfactors;
run;

%let vars=1;
%let names=;
%do %while (%Qscan(&riskfactors,&vars) ne );
%let names=&&names %Qscan(&riskfactors,&vars);
%let vars=%eval(&vars+1);
%end;
%let vars=%eval(&vars-1);

%put &names;

data _design_;
set _design_;
if nmiss(of &riskfactors)=0 and nmiss(of &out)=0;
run;

data _indices_;
drop i perms;
array x (&vars) (1:&vars);

perms=fact(&vars);
/*vars=4 and perms=24*/
do i=1 to perms;
call allperm(i, of x(*));
output;
end;
run;

proc iml;
use _design_;
read all var {intercept &riskfactors} into X;
close _design_;
use _model_;
read var {intercept &riskfactors} into beta;
close _model_;
use _indices_;
read all var _num_ into indices;
close _indices_;

start pp(pcases,X,beta);
pcases=sum(1/(1+exp(-X*beta`)));
finish;

run pp(pcases,X,beta);

print (pcases);

pred_cases_m=shape(.,nrow(indices),&vars); /*24 rows , 4 columns*/
prev_cases_m=shape(.,nrow(indices),&vars); /*24 rows , 4 columns*/

X1=X;
do i=1 to nrow(indices); /*24*/
do k=1 to &vars; /*4*/
print (indices[i,k]);
X[,indices[i,k]+1]=shape(0,nrow(X1),1);
pred_cases_m[i,k]=sum(1/(1+exp(-X*beta`)));
print (pred_cases_m[i,k]);
end;
X=X1;
end;

[IanWakeling]

The code does not make much sense to me.  The SHAPE function is being used to generate a vector of zeros and then that vector is being copied repeateadly into columns 2 to 5 of X.  I believe the one line

X [ , 2:5] = 0;

would have the same effect as the double loop.   But you need to test this out - try printing the matrix X before and after.

[IanWakeling]

I notice that the line "X = X1" is missing from the first code example that you posted. With this line the code makes more sense as X is reset after each iteration of the outermost loop.  The result is that the columns of X are set to zero in every possible order.

[Rick_SAS]

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