As long as you have confidence on what you are doing, it doesn't matter how you do it. There are always many ways of getting the same result when programming.
I agree w/ Nick that Stata offers better ways of creating the identifier. However, the old method I reminded will do the trick w/ any programming language - OK, this is a Stata list but if you are new to it, it may be useful to understand the basic algorithm to perform the task.
The e-mail I sent got stuck in my outgoing server since last week. I apologize for it being sent so late to the list.
Rafael G. Osorio
-----Original Message-----
From: [email protected] [mailto:[email protected]] On Behalf Of n j cox
Sent: Monday, July 30, 2007 12:22 PM
To: [email protected]
Subject: st: RE: RE: Identifier from three variables
This isn't another way of doing it. It is a variation
on a way already discussed in this thread by Svend Juul
and indeed by myself (#3 in the post to which this
post replies).
Generating a numeric identifier by this method has no
real advantages that I can see and at least five disadvantages
compared with concatenation of strings:
1. You need to check the limits of each identifier,
as Rafael points out, as the result will be legal to Stata.
2. If you forget #1 or make a mistake on #1, the resulting
problems may not be obvious.
3. Here the default variable type is a -float-. Using
that to hold (very large) integers could run into precision
problems. Rafael is careful to recommend using a -double-
if needed but the problem is that everyone has to remember to do
that.
4. Concatenation can make use of separators, as in
egen id = concat(house family individual), p("_")
5. The reverse engineering from composite to
individual identifiers is easier with -split-
(provided you follow #4) than it is usually is
with composite integers.
Rafael's method, which is quite often used, is relatively
simple and can be unproblematic, but it is not a good
general method in my view.
Nick
[email protected]
Rafael Osorio
Other way of doing it, supposing each identifier is <= 99:
gen id = house*1000+family*100+individual
By doing it this way you can tell to which house and family a specific
person belongs to. If your data was not sorted by the identifiers, if
you sort by this new identifier the result will be equal to sorting it
by all identifiers. If your final id is large: gen double id...
If your identifiers are greater than 99, adjust the multipliers.
Nick Cox
1. Don't do that. Use -egen, group()- with the -label- option.
2. See also
FAQ . . . . . . . . . . . . . . . . . . . . . . Creating group
identifiers
3/01 How do I create individual identifiers numbered
from 1 upwards?
http://www.stata.com/support/faqs/data/group.html
3. If you really must, look into -egen, concat()-, including
its options.
Nick
[email protected]
N�dia N. Sim�es
>
> How can I generate an identifier from three variables?
> In my data I have a column for the house, one for the family
> and one for the individual
>
> example:
>
> house family individual
> 1 1 1
> 1 1 2
> 1 1 3
> 2 1 1
> 2 1 2
> ...
>
> and I would like to know how can I create an identifier per
> individual such as:
>
> individual
> 010101
> 010102
> 010103
> 020101
> ...
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