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st: generate new distinct individual ids by organizational id and year
From
Erik Aadland <[email protected]>
To
"[email protected]" <[email protected]>
Subject
st: generate new distinct individual ids by organizational id and year
Date
Tue, 7 May 2013 08:18:39 +0000
Dear Statalist.
I have an unbalanced panel dataset containing organizations with distinct org_id, and employees within these organizations with distinct ind_id. Some employees are missing. I have calculated the total number of missing employees per org_id per year, called "missing_employees".
Now, by each org_id and year I need to generate a set of distinct ind_ids corresponding to the number of missing employees. The new_ind_ids generated for a distinct org_id and year should preferably be reused in later years.
In laymans terms, I need to generate new ind_id observations for each org_id and year so that I have an ind_id for all employees in each org_id and year.
Example:
year org_id ind_id missing_employees
2000 1 1 1
2000 1 2 1
2000 2 3 2
2000 2 3 2
2000 2 4 2
2000 3 5 0
2001 1 2 2
2001 1 7 2
Given such a structure, I need to generate distinct new_ind_ids e.g. starting after the last distinct ind_id I have in my dataset (e.g. if last ind_id==100, new_ind_id starts at 101):
year org_id ind_id missing_employees new_ind_ids
2000 1 1 1
2000 1 2 1
2000 1 101
2000 2 3 2
2000 2 3 2
2000 2 4 2
2000 2 102
2000 2 103
2000 3 5 0
2001 1 2 2
2001 1 7 2
2001 1 101
2001 1 104
Note that for org_id==1 in 2000, I generate 1 new_ind_id == 101 since one employee was missing. In 2001 I have 2 employees missing for org_id == 1. Since I have already generated new_ind_id == 101 for org_id==1, this can be used again in 2001 and only one new_ind_id needs to be generated; new_ind_id == 104.
If the number of missing employees for a distinct org_id goes down from one year to the next, I would like to drop the corresponding number of new_ind_ids for that year (i.e. remove some of the reused new_ind_ids).
Any input on these problems would be greatly appreciated.
Kind regards,
Erik.
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