How can I code proc sql not in select ... in sas merge? - sas

Question

Good Afternoon,
So, I started to learn code in SAS.
I want to have exact same result of
proc sql;
create table award_print_new as
select *
from awards_try2
where BOR_ITEM_TYPE NOT IN (SELECT BOR_ITEM_TYPE from FADSFUND);
run;
I though this is the exact code, but the results are different, so I have been wrong.
proc sort data=awards_try2;
by BOR_ITEM_TYPE;
run;
proc sort data=FADSFUND;
by BOR_ITEM_TYPE;
run;
data award_print;
set awards_try2 (in=a) FADSFUND (in=b);
by BOR_ITEM_TYPE;
if a and not b;
run;
Like below there are 9525 observations instead of 681. how can I get 681 in SAS code?
1665 data award_print;
1666 set awards_try2 (in=a) FADSFUND (in=b);
1667 by BOR_ITEM_TYPE;
1668 if a and not b;
1669 run;
NOTE: There were 9525 observations read from the data set WORK.AWARDS_TRY2.
NOTE: There were 1226 observations read from the data set WORK.FADSFUND.
NOTE: The data set WORK.AWARD_PRINT has 9525 observations and 22 variables.
NOTE: DATA statement used (Total process time):
real time 0.01 seconds
cpu time 0.01 seconds
1670
1671 proc sql;
1672 create table award_print_new as
1673 select *
1674 from awards_try2
1675 where BOR_ITEM_TYPE NOT IN (SELECT BOR_ITEM_TYPE from FADSFUND);
NOTE: Table WORK.AWARD_PRINT_NEW created, with 681 rows and 15 columns.
1676 run;

Answer 1

You didn't merge the datasets. Because you used a SET statement you merely interleaved them. You can tell from the notes:
NOTE: There were 9525 observations read from the data set WORK.AWARDS_TRY2.
NOTE: There were 1226 observations read from the data set WORK.FADSFUND.
NOTE: The data set WORK.AWARD_PRINT has 9525 observations and 22 variables.
You are just writing out the observations read from the first dataset.
data award_print;
merge awards_try2 (in=a) FADSFUND (in=b);
by BOR_ITEM_TYPE;
if a and not b;
run;
Note you don't need the first part of the IF condition. If you are merging only two datasets any observation that is not in B must have come from A.
Other source of differences could when B has more copies of a particular value of the BY variable(s) than A. Also if B has other non BY variables their values could overwrite the values of the same variables read from A. In this case however no data will be overwritten since you are not writing the observations where B contributed data. However if B has variables that are not in A then they will be added to the output dataset, but with missing values on all observations.