Whether a variable is an identifier can be asked directly using -isid- without the need for installing user-written add-ons. That's not quite the main issue here, as there is no problem working with -Case Person- together.
That's said, let's spell out Jochen's good solution in a little more detail.
In terms of Jim's original question, the implication is that
var1 var2 var3
each have at most one non-missing value for each distinct combination of -Case- and -Person-. Otherwise what Jim wants to do is a bad idea.
That's directly checkable:
foreach v in var1 var2 var3 {
bysort Case Person (`v') : assert missing(`v'[2])
}
Consider -var1- as an example.
After -sort Case Person var1-, any single non-missing value of -var1- will be first within groups defined by -Case Person-. The second value should be missing. If it's not, the assumption is incorrect, i.e. there are at least 2 non-missing values.
Assuming good health in this sense, Jochen's solution is then to -collapse-.
collapse var1 var2 var3, by(Case Person)
which will work fine as the mean of one non-missing value is always the same value.
Nick
[email protected]
Jochen Späth
I am wondering if case and person are actually the only identifiers in your data set because they do not uniquely identify your observations, maybe you should check this with the -unique- command first (if not installed type
-ssc install unique-). Anyway, if they are the only identifiers as you stated in your question, you could try -collape var1 var2 var3, by(case person)-. This, however, will only help if per case and person you have really only one non-missing observation in each variable to be collapsed, as showed in your example. Otherwise, the collapsed mean would be false.
Jim O'Grady
I have a dataset like this:
Case Person var1 var2 var3
345 01 1 . .
345 01 . 1 .
345 02 . 1 .
345 02 . . 1
I want to merge the observations which have the same identifiers (case
and person) to get
Case Person var1 var2 var3
345 01 1 1 .
345 02 . 1 1
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