Sorry, I wasn't sure what you meant. This way will return more than one
answer, right?
N<-c(1,2,1,3)
R<-c(1.75,3.5,1.75,1.3125)
## get all 126 combinations of five 0's and four 1's for matrix
cbn<-as.matrix(expand.grid( rep( list(0:1), 9)))
cbn<- cbn[rowSums(cbn)==4,]
ans<-list()
ctr<-0
## loop through each combination
for (i in 1:126){
x<-cbn[i,]
## replace 1's with N
x[which(x==1)]<-N
## create matrix
dim(x)<-c(3,3)
## calculate y and new R
y <-sum(x) * x / (rowSums(x)%o%colSums(x))
R1<-y[y[1:3,]>0]
# check if equal to original R
if(identical(R, R1)) ans[[ctr<-ctr+1]]<-x
}
ans
[[1]]
[,1] [,2] [,3]
[1,] 0 0 1
[2,] 1 0 3
[3,] 0 2 0
[[2]]
[,1] [,2] [,3]
[1,] 0 0 1
[2,] 0 2 0
[3,] 1 0 3
[[3]]
[,1] [,2] [,3]
[1,] 0 2 0
[2,] 0 0 1
[3,] 1 0 3
Chris
francogrex wrote:
>
> Hi, thanks but the way you are doing it is to assign the values of N in
> the x matrix, knowing from the example I have given where they are
> supposed to be. While the assumption is, you ONLY have values of N and R
> and do NOT know where they would be placed in the x and y matrix a-priori,
> but their position has to be derived from only the (N and R) dataframe you
> have.
>
>
>
>
--
View this message in context:
http://www.nabble.com/Generating-these-matrices-going-backwards-tf4807447.html#a13782676
Sent from the R help mailing list archive at Nabble.com.
______________________________________________
R-help@r-project.org mailing list
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.
