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RE: st: how to construct a mean squared Euclidean measure?


From   "chanchal.balachandran@usi.ch" <chanchal.balachandran@usi.ch>
To   "statalist@hsphsun2.harvard.edu" <statalist@hsphsun2.harvard.edu>
Subject   RE: st: how to construct a mean squared Euclidean measure?
Date   Thu, 17 Feb 2011 18:53:49 +0000

Hi David,

I got a solution from another statalist user, but I think I will check it using the method you suggested as well.

thanks for your kind reply.

Chanchal

________________________________________
From: owner-statalist@hsphsun2.harvard.edu [owner-statalist@hsphsun2.harvard.edu] on behalf of David Kantor [kantor.d@att.net]
Sent: 17 February 2011 18:56
To: statalist@hsphsun2.harvard.edu
Subject: Re: st: how to construct a mean squared Euclidean measure?

Hi.

I don't follow all of what's happening here, but
what you want looks like a Mahalanobis distance measure.
Check
  ssc desc mahapick

If that fits, you would use -mahascores-,
possibly with the -euclidean- option (or maybe not).

HTH
--David

At 12:15 PM 2/17/2011, Mike Lacy replied to Chanchal:

> >
> >Date: Wed, 16 Feb 2011 16:24:57 +0000
> >From: "chanchal.balachandran@usi.ch" <chanchal.balachandran@usi.ch>
> >Subject:
> >
> >Hi all,
> >
> >I am using Stata 10.1. I want to create a variable which captures the
> >dissimilarity measure of a member who exit from a work group, sorted by firm
> >and year. The dissimilarity is calculated in terms of the member¡¯s job
> >tenure in the group with respect to other members¡¯ job tenures. More
> >precisely the measure I am trying to construct is the mean squared Euclidean
> >distance of a focal member i  from each incumbent team member j  and is
> >given as: ¡Ì( ¡Æ(Xi ¨C Xj)2/(n-1)), where Xi is the tenure of the focal
> >individual i, Xj is the tenure of incumbent j, with i not equal to j and n
> >is the number of group members. Does anyone have an idea how to construct
> >this measure using Stata? For more clarity, let me describe the data
> >structure as below.
> >
> >
> >
>If I understand Chanchal correctly, I think this will work.  It is
>probably not the most efficient approach, as it involves looping over cases,
>but it ran in a few seconds on my puny machine even when I expanded
>the data by 1000. A little byable program would probably be faster.
>
>clear
>input ObservationID Year FirmID MemberID Tenure
>  1 1960 1 1 4
>  2 1960 1 2 2
>  3 1960 1 3 1
>  4 1960 2 4 2
>  5 1960 2 5 1
>  6 1961 1 1 4
>  7 1961 1 3 2
>  8 1961 2 2 0
>  9 1961 2 4 3
>  10 1961 2 5 2
>end
>//
>egen group = group(FirmID Year)  // I think this is what is intended
>sort group
>egen case1 = min(_n), by(group)
>egen caseN = max(_n), by(group)
>gen dij2 = 0
>forval i = 1/`=_N' {
>    quiet replace dij2 = dij2 + (Tenure- Tenure[`i'] )^2 ///
>          if inrange(`i', case1, caseN)  // restrict to group members
>}
>// Inspect results before dividing, easier to check by eyeball
>by group: list FirmID Year Tenure dij2
>//
>quiet replace dij2 = dij2/(caseN - case1) // n-1 within group
>
>Regards,
>=-=-=-=-=-=-=-=-=-=-=-=-=
>Mike Lacy, Assoc. Prof.
>Soc. Dept., Colo. State. Univ.
>Fort Collins CO 80523 USA
>(970)-491-6721


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