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RE: st: RE: Re: local macro with if qualifier

From   "Nick Cox" <>
To   <>
Subject   RE: st: RE: Re: local macro with if qualifier
Date   Wed, 8 Apr 2009 16:55:07 +0100

Thanks for the explanation. You might find the discussion in 

SJ-9-1  pr0046  . . . . . . . . . . . . . . . . . . .  Speaking Stata:
>  Rowwise
        (help rowsort, rowranks if installed) . . . . . . . . . . .  N
> . J. Cox
        Q1/09   SJ 9(1):137--157
        shows how to exploit functions, egen functions, and Mata
        for working rowwise; rowsort and rowranks are introduced

of use or interest, at least for related problems. 


Sabrina Carrossa

My object is to find the best friend among a list of friends.
I have a list of  "i" occupation (in the db they are 20, in the
example just 3). For each occupation in the db I have the variables:
- d31b_i : do you have a friend that have the occupation "i"? (yes/no)
- d31c_i: how many time do you spend with this person in the
occupation "i"? (0/6, with 6 a lot of time)
-  d31d_i: how do you feel close with this person in the occupation
"i"? (0/10, with 10 really close)

Than, my idea is that the best friend is a friend with:
- d31b_i=1
- d31c_i= max (among each i)
- d31d_i=max (among each i)
 And, if I have more friends with the same value for these three vars,
the best friend is the friend with higher occupation (max i)

* Stata Code:

* compute n vars as sum of the information on d31b d31c d31d and
occupation (i)
forval i=1/3 {
	gen d31_`i'=0
	replace d31_`i'= (d31b_`i'*100000) + (d31c_`i'*10000) +
(d31d_`i'*100) + `i'

* in the first step i assign to friend the value of the friend in the
first occupation (i=1)
* than i compare the value of this first friend with the values of the
other friends
* if the values of the other friends are highest (more contact and
more closeness) I replace the value of friend
gen friend=d31_1
gen f=1
forval y=2/3 {
	replace f=`y' if confront<d31_`y'
	replace friend= d31_`y' if confront<d31_`y'

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