CODEWITHSUNDEEP
sas
user634455
user634455

Reputation: 115

SAS computation using double loops

I am trying to compute using two loops. But I am not very familiar with loop elements.

Here is my data:

data try;
 input rs t a b c;
 datalines;
 0    600    
 1    600       0.02514  667.53437  0.1638
 2    600       0.2766   724.60233  0.30162
 3    610       0.01592  792.34628  0.21354
 4    615.2869  0.03027  718.30377  0.22097
 5    636.0273  0.01967  705.45965  0.16847
 ;
run;

What I am trying to compute is that for each 'T' value, all elements of a, b, and c need to be used for the equation. Then I create varaibles v1-v6 to put results of the equation for each T1-T6. After that, I create CS to sum all the elements of v.

So my result dataset will look like this:

rs    T           a          b          c        v1       v2        v3      v4  v5      v6    CS 
0    600                                                                                     sum of v1
1    600       0.02514   667.53437    0.1638                                                 sum of v2
2    600       0.2766    724.60233    0.30162                                                sum of v3
3    610       0.01592   792.34628    0.21354                                                sum of v4
4    615.2869  0.03027   718.30377    0.22097                                                sum of v5
5    636.0273  0.01967   705.45965    0.16847                                                sum of v6

I wrote a code below to do this but got errors. Mainly I am not sure how to use i and j properly to link all elements of variables. Can someone point out what i did not think correct? I am aware that myabe I should not use sum function to cum up elements of a variable but not sure which function to use. 

data try3;
 set try;
 retain v1-v6;
 retain t a b c;
 array v(*) v1-v6;
 array var(*) t a b c;
 cs=0;
 do i=1 to 6;
  do j=1 to 6;
   v[i,j]=(2.89*(a[j]**2*(1-c[j]))/
    ((c[j]+exp(1.7*a[j]*(t[i]-b[j])))*
    ((1+exp(-1.7*a[j]*(t[i]-b[j])))**2));
   cs[i]=sum(of v[i,j]-v[i,j]);
  end;
 end;
run;

Forexample, v1 will be computed like v[1,1] =0 because there is no values for a b c.

For v[1,2]=(2.89*0.02514**2(1-0.1638))/((0.1638+exp(1.7*0.02514*600-667.53437)))*((1+exp(-1.7*0.02514*(600-667.5347)))**2)).

v[1,3]]=(2.89*0.2766**2(1-0.30162))/((0.30162+exp(1.7*0.2766*600-724.60233)))*((1+exp(-1.7*0.2766*(600-724.60233)))**2)).

v[1,4] will be using the next line values of a b c but the t will be same as the t[1]. and do this until the last row. And that will be v1. And then I need to sum all the elements of v1 like v1{1,1] +v1[1,2]+ v1{1,3] ....v1[1,6] to make cs[1,1].

Upvotes: 0

Views: 1705

Answers (1)

Tom Quarendon
Tom Quarendon

Reputation: 5708

The SAS language isn't that good at doing these kinds of things, which are essentially matrix calculations. The DATA step normally processes one observation at a time, though you can carry calculations over using the RETAIN statement. It is possible that you could get a cleaner result than this if you had access to PROC IML (which does matrix calculations natively), but assuming that you don't have access to IML, you need to do something like the following. I'm not 100% sure that it is what you need, but I think it is along the right lines:

    data try;
     infile cards missover;
     input rs t a b c;
     datalines;
     0    600    
     1    600       0.02514  667.53437  0.1638
     2    600       0.2766   724.60233  0.30162
     3    610       0.01592  792.34628  0.21354
     4    615.2869  0.03027  718.30377  0.22097
     5    636.0273  0.01967  705.45965  0.16847
     ;
    run;

    data try4(rename=(aa=a bb=b cc=c css=cs tt=t vv1=v1 vv2=v2 vv3=v3 vv4=v4 vv5=v5 vv6=v6));
    * Construct arrays into which we will read all of the records;
    array t(6);
    array a(6);
    array b(6);
    array c(6);
    array v(6,6);
    array cs(6);

    * Read all six records;
    do i=1 to 6;
      set try(rename=(t=tt a=aa b=bb c=cc));
      t[i] = tt;
      a[i] = aa;
      b[i] = bb;
      c[i] = cc;
    end;

    * Now do the calculation, which involves values from each
      row at each iteration;
    do i=1 to 6;
      cs[i]=0;
      do j=1 to 6;
        v[i,j]=(2.89*(a[j]**2*(1-c[j]))/
           ((c[j]+exp(1.7*a[j]*(t[i]-b[j])))*
           ((1+exp(-1.7*a[j]*(t[i]-b[j])))**2)));
        cs[i]+v[i,j];
      end;

      * Then output the values for this iteration;
      tt=t[i];
      aa=a[i];
      bb=b[i];
      cc=c[i];
      css=cs[i];
      vv1=v[i,1];
      vv2=v[i,2];
      vv3=v[i,3];
      vv4=v[i,4];
      vv5=v[i,5];
      vv6=v[i,6];
      keep tt aa bb cc vv1-vv6 css;
      output try4;
     end;

Note that I have to construct arrays of known size, that is you have to know how many input records there are.

The first half of the DATA step constructs arrays into which the values from the input data set are read. We read all of the records, and then we do all of the calculations, since we have all of the values in memory in the matricies. There is some fiddling with RENAMES so that you can keep the array names t, a, b, c etc but still have variables named a, b, c etc in the output data set.

So hopefully that might help you along a bit. Either that or confuse you because I've misunderstood what you're trying to do!

Upvotes: 4

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