CODEWITHSUNDEEP
rsamplingsimulationmatrix
Bstat
Bstat

Reputation: 121

How to replace an element of a symmetric matrix randomly?

Suppose I have a matrix like so:

data=matrix(c(1,0,0,0,0,0,1,0,0.6583,0,0,0,1,0,0,0,0.6583,0,1,0,0,0,0,0,1),nrow=5,ncol=5)

     [,1]   [,2] [,3]   [,4] [,5]
[1,]    1 0.0000    0 0.0000    0
[2,]    0 1.0000    0 0.6583    0
[3,]    0 0.0000    1 0.0000    0
[4,]    0 0.6583    0 1.0000    0
[5,]    0 0.0000    0 0.0000    1

How do I create another matrix, say "data2", such that it has the same number of off-diagonal nonzero elements as "data" but in another location other than the one in data? The randomly simulated data will be uniform (so runif).

Upvotes: 2

Views: 499

Answers (2)

MånsT
MånsT

Reputation: 974

Here is a somewhat clumsy way to do this. It works well for small matrices but would be too slow if you're going to use this for some very high-dimensional problems.

# Current matrix:
data=matrix(c(1,0,0,0,0,0,1,0,0.6583,0,0,0,1,0,0,0,0.6583,0,1,0,0,0,0,0,1),nrow=5,ncol=5)

# Number of nonzero elements in upper triangle:
no.nonzero<-sum(upper.tri(data)*data>0)

# Generate same number of new nonzero correlations:
new.cor<-runif(no.nonzero,-1,1)

# Create new diagonal matrix:
p<-dim(data)[1]
data2<-diag(1,p,p)

### Insert nonzero correlations: ###

# Step 1. Identify the places where the nonzero elements can be placed:

pairs<-(p^2-p)/2 # Number of element in upper triangle
combinations<-matrix(NA,pairs,2) # Matrix containing indices for those elements (i.e. (1,2), (1,3), ... (2,3), ... and so on)

k<-0
for(i in 1:(p-1))
{
    for(j in {i+1}:p)
    {
        k<-k+1
        combinations[k,]<-c(i,j)
    }
}

# Step 2. Randomly pick indices:

places<-sample(1:k,no.nonzero)

# Step 3. Insert nonzero correlations:

for(i in 1:no.nonzero)
{
    data2[combinations[places[i],1],combinations[places[i],2]]<-data2[combinations[places[i],2],combinations[places[i],1]]<-new.cor[i]
}

Upvotes: 1

djhurio
djhurio

Reputation: 5536

Not really understood the question. There are two off-diagonal and non-zero elements (0.6583) in the example, right? Is matrix with two elements the result you want in this case?

data=matrix(c(1,0,0,0,0,0,1,0,0.6583,0,0,0,1,0,0,0,0.6583,0,1,0,0,0,0,0,1),nrow=5,ncol=5)

# Convert to vector
data2 <- as.numeric(data)

# Remove diagonal
data2 <- data2[as.logical(upper.tri(data) | lower.tri(data))]

# Remove 0 elements
data2 <- data2[data2 != 0]

data2

Upvotes: 0

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