Not a clear question about the drop option and drop statement ?

[ResoluteCarbon]

Hi SAS Instructor,

Today I went through a question 1.07 in Programming 2. The question and answer are as below

First, to me, it is not the easiest way, and there is no way for us to judge if it is the easiest one, for me, I can do by setting the option drop right at the set statement ( it is easier because I do not need to create a new data statement for drop.

Secondly, if I were the instructors, I would changed the question to :how to drop the EndDate in three implicitly output datasets but this available is still calculable (that is when we can use the data statement drop EndDate; ) or how to drop......and this variable is non-calculable during the datatstep (at that time we set an option drop right at the set statement).

Please let me know if I fall into any fallacy.

Thank you.

[japelin · Accepted solution]

Since we are being asked how to most easiest way to drop from three tables in a data step, we will consider the following three methods.
1st. Use the drop= dataset option in the data statement.
2nd. Use the drop= dataset option in a set statement.
3rd. Use the drop statement.

Of these, 2nd is not suitable because it will not be able to calculate the EndDate - StartDate.
1st. and 3rd. are almost the same, but with drop= dataset option, you need to write three, whereas with the drop statement, you only need to write one, so the answer is the drop statement.

I think it is clear question.

[Shmuel]

In this case you can't drop ENDDATE from input because you need it to calculate StormLength.

If you use DROP statement it relates to all output data sets, and you need not enter the DROP option for each output data set.

[ballardw]

If your source data is not clean you may have to use a dataset option to drop one or more variables because they are of a different type in some data sets.

data junk1;
   x='1234';
   y=42;
run;
data junk2;
  x=123;
  y= 56;
run;

/* this fails because X is both numeric and character*/
data combined;
   set junk1
       junk2
   ;
   drop x;
run;
/* this works because only one type of x makes it into
   the data vector
*/
data combined2;
   set junk1 (drop=x)
       junk2
   ;
   drop x;
run;