Add the index of the column with the maximum value

问题:

My question is simple. When data is as below,

var1    var2    var3
10      40      60
15      10      5

I want to add a new column MaxValueVar that returns index of a column that has maximum value among var1, var2 and var3. That is, I want to make a table as below.

var1    var2    var3    MaxValueVar
10      40      60      3
15      10      5       1

In R I would use:

apply(vector, 1, which.max)

How can I accomplish this using SAS?

回答1:

One Solution for your reference according to the sample you provide here. You didn't mention how to deal with ties. Here for ties, the first occurrence is fetched.

data test;
input var1 var2 var3;
datalines;
10      40      60
15      10      5
run;

data test;
 set test;
 maxvalue=max(of var1-var3);
 maxvaluevar=0;
  array vars (*) var1-var3;
    do i=1 to dim(vars);
     if maxvaluevar=0 then maxvaluevar=i*(maxvalue=vars(i))+maxvaluevar;
    end;
 drop i maxvalue;
run;

回答2:

Here is a similar solution to that of Robbie's. It uses the vname() function to get the variable name of the first maximum in addition to the index in the array.

data maxarr (drop=i);
input var1 var2 var3;
 array vars(*) var1-var3;
 max=max(of vars(*));
   do i=1 to dim(vars);
     if vars(i)=max then do;
          MaxValueIndx=i;
          MaxValueVar=vname(vars(i));
          leave;
     end;
   end;
datalines;
10      40      60
15      10      5
;
run;

proc print data=maxarr noobs;run;

回答3:

The code below should work as intended, as well as creating more columns in the instance of there being ties. I know you said not to bother, but it was bugging me!

Input test data:

data test;
    input var1 var2 var3;
    cards;
10 40 60
15 10 5
7 8 9
13 13 10
5 7 6
10 11 12
10 10 10
1 3 2
3 3 1
;
run;

Code to check for maximum values:

data test (drop = i howmanymax);
    set test;
    retain howmanymax;

    howmanymax=0;

    array varlist[*] var1--var3;
    array maxnum[3];

    do i = 1 to DIM(varlist);
        if varlist[i] = max(of var1--var3)
            then do;
                howmanymax+1;
                maxnum[howmanymax] = i;
            end;
    end;

run;

Output looks like:

  var1  var2  var3  nummax1 nummax2 nummax3
    10    40    60        3       .       .
    15    10     5        1       .       .
     7     8     9        3       .       .
    13    13    10        1       2       .
     5     7     6        2       .       .
    10    11    12        3       .       .
    10    10    10        1       2       3
     1     3     2        2       .       .
     3     3     1        1       2       .