do - end structure and if structure interaction

[emaguin]

Again, i want to emphasize i come from spss and i'm very new to sas. I'm looking for sas equivalents.

Here's some spss code.

vector xvec=x1 to x10. * same as array xvec{10} x1-x10;

loop i=1 to 10; * same as do i=1 to 10;

do if (z eq 12).

compute x(i)=x(i)+2.

else if (z eq 15).

compute x(i)=x(i)+5.

else.

compute x(i)=x(i)+15.

end if.

end loop.

An alternative (with a variation) would-could be

vector xvec=x1 to x10. 

do if (z eq 12).

loop i=1 to 10; 

compute x(i)=x(i)+2.

end loop.

else if (z eq 15).

loop i=1 to 10; 

compute x(i)=x(i)+15.

end loop.

else.

compute w=x1*x3.

end if.

end loop.

I know (courtesy of Ron Cody's book) that sas has a if-else if structure but  my (very limited) understanding is that if-else if can't interact with a do-end structure because the if-else if must have a 'then' result. 

I assume sas can do this; but how?

Thanks, Gene Maguin

PS. I'm interested in suggestions of sas books that build on Ron's book.

[PGStats]

I don't know SPSS syntax but your code looks like

array x{10};
do i = 1 to dim(x);
    if z = 12 then x{i} = x{i} + 2;
    else if z = 15 then x{i} = x{i} + 5;
    else x{i} = x{i} + 15;
    end;

[ballardw]

If I remember the SPSS correctly then

vector xvec=x1 to x10. 
do if (z eq 12).
loop i=1 to 10; 
compute x(i)=x(i)+2.
end loop.
else if (z eq 15).
loop i=1 to 10; 
compute x(i)=x(i)+15.
end loop.
else.
compute w=x1*x3.
end if.
end loop.
 
would be:

array x{10};
if z = 12 then do i = 1 to dim(x);
   x{i} = x{i} + 2;
end;
else if z = 15 then do i = 1 to dim(x);
   x{i} = x{i} + 5;
end;
else  w=x1*x3;

You could use the same approach that @PGStats with  the W=x1*x3 calculation instead of the x{I}=x{I}+15 , it would just be done 10 times with identical results.

Or

select (z);
   when (12) do i = 1 to dim(x);
               x{i} = x{i} + 2;
            end;
   when (15) do i = 1 to dim(x);
               x{i} = x{i} + 5;
            end;
   otherwise w=x1*x3;
end;

The select statement/when/block becomes more attractive when there are more than 2 or 3 "elses" involved.

And sometimes you get semi-slick values and

if z in (12 15) then do i= 1 to dim(x);
   x[i] = x[i] + (z-10);
end;
else  w=x1*x3;

[ChrisNZ]

loop i=1 to 10; * same as do i=1 to 10;

do if (z eq 12).

compute x(i)=x(i)+2.

else if (z eq 15).

compute x(i)=x(i)+5.

else.

compute x(i)=x(i)+15.

end if.

end loop.

Another way::

do I= 1 to dim(X);
   X[I] = X[I] + 2*(Z=12) + 5*(Z=15) + 15*(Z ^in (12 15));
end;