Random id for same mother

[malena]

Hello,

I am trying to assign a random id to mothers in a dataset for 4 years. Of course there are many mothers who give birth more than once in this dataset so I will like to use the same random id if that is the case. I have close to 200,000 records and a key using last name, first name and DOB of the mothers.

thanks!

[malena]

Hello,

I am trying to assign a random id to mothers in a dataset for 4 years. Of course there are many mothers who give birth more than once in this dataset so I will like to use the same random id if that is the case. I have close to 200,000 records and a key using last name, first name and DOB of the mothers.

thanks!

[malena]

I am trying to assign a random id to mothers in a dataset for 4 years. Of course there are many mothers who give birth more than once in this dataset so I will like to use the same random id if that is the case. I have close to 200,000 records and a key using last name, first name and DOB of the mothers.

thanks!

[LinusH]

What will be the use of the random id?

[TomKari]

People frequently use UUIDs for this. Check out the UUIDGEN function.

Tom

[PGStats]

To get random ids going from 1 to the number of distinct mothers, do:

proc sort data=mothers; by lastName firstName dob; run;

data mothers2;
call streaminit(7568);
retain id;
set mothers; by lastName firstName dob;
if first.dob then id = rand("UNIFORM");
run;

proc rank data=mothers2 out=mothersId ties=dense;
var id;
run;

(untested)

[PGStats · Accepted solution]

Or, if you just want large non repeating numbers as ids :

proc sort data=mothers; by lastName firstName dob; run;

data mothersId;
call streaminit(7566);
retain id;
set mothers; by lastName firstName dob;
if first.dob then id = int(1e12*rand("UNIFORM"));
format id z12.0;
run;

[malena]

thank you! this works perfectly

[Ksharp]

Hash Table

data have;
input first $ last $ dob : date9.;
format dob date9.;
cards;
a b 03jun1991
b c 09jan1980
b c 09jan1980
a b 03jun1991
;
run;
data want;
 if _n_=1 then do;
  if 0 then set have;
  declare hash h();
  h.definekey('first','last','dob');
  h.definedata('id');
  h.definedone();
 end;
 set have;
 rc=h.find();
 if rc ne 0 then do;n+1;id=n;h.add();end;
 drop rc n;
run;

[LinusH]

Not knowing the use case, but perhaps hashing the concatenated business key values could be one solution.
The id values will distributed, it's quite safe if privacy is required and it's simple to use: given the same input always the same hash is produced. Check out the md5() function.

[Reeza]

Hashing isn't a good idea...

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