如何隨機替換對稱矩陣的元素？

Question

假設我有一個矩陣，像這樣：

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 

如何創建另一個矩陣，說「數據2」，使得其具有相同數量的非對角線非零元素爲「數據」，但在其他位置除了數據之外呢？隨機模擬的數據將是統一的（如此runif）。

Answer 1

沒有真正理解這個問題。在這個例子中有兩個非對角元素和非零元素（0.6583），對嗎？在這種情況下，矩陣是否包含兩個元素？

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 

Answer 2

這是一個有點笨拙的做法。它適用於小型矩陣，但如果要將其用於某些非常高維的問題，速度會太慢。

# 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] 
}