The number of obs in datasets do not add up after MERGE

[somebody]

This might be a simple task but I could not figure it out. I am trying to merge 3 datasets together. if they dont merge by the variables, then split out to a nomatch dataset. Theoretically, the matches and no-matches number should add up to the number of obs in the datasets. Let's look at my code for better understanding 

data MATCH
	nomatch_A
	nomatch_B
	nomatch_C;

	merge data_A(in=A) 
		data_B(in=B)
		data_C(in=C);
	by var1 var2 var3 var4;
	if A and B and C then output MATCH;
	else if A and (not B or not C) then output nomatch_A;
	else if B and (not C or not A)  then output nomatch_B;
	else if C and (not A or not B) then output nomatch_C;
run;

so if A has 1,000 obs and there are 800 obs in the dataset MATCH, then nomatch_A should have 200 obs. However, I got a number less than 200. Why? where did the other go?

[gamotte]

Hello,

I think there is a problem with the logic behind datasets nomatch_B and nomatch_C in that they are only populated

if the preceeding conditions are not matched.

For example, if an observation is in A and B but not in C, then it will not appear in nomatch_B.

Try (no data => not tested) :

if A and B and C then output MATCH;
else do;
    if A then output nomatch_A;
    if B then output nomatch_B;
    if C then output nomatch_C;
end;

[Astounding]

I would recommend adjusting how you think about this problem.  Let's look at an approach first, and discuss a couple of the pitfalls later.  Consider:

data want;
	merge data_A(in=A) 
		data_B(in=B)
		data_C(in=C);
	by var1 var2 var3 var4;
	match_pattern = '---';
	if A then substr(match_pattern, 1, 1) = 'A';
	if B then substr(match_pattern, 2, 1) = 'B';
	if C then substr(match_pattern, 3, 1) = 'C';
run;

proc freq data=want;
   tables match_pattern;
run;

This creates a single variable that shows, for each observation, where it came from.  That should be sufficient for you to determine what should happen next.

Some of the issues may involve your expectations of what the number of observations should be.  To explore that, consider a simplified case where DATA_A and DATA_B both contain 10 observations, and we're merging by just one variable named ID.

data test1;
   merge data_A data_B;
   by ID;
run;

The number of observations generated could be anywhere from 10 to 20, depending on the number of matches and mismatches.  If you allow that either data set might contain duplicate values for the same ID, the complexity increases:

data test2;
   merge data_A (in=A) data_B;
   by ID;
   if A;
run;

Now the number of observations generated must be at least 10, but could be as many as 19.  The 19 would occur of DATA_B contains just a single ID, repeated 10 times, that matches one of the IDs in DATA_A.

If both DATA_A and DATA_B could contain duplicate observations for the same ID, the problem gets so complex that SAS issues a warning in the log ... something along the lines of "more than one data set contains multiple observations for ID".  So be sure to check the log for that type of message.

At any rate, the first program should at least give you a chance to figure this one out.

[mkeintz]

BTW, this is one of the nice features of the SUBSTR function - it can be put on the left of the equals sign in an assignment statement.  Just the ticket here.

[somebody]

thank you for the detailed response. I have removed all duplicates so there would not be any obs with more than 1 match.

Your first set of code helps to see where the obs come from, but I don't think it show the ones that are missing (because the numbers don't add up)

[Astounding]

I can't see the numbers that you are looking at. If you would like feedback ...

Post the number of observations you start with from each source (after you had removed duplicates), and the PROC FREQ results.

[ballardw]

I would suggest creating non-temporary variables with the values of the IN variables and you can examine explicitly which combinations of values are appearing "incorrectly" in which set and adjust logic accordingly.

[somebody]

can you please explain a bit more on the "non-temporary variables"? what should I create?

[smantha]

Refer to this regionlook at the image in your case if A and B and C is A intersection B Intersection C. A and (not B or Not C) will be the pink region and the other two cases are represented by Blue non overlapping parts of B and Green non overlapping parts of C. That means you will be missing intersection parts of A,B; A,C and BC respectively

[smantha]

The answer lies in how you set up your clauses

if A and (not B or not C)  translates to if A =1 and (B=0 or C=0)

      possible outcomes

                          A=1 B=1 C=1

                         A=1 B=0 C=1

                         A=1 B=1 C=0

                         A=1 B=0 C=0

if your objective is to truly capture the fourth case then the condition becomes A=1 and (B=0 and C=0) and the counts will add up

The second and third conditions are captured by 

if B and (Not A or Not C), if C and (Not A or Not B) so the datasets are not mutually exclusive and hence the row count does not add up.

[Tom]

The are 8 possible ways to populate 3 binary variables.  One combination is impossible (all false).  Your program is only writing out 4.   So either you skipped some of them or included them into multiple datasets.

Here is how to get all 7 combinations.

data 
  MATCH
  A_ONLY
  B_ONLY
  C_ONLY
  AB_ONLY
  AC_ONLY
  BC_ONLY
;
  merge data_A(in=A)  data_B(in=B) data_C(in=C);
  by var1 var2 var3 var4;
  if A and B and C then output MATCH;
  else if A and not (B or C) then output A_ONLY;
  else if B and not (C or A)  then output B_ONLY;
  else if C and not (A or B) then output C_ONLY;
  else if A and B then output AB_ONLY;
  else if A and C then output AC_ONLY;
  else if B and C then output BC_ONLY;
  else put 'This combination is impossible. ' (_n_ a b c) (=);
run;

Checking your logic it seems you do account for all 7 combinations. 

data test;
 do a=1,0; do b=1,0; do c=1,0;
  length m1 m2 $20;
  if A and B and C then m1='MATCH';
  else if A and (not B or not C) then m1='nomatch_A';
  else if B and (not C or not A)  then m1='nomatch_B';
  else if C and (not A or not B) then m1='nomatch_C';
  else m1='none';
  if a and b and c then m2='MATCH';
  else if a and b and not c then m2='AB_ONLY';
  else if a and not b and c then m2='AC_ONLY';
  else if not a and b and c then m2='BC_ONLY';
  else if a then m2='A_ONLY';
  else if b then m2='B_ONLY';
  else if c then m2='C_ONLY';
  else m2='none';
  output;

 end; end; end;
run;

Not clear if this is exactly what you wanted.

Obs    a    b    c    m1           m2

 1     1    1    1    MATCH        MATCH
 2     1    1    0    nomatch_A    AB_ONLY
 3     1    0    1    nomatch_A    AC_ONLY
 4     1    0    0    nomatch_A    A_ONLY
 5     0    1    1    nomatch_B    BC_ONLY
 6     0    1    0    nomatch_B    B_ONLY
 7     0    0    1    nomatch_C    C_ONLY
 8     0    0    0    none         none