# [R] sampling and gini index

Achim Zeileis zeileis at ci.tuwien.ac.at
Mon Mar 10 18:03:50 CET 2003

```On Monday 10 March 2003 16:34, Carlos Ortega wrote:

> Cesar,
>
> For the first part, please check the function included. For the
>
> Regards,
> Carlos.
>
> g.index<-function(y) {
> 	sum.res<-0
> 	y.lg<-length(y)
> 	y.mean<-mean(y)
>
> 	for (i in 1:y.lg) {
>   		for (j in 1:y.lg) {
>     			ratio.res<-abs(y[i]-y[j]) / (2 * y.lg^2 * y.mean)
>     			sum.res<-sum.res+ratio.res
>   		}
> 	}
>   	return(sum.res)
>  }
>
>  y<-rnorm(284)
>  g.index(y)
>

A more efficient implementation of the Gini index can be found in the
ineq package. See
help(Gini)
help(ineq)
Z

>
>
>
> -----Mensaje original-----
> De: r-help-bounces at stat.math.ethz.ch
> [mailto:r-help-bounces at stat.math.ethz.ch]En nombre de Cesar Ortega
> Enviado el: lunes, 10 de marzo de 2003 12:17
> Para: r-help at stat.math.ethz.ch
> CC: r-help at stat.math.ethz.ch
> Asunto: [R] sampling and gini index
>
>
> Hi there,
>
>
> I am new in R, and I was wondering if I could do the following in R,
> since I have tried in SPSS and I have only done part of it.  I
> would appreciate any help to build this routine in R, if possible:
>
>
>
> I have a column of 284 elements Y, [...]
>
> > Iam reading these as 284 cases with a single
> > variable, which I will call Y
> >
> > >  where first I need to calculate:
> > >
> > >GINP=[SUMi SUMj {|Yi - Yj|}]/[2(N**2)*MEAN(Y)], where Yi and Yj
> >
> > are
> >
> > >the 284 elements, 0<Y1<=Y2...<=Y284.  j: is the next position of
> > > i.
> >
> > Here, I am doing a calculation to yield a single number.
> > Calculating
>
>  (I assume that N=284)
>
> > >Next, I need to take the 284 numbers and resampling them in
> >
> > [groups
> >
> > >of] n data (like 3, 4, 5,  etc) for M number of samples ( like
> >
> > 1000,
> >
> > >2000, etc, one at a time) in 3 sampling methods:
> > >
> > >1. Simple sampling without replacing.
> > >2. Fixed systematic sampling.
> > >3. Proportional sampling Madow.
>
> SIMPLE SAMPLING WITHOUT REPLACING THE POSITION VALUE, FOR INTANCE 3,
> IF WE HAVE THE POSITION 3,7,9 WITH ITS VALUES, WE COULD HAVE 3,7,10,
> BUT WE COULD NOT HAVE THE POSITIONS 9,3,7 AGAIN.  SO THIS IS A
> COMBINATION, AND IF WE TAKE 3 COMBINATIONS OUT 284 WE HAVE
> 284!/(3!*281!). AND WE NEED TO START FROM 1000 GROUPS OF
> COMBINATIONS. IT IS LIKE
> IF WE HAVE 1,2,3,4,5 AND WE WANT COMBINATIONS OF 2, WE COULD HAVE IN
> 1,2 AND 1,3 AND 1,4 AND 1,5 AND 2,3 AND 2,4, ETC, BUT IN RANDOM
> ORDER.
>
> > >After I have the M(1000, 2000) samples of n ( 3,4, 5)elements in
> >
> > a column,
> >
> > >I need to take one by one each of the 1000 samples of each n
> >
> > elements
> >
> > >and calculate:
> > >
> > >PI(i)=n/N in simple and fixed sampling and i belogns to the
> >
> > sample. In
> >
> > >MADOW  PI(i) is proprortional to an auxiliary variable Xi:PI(i)=
> >
> > RXi,
> >
> > >where r= MOD(TN,R), and TN=SUMi(Xi), i to N.
>
> WE KNOW r, AND TN, BUT WE NEED TO FIND R FROM THE MODAL AND r.
>
> AND CALCULATE:
> > >ESTIMATED N= SUM {1/PI(i)};
> > >ESTIMATED T(Y)= SUM {Yi/PI(i)};  i belongs to the sample.
>
> GINMi=[SUMi{(Yi/PI(i))*[1/PI(i)+SUMj(2/PI(j)]}]/[(2*ESTIMATED
> N*ESTIMATED T)].     i belongs to  M (1000) times, WHERE j=i+1
>
> WE HAVE ONLY ONE VARIABLE Y, BUT WITH THE FIRST POSITION i AND THE
> NEXT j, IN THE ASCENDING ORDER.
>
> > >And last to get the errors as:
> > >
> > >E1=  [SUMi {|GINP-GINMi |}]/[M] ; i belongs M
> > >
> > >E2= SQRT[SUMi {|GINP-GINMi |}]/[M]; i belongs TO M
>
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```