Thanks Nick!
Maria Ana
On Feb 10, 2010, at 9:05 AM, "Nick Cox" <[email protected]> wrote:
> See also for related technique
>
> SJ-8-4 dm0043 . Tip 71: The problem of split identity, or how to
> group
> dyads
> . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . N.
> J. Cox
> Q4/08 SJ 8(4):588--591 (no
> commands)
> tip on how to handle dyadic identifiers
>
> Nick
> [email protected]
>
>
> Martin Weiss
>
> clear*
>
> input byte(Person1_id ClosestPerson Person1_age)
> 1 2 10
> 2 1 20
> 3 1 30
> 4 3 40
> end
>
> gen lower=min( Person1_id, ClosestPerson)
> gen upper=max( Person1_id, ClosestPerson)
> bys lower upper ( Person1_age): gen byte dup=!(_N==1)
> by lower upper ( Person1_age): replace dup=2 if _n==_N &dup!=0
> l, noo
>
> Vitorino, Maria Ana
>
> I was wondering if there was a relatively simple way to achieve the
> following:
>
> My data looks like this:
>
> Person1_id ClosestPerson Person1_age
> 1 2 10
> 2 1 20
> 3 1 30
> 4 3 40
> ...
>
> What I would like to do is to signal "duplicate" rows for which
> Person1
> is
> the oldest. Since this is hard to explain, the resulting data would be
> for
> the example above:
>
> Person1_id ClosestPerson Person1_age dup
> 1 2 10
> 1
> 2 1 20
> 2
> 3 1 30
> 0
> 4 3 40
> 0
> ...
>
> Note that the column dup has the following coding:
> dup=1 if there is another pair of ppl identical to the one in this row
> and
> Person1 is the youngest of the pair
> dup=2 if there is another pair of ppl identical to the one in this row
> and
> Person1 is the oldest of the pair
> dup=0 if there is no other pair of ppl identical to the one in this
> row
>
> *
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*
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