Find Min and Max of five variables with min above zero for each row

[jbird]

Hello, 

I have a data set with 6 variables: ID, IR00, IR10, IR20, IR30, IR40.  ID is an 8 digit numeric ID.  IR00 through IR40 are each continous variable from 0 to 1.  I need to find the min and max for each ID across the IR00 to IR40 variables. The min has to be above 0.  

I have found the max using:

data new;

set old;

MAX_IR = max(of IR00 - IR40) ;

output;

keep ID IR00 IR10 IR20 IR30 IR40 MAX_IR; 

run; 

But I cannot figure out how to find the min value above zero.  Any help would be appreciated. 

[Patrick]

Can't think of anything "shorthand". Below should work.

data new;
  set old;
  array vars{*} IR00 - IR40;
  do i=1 to dim(vars);
    if vars[i]>0 then min_ir=min(min_ir,vars[i]);
  end;

  MAX_IR = max(of IR00 - IR40);
  output;
  keep ID IR00 IR10 IR20 IR30 IR40 MAX_IR min_ir;
run;

[SASKiwi]

Try this. The SMALLEST function ignores missing values:

data new;
  set old;
  MAX_IR = max(of IR00 - IR40);
  if IR00 = 0 then IR00 = .;
  if IR10 = 0 then IR10 = .;
  if IR20 = 0 then IR20 = .;
  if IR30 = 0 then IR30 = .;
  if IR40 = 0 then IR40 = .;
  MIN_IR = smallest(of IR00 - IR40);
run; 

[Patrick]

@SASKiwi Also the min() function skips missings but I believe the real "challenge" is: "The min has to be above 0. "

if IR00 <= 0 then IR00 = .;

[yabwon]

With SAS Arrays?

data old;
  call streaminit(42);
  do id = 1 to 1e2;
    array ir IR00 IR10 IR20 IR30 IR40;
    do over IR;
      IR = rand("integer",0,9)/10;
    end;
  output;
  end;
run;

data new;
  set old;

  array IR IR:;
  do over IR;
               MAX_IR = max(MAX_IR,IR);
    if IR then MIN_IR = min(MIN_IR,IR);
  end;

  keep ID IR: M:; 
run; 
proc print;
run;

Bart

P.S. Just for fun

data new;
  set old;

  array IR IR:;
  do over IR;
    MAX_IR = MAX_IR<>IR;
    MIN_IR = min(MIN_IR,IR+(0=IR));
  end;

  keep ID IR: M:; 
run;

[yabwon]

One more just for fun:

data new;
  set old;

  array IR IR:; /* make 0 to . */
  do over IR;
    IR = ifn(IR,IR,.);
  end;
  MAX_IR = max(of IR[*]);
  MIN_IR = min(of IR[*]);

  set old;

  keep ID IR: M:; 
run; 
proc print;
run;

[mkeintz]

You may have to protect against all missing IR values or all zeroes.  If so, then:

data want (drop=_:);
  set have ;
  if n(of IR:)>0 then max=max(of IR:);
  if max=0 then max=.;

  if max>0 then do _r=1 to n(of IR:) until(min>0);
    min=smallest(_r,of IR:);
  end;
run;

The "do ... UNTIL" using smallest is a way to avoid min=0.