IF-THEN statement is SAS IML when there are zeros in matrix

[pari2189]

I'm trying to perform logical operation through if-then statement but since my matrix has some value sas zero it fails to perform the operation correctly

Start M_1(y);
p=13;
outarray_M=j(nrow(y),ncol(y),.);
outarray_01=j(nrow(y),ncol(y),.);
do i=14 to (nrow(y)-1);
idx= max(2,i-p+1):i;
unit= mad(y[idx,],"nmad");
outarray_M[i,]=unit;
if all(outarray_M[i,] ^= 0) then outarray_01[i+1,] = abs((y[i+1,]-median(y[idx,])))/mad(y[idx,],"nmad");
else outarray_01[i+1,]= choose(y[i+1,] =0, 0, 9999);
end;
return(outarray_01);
finish;

[PaigeMiller]

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@pari2189 wrote:

I'm trying to perform logical operation through if-then statement but since my matrix has some value sas zero it fails to perform the operation correctly

Can you show us the SASLOG that contains the error?

[pari2189]

There is no error but the results are incorrect. 

So what I want is if it is a non-zero value perform the calculation as per the formula and if it is zero go to the choose condition. But since  if-then statement returns false because some values in matrix are zero then it never performs the calculation and goes to choose condition directly.

[ballardw]

If

 mad(y[idx,],"nmad") returns 0

then you have a division by 0 which is double-plus-ungood.

[Rick_SAS]

I think ballardw mailed it: The MAD function will return 0 when more than half of the values are the same value.

In addition, I'd like to point out that the outarray_M matrix is not necessary in your code. Hopefully, that code is left over from a more complicated example (thanks for supplying a short module for us to view!). If not, you can use 

if unit ^= 0 then...