Hash join question

[evp000]

Hi,

In Art Carpenter's Innovative SAS Techniques, Chapter 3, section 3.3.7 there is the following code:

data trans;

     length lname $10 fname $6 sex $1;

     fname = 'Mary'; lname = 'Adams'; sex = 'N'; output;

     fname = 'Joan'; lname = 'Adamson'; sex = 'x'; output;

     fname = 'Peter'; lname = 'Anla'; sex = 'A'; output;

run;

data newdemog (drop = rc);

     declare hash upd(hashexp:10);

     upd.definekey('lname','fname');

     upd.definedata('sex');

     upd.definedone();

     do until(lasttrans);

          set trans end=lasttrans;

          rc=upd.add();

     end;

     do until(lastdemog);

          set advrtp.demog end=lastdemog;

          rc=upd.find();

          output newdemog;

     end;

     stop;

run;

Was it necessary to have the first loop to populate the hash table, or would it have been the same to just declare it with

declare hash upd (dataset: 'trans');  ?

Does this code do the same thing (without do loops)?

data newdemog (drop = rc);

     if 0 then set advrtp.demog trans ;

     if _n_ = 1 then

     do;

          declare hash upd(hashexp:10 dataset: 'trans' );

          upd.definekey('lname','fname');

          upd.definedata('sex');

          upd.definedone();

     end;

     set advrtp.demog;

     call missing(sex);

     rc = upd find();

     output;

run;

Thanks for your input.

[Haikuo]

When asking if they are the same, you are implying several things at the same time, for example:

1. Will they get the same outcome? Yes, I believe so.

2.Will they get the same/similar performance/efficiency? I have no clue. You will have to benchmark it yourself .

in between REPLACE() and ADD() methods, REPLACE() is the one that will produce different outcome if you have duplicated keys.

Haikuo

[evp000]

Ok, thanks.  Yes, I was wondering about the output, not the performance. 

[gergely_batho]

Yes, the two codes produce the same output (though I have not tested).

I think in older versions of SAS there was no dasaset argument tag.

[evp000]

Thanks.