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serrld113
Obsidian | Level 7

 

Hello,

 

I need to find the ID where the first 3 characters of the field are between (‘A’) and (‘Z’)...So if I have this list:

 

5021060
A310603
D1A5421
5022000
5001050
5002000
5010020
5010045
5010045
50A5421
5012060
5013100

then the SAS program would output only these:

 

A310603

D1A5421

50A5421

 

Thank you!

1 ACCEPTED SOLUTION

Accepted Solutions
kiranv_
Rhodochrosite | Level 12

something like this

data have;
input string $;
datalines;
5021060
A310603
D1A5421
5022000
5001050
5002000
5010020
5010045
5010045
50A5421
5012060
5013100
501310D
;

data want;
set have;
if anyalpha(substr(string,1,3)) gt 0;
run;

View solution in original post

2 REPLIES 2
kiranv_
Rhodochrosite | Level 12

something like this

data have;
input string $;
datalines;
5021060
A310603
D1A5421
5022000
5001050
5002000
5010020
5010045
5010045
50A5421
5012060
5013100
501310D
;

data want;
set have;
if anyalpha(substr(string,1,3)) gt 0;
run;
RW9
Diamond | Level 26 RW9
Diamond | Level 26

Post test data in the form of a datastep!

Something like (and not tested as not typing all that in):

data want;
  set have;
  if lengthn(compress(substr(yourvar,1,3),"","ka")) > 0 then output;
run;

So take the first 3 chars, compress down to only a-z and if length > 0 at least one is present

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