Pfthb

I am a bit puzzled why the output is a certain way on the following program.

 data a;
 input name$ lv @@;
 cards;
 Frank 1 Joan 2 Sui 3 Burt 4 Kelly . Juan 1
 ;
 data b;
 set a;
 if lv=.
 then expertise='?';
 else if lv=1 
 then expertise='L';
 else if lv=2 or 3 
 then expertise ='M';
 else expertise ='H';
 run;
 proc print data=b;
 run;

In the program above, I am expecting that the output for the observation containing Burt has expertise of value H, but for some reason it is M.

I was thinking that the if statement should be lv=(2 or 3) but when I do that now those that I am thinking that the lv= 2 and lv=3 whose expertise should equal M now becomes H as well.

*This is probably explained by the fact that the syntax is inappropriate, and lv is never (2 or 3) /I am not sure why this does not cause an error/ thus the ELSE statement is executed.

I have a feeling that I am not understanding how the ELSE is really working.

However, according to that logic then if the lv=2 part is recognized but the others are not, I am not sure why the lv= 3 and lv=4 does not produce an H.

My goal is to understand why the program runs as if there is no syntax error and why the output is what it is.

Your mistake is here

If lv = 4 or 0 

You think this means

If lv = 4 or lv = 0

But it doesn’t. What actually happens is

If (lv = 4) or 0

In SAS 0 is false and any number greater than 1 is true. Not sure about decimals. Anyways. The or condition makes this one false. But in your first example, it makes it always true.

Your syntax is incorrect, but not illegal, so SAS did what you asked for instead of what you wanted. The condition lv=2 or 3 is evaluated from left to right. So lv=2 will result in either 1 or 0 depending whether the value of lv is 2 or not. But both 1 or 3 and 0 or 3 will be true since 3 is always considered true.

What you really want is the in operator.

if missing(lv) then expertise='?';
else if lv=1 then expertise='L';
else if lv in (2 3) then expertise ='M';
else expertise ='H';