Maybe something like this?
N<-c(1,2,1,3)
## create empty matrix
x<-diag(0,3)
## fill off diagonal
x[row(x)==col(x)+1]<-N[1:2]
# fill 3rd column
x[1:2,3]<-N[3:4]
Or create a function and return both x and y
mat3<-function(N)
{
x<-diag(0,3)
# fill each element separately
x[2,1]<-N[1]
x[3,2]<-N[2]
x[1,3]<-N[3]
x[2,3]<-N[4]
dimnames(x)<-list(c("D1", "D2", "D3"), c("E1", "E2", "E3"))
y =sum(x) * x / (rowSums(x)%o%colSums(x))
list(x=x,y=y)
}
mat3(N)
$x
E1 E2 E3
D1 0 0 1
D2 1 0 3
D3 0 2 0
$y
E1 E2 E3
D1 0.00 0.0 1.7500
D2 1.75 0.0 1.3125
D3 0.00 3.5 0.0000
francogrex wrote:
>
> I have generated the following:
>
> x=
> E1 E2 E3
> D1 0 0 1
> D2 1 0 3
> D3 0 2 0
>
> y=
> E1 E2 E3
> D1 0 0 1.75
> D2 1.75 0 1.3125
> D3 0 3.5 0
>
> Where x and y are linked by:
> y =sum(x) * x / (rowSums(x)%o%colSums(x))
>
> N=x[x[1:3,]>0]
> R=y[y[1:3,]>0]
>
> Now suppose I ONLY have N and R linked in this way below where each N
> corresponds to an R
>
> N R
> 1 1.7500
> 2 3.5000
> 1 1.7500
> 3 1.3125
>
> Is there a way to generate matrix "x" and matrix "y" having only the N
> and the R as above?
>
>
>
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