BookmarkSubscribeRSS Feed
chelm24
Calcite | Level 5

Hello,

 

I require assistance with comparing two variables in SAS and determining if there are any partial record matches.

data have;
length VAR1 $100 VAR2 $100;
input VAR1 $ VAR2 $;
infile datalines dlm='|';
datalines;
1.DR. MORRISON|
1.MORRISON| ABCFG MORRISON
1.DR. MORRISON| MORRISON
1.DR. MORRISON| DR. WRIGHT
1. LA HOSPITAL| SAN DIEGO
;

data want / Expected Result

VAR1VAR2PARTIAL MATCH?
1.DR. MORRISON NO
1.MORRISONABCFG MORRISONYES
1.DR. MORRISONMORRISONYES
1.DR. MORRISONDR.WRIGHTNO
1. LA HOSPITALSAN DIEGONO
5 REPLIES 5
ballardw
Super User

Define your criteria for "partial match". 4 letters the same in sequence? 5 ? 6? Some other rule?

 

There are several SAS functions, COMPGED, SPEDIS and COMPLEV that will provide scores of spelling "distance", or a measure of similarity. I would try all three, and read the documentation, to select which seems to fit your data and need best. The lower the score returned the more similar two variables are.

 

data have;
length VAR1 $100 VAR2 $100;
input VAR1 $ VAR2 $;
infile datalines dlm='|';
Compgedscore = compged(var1, var2);
Complevscore = complev(var1, var2);
Spedisscore  = spedis(var1, var2);
datalines;
1.DR. MORRISON|
1.MORRISON| ABCFG MORRISON
1.DR. MORRISON| MORRISON
1.DR. MORRISON| DR. WRIGHT
1. LA HOSPITAL| SAN DIEGO
;
chelm24
Calcite | Level 5

@ballardw , I need to determine the words that match between 2 variables and not by score. Partial match at least >= 4 letters the same in sequence.

 

VAR1VAR2PARTIAL MATCH?MATCH
1.DR. MORRISON NO 
1.MORRISONABCFG MORRISONYESMORRISON
1.DR. MORRISONMORRISONYESMORRISON
1.DR. MORRISONDR.WRIGHTNO 
1. LA HOSPITALSAN DIEGONO 

 

 

Patrick
Opal | Level 21

@chelm24 wrote:

@ballardw , I need to determine the words that match between 2 variables and not by score. Partial match at least >= 4 letters the same in sequence.


Exact match of WORD OR exact match of any string of 4 characters within the same WORD. 

Above two options should be doable BUT if you go for the option with 4 characters it could then be any two words as long as they share a sequence of 4 identical characters. 

Tom
Super User Tom
Super User

Just test each word.

data want ;
  set have;
  do i=1 to countw(var1,' ,.()-') until(found);
    word=scan(var1,i,' ,.()-');
    if length(word)>3 then found = 0<findw(var2,word,' ,.()-','it');
  end;
  if not found then do; 
     word=' ';
     i=0;
  end;
run;
Obs         VAR1         VAR2              i    found      word

 1     1.DR. MORRISON                      0      0
 2     1.MORRISON        ABCFG MORRISON    2      1      MORRISON
 3     1.DR. MORRISON    MORRISON          3      1      MORRISON
 4     1.DR. MORRISON    DR. WRIGHT        0      0
 5     1. LA HOSPITAL    SAN DIEGO         0      0
Ksharp
Super User
data have;
length VAR1 $100 VAR2 $100;
input VAR1 $ VAR2 $;
infile datalines dlm='|';

if find(compress(var1,,'ka'),compress(var2,,'ka'),'i') or 
   find(compress(var2,,'ka'),compress(var1,,'ka'),'i') then MATCH='Yes' ;
  else MATCH='No ' ;

datalines;
1.DR. MORRISON|
1.MORRISON| ABCFG MORRISON
1.DR. MORRISON| MORRISON
1.DR. MORRISON| DR. WRIGHT
1. LA HOSPITAL| SAN DIEGO
;

hackathon24-white-horiz.png

The 2025 SAS Hackathon Kicks Off on June 11!

Watch the live Hackathon Kickoff to get all the essential information about the SAS Hackathon—including how to join, how to participate, and expert tips for success.

YouTube LinkedIn

What is Bayesian Analysis?

Learn the difference between classical and Bayesian statistical approaches and see a few PROC examples to perform Bayesian analysis in this video.

Find more tutorials on the SAS Users YouTube channel.

SAS Training: Just a Click Away

 Ready to level-up your skills? Choose your own adventure.

Browse our catalog!

Discussion stats
  • 5 replies
  • 1662 views
  • 0 likes
  • 5 in conversation