topic creating new varibles within a range of values of 6 different variables in SAS Programming

creating new varibles within a range of values of 6 different variables

lalohg — Thu, 16 Feb 2017 19:15:00 GMT

Hello there, Among other variables, I have a data set with 6 character variables (lenght 4) like this:

Obs	Var1	Var2	Var3	Var4	Var5	Var6
1	O809	O809	Z370	Z301	Z390	A20X
2	B171	B172	K746	I10X	I519	K546
3	O809	O809	Z370	Z302	Z390
4	X101	X102	I12X

and would like:

1.- to get rid of the "X" but only for entries with the "X" at the end so de data would look like:

Obs	Var1	Var2	Var3	Var4	Var5	Var6
1	O809	O809	Z370	Z301	Z390	A20
2	B171	B172	K746	I10	I519	K546
3	O809	O809	Z370	Z302	Z390
4	X101	X102	I12

then

2.- I would like to create new variables that are within a specific range, for example if I want a new variable Z to select any entry of Var1 to Var 6 between Z301 and Z370, the new variable would be like the example below, in this case "VariableZ", if I want another new varible to select any value of Var1 to Var6 to be between B171 and B172 or between X101 and X102 the new variables would be respectivelly like "VariableB" and "VariableX" below and so on

New data set
Obs	Var1	Var2	Var3	Var4	Var5	Var6	VariableZ	VariableB	VariableX
1	O809	O809	Z370	Z301	Z390	A20	Z301-Z370
2	B171	B172	K746	I10	I519	K546		B171-172
3	O809	O829	Z370	Z302	Z390		Z301-Z370
4	X101	D109	I12						X101-X102

all your help will be appreciated

Thanks

EHG.

Re: creating new varibles within a range of values of 6 different variables

Astounding — Thu, 16 Feb 2017 19:49:44 GMT

Part 1 is pretty easy. In a DATA step:

array var {6};

do i=1 to 6;

if substr(var{i}, 4, 1) = 'X' then substr(var{i}, 4, 1) = ' ';

end;

For part 2, I'm not really sure what you are trying to achieve here. But I would warn you about making character comparisons. For example, as character strings, "Z32" falls within the range of "Z301" through "Z370".

Re: creating new varibles within a range of values of 6 different variables

art297 — Thu, 16 Feb 2017 20:17:33 GMT

I'm sure that someone (Hi @Reeza) will complain about my using DO OVER, the following is an easy way to accomplish both tasks:

data have;
  input (Var1-Var6) ($);
  cards;
O809 O809 Z370 Z301 Z390 A20X
B171 B172 K746 I10X I519 K546
O809 O809 Z370 Z302 Z390 . 
X101 X102 I12X  .  .  .
;
data want;
  set have;
  array stuff var1-var6;
  do over stuff;
    if substr(stuff,length(stuff),1) eq 'X' then
      substr(stuff,length(stuff),1) = '';
    if substr(stuff,1,2) eq 'Z3' and
      1<=input(substr(stuff,3,2),8.)<=70  then variablez='Z301-Z370';
    if stuff in ('B171','B172') then variableb='B171-B172';
  end;
run;

HTH,

Art, CEO, AnalystFinder.com

Re: creating new varibles within a range of values of 6 different variables

mkeintz — Thu, 16 Feb 2017 22:42:52 GMT

This is a good task to show off the benefits of the CHAR function (extracts a single character substring), the INPUT function, and especially the SELECT statement:

data have;
  input (Var1-Var6) ($);
  cards;
O809 O809 Z370 Z301 Z390 A20X
B171 B172 K746 I10X I519 K546
O809 O809 Z370 Z302 Z390 . 
X101 X102 I12X  .  .  .
;
data want;
  set have;
  array v {*} var1-var6;
  do I=1 to dim(v);
    if char(var6,4)='X' then substr(var6,4,1)=' ';
    select (char(v{I},1));
      when ('Z') if 301 <= input(substr(v{I},2),best32.) <= 370 then varz='Z301-Z370';
      when ('B') if 171 <= input(substr(v{I},2),best32.) <= 172 then varb='B171-B172';
      when ('X') if 101 <= input(substr(v{I},2),best32.) <= 102 then varx='X101-X102';
      otherwise ;
    end;
  end;
run;

The INPUT function is set to an informat of BEST32. as an insurance policy. It means you generally don't have to worry about the length of the character variables being INPUTed.

Re: creating new varibles within a range of values of 6 different variables

lalohg — Mon, 20 Feb 2017 13:17:16 GMT

Thank you everybody,

the code worked just fine

your help is very much appreciated.