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Re: st: ranking with weights


From   Steven Samuels <[email protected]>
To   [email protected]
Subject   Re: st: ranking with weights
Date   Tue, 2 Dec 2008 13:53:40 -0500

Cindy, What are the analytic units (people? regions?). What are the "weights"? What is "expenditure"? How is it measured. What do you mean that some regions are "less sampled" than others. It's not clear, for example, if this is a sample, and if so, of what? So, please describe the study design in detail. Last question: what is the purpose of the ranking?

-Steve

On Dec 2, 2008, at 12:54 PM, Cindy Gao wrote:

Hello

I am trying to find a way to rank weighted data (since the egen function -rank- does not work with weights). A simple way would be order the data in terms of variable that I have interest in (monthly expenditure) and then create a new variable like -g rank1=sum(weight)-. But, there is problem. Some of my observations are "tied" as they have the same level of expenditure. Using the simple method I mention means that some observations are ranked above others even though they have same level of expenditure. This is a problem as the weights are large so you find that 2 observations are ranked with bug gap in between even though same level of expenditure. It is even bigger problem because the weights might be correlated with some other variables I am interested in (like region, since some regions are less sampled than other). I also try multiplying the expenditure ranking by the weight, but this gives wrong results (for example they do not add up to weighted total). Can anyone help? In other words, I would like for all observations with same expenditure to have same rank, which I assume would be some average of all the weighted observations having that same expenditure. I include a sample dataset below:

expenditure       weighting        rank       rank1      weighted_rank
10 341 1 341 341
12                          1065          2.5        1406         ???
12                          98             2.5        1504
15                          254            4          1758
.......

thanks,

Cindy






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