id lowercase remove

[pavank]

data dsn;
input ID $ ;
datalines;
aBC.L
ABCa.L
cDE.L
BDEna.L
bNE.L
HDF.L
;
run;


data dsn_output;
    set dsn;
new_id=compress(id,,'ku');
 proc print noobs;
   run;

without REGEX  Method required output

ABC.L
ABC.L
CDE.L
BDE.L
BNE.L
HDF.L

[PaigeMiller]

You are not using REGEX. I don't understand what you want.

[pavank]

Hi @PaigeMiller  

I am trying below output typically using functions

ABC.L
ABC.L
CDE.L
BDE.L
BNE.L
HDF.L

[PaigeMiller]

I showed in your last post how to pull strings apart and make some of it uppercase. It is essentially the same here, except you seem to want (but you haven't explicitly said so) to find lowercase letters. How can you do that? You would use the FINDC function with the modifier 'L'. See if you can put together what I did in your last post, with the FINDC function.

With regards to my phrase above "but you haven't explicitly said so" — I'd like to see you explain more of the problem, rather than writing as few words as possible. Instead, write complete clear explanations. Be generous with words, be generous with your explanations, be generous with information. Its not enough to show us the input and output that you want, you need to explain in words what you want. Please start doing that.

Still unexplained is: are there always 3 letters that will be capitalized, then a dot, and then another capital letter? Or can there be more than 3 letters before the dot? Please explain that part as well. 

[Kurt_Bremser · Accepted solution]

Make sure that the first three characters are uppercase, then use COMPRESS with the correct modifier:

data dsn;
input ID $;
new_id = compress(upcase(substr(id,1,3)) !! substr(id,4),,'kup');
datalines;
aBC.L
ABCa.L
cDE.L
BDEna.L
bNE.L
HDF.L
;

[Tom]

What is the rule?  This rule works for your example data.  Take the first three characters, uppercase them and append .L.

data dsn;
  input ID $ ;
  new_id=catx('.',upcase(substr(id,1,3)),'L');
datalines;
aBC.L
ABCa.L
cDE.L
BDEna.L
bNE.L
HDF.L
;

Result

OBS      ID       new_id

 1     aBC.L      ABC.L
 2     ABCa.L     ABC.L
 3     cDE.L      CDE.L
 4     BDEna.L    BDE.L
 5     bNE.L      BNE.L
 6     HDF.L      HDF.L

[pavank]

Hi @Tom  

Thank you very much for your solution