Concatnation of character and number into one variable

[sgarg]

Hi, I need to create a new variable from three variables. example

a = 'S'  -->   of size 1 character

b = 'XYZ' --->  of size 3 character. Here value of b can be  'XY'  or 'X'  This means  size of variable is 3 but can have 2 character or 1 character value.

c = 11 ---->  this is 2 digit numeric  and can have values from 1 to 20  This means size of variable is 2 but can have 1 digit or 2 digit.

I need to combined all three variable and  create new variable  Z  of size $7.

So as per Example

if

a = 'S'   b = 'XY'  c = 1   then  Z = 'SXY 01'  ( 1 space in between)

a = 'S'   b = 'X'    c = 5   then  Z = 'SX  05'  ( two spaces in between)

I tried CAT, CATX, CATX  functions but not getting results.

Please Advice what i can do...

Thank you.

[novinosrin]

HI @sgarg

Z  of size $7. or size $6(3+2+1)?

[novinosrin]

data have;
length a $1 b $3 ;
a = 'S'  ; b ='XY' ; c = 1;
output;
a = 'S' ;  b = 'X'   ; c = 5 ;
output;
run;

data want;
set have;
length _c $2 z $6;
_c=put(c,z2.);
z=peekc(addr(a),1)||peekc(addr(b),3)||peekc(addr(_c),2);
run;

NOTES:

1. We do not know your real data whether those variables are grouped contiguous. So the need for peekc *3 times. If it is contiguous, just extracting the left most var's addr of the leftmost byte with vlength combination of a,b,c will get the required in one shot.

2. APP functions use is highly dependent on knowing data first.

3. While for the description that you gave, that answered.

4. Use the long brethren if  you are working on 64 bit machine i.e substitute with peekclong, addrlong accordingly

5. If you are not used to APP data management functions, ignore my post

[sgarg]

@novinosrin  Thanks for your reply. I am using SAS Enterprise guide and got error see NOTE section below.

27         data have;
28         length a $1 b $3 ;
29         a = 'S'  ; b ='XY' ; c = 1;
30         output;
31         a = 'S' ;  b = 'X'   ; c = 5 ;
32         output;
33         run;

NOTE: The data set WORK.HAVE has 2 observations and 3 variables.
NOTE: DATA statement used (Total process time):
      real time           0.00 seconds
      cpu time            0.00 seconds
      

34         
35         data want;
36         set have;
37         length _c $2 z $6;
38         _c=put(c,z2.);
39         z=peekc(addr(a),1)||peekc(addr(b),3)||peekc(addr(_c),2);
40         run;

NOTE: ADDR function cannot be used on this platform. Use ADDRLONG instead.
NOTE: PEEKC function cannot be used on this platform. Use PEEKCLONG instead.
NOTE: There were 2 observations read from the data set WORK.HAVE.
NOTE: The data set WORK.WANT has 2 observations and 5 variables.
NOTE: DATA statement used (Total process time):
      real time           0.00 seconds
      cpu time            0.00 seconds
     

[sgarg]

@novinosrin    Correct Z is of size  $6.  lol.. Thanks.

[novinosrin · Accepted solution]

Forget APP functions. How about CAT?

data have;
length a $1 b $3 ;
a = 'S'  ; b ='XY' ; c = 1;
output;
a = 'S' ;  b = 'X'   ; c = 5 ;
output;
run;

data want;
set have;
length _c $2 z $6;
_c=put(c,z2.);
z=cat(a,b,_c);
run;

[sgarg]

@novinosrin  Yes it worked. Thank you Thank you so much..

[Allaluiah]

@sgarg  Mark the post as answered having acknowledged the help. That completes and closes it.

[novinosrin]

Finally for fun and completeness,

My Notes above point 4. emphasizes LONG APP' for the 64 bit machine, therefore the code changes

data have;
length a $1 b $3 ;
a = 'S'  ; b ='XY' ; c = 1;
output;
a = 'S' ;  b = 'X'   ; c = 5 ;
output;
run;

data want;
set have;
length _c $2 z $6;
_c=put(c,z2.);
z=peekclong(addrlong(a),1)||peekclong(addrlong(b),3)||peekclong(addrlong(_c),2);
run;