Dear All, I cannot unfortunately provide any R code, otherwise I would not need to post this in the first place. I need to generate a sample of N positive non-zero integers such that their mean is *exactly* 2(N-1)/N, i.e. the mean depends on the length of the sample. For a start, we can assume that every integer in my sample can assume only the values 1, 2 and 3. Any suggestion is appreciated. Cheers Lorenzo
n_1 + ... + n_N = 2(N-1) is requested for integers n_i >= 1. What about c(rep(2, N-2), 1, 1))? but I'm afraid that this was not what you really wanted. ;-) However, you didn't say if your sample should be random. :-) Best regards -- Gerrit --------------------------------------------------------------------- Dr. Gerrit Eichner Mathematical Institute, Room 212 gerrit.eichner at math.uni-giessen.de Justus-Liebig-University Giessen Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany http://www.uni-giessen.de/eichner --------------------------------------------------------------------- Am 01.08.2019 um 12:27 schrieb Lorenzo Isella:> Dear All, > I cannot unfortunately provide any R code, otherwise I would not need to > post this in the first place. > I need to generate a sample of N positive non-zero integers such that > their mean is *exactly* 2(N-1)/N, i.e. the mean depends on the length of > the sample. > For a start, we can assume that every integer in my sample can assume > only the values 1, 2 and 3. > Any suggestion is appreciated. > Cheers > > Lorenzo > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.
Yes, you are right (and yours is one of the possible cases).
I think this works (I resorted to pen and paper for once)
generate_k <- function(N, n3){
n1 <- 2+n3
n2 <- N-n1-n3
out <- c(rep(1, n1), rep(2, n2), rep(3, n3))
return(out)
}
where N is the length of the sample, n3 is the number of times I have the number
3 in the sample (n1 and n2 are the same for 1 and 2, respectively).
So far it holds!
L.
On Thu, Aug 01, 2019 at 12:37:47PM +0200, Gerrit Eichner
wrote:>n_1 + ... + n_N = 2(N-1) is requested for integers n_i >= 1.
>
>What about c(rep(2, N-2), 1, 1))?
>
>but I'm afraid that this was not what you really wanted. ;-)
>However, you didn't say if your sample should be random. :-)
>
> Best regards -- Gerrit
>
>---------------------------------------------------------------------
>Dr. Gerrit Eichner Mathematical Institute, Room 212
>gerrit.eichner at math.uni-giessen.de Justus-Liebig-University Giessen
>Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
>http://www.uni-giessen.de/eichner
>---------------------------------------------------------------------
>
>Am 01.08.2019 um 12:27 schrieb Lorenzo Isella:
>>Dear All,
>>I cannot unfortunately provide any R code, otherwise I
>>would not need to post this in the first place.
>>I need to generate a sample of N positive non-zero
>>integers such that their mean is *exactly* 2(N-1)/N,
>>i.e. the mean depends on the length of the sample.
>>For a start, we can assume that every integer in my
>>sample can assume only the values 1, 2 and 3.
>>Any suggestion is appreciated.
>>Cheers
>>
>>Lorenzo
>>
>>______________________________________________
>>R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>https://stat.ethz.ch/mailman/listinfo/r-help
>>PLEASE do read the posting guide
>>http://www.R-project.org/posting-guide.html
>>and provide commented, minimal, self-contained, reproducible code.
2(N-1)/N = 2 - 2/N.
So one way to get exactly that mean is to make all the numbers
2 except for two of them which are 1.
N < 2 : can't be done.
N = 2 : only [1,1] does the job.
N = 3 : the sum of the three numbers must be 4, so none of them
can be 3, so [1,1,2] [1,2,1] [2,1,1] are the only
possibilities.
N = 4 : [1,1,1,3] [1,1,2,2] [1,1,3,1] [1,2,1,2] [1,2,2,1] [1,3,1,1]
[2,1,1,2] [2,1,2,1] [2,2,1,1] [3,1,1,1]
Is there a pattern here?
Yes. There must be an integer k such that 0 <= 2k <= N-2
and then you have a rearrangement of (k+2) 1s, (N-2-2k) 2s, and k 3s.
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Hello, On Thu, Aug 01, 2019 at 11:17:30PM +1200, Richard O'Keefe wrote:>2(N-1)/N = 2 - 2/N. >So one way to get exactly that mean is to make all the numbers >2 except for two of them which are 1. > >N < 2 : can't be done. >N = 2 : only [1,1] does the job. >N = 3 : the sum of the three numbers must be 4, so none of them > can be 3, so [1,1,2] [1,2,1] [2,1,1] are the only > possibilities. >N = 4 : [1,1,1,3] [1,1,2,2] [1,1,3,1] [1,2,1,2] [1,2,2,1] [1,3,1,1] >[2,1,1,2] [2,1,2,1] [2,2,1,1] [3,1,1,1] > >Is there a pattern here? >Yes. There must be an integer k such that 0 <= 2k <= N-2 >and then you have a rearrangement of (k+2) 1s, (N-2-2k) 2s, and k 3s.This is what I did by solving a set of equations n1+n2+n3=N (i.e. the number of 1, 2 and 3 equals N, the length of my sample) and n2+2*n2+3*n3=2*(N-1) without delving into the details, you are composing (fractal) structures of N spheres in 3D, with an algorithm which establishes that every sphere has on average (2*N-1)/N neighbours. One solution is a simple chain, without bifurcations, where every sphere has two neighbours apart from the two terminal spheres, each of which has just one neighbour. Whenever a bifurcation arises, one sphere will have three neighbors. L.