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: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] nombre de Cesar Ortega > Enviado el: lunes, 10 de marzo de 2003 12:17 > Para: [EMAIL PROTECTED] > CC: [EMAIL PROTECTED] > 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 > > ______________________________________________ > [EMAIL PROTECTED] 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 [EMAIL PROTECTED] for details. > > > _____ > The information in this email is confidential and it may not be\... > [[dropped]] > > ______________________________________________ > [EMAIL PROTECTED] mailing list > https://www.stat.math.ethz.ch/mailman/listinfo/r-help ______________________________________________ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help