A solution amongst others.   condition 1 -> And change each time i run the code. condition 2 -> Any numbers, but without repeating them.   data t_a(keep=zRandom);    array xRand(6);    x = 0;    do until (x=6);       y = ceil(60*ranuni(-3));      *** covers condition 1 negative argument ***;       if not(y in xRand) then do; *** covers condition 2 ***;          x+1;          xRand(x)=y;       end;    end;    do  i = 1 to 6;      zRandom = xRand(i);      output;    end; run; ... View more

for the results you want , you have a type in   8310 no 30   It should be   8210 no 30.   A solution amongst others (I assume the input dataset is sorted w/r/t the variable id):   data have; input Id dx $ test; datalines; 8210 no 1 8210 yes 20 8210 no 30 8321 yes 11 8321 no 2 ; run; 　 data want; do until (last.id); set have; by id; output; end; do until (last.id); set have; by id; output; end; do until (last.id); set have; by id; output; end; run;   proc print data=want; run; ... View more

You can use %let value = %nrbquote(ABC","123","TEST","HELLO","WHAT); and then the data step works: data test; value = "&value"; run; ... View more

A pedestrian view of the question. The function (F) in question has 26 parameters and thus a clear view is complicated. The function (F) has poles at x=0 and x=-1 so the roots of F will not include 0 and -1. First, assign some values to the said 26 parameters. Second, find at least 1 root of F within either in the (0 to 1) or (0 to -1) range. Assumption about the parameters:    1- p13 and p26 are greater than 1, with p26> p13.    2- besides p13 and p26, the other parameters are positive and less than 1. I am using a crude and pedestrian method to find the roots. /************************************************/ /**** sample random sets of parameters p(26) ****/ /************************************************/ data t_a(keep=i p:);    array p(26);    do i = 1 to 10;    do j = 1 to 26;            if (j=13) then p(j)= ceil(30*ranuni(5));       else if (j=26) then p(j)= p(13)+ ceil(20*ranuni(7));       else p(j) =  ceil(10000*ranuni(11))/10000;    end;    output;    end; run; /************************************************/ /**** for each set of parameters find a root ****/ /************************************************/ data t_b(keep=i p: aZero aChng aRoot bSign);    array p(26);    set t_a;    aSign=0;    aZero=0;    aFunc=0;    aChng=0;    aSteps=1000;    bSign=1;    do j=0 to aSteps;       aZero=bSign*j*(1/aSteps);       aFlag=aFunc;       do k=1 to 14;               if (k=1)  then aFunc+(p(13)-p(26))*aZero*((1+aZero)**12);          else if (k=14) then aFunc+(p(25)*(p(12)-aZero));                         else aFunc+(p(11+k)*(p(k-1)-aZero)*aZero*((1+aZero)**(13-k)));       end;       aFunc= round(aFunc,0.00001);            if (j=0) then aSign= sign(aFunc);       else if not(sign(aFunc)=aSign) then do;               aChng=1;               aRoot= aZero-bSign*(1/aSteps)*(abs(aFunc)/(abs(aFunc)+abs(aFlag)));               aRoot=round(aRoot,0.000001);               leave;            end;    end;    output; run; proc print data=t_b;   var i bSign aZero aRoot aChng;   format aRoot 9.6; run; /********************/ /*** some results ***/ /********************/ When looking at the (0 to 1) range (bSign=1) one finds at least 1 root (aRoot). aChng=1 -> the function (aFunc) changed signs. i   bSign   aZero    aRoot   aChng =================================== 1   1       0.059   0.058985   1 2   1       0.431   0.430937   1 3   1       0.024   0.023931   1 4   1       0.007   0.006165   1 5   1       0.024   0.023457   1 6   1       0.012   0.011116   1 7   1       0.014   0.01384    1 8   1       0.127   0.126725   1 9   1       0.108   0.107417   1 10  1       0.025   0.024228   1 When looking at the (0 to -1) range (bSign=-1), not all sets of p(26) have a root in the (0 to -1) range. No root (aRoot=.) when there is no change in sign (aChng=0). i   bSign   aZero    aRoot      aChng ====================================== 1   -1      -1         .         0 2   -1      -0.057   -0.056783   1 3   -1      -0.144   -0.143403   1 4   -1      -1         .         0 5   -1      -0.256   -0.255269   1 6   -1      -0.056   -0.055367   1 7   -1      -1         .         0 8   -1      -0.416   -0.415036   1 9   -1      -0.042   -0.041868   1 10  -1      -0.279   -0.278255   1 ... View more

The right-hand-side of the 2 examples is the ratio of 1 polynomial over another polynomial. You write: "the actual polynomial have 12 terms". Do both polynomials (numerator and denominator) have 12 terms? Or does the equation have a grand total of 12 parameters? ... View more

A proposed solution. Some of the hints are already given in this forum. I am looking to match pairs with the smallest score differences. One can at some later point remove the pairs with a score difference greater than the minimum wanted. Same sample datasets. /*********************/ /***** DataSet 1 *****/ /*********************/ data t_a(keep=id1 zScore);   do id = 1 to 10000;      id1 = compress('A'||put(id, z5.));      zScore=  ceil(100000*ranuni(3));      output;   end; run; /*********************/ /***** DataSet 2 *****/ /*********************/ data t_b(keep=id2 zScore);   do id = 1 to 50000;      id2 = compress('B'||put(id, z5.));      zScore=  ceil(200000*ranuni(3));      output;   end; run; /**************************************/ /** One can do a full Cartesian Join **/ /**    or one can adjust the range   **/ /**************************************/ PROC SQL;    create table t_c as    select a.id1, b.id2, a.zScore as zS1, b.zScore as zS2,           (a.zScore-b.zScore) as difx,            abs(a.zScore-b.zScore) as difa    from   t_a a, t_b b    where (-100 <= (a.zScore-b.zScore) <=100);    /*********************************/    /** check if all id1 are in t_c **/    /*********************************/    select count(distinct id1) as c_id1 from t_c; QUIT; proc sort data=t_c; by difa difx; run; /*****************************/ /**** the 1-to-1 matching ****/ /*****************************/ data t_want(keep=id1 id2 difa difx zS1 zS2);    if _N_=1 then do;       declare hash ha(multidata:'N');          ha.definekey('mbr1');          ha.definedata('mbr1');          ha.definedone();       declare hash hb(multidata:'N');          hb.definekey('mbr2');          hb.definedata('mbr2');          hb.definedone();    end;    do until (aDone);       set t_c;       mbr1=id1;       mbr2=id2;       k1=ha.find();       k2=hb.find();       if (k1 and k2) then do;          ha.add();          hb.add();          output;       end;    end; run; /***************************/ /**** some idiot checks ****/ /***************************/ proc sql;    select count(distinct id1) as c_id1, count(distinct id2) as c_id2    from t_want;    select difa,difx, sum(1) as c_pairs    from t_want    group by difa, difx; quit; ... View more