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Cesar Ortega

2003-Mar-10 11:17 UTC

[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

Carlos Ortega

2003-Mar-10 15:34 UTC

head link

[R] sampling and gini index

Cesar,

For the first part, please check the function included. For the sampling,
please check "?sample".

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)





-----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
______________________________________________
R-help at stat.math.ethz.ch mailing list
https://www.stat.math.ethz.ch/mailman/listinfo/r-help

###  This email has been checked for all known viruses by the
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partha_bagchi@hgsi.com

2003-Mar-10 16:28 UTC

head link

[R] sampling and gini index

I think you can compute the sum using 
> sum(abs( outer(y, y, FUN= "-")))





"Carlos Ortega" <carlos.ortega@minorplanet.com>
Sent by: r-help-bounces@stat.math.ethz.ch
03/10/2003 10:34 AM
Please respond to carlos.ortega

 
        To:     "Cesar Ortega" <cortega@unitec.edu>,
<r-help@stat.math.ethz.ch>
        cc: 
        Subject:        RE: [R] sampling and gini index


Cesar,

For the first part, please check the function included. For the sampling,
please check "?sample".

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)





-----Mensaje original-----
De: r-help-bounces@stat.math.ethz.ch
[mailto:r-help-bounces@stat.math.ethz.ch]En nombre de Cesar Ortega
Enviado el: lunes, 10 de marzo de 2003 12:17
Para: r-help@stat.math.ethz.ch
CC: r-help@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
______________________________________________
R-help@stat.math.ethz.ch mailing list
https://www.stat.math.ethz.ch/mailman/listinfo/r-help

###  This email has been checked for all known viruses by the
###  Firstnet anti-virus system - http://www.firstnet.net.uk
###  Please email fav@firstnet.net.uk for details.


_____
The information in this email is confidential and it may not be\... [[dropped]]

Achim Zeileis

2003-Mar-10 17:03 UTC

head link

[R] sampling and gini index

On Monday 10 March 2003 16:34, Carlos Ortega wrote:
> Cesar,
>
> For the first part, please check the function included. For the
> sampling, please check "?sample".
>
> 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
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://www.stat.math.ethz.ch/mailman/listinfo/r-help
>
> ###  This email has been checked for all known viruses by the
> ###  Firstnet anti-virus system - http://www.firstnet.net.uk
> ###  Please email fav at firstnet.net.uk for details.
>
>
> _____
> The information in this email is confidential and it may not be\...
> [[dropped]]
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://www.stat.math.ethz.ch/mailman/listinfo/r-help

Cesar Ortega

2003-Mar-10 17:41 UTC

head link

[R] sampling and gini index

Thanks for you input,

However, I still need to do the following:

1. How can I work on the other 2 methods:Fixed systematic sampling and.
Proportional sampling Madow.

2. After I have done the 3 samplings methods, obtaining M(1000,or 2000, 
or more, etc. for each sampling method) samples of n ( 3,4, 5, etc)
elements in a column,I need to take one by one each of the 1000, 2000, 
etc. samples of each n elements and calculate in the 3 sampling methods:

a. PI(i)=n/N in simple and fixed sampling and i belogns to the sample. 
PI(i) In MADOW 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.

ALSO 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,etc.) 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

Thanks for your kind response,

Cesar
----- Original Message -----
From: Achim Zeileis <zeileis at ci.tuwien.ac.at>
Date: Monday, March 10, 2003 6:03 pm
Subject: Re: [R] sampling and gini index
> On Monday 10 March 2003 16:34, Carlos Ortega wrote:
> 
> > Cesar,
> >
> > For the first part, please check the function included. For the
> > sampling, please check "?sample".
> >
> > 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
> >
> > ______________________________________________
> > R-help at stat.math.ethz.ch mailing list
> > https://www.stat.math.ethz.ch/mailman/listinfo/r-help
> >
> > ###  This email has been checked for all known viruses by the
> > ###  Firstnet anti-virus system - http://www.firstnet.net.uk
> > ###  Please email fav at firstnet.net.uk for details.
> >
> >
> > _____
> > The information in this email is confidential and it may not be\...
> > [[dropped]]
> >
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R help - Mar 2003 - sampling and gini index

[R] sampling and gini index

[R] sampling and gini index

[R] sampling and gini index

[R] sampling and gini index

[R] sampling and gini index