topic Re: conditional combinations in SAS Programming

conditional combinations

cminard — Mon, 19 Apr 2021 15:22:29 GMT

Suppose that you have the following data set:

data temp;
	do set = 1 to 4;
		do count = 0 to 5;
			output;
		end;
	end;
run;

Now, you want to get all possible combinations of sets from this data set such that the sum of the "count" is 5. You have to include each "set" one time.

For example, I want:

combination	set	count
1	1	5
1	2	0
1	3	0
1	4	0
2	1	4
2	2	1
2	3	0
2	4	0
3	1	4
3	2	0
3	3	1
3	4	0
etc	etc	etc

Any suggestions on how to accomplish this would be appreciated. TIA!

Re: conditional combinations

ballardw — Mon, 19 Apr 2021 15:40:59 GMT

I am not seeing how your "want" complies with " sum of the "count" is 5." . You likely have to go into a bit more detail of exactly what "You have to include each "set" one time." means as well.

If you mean that you have other data than your example data set it is quite possible the "sum of count=5" may not occur for "set". So what would be the rule for selecting the one that does not match.

Re: conditional combinations

cminard — Mon, 19 Apr 2021 15:56:36 GMT

In the example data set I have:

Set 1: 5+0+0+0=5

Set 2: 4+1+0+0=5

Set 3: 4+0+1+0=5

No, there is no other data.

Re: conditional combinations

FreelanceReinh — Mon, 19 Apr 2021 17:32:23 GMT

Hello @cminard,

You can create a Cartesian product (view) with PROC SQL and then transpose the result with a DATA step.

proc sql;
create view _tmp as
select t1.count as a1,
       t2.count as a2,
       t3.count as a3,
       t4.count as a4
from temp(where=(set=1)) t1,
     temp(where=(set=2)) t2,
     temp(where=(set=3)) t3,
     temp(where=(set=4)) t4
having a1+a2+a3+a4=5
order by 1 desc, 2 desc, 3 desc, 4 desc;
quit;

data want(drop=a:);
set _tmp;
array a[*] a:;
combination=_n_;
do set=1 to dim(a);
  count=a[set];
  output;
end;
run;

Re: conditional combinations

Ksharp — Tue, 20 Apr 2021 14:04:46 GMT

It is OR things. Try post it at OR forum and calling @RobPratt

proc optmodel;
num numLists=4;
set LIST{1..numLists};
LIST[1]=0..5;
LIST[2]=0..5;
LIST[3]=0..5;
LIST[4]=0..5;

num target=5;
var x{i in 1..numLIsts,j in LIST[i]} binary;
con oneitemperlist{i in 1..numLists} : sum{j in LIST[i]} x[i,j]=1;
con sumtotarget:sum{i in 1..numLists,j in LIST[i]} j*x[i,j]=target;
solve with clp/findallsolns;
set SUPPORT{s in 1.._NSOL_}={i in 1..numLists,j in LIST[i]:x[i,j].sol[s]>0.5};
put SUPPORT[*];
quit;

{<1,5>,<2,0>,<3,0>,<4,0>} {<1,4>,<2,0>,<3,0>,<4,1>} {<1,4>,<2,1>,<3,0>,<4,0>} {<1,4>,<2,0>,<3,1>,<4,0>} {<1,3>,<2,1>,<3,0>,<4,1>} {<
 1,3>,<2,0>,<3,1>,<4,1>} {<1,3>,<2,1>,<3,1>,<4,0>} {<1,3>,<2,0>,<3,0>,<4,2>} {<1,3>,<2,2>,<3,0>,<4,0>} {<1,3>,<2,0>,<3,2>,<4,0>} {<1,
 2>,<2,1>,<3,1>,<4,1>} {<1,2>,<2,1>,<3,0>,<4,2>} {<1,2>,<2,0>,<3,1>,<4,2>} {<1,2>,<2,2>,<3,0>,<4,1>} {<1,2>,<2,2>,<3,1>,<4,0>} {<1,2>
 ,<2,0>,<3,2>,<4,1>} {<1,2>,<2,1>,<3,2>,<4,0>} {<1,2>,<2,0>,<3,3>,<4,0>} {<1,2>,<2,3>,<3,0>,<4,0>} {<1,2>,<2,0>,<3,0>,<4,3>} {<1,1>,<
 2,1>,<3,1>,<4,2>} {<1,1>,<2,2>,<3,0>,<4,2>} {<1,1>,<2,2>,<3,1>,<4,1>} {<1,1>,<2,0>,<3,2>,<4,2>} {<1,1>,<2,2>,<3,2>,<4,0>} {<1,1>,<2,
 1>,<3,2>,<4,1>} {<1,1>,<2,0>,<3,3>,<4,1>} {<1,1>,<2,1>,<3,3>,<4,0>} {<1,1>,<2,3>,<3,0>,<4,1>} {<1,1>,<2,3>,<3,1>,<4,0>} {<1,1>,<2,0>
 ,<3,1>,<4,3>} {<1,1>,<2,1>,<3,0>,<4,3>} {<1,1>,<2,0>,<3,4>,<4,0>} {<1,1>,<2,0>,<3,0>,<4,4>} {<1,1>,<2,4>,<3,0>,<4,0>} {<1,0>,<2,2>,<
 3,1>,<4,2>} {<1,0>,<2,1>,<3,2>,<4,2>} {<1,0>,<2,2>,<3,2>,<4,1>} {<1,0>,<2,0>,<3,3>,<4,2>} {<1,0>,<2,1>,<3,3>,<4,1>} {<1,0>,<2,2>,<3,
 3>,<4,0>} {<1,0>,<2,3>,<3,0>,<4,2>} {<1,0>,<2,3>,<3,1>,<4,1>} {<1,0>,<2,3>,<3,2>,<4,0>} {<1,0>,<2,2>,<3,0>,<4,3>} {<1,0>,<2,0>,<3,2>
 ,<4,3>} {<1,0>,<2,1>,<3,1>,<4,3>} {<1,0>,<2,1>,<3,4>,<4,0>} {<1,0>,<2,0>,<3,4>,<4,1>} {<1,0>,<2,0>,<3,1>,<4,4>} {<1,0>,<2,1>,<3,0>,<
 4,4>} {<1,0>,<2,4>,<3,1>,<4,0>} {<1,0>,<2,4>,<3,0>,<4,1>} {<1,0>,<2,0>,<3,0>,<4,5>} {<1,0>,<2,0>,<3,5>,<4,0>} {<1,0>,<2,5>,<3,0>,<4,
 0>}

解汇总
Solver	CLP
变量选择	MINR
Objective Function	(0)
Solution Status	All Solutions Found
Objective Value	0
 	 
Solutions Found	56
Presolve Time	0.00
Solution Time	0.01

Re: conditional combinations

RobPratt — Tue, 20 Apr 2021 20:47:00 GMT

That works, or you can use 4 integer variables and 1 linear constraint:

proc optmodel;
   num numLists = 4;
   num target = 5;
   var X {1..numLists} >= 0 <= target integer;
   con SumToTarget: sum {i in 1..numLists} X[i] = target;
   solve with clp / findallsolns;
   create data want from [combination set]={s in 1.._NSOL_, i in 1..numLists} count=X[i].sol[s];
quit;
proc sort data=want;
   by combination set;
run;

Re: conditional combinations

mkeintz — Wed, 21 Apr 2021 06:51:10 GMT

Just to add a data step solution.

First, if you really have count=0 through 5 for each and every set, then you don't need data set TEMP (as in the proc optmodel solutions):

data want (keep=set1-set4);
  sum0=0;
  do set1=0 to 5;
    sum1=set1+sum0; 
    if sum1>5 then continue;
    do set2=0 to 5;
      sum2=sum1+set2;
      if sum2>5 then continue;
      do set3=0 to 5;
        sum3=sum2+set3; 
        if sum3>5 then continue;
        do set4=0 to 5;
          sum4=sum3+set4;  
          if sum4>5 then continue;
          if sum4=5 then output;
        end;
      end;
    end;
  end;
run;

As I said, this solution doesn't require data set TEMP. But let's say data set TEMP has some "holes", i.e. not every SET has every COUNT from 0 through 5: Then you need to read TEMP to determine which combinations to skip (the if x{n,setn}=. then continue; statements below):

data want (keep=set1-set4);
  array x {4,0:5} _temporary_;
  set temp end=end_of_temp;
  x{set,count}=1;
  if end_of_temp;

  sum0=0;
  do set1=0 to 5;
    if x{1,set1}=. then continue;
    sum1=set1+sum0; 
    if sum1>5 then continue;
    do set2=0 to 5;
      if x{2,set2}=. then continue;
      sum2=sum1+set2;
      if sum2>5 then continue;
      do set3=0 to 5;
        if x{3,set3}=. then continue;
        sum3=sum2+set3; 
        if sum3>5 then continue;
        do set4=0 to 5;
          if x{4,set4}=. then continue;
          sum4=sum3+set4;  
          if sum4>5 then continue;
          if sum4=5 then output;
        end;
      end;
    end;
  end;
run;

Conversion to macro code can shorten the above. And it would be more amenable to a larger number of sets (just change the %let nsets=4; statement below), or a change in the count range, or target value:

%macro test;
  %let nsets=4;
  %let countrange=0:5;
  %let target=5;

  data want (keep=set1-set&nsets);
    array x {&nsets,&countrange} _temporary_;
    set temp end=end_of_temp;
    x{set,count}=1;
    if end_of_temp;

    sum0=0;
    %do s=1 %to &nsets;
      do set&s = lbound(x,2) to hbound(x,2);
        if x{&s,set&s}=. then continue;
        sum&s=set&s + sum%eval(&s-1);
        if sum&s>&target then continue;
    %end;

      if sum&nsets=&target then output;

    %do s=1 %to &nsets;
      end;
    %end;
  run;
%mend test;
%test;

Note

the statement

  sum&s=set&s + sum%eval(&s-1);

is converted to

   sum1=set1 + sum0;
   sum2=set2 + sum1;

etc.